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| subject
stringclasses 1
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stringclasses 32
values | topic
stringclasses 178
values | question
stringlengths 26
9.64k
| options
stringlengths 2
1.63k
| correct_option
stringclasses 5
values | answer
stringclasses 293
values | explanation
stringlengths 13
9.38k
| question_type
stringclasses 3
values | paper_id
stringclasses 149
values |
|---|---|---|---|---|---|---|---|---|---|---|
1l6dxf7ak
|
maths
|
hyperbola
|
tangent-to-hyperbola
|
<p>Let the equation of two diameters of a circle $$x^{2}+y^{2}-2 x+2 f y+1=0$$ be $$2 p x-y=1$$ and $$2 x+p y=4 p$$. Then the slope m $$ \in $$ $$(0, \infty)$$ of the tangent to the hyperbola $$3 x^{2}-y^{2}=3$$ passing through the centre of the circle is equal to _______________.</p>
|
[]
| null |
2
|
$$
\begin{aligned}
&2 p+f-1=0 \quad\dots(1)\\\\
&2-p f-4 p=0 \quad\dots(2)\\\\
&2=p(f+4) \\\\
&p=\frac{2}{f+4} \\\\
&2 p=1-f \\\\
&\frac{4}{f+4}=1-f \\\\
&f^2+3 f=0 \\\\
&f=0 \text { or }-3
\end{aligned}
$$<br/><br/>
Hyperbola $3 x^2-y^2=3, x^2-\frac{y^2}{3}=1$<br/><br/>
$$
\mathrm{y}=\mathrm{mx} \pm \sqrt{\mathrm{m}^2-3}
$$<br/><br/>
It passes $(1,0)$ <br/><br/>$o=m \pm \sqrt{m^2-3}$<br/><br/>
$m$ tends $\infty$<br/><br/>
$$
\begin{aligned}
&\text {It passes }(1,3) \\\\
&3=m \pm \sqrt{m^2-3} \\\\
&(3-m)^2=m^2-3 \\\\
&m=2
\end{aligned}
$$
|
integer
|
jee-main-2022-online-25th-july-morning-shift
|
1ldo4xfyk
|
maths
|
hyperbola
|
tangent-to-hyperbola
|
<p>Let $$\mathrm{P}\left(x_{0}, y_{0}\right)$$ be the point on the hyperbola $$3 x^{2}-4 y^{2}=36$$, which is nearest to the line $$3 x+2 y=1$$. Then $$\sqrt{2}\left(y_{0}-x_{0}\right)$$ is equal to :</p>
|
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "$$-$$9"}, {"identifier": "C", "content": "$$-$$3"}, {"identifier": "D", "content": "9"}]
|
["B"]
| null |
If $\left(x_0, y_0\right)$ is point on hyperbola then tangent at $\left(x_0, y_0\right)$ is parallel to $3 x+2 y=1$
<br/><br/>Equation of tangent $= \frac{x x_0}{12}-\frac{y y_0}{9}=2$
<br/><br/>Slope of tangent $=\frac{-3}{2}$
<br/><br/>Equation of tangent in slope form
<br/><br/>$y=\frac{-3}{2} x \pm \sqrt{12 \cdot \frac{9}{4}-9}$
<br/><br/>$y=\frac{-3}{2} x \pm 3 \sqrt{2}$
<br/><br/>or $3 x+2 y=6 \sqrt{2}$
<br/><br/>Comparing
<br/><br/>$$
\begin{aligned}
& \frac{\frac{x_0}{12}}{3}=\frac{\frac{-y_0}{9}}{2}=\frac{1}{6 \sqrt{2}} \\\\
& x_0=3 \sqrt{2}, y_0=\frac{-3}{\sqrt{2}} \\\\
& \sqrt{2}\left(y_0-x_0\right)=-3-6=-9 \\\\
&
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-1st-february-evening-shift
|
1lgoy09kt
|
maths
|
hyperbola
|
tangent-to-hyperbola
|
<p>The foci of a hyperbola are $$( \pm 2,0)$$ and its eccentricity is $$\frac{3}{2}$$. A tangent, perpendicular to the line $$2 x+3 y=6$$, is drawn at a point in the first quadrant on the hyperbola. If the intercepts made by the tangent on the $$\mathrm{x}$$ - and $$\mathrm{y}$$-axes are $$\mathrm{a}$$ and $$\mathrm{b}$$ respectively, then $$|6 a|+|5 b|$$ is equal to __________</p>
|
[]
| null |
12
|
<p>Given that the foci are at $(\pm 2, 0)$, the distance between the foci is $2ae = 2\times2 = 4$. <br/><br/>So, $a = \frac{4}{3}$. </p>
<p>The eccentricity of the hyperbola is given as $\frac{3}{2}$. Thus, $e = \frac{3}{2}$. </p>
<p>For a hyperbola, $e^2 = 1 + \frac{b^2}{a^2}$, which gives $b^2 = a^2(e^2 - 1)$.<br/><br/> Substituting the given values, we get $b^2 = \frac{16}{9} - \frac{16}{9} = \frac{20}{9}$. </p>
<p>The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, or $\frac{9x^2}{16} - \frac{9y^2}{20} = 1$.</p>
<p>The slope of the tangent line, which is perpendicular to the given line $2x + 3y = 6$, is $\frac{3}{2}$.</p>
<p>The equation of the tangent to a hyperbola is $y = mx \pm \sqrt{a^2 m^2 - b^2}$. Substituting the given values, we get $y = \frac{3x}{2} \pm \sqrt{\frac{16}{9}\times\frac{9}{4} - \frac{20}{9}} = \frac{3x}{2} \pm \frac{4}{3}$.</p>
<p>The $x$-intercept occurs when $y = 0$, so $|a| = \frac{8}{9}$, and the $y$-intercept occurs when $x = 0$, so $|b| = \frac{4}{3}$.</p>
<p>Finally, $|6a| + |5b| = 6\frac{8}{9} + 5\frac{4}{3} = \frac{48}{9} + \frac{60}{9} = \frac{108}{9} = 12$.</p>
|
integer
|
jee-main-2023-online-13th-april-evening-shift
|
1lgq0o0y8
|
maths
|
hyperbola
|
tangent-to-hyperbola
|
<p>Let $$m_{1}$$ and $$m_{2}$$ be the slopes of the tangents drawn from the point $$\mathrm{P}(4,1)$$ to the hyperbola $$H: \frac{y^{2}}{25}-\frac{x^{2}}{16}=1$$. If $$\mathrm{Q}$$ is the point from which the tangents drawn to $$\mathrm{H}$$ have slopes $$\left|m_{1}\right|$$ and $$\left|m_{2}\right|$$ and they make positive intercepts $$\alpha$$ and $$\beta$$ on the $$x$$-axis, then $$\frac{(P Q)^{2}}{\alpha \beta}$$ is equal to __________.</p>
|
[]
| null |
8
|
Equation of tangent to the hyperbola $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
<br/><br/>$$
y=m x \pm \sqrt{a^2-b^2 m^2}
$$
<br/><br/>Given the hyperbola $H: \frac{y^2}{25} - \frac{x^2}{16} = 1$, the equation of the tangent to this hyperbola can be written as :
<br/><br/>$$y=mx \pm \sqrt{25 - 16m^2}$$
<br/><br/>We know that the tangents pass through the point $P(4, 1)$, which gives us the equation :
<br/><br/>$$1 = 4m \pm \sqrt{25 - 16m^2}$$
<br/><br/>Squaring both sides to get rid of the square root, we obtain :
<br/><br/>$$(4m - 1)^2 = 25 - 16m^2$$
<br/><br/>which simplifies to the quadratic equation :
<br/><br/>$$4m^2 - m - 3 = 0.$$
<br/><br/>Solving this equation, we find the roots $m_1 = 1$ and $m_2 = -\frac{3}{4}$, which are the slopes of the tangents.
<br/><br/>Given that we are interested in the positive values of the slopes, we consider $|m_1| = 1$ and $|m_2| = \frac{3}{4}$.
<br/><br/>The equations of the tangents are then :
<br/><br/>1) $y = x - 3$ and
<br/><br/>2) $y = \frac{3}{4}x - 4$.
<br/><br/>The x-intercepts of these lines (when $y = 0$) are given by $x = \alpha = 3$ for the first line and $x = \beta = \frac{16}{3}$ for the second line.
<br/><br/>The intersection point $Q$ of these tangents is found by solving the system of equations $y = x - 3$ and $y = \frac{3}{4}x - 4$, which gives $Q(-4, -7)$.
<br/><br/>The square of the distance $PQ$ is then $(-4-4)^2 + (-7-1)^2 = 128$.
<br/><br/>Therefore,
$$\frac{(PQ)^2}{\alpha \beta} = \frac{128}{3 \times \frac{16}{3}} = 8$$
|
integer
|
jee-main-2023-online-13th-april-morning-shift
|
lv2er3wn
|
maths
|
hyperbola
|
tangent-to-hyperbola
|
<p>Consider a hyperbola $$\mathrm{H}$$ having centre at the origin and foci on the $$\mathrm{x}$$-axis. Let $$\mathrm{C}_1$$ be the circle touching the hyperbola $$\mathrm{H}$$ and having the centre at the origin. Let $$\mathrm{C}_2$$ be the circle touching the hyperbola $$\mathrm{H}$$ at its vertex and having the centre at one of its foci. If areas (in sq units) of $$C_1$$ and $$C_2$$ are $$36 \pi$$ and $$4 \pi$$, respectively, then the length (in units) of latus rectum of $$\mathrm{H}$$ is</p>
|
[{"identifier": "A", "content": "$$\\frac{28}{3}$$\n"}, {"identifier": "B", "content": "$$\\frac{11}{3}$$\n"}, {"identifier": "C", "content": "$$\\frac{14}{3}$$\n"}, {"identifier": "D", "content": "$$\\frac{10}{3}$$"}]
|
["A"]
| null |
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwhfif9k/1a3e3d85-d2e7-4099-8558-58b863f1c6ee/866ac480-1802-11ef-b156-f754785ad3ce/file-1lwhfif9l.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwhfif9k/1a3e3d85-d2e7-4099-8558-58b863f1c6ee/866ac480-1802-11ef-b156-f754785ad3ce/file-1lwhfif9l.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 4th April Evening Shift Mathematics - Hyperbola Question 4 English Explanation"></p>
<p>$$\begin{aligned}
& C_1: x^2+y^2=a^2 \Rightarrow \text { area }=\pi a^2=36 \pi \Rightarrow a=6 \\
& C_2: x^2+(y-a e)^2=(a e-a)^2 \\
& \therefore \quad \pi(a e-a)^2=4 \pi \\
& \Rightarrow 36(e-1)^2=4 \\
& \Rightarrow e-1=\frac{1}{3} \Rightarrow e=\frac{4}{3} \\
& \Rightarrow b^2=28 \\
& \therefore \quad L R=\frac{2 b^2}{a}=\frac{2 \times 28}{6}=\frac{28}{3} \text { units }
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-4th-april-evening-shift
|
lvb2951p
|
maths
|
hyperbola
|
tangent-to-hyperbola
|
<p>The length of the latus rectum and directrices of hyperbola with eccentricity e are 9 and $$x= \pm \frac{4}{\sqrt{3}}$$, respectively. Let the line $$y-\sqrt{3} x+\sqrt{3}=0$$ touch this hyperbola at $$\left(x_0, y_0\right)$$. If $$\mathrm{m}$$ is the product of the focal distances of the point $$\left(x_0, y_0\right)$$, then $$4 \mathrm{e}^2+\mathrm{m}$$ is equal to _________.</p>
|
[]
| null |
61
|
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwaskftu/d526b7cc-575f-4620-8047-7c036e1ade54/189a8020-145c-11ef-a76a-45e5478b32e1/file-1lwaskftv.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwaskftu/d526b7cc-575f-4620-8047-7c036e1ade54/189a8020-145c-11ef-a76a-45e5478b32e1/file-1lwaskftv.png" loading="lazy" style="max-width: 100%;height: auto;display: block;margin: 0 auto;max-height: 40vh;vertical-align: baseline" alt="JEE Main 2024 (Online) 6th April Evening Shift Mathematics - Hyperbola Question 1 English Explanation"></p>
<p>$$y=m x \pm \sqrt{a^2 m^2-b^2}$$ is the tangent</p>
<p>$$\begin{aligned}
& \Rightarrow m=\sqrt{3} \Rightarrow a^2 m^2-b^2=3 \\
& \Rightarrow 3 a^2-b^2=3
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \frac{2 b^2}{a}=9 \Rightarrow b^2=\frac{9 a}{2} \Rightarrow 3 a^2-\frac{9 a}{2}=3 \\
& \Rightarrow a^2-\frac{3}{2} a-1=0 \\
& \Rightarrow a=2 \text { or }-0.5 \text { (ignore) } \\
& \Rightarrow b=3 \\
& \Rightarrow \frac{x^2}{4}-\frac{y^2}{9}=1 \\
& \Rightarrow \text { Solving hyperbola and tangent } y=\sqrt{3 x}-\sqrt{3} \\
& x_0=4, y_0=3 \sqrt{3}, e=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2} \\
& P F_1 \cdot P F_2= \\
& \sqrt{(4-\sqrt{13})^2+(3 \sqrt{3})^2} \sqrt{(4+\sqrt{13})^2+(3 \sqrt{3})^2} \\
& =\sqrt{(56-8 \sqrt{13})(56+8 \sqrt{13})} \\
& =\sqrt{2304}=48=m \\
& \Rightarrow 4 e^2+m=13+48=61
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-6th-april-evening-shift
|
1ktgqhz9h
|
maths
|
indefinite-integrals
|
integration-by-partial-fraction
|
If $$\int {{{2{e^x} + 3{e^{ - x}}} \over {4{e^x} + 7{e^{ - x}}}}dx = {1 \over {14}}(ux + v{{\log }_e}(4{e^x} + 7{e^{ - x}})) + C} $$, where C is a constant of integration, then u + v is equal to _____________.
|
[]
| null |
7
|
$$2{e^x} + 3{e^{ - x}} = A(4{e^x} + 7{e^{ - x}}) + B(4{e^x} - 7{e^{ - x}}) + \lambda $$<br><br>2 = 4A + 4B ; 3 = 7A $$-$$ 7B ; $$\lambda$$ = 0<br><br>$$A + B = {1 \over 2}$$<br><br>$$A - B = {3 \over 7}$$<br><br>$$A = {1 \over 2}\left( {{1 \over 2} + {3 \over 7}} \right) = {{7 + 6} \over {28}} = {{13} \over {28}}$$<br><br>$$B = A - {3 \over 7} = {{13} \over {28}} - {3 \over 7} = {{13 - 12} \over {28}} = {1 \over {28}}$$<br><br>$$\int {{{13} \over {28}}dx + {1 \over {28}}\int {{{4{e^x} - 7{e^{ - x}}} \over {4{e^x} + 7{e^{ - x}}}}} dx} $$<br><br>= $${{13} \over {28}}x + {1 \over {28}}\ln |4{e^x} + 7{e^{ - x}}| + C$$<br><br>$$u = {{13} \over 2};v = {1 \over 2}$$<br><br>$$\Rightarrow$$ u + v = 7
|
integer
|
jee-main-2021-online-27th-august-evening-shift
|
1ktke5n0m
|
maths
|
indefinite-integrals
|
integration-by-partial-fraction
|
If $$\int {{{\sin x} \over {{{\sin }^3}x + {{\cos }^3}x}}dx = } $$
<br/><br/>$$\alpha {\log _e}|1 + \tan x| + \beta {\log _e}|1 - \tan x + {\tan ^2}x| + \gamma {\tan ^{ - 1}}\left( {{{2\tan x - 1} \over {\sqrt 3 }}} \right) + C$$, when C is constant of integration, then the value of $$18(\alpha + \beta + {\gamma ^2})$$ is ______________.
|
[]
| null |
3
|
$$ = \int {{{{{\sin x} \over {{{\cos }^3}x}}} \over {1 + {{\tan }^3}x}}dx = \int {{{\tan x.{{\sec }^2}x} \over {(\tan x + 1)(1 + {{\tan }^2}x - \tan x)}}dx} } $$<br><br>Let $$\tan x = t \Rightarrow {\sec ^2}x.\,dx = dt$$<br><br>$$ = \int {{t \over {(t + 1)({t^2} - t + 1)}}dt} $$<br><br>$$ = \int {\left( {{A \over {t + 1}} + {{B(2t - 1)} \over {{t^2} - t + 1}} + {C \over {{t^2} - t + 1}}} \right)dx} $$<br><br>$$ \Rightarrow A({t^2} - t + 1) + B(2t - 1)({t^2} - t + 1) + C(t + 1) = t$$<br><br>$$ \Rightarrow {t^2}(A + 2B) + t( - A + B + C) + A - B + C = 1$$<br><br>$$\therefore$$ $$A + 2B = 0$$ ..... (1)<br><br>$$ - A + B + C = 1$$ ... (2)<br><br>$$A - B + C = 0$$ ... (3)<br><br>$$ \Rightarrow C = {1 \over 2} \Rightarrow A - B = - {1 \over 2}$$ ... (4)<br><br>$$A + 2B = 0$$<br><br>$$A - B = - {1 \over 2}$$<br><br>$$\Rightarrow$$ $$3B = {1 \over 2} \Rightarrow B = {1 \over 6}$$<br><br>$$A = - {1 \over 3}$$<br><br>$$I = - {1 \over 3}\int {{{dt} \over {1 + t}} + {1 \over 6}\int {{{2t - 1} \over {{t^2} - t + 1}}dt + {1 \over 2}\int {{{dt} \over {{t^2} - t + 1}}} } } $$<br><br>$$ = - {1 \over 3}\ln |(1 + \tan x)| + {1 \over 6}\ln |{\tan ^2}x - \tan x + 1| + {1 \over 2}.{2 \over {\sqrt 3 }}{\tan ^{ - 1}}\left( {{{\left( {\tan x - {1 \over 2}} \right)} \over {{{\sqrt 3 } \over 2}}}} \right)$$<br><br>$$ = - {1 \over 3}\ln |(1 + \tan x)| + {1 \over 6}\ln |{\tan ^2}x - \tan x + 1| + {1 \over {\sqrt 3 }}{\tan ^{ - 1}}\left( {{{2\tan x - 1} \over {\sqrt 3 }}} \right) + C$$<br><br>$$\alpha = - {1 \over 3},\beta = {1 \over 6},\gamma = {1 \over {\sqrt 3 }}$$<br><br>$$18(\alpha + \beta + {\gamma ^2}) = 18\left( { - {1 \over 3} + {1 \over 6} + {1 \over 3}} \right) = 3$$
|
integer
|
jee-main-2021-online-31st-august-evening-shift
|
Uw47dOEvkjyfJj7r
|
maths
|
indefinite-integrals
|
integration-by-parts
|
The integral $$\int {\left( {1 + x - {1 \over x}} \right){e^{x + {1 \over x}}}dx} $$ is equal to
|
[{"identifier": "A", "content": "<img class=\"question-image\" src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734264772/exam_images/je7jltiyhhqoek98uddy.webp\" loading=\"lazy\" alt=\"JEE Main 2014 (Offline) Mathematics - Indefinite Integrals Question 68 English Option 1\"> "}, {"identifier": "B", "content": "<img class=\"question-image\" src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734263643/exam_images/iyvs2mdeuwsvllarbgrw.webp\" loading=\"lazy\" alt=\"JEE Main 2014 (Offline) Mathematics - Indefinite Integrals Question 68 English Option 2\"> "}, {"identifier": "C", "content": "<img class=\"question-image\" src=\"https://gateclass.cdn.examgoal.net/Vr0Zu3a3X1vIGgHzv/yYQIuzwacMH8vGTwCrmF1Y4FuHgaa/wr2jyH6V0lm8gcfeGXBHmN/uploadfile.jpg\" loading=\"lazy\" alt=\"JEE Main 2014 (Offline) Mathematics - Indefinite Integrals Question 68 English Option 3\"> "}, {"identifier": "D", "content": "<img class=\"question-image\" src=\"https://res.cloudinary.com/dckxllbjy/image/upload/v1734266860/exam_images/c5mc2bmicbb59eklgten.webp\" loading=\"lazy\" alt=\"JEE Main 2014 (Offline) Mathematics - Indefinite Integrals Question 68 English Option 4\"> "}]
|
["D"]
| null |
Let $$I = \int {\left( {1 + x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle x$}}}}dx$$
<br><br>$$ = \int {{e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle x$}}}}} dx + \int {\left( {x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle x$}}}}dx$$
<br><br>$$ = x.{e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle x$}}}} - \int {x\left( {1 - {1 \over {{x^2}}}} \right)} {e^{x+{\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle x$}}}}dx$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \int {\left( {x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle x$}}}}dx$$
<br><br>$$ = x.{e^{x + {1 \over x}}} - \int {\left( {x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle x$}}}}dx$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \int {\left( {x - {1 \over x}} \right)} {e^{x + {\raise0.5ex\hbox{$\scriptstyle 1$}
\kern-0.1em/\kern-0.15em
\lower0.25ex\hbox{$\scriptstyle x$}}}}dx$$
<br><br>$$ = x{e^{x + {1 \over x}}} + C$$
|
mcq
|
jee-main-2014-offline
|
k0vrHx9KC2v0TCKPU1nFZ
|
maths
|
indefinite-integrals
|
integration-by-parts
|
The integral $$\int \, $$cos(log<sub>e</sub> x) dx is equal to : (where C is a constant of integration)
|
[{"identifier": "A", "content": "$${x \\over 2}$$[sin(log<sub>e</sub> x) $$-$$ cos(log<sub>e</sub> x)] + C"}, {"identifier": "B", "content": "x[cos(log<sub>e</sub> x) + sin(log<sub>e</sub> x)] + C"}, {"identifier": "C", "content": "$${x \\over 2}$$[cos(log<sub>e</sub> x) + sin(log<sub>e</sub> x)] + C"}, {"identifier": "D", "content": "x[cos(log<sub>e</sub> x) $$-$$ sin(log<sub>e</sub> x)] + C"}]
|
["C"]
| null |
$${\rm I} = \int {\cos \left( {\ell nx} \right)} dx$$
<br><br>$${\rm I} = \cos (\ln x).x + \int {\sin \left( {\ell nx} \right)dx} $$
<br><br>$${\rm I} = \cos \left( {\ell nx} \right)x + \left[ {\sin \left( {\ell nx} \right).x - \int {\cos \left( {\ell nx} \right)dx} } \right]$$
<br><br>$${\rm I} = {x \over 2}\left[ {\sin \left( {\ell nx} \right) + \cos \left( {\ell nx} \right)} \right] + C$$
|
mcq
|
jee-main-2019-online-12th-january-morning-slot
|
1l6re7z1v
|
maths
|
indefinite-integrals
|
integration-by-parts
|
<p>For $$I(x)=\int \frac{\sec ^{2} x-2022}{\sin ^{2022} x} d x$$, if $$I\left(\frac{\pi}{4}\right)=2^{1011}$$, then</p>
|
[{"identifier": "A", "content": "$$3^{1010} I\\left(\\frac{\\pi}{3}\\right)-I\\left(\\frac{\\pi}{6}\\right)=0$$"}, {"identifier": "B", "content": "$$3^{1010} I\\left(\\frac{\\pi}{6}\\right)-I\\left(\\frac{\\pi}{3}\\right)=0$$"}, {"identifier": "C", "content": "$$3^{1011} I\\left(\\frac{\\pi}{3}\\right)-I\\left(\\frac{\\pi}{6}\\right)=0$$"}, {"identifier": "D", "content": "$$3^{1011} I\\left(\\frac{\\pi}{6}\\right)-I\\left(\\frac{\\pi}{3}\\right)=0$$"}]
|
["A"]
| null |
<p>Given,</p>
<p>$$I(x) = \int {{{{{\sec }^2}x - 2022} \over {{{\sin }^{2022}}x}}dx} $$</p>
<p>$$ = \int {{{{{\sec }^2}x} \over {{{\sin }^{2022}}x}}dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} } $$</p>
<p>$$ = \int {{1 \over {{{\sin }^{2022}}x}}\,.\,{{\sec }^2}x\,dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} } $$</p>
<p>$$ = {1 \over {{{\sin }^{2022}}x}}\,.\,\tan x - \int {\left( {{{ - 2022} \over {{{\sin }^{2023}}x}}\,.\,\cos x\,.\,\tan x} \right)dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx + C} } $$</p>
<p>$$ = {{\tan x} \over {{{\sin }^{2022}}x}} + \int {\left( {{{2022} \over {{{\sin }^{2023}}x}}\,.\,\cos x\,.\,{{\sin x} \over {\cos x}}} \right)dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx + C} } $$</p>
<p>$$ = {{\tan x} \over {{{\sin }^{2022}}x}} + \int {{{2022} \over {{{\sin }^{2022}}x}}dx - \int {{{2022} \over {{{\sin }^{2022}}x}}dx} } $$</p>
<p>$$ = {{\tan x} \over {{{\sin }^{2022}}x}} + C$$</p>
<p>Given, $$I\left( {{\pi \over 4}} \right) = {2^{1011}}$$</p>
<p>$$\therefore$$ $$I\left( {{\pi \over 4}} \right) = {{\tan \left( {{\pi \over 4}} \right)} \over {{{\left( {\sin {\pi \over 4}} \right)}^{2022}}}} + C$$</p>
<p>$$ \Rightarrow {2^{1011}} = {1 \over {{{\left( {{1 \over {\sqrt 2 }}} \right)}^{2022}}}} + C$$</p>
<p>$$ \Rightarrow C = {2^{1011}} - {2^{1011}} = 0$$</p>
<p>$$\therefore$$ $$I(x) = {{\tan x} \over {{{\sin }^{2022}}x}}$$</p>
<p>$$\therefore$$ $$I\left( {{\pi \over 3}} \right) = {{\tan {\pi \over 3}} \over {{{\left( {\sin {\pi \over 3}} \right)}^{2022}}}} = {{\sqrt 3 } \over {{{\left( {{{\sqrt 3 } \over 2}} \right)}^{2022}}}}$$</p>
<p>$$I\left( {{\pi \over 6}} \right) = {{{1 \over {\sqrt 3 }}} \over {{{\left( {{1 \over 2}} \right)}^{2022}}}} = {1 \over {\sqrt 3 }} \times {(2)^{2022}}$$</p>
<p>From option (A),</p>
<p>$${3^{1010}}\,.\,I\left( {{\pi \over 3}} \right) - I\left( {{\pi \over 6}} \right)$$</p>
<p>$$ = {3^{1010}}\,.\,\sqrt 3 \,.\,{\left( {{2 \over {\sqrt 3 }}} \right)^{2022}} - {{{{(2)}^{2022}}} \over {\sqrt 3 }}$$</p>
<p>$$ = {3^{1010}}\,.\,\sqrt 3 \times {{{2^{2022}}} \over {{3^{1011}}}} - {{{2^{2022}}} \over {\sqrt 3 }}$$</p>
<p>$$ = {{{2^{2022}}} \over {\sqrt 3 }} - {{{2^{2022}}} \over {\sqrt 3 }} = 0$$</p>
|
mcq
|
jee-main-2022-online-29th-july-evening-shift
|
1lgxgz9mo
|
maths
|
indefinite-integrals
|
integration-by-parts
|
<p>If $$I(x) = \int {{e^{{{\sin }^2}x}}(\cos x\sin 2x - \sin x)dx} $$ and $$I(0) = 1$$, then $$I\left( {{\pi \over 3}} \right)$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$ - {e^{{3 \\over 4}}}$$"}, {"identifier": "B", "content": "$$ - {1 \\over 2}{e^{{3 \\over 4}}}$$"}, {"identifier": "C", "content": "$${e^{{3 \\over 4}}}$$"}, {"identifier": "D", "content": "$${1 \\over 2}{e^{{3 \\over 4}}}$$"}]
|
["D"]
| null |
$$
\begin{aligned}
& \text { Given, } I(x)=\int e^{\sin ^2 x}(\cos x \sin 2 x-\sin x) d x \\\\
& =\int e^{\sin ^2 x} \cdot \cos x \cdot \sin 2 x d x-\int \sin x e^{\sin ^2 x} d x \\\\
& =\int \frac{\cos x}{\mathrm{I}} \cdot \frac{e^{\sin ^2 x} \cdot \sin 2 x}{\mathrm{II}} d x-\int \sin x \cdot e^{\sin ^2 x} d x \\\\
& =\cos x \cdot e^{\sin ^2 x}-\int(-\sin x) e^{\sin ^2 x} d x-\int \sin x e^{\sin ^2 x} d x \\\\
& =\cos x \cdot e^{\sin ^2 x}+\int \sin x e^{\sin ^2 x} \cdot d x-\int \sin x \cdot e^{\sin ^2 x} d x+C
\end{aligned}
$$
<br/><br/>$I(x)=e^{\sin ^2 x} \cdot \cos x+C$
<br/><br/>Given, $I(0)=1 \Rightarrow C=0$
<br/><br/>So, $ I(x)=e^{\sin ^2 x} \cdot \cos x$
<br/><br/>$$
\Rightarrow I(\pi / 3)=e^{3 / 4} \cdot \frac{1}{2}=\frac{e^{3 / 4}}{2}
$$
|
mcq
|
jee-main-2023-online-10th-april-morning-shift
|
1lh21fmey
|
maths
|
indefinite-integrals
|
integration-by-parts
|
<p>Let $$I(x)=\int \frac{x^{2}\left(x \sec ^{2} x+\tan x\right)}{(x \tan x+1)^{2}} d x$$. If $$I(0)=0$$, then $$I\left(\frac{\pi}{4}\right)$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$\\log _{e} \\frac{(\\pi+4)^{2}}{32}-\\frac{\\pi^{2}}{4(\\pi+4)}$$"}, {"identifier": "B", "content": "$$\\log _{e} \\frac{(\\pi+4)^{2}}{16}-\\frac{\\pi^{2}}{4(\\pi+4)}$$"}, {"identifier": "C", "content": "$$\\log _{e} \\frac{(\\pi+4)^{2}}{16}+\\frac{\\pi^{2}}{4(\\pi+4)}$$"}, {"identifier": "D", "content": "$$\\log _{e} \\frac{(\\pi+4)^{2}}{32}+\\frac{\\pi^{2}}{4(\\pi+4)}$$"}]
|
["A"]
| null |
We have,
<br/><br/>$$
\begin{aligned}
I(x)= & \int \frac{x^2\left(x \sec ^2 x+\tan x\right)}{(x \tan x+1)^2} d x \\\\
= & x^2 \int \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2} d x \\\\
& \quad-\int\left\{\frac{d}{d x}\left(x^2\right) \int \frac{x \sec ^2 x+\tan x}{(x \tan x+1)^2} d x\right\} d x \text { (integration by parts) }
\end{aligned}
$$
<br/><br/>$$
=x^2\left(\frac{-1}{x \tan x+1}\right)+\int \frac{2 x}{x \tan x+1} d x
$$
<br/><br/>Now, let
<br/><br/>$$
\begin{aligned}
I_1 & =2 \int \frac{x}{x \tan x+1} d x \\\\
& =2 \int \frac{x \cos x}{x \sin x+\cos x} d x
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \text { On putting } x \sin x+\cos x=t \\\\
& \Rightarrow (x \cos x+\sin x-\sin x) d x=d t \\\\
& \Rightarrow x \cos x d x=d t
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
&\therefore I_1 =2 \int \frac{d t}{t}=2 \log t+c \\\\
& =2 \log (x \sin x+\cos x)+c
\end{aligned}
$$
<br/><br/>$$
\begin{gathered}
\therefore I(x)=\frac{-x^2}{x \tan x+1}+2 \log(x \sin x+\cos x)+c
\end{gathered}
$$
<br/><br/>When, $x=0$, then
<br/><br/>$$
\begin{array}{ll}
& I(0)=0+2 \log (1)+c=0 \\\\
&\Rightarrow c=0 \\\\
&\therefore I(x)=\frac{-x^2}{x \tan x+1}+2 \log (x \sin x+\cos x)
\end{array}
$$
<br/><br/>$$
\begin{aligned}
&\therefore I(x) =\frac{-x^2}{x \tan x+1}+2 \log (x \sin x+\cos x) \\\\
&\Rightarrow I\left(\frac{\pi}{4}\right) =\frac{-\frac{\pi^2}{16}}{\frac{\pi}{4}+1}+2 \log \left(\frac{\pi}{4} \times \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\right) \\\\
& =\log \left(\frac{(\pi+4)^2}{32}\right)-\frac{\pi^2}{4(\pi+4)}
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-6th-april-morning-shift
|
lv2eqxr8
|
maths
|
indefinite-integrals
|
integration-by-parts
|
<p>If $$\int \operatorname{cosec}^5 x d x=\alpha \cot x \operatorname{cosec} x\left(\operatorname{cosec}^2 x+\frac{3}{2}\right)+\beta \log _x\left|\tan \frac{x}{2}\right|+\mathrm{C}$$
where $$\alpha, \beta \in \mathbb{R}$$ and $$\mathrm{C}$$ is the constant of integration, then the value of $$8(\alpha+\beta)$$ equals _________.</p>
|
[]
| null |
1
|
<p>$$\begin{aligned}
& I=\int(\operatorname{cosec} x)^5 d x=\int(\operatorname{cosec} x)^3(\operatorname{cosec} x)^2 d x \\
& =(\operatorname{cosec} x)^3 \int \operatorname{cosec}^2 x d x- \\
& \int\left(\frac{d}{d x}(\operatorname{cosec} x)^3 \int \operatorname{cosec}^2 x d x\right) d x
\end{aligned}$$</p>
<p>$$\begin{aligned}
& =-\cot x(\operatorname{cosec} x)^3-\int 3 \operatorname{cosec}^2 x \cdot(-\operatorname{cosec} x \cot x)(-\cot x) d x \\
& =-\cot x(\operatorname{cosec} x)^3-\int 3 \operatorname{cosec}^3 x \cot ^2 x d x \\
& =-\cot x(\operatorname{cosec} x)^3-3 \int(\operatorname{cosec} x)^3\left(\operatorname{cosec}^2 x-1\right) d x \\
& I=-\cot x(\operatorname{cosec} x)^3-3 I+3 \int(\operatorname{cosec} x)^3 d x \\
& 4 I=-\cot x(\operatorname{cosec} x)^3+3 \int(\operatorname{cosec} x)^3 d x \\
& =-\cot x(\operatorname{cosec} x)^3+3 I_1 \\
& I_1=\int \operatorname{cosec} x \cdot \operatorname{cosec}^2 x d x=\operatorname{cosec} x(-\cot x)- \\
& \qquad \int(-\operatorname{cosec} x \cot x)(-\cot x) d x
\end{aligned}$$</p>
<p>$$\begin{aligned}
& I_1=-\operatorname{cosec} x \cot x-\int \operatorname{cosec} x\left(\operatorname{cosec}^2 x-1\right) d x \\
& I_1=-\operatorname{cosec} x \cot x-I+\int \operatorname{cosec} x d x \\
& 2 l_1=-\operatorname{cosec} x \cot x+\ln \left|\tan \frac{x}{2}\right| \\
& \Rightarrow \quad 4 l=-\cot x(\operatorname{cosec} x)^3-\frac{3}{2} \operatorname{cosec} x \cot x +\frac{3}{2} \ln \left|\tan \frac{x}{2}\right|+C
\end{aligned}$$</p>
<p>$$\begin{aligned}
& I=-\frac{1}{4} \operatorname{cosec} x \cot x\left(\operatorname{cosec}^2 x+\frac{3}{2}\right)+\frac{3}{8} \ln \left|\tan \frac{x}{2}\right|+C \\
& \Rightarrow \alpha=-\frac{1}{4}, \beta=\frac{3}{8} \Rightarrow 8(\alpha+\beta)=1
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-4th-april-evening-shift
|
hKe3lNduHiApUmZ8
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
$$\int {{{\left\{ {{{\left( {\log x - 1} \right)} \over {1 + {{\left( {\log x} \right)}^2}}}} \right\}}^2}\,\,dx} $$ is equal to
|
[{"identifier": "A", "content": "$${{\\log x} \\over {{{\\left( {\\log x} \\right)}^2} + 1}} + C$$ "}, {"identifier": "B", "content": "$${x \\over {{x^2} + 1}} + C$$ "}, {"identifier": "C", "content": "$${{x{e^x}} \\over {1 + {x^2}}} + C$$ "}, {"identifier": "D", "content": "$${x \\over {{{\\left( {\\log x} \\right)}^2} + 1}} + C$$ "}]
|
["D"]
| null |
$$\int {{{{{\left( {\log x - 1} \right)}^2}} \over {{{\left( {1 + {{\left( {\log x} \right)}^2}} \right)}^2}}}} dx$$
<br><br>$$ = \int {{{1 + {{\left( {\log x} \right)}^2} - 2\log x} \over {{{\left[ {1 + {{\left( {\log x} \right)}^2}} \right]}^2}}}} $$
<br><br>$$ = \int {\left[ {{1 \over {\left( {1 + {{\left( {\log x} \right)}^2}} \right)}} - {{2\log x} \over {{{\left( {1 + {{\left( {\log x} \right)}^2}} \right)}^2}}}} \right]} dx$$
<br><br>$$ = \int {\left[ {{{{e^t}} \over {1 + {t^2}}} - {{2t\,{e^t}} \over {{{\left( {1 + {t^2}} \right)}^2}}}} \right]} dt$$
<br><br>put $$\log x = t \Rightarrow dx = {e^t}\,dt$$
<br><br>$$ = \int {{e^t}} \left[ {{1 \over {1 + {t^2}}} - {{2t} \over {{{\left( {1 + {t^2}} \right)}^2}}}} \right]dt$$
<br><br>$$\left[ \, \right.$$ which is of the form
<br><br>$$\left. {\int {{e^x}\left( {f\left( x \right) + f'\left( x \right)dx} \right)} } \right]$$
<br><br>$$ = {{{e^t}} \over {1 + {t^2}}} + c = {x \over {1 + {{\left( {\log x} \right)}^2}}} + c$$
|
mcq
|
aieee-2005
|
zX0bJHNzNUnkDwG7
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
The value of $$\sqrt 2 \int {{{\sin xdx} \over {\sin \left( {x - {\pi \over 4}} \right)}}} $$ is
|
[{"identifier": "A", "content": "$$\\,x + \\log \\,\\left| {\\,\\cos \\left( {x - {\\pi \\over 4}} \\right)\\,} \\right| + c$$"}, {"identifier": "B", "content": "$$\\,x - \\log \\,\\left| {\\,\\sin \\left( {x - {\\pi \\over 4}} \\right)\\,} \\right| + c$$"}, {"identifier": "C", "content": "$$\\,x + \\log \\,\\left| {\\,\\sin \\left( {x - {\\pi \\over 4}} \\right)\\,} \\right| + c$$ "}, {"identifier": "D", "content": "$$\\,x - \\log \\,\\left| {\\,\\cos \\left( {x - {\\pi \\over 4}} \\right)\\,} \\right| + c$$"}]
|
["C"]
| null |
Let $$I = \sqrt 2 \int {{{\sin \,xdx} \over {\sin \left( {x - {\pi \over 4}} \right)}}} $$
<br><br>Put $$x - {\pi \over 4} = t$$
<br><br>$$ \Rightarrow dx = dt$$
<br><br>$$ \Rightarrow I = \sqrt 2 \int {{{\sin \left( {t + {\pi \over 4}} \right)} \over {\sin \,t}}} dt$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\, = {{\sqrt 2 } \over {\sqrt 2 }}\int {\left( {{{\sin t + \cos t} \over {\sin t}}} \right)} \,\,dt$$
<br><br>$$ \Rightarrow I = \int {\left( {1 + \cot \,t} \right)} dt$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\, = t + \log \left| {\sin t} \right| + {c_1}$$
<br><br>$$ = x - {\pi \over 4} + \log \left| {\sin \left( {x - {\pi \over 4}} \right)} \right| + {c_1}$$
<br><br>$$ = x + \log \left| {\sin \left( {x - {\pi \over 4}} \right)} \right| + c$$
<br><br>$$\left( \, \right.$$ where $${c = {c_1} - {\pi \over 4}}$$ $$\left. \, \right)$$
|
mcq
|
aieee-2008
|
2ACvJSdqk8b5slgg
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If the $$\int {{{5\tan x} \over {\tan x - 2}}dx = x + a\,\ln \,\left| {\sin x - 2\cos x} \right| + k,} $$ then $$a$$ is
<br/>equal to :
|
[{"identifier": "A", "content": "$$-1$$ "}, {"identifier": "B", "content": "$$-2$$ "}, {"identifier": "C", "content": "$$1$$ "}, {"identifier": "D", "content": "$$2$$ "}]
|
["D"]
| null |
$$\int {{{5\tan x} \over {\tan x - 2}}} dx$$
<br><br>$$ = \int {{{5{{\sin x} \over {\cos x}}} \over {{{\sin x} \over {\cos x}} - 2}}} \,dx$$
<br><br>$$ = \int {\left( {{{5\sin x} \over {\cos x}} \times {{\cos x} \over {\sin x - 2\cos x}}} \right)} \,dx$$
<br><br>$$ = \int {{{5\,\sin \,x\,dx} \over {\sin x - 2\,\cos x}}} $$
<br><br>$$ = \int {\left( {{{4\sin x + \sin x + 2\cos x - 2\cos x} \over {\sin x - 2\cos x}}} \right)} \,dx$$
<br><br>$$ = \int {{{\left( {\sin x - 2\cos x} \right) + \left( {4\sin x + 2\cos x} \right)} \over {\sin x - 2\cos x}}} \,dx$$
<br><br>$$ = \int {{{\left( {\sin x - 2\cos x} \right) + 2\left( {\cos x + 2\sin x} \right)} \over {\left( {\sin x - 2\cos x} \right)}}} \,dx$$
<br><br>$$ = \int {{{\sin x - 2\cos x} \over {\sin x - 2\cos x}}dx + 2\int {\left( {{{\cos x + 2\sin x} \over {\sin x - 2\cos x}}} \right)} } \,dx$$
<br><br>$$ = \int {dx + 2\int {{{\cos x + 2\sin x} \over {\sin x - 2\cos \,x}}} } \,dx$$
<br><br>$$ = {I_1} + {I_2}$$ where $${I_1} = \int {dx} $$ and
<br><br>$${I_2} = 2\int {{{\cos x + 2\sin x} \over {\sin x - 2\cos x}}\,dx} $$
<br><br>Put $$\sin x - 2\cos x = t$$
<br><br>$$ \Rightarrow \left( {\cos x + 2\sin x} \right)dx = dt$$
<br><br>$$\therefore$$ $${I_2} = 2\int {{{dt} \over t}} = 2\ln \,t + C$$
<br><br>$$ = 2\,\ln \left( {\sin \,x - 2\cos \,x} \right) + C$$
<br><br>Hence, $${I_1} + {I_2}$$
<br><br>$$ = \int {dx + 2\ln \left( {\sin x - 2\cos x} \right) + c} $$
<br><br>$$ = x + 2\ln \left| {\left( {\sin x - 2\cos x} \right)} \right| + k$$
<br><br>$$ \Rightarrow a = 2$$
|
mcq
|
aieee-2012
|
7Kbr9AwkQApiexVy
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$\int {f\left( x \right)dx = \psi \left( x \right),} $$ then $$\int {{x^5}f\left( {{x^3}} \right)dx} $$ is equal to
|
[{"identifier": "A", "content": "$${1 \\over 3}\\left[ {{x^3}\\psi \\left( {{x^3}} \\right) - \\int {{x^2}\\psi \\left( {{x^3}} \\right)dx} } \\right] + C$$ "}, {"identifier": "B", "content": "$${1 \\over 3}{x^3}\\psi \\left( {{x^3}} \\right) - 3\\int {{x^3}\\psi \\left( {{x^3}} \\right)dx} + C$$ "}, {"identifier": "C", "content": "$${1 \\over 3}{x^3}\\psi \\left( {{x^3}} \\right) - \\int {{x^2}\\psi \\left( {{x^3}} \\right)dx} + C$$ "}, {"identifier": "D", "content": "$${1 \\over 3}\\left[ {{x^3}\\psi \\left( {{x^3}} \\right) - \\int {{x^3}\\psi \\left( {{x^3}} \\right)dx} } \\right] + C$$ "}]
|
["C"]
| null |
Let $$\int {f\left( x \right)dx = \psi \left( x \right)} $$
<br><br>Let $$I = \int {{x^5}} f\left( {{x^3}} \right)dx$$
<br><br>put $${x^3} = t \Rightarrow 3{x^2}dx = dt$$
<br><br>$$I = {1 \over 3}\int {3.{x^2}} .{x^3}.f\left( {{x^3}} \right).dx$$
<br><br>$$ = {1 \over 3}\int {tf} \left( t \right)dt$$
<br><br>$$ = {1 \over 3}\left[ {t\int {f\left( t \right)dt - \int {f\left( t \right)dt} } } \right]$$
<br><br>$$ = {1 \over 3}\left[ {t\psi \left( t \right) - \int {\psi \left( t \right)dt} } \right]$$
<br><br>$$ = {1 \over 3}\left[ {{x^3}\psi \left( {{x^3}} \right) - 3\int {{x^2}\psi \left( {{x^3}} \right)dx} } \right] + C$$
<br><br>$$ = {1 \over 3}{x^3}\psi \left( {{x^3}} \right) - \int {{x^2}} \psi \left( {{x^3}} \right)dx + C$$
|
mcq
|
jee-main-2013-offline
|
nZ3cEdvpgAwa9myY
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
The integral $$\int {{{dx} \over {{x^2}{{\left( {{x^4} + 1} \right)}^{3/4}}}}} $$ equals :
|
[{"identifier": "A", "content": "$$ - {\\left( {{x^4} + 1} \\right)^{{1 \\over 4}}} + c$$"}, {"identifier": "B", "content": "$$ - {\\left( {{{{x^4} + 1} \\over {{x^4}}}} \\right)^{{1 \\over 4}}} + c$$ "}, {"identifier": "C", "content": "$$ {\\left( {{{{x^4} + 1} \\over {{x^4}}}} \\right)^{{1 \\over 4}}} + c$$ "}, {"identifier": "D", "content": "$$ {\\left( {{x^4} + 1} \\right)^{{1 \\over 4}}} + c$$"}]
|
["B"]
| null |
$$1 = \int {{{dx} \over {{x^2}{{\left( {{x^4} + 1} \right)}^{3/4}}}}} $$
<br><br>$$ = \int {{{dx} \over {{x^3}{{\left( {1 + {x^{ - 4}}} \right)}^{3/4}}}}} $$
<br><br>Let $${x^{ - 4}} = y$$
<br><br>$$ \Rightarrow - 4{x^{ - 3}}\,dx = dy$$
<br><br>$$ \Rightarrow dx = {{ - 1} \over 4}{x^3}dy$$
<br><br>$$\therefore$$ $$I = {{ - 1} \over 4}\int {{{{x^3}dy} \over {{x^3}{{\left( {1 + y} \right)}^{3/4}}}}} $$
<br><br>$$ = {{ - 1} \over 4}\int {{{dy} \over {{{\left( {1 + y} \right)}^{3/4}}}}} $$
<br><br>$$ = {{ - 1} \over 4} \times 4{\left( {1 + y} \right)^{1/4}}$$
<br><br>$$ = - 1{\left( {1 + {x^{ - 4}}} \right)^{1/4}} + C$$
<br><br>$$ = - {\left( {{{{x^4} + 1} \over {{x^4}}}} \right)^{1/4}} + C$$
|
mcq
|
jee-main-2015-offline
|
kUOFTa2sNbMr1KkF
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
The integral $$\int {{{2{x^{12}} + 5{x^9}} \over {{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx$$ is equal to :
|
[{"identifier": "A", "content": "$${{{x^5}} \\over {2{{\\left( {{x^5} + {x^3} + 1} \\right)}^2}}} + C$$ "}, {"identifier": "B", "content": "$${{ - {x^{10}}} \\over {2{{\\left( {{x^5} + {x^3} + 1} \\right)}^2}}} + C$$ "}, {"identifier": "C", "content": "$${{{-x^5}} \\over {{{\\left( {{x^5} + {x^3} + 1} \\right)}^2}}} + C$$ "}, {"identifier": "D", "content": "$${{ {x^{10}}} \\over {2{{\\left( {{x^5} + {x^3} + 1} \\right)}^2}}} + C$$ "}]
|
["D"]
| null |
$$\int {{{2{x^{12}} + 5{x^9}} \over {{{\left( {{x^5} + {x^3} + 1} \right)}^3}}}} dx$$
<br><br>Dividing by $${x^{15}}$$ in numerator and denominator
<br><br>$$\int {{{{2 \over {{x^3}}} + {5 \over {{x^6}}}dx} \over {{{\left( {1 + {1 \over {{x^2}}} + {1 \over 5}} \right)}^3}}}} $$
<br><br>Substitute $$1 + {1 \over {{x^2}}} + {1 \over {{x^5}}} = t$$
<br><br>$$ \Rightarrow \left( {{{ - 2} \over {{x^3}}} - {5 \over {{x^6}}}} \right)dx = dt$$
<br><br>$$ \Rightarrow \left( {{2 \over {{x^3}}} + {5 \over {{x^6}}}} \right)dx = - dt$$
<br><br>This gives,
<br><br>$$\int {{{{2 \over {{x^3}}} + {5 \over {{x^6}}}dx} \over {{{\left( {1 + {1 \over {{x^2}}} + {1 \over {{x^5}}}} \right)}^3}}}} $$
<br><br>$$ = \int {{{ - dt} \over {{t^3}}}} = {1 \over {2{t^2}}} + C$$
<br><br>$$ = {1 \over {2{{\left( {1 + {1 \over {{x^2}}} + {1 \over {{x^5}}}} \right)}^2}}} + C$$
<br><br>$$ = {{{x^{10}}} \over {2{{\left( {{x^5} + {x^3} + 1} \right)}^2}}} + C$$
|
mcq
|
jee-main-2016-offline
|
LQe1HRHuBYA8dRTJJfiHH
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
The integral $$\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}} $$ is equal to :
<br/><br/>(where C is a constant of integration.)
|
[{"identifier": "A", "content": "$$ - 2\\sqrt {{{1 + \\sqrt x } \\over {1 - \\sqrt x }}} + C$$ "}, {"identifier": "B", "content": "$$ - 2\\sqrt {{{1 - \\sqrt x } \\over {1 + \\sqrt x }}} + C$$"}, {"identifier": "C", "content": "$$ - \\sqrt {{{1 - \\sqrt x } \\over {1 + \\sqrt x }}} + C$$ "}, {"identifier": "D", "content": "$$2\\sqrt {{{1 + \\sqrt x } \\over {1 - \\sqrt x }}} + C$$ "}]
|
["B"]
| null |
I = $$\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt {x - {x^2}} }}} $$
<br><br>= $$\int {{{dx} \over {\left( {1 + \sqrt x } \right)\sqrt x \sqrt {1 - x} }}} $$
<br><br>Let 1 + $$\sqrt x $$ = t
<br><br>$$ \Rightarrow $$$$\,\,\,$$$${1 \over {2\sqrt x }}\,dx$$ = dt
<br><br>I = $$\int {{{2dt} \over {t\sqrt {2t - {t^2}} }}} $$
<br><br>Again let t = $${1 \over z}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ dt = $$-$$ $${1 \over {{z^2}}}dz$$
<br><br>$$\therefore\,\,\,$$ I = 2 $$\int {{{ - {1 \over {{z^2}}}dz} \over {{1 \over z}\sqrt {{2 \over z} - {1 \over {{z^2}}}} }}} $$
<br><br>= 2$$\int {{{ - dz} \over {\sqrt {2z - 1} }}} $$
<br><br>= $$ - 2\sqrt {2z - 1} + C$$
<br><br>= $$ - 2\sqrt {{2 \over t} - 1} + C$$
<br><br>= $$- 2\sqrt {{{2 - t} \over t}} + C$$
<br><br>= $$ - 2\sqrt {{{1 - \sqrt x } \over {1 + \sqrt x }}} + C$$
|
mcq
|
jee-main-2016-online-10th-april-morning-slot
|
7Gnzjnba9fRxVw93NEa9p
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}} = {\left( {\tan x} \right)^A} + C{\left( {\tan x} \right)^B} + k,$$
<br/><br/>where k is a constant of integration, then A + B +C equals :
|
[{"identifier": "A", "content": "$${{21} \\over 5}$$ "}, {"identifier": "B", "content": "$${{16} \\over 5}$$"}, {"identifier": "C", "content": "$${{7} \\over 10}$$"}, {"identifier": "D", "content": "$${{27} \\over 10}$$"}]
|
["B"]
| null |
$$\int {{{dx} \over {{{\cos }^3}x\sqrt {2\sin 2x} }}} $$
<br><br>= $$\int {{{dx} \over {{{\cos }^3}x\sqrt {4\sin x\cos x} }}} $$
<br><br>= $$\int {{{dx} \over {2{{\cos }^4}x\sqrt {\tan x} }}} $$
<br><br>Let tan x = t<sup>2</sup>
<br><br>$$ \Rightarrow $$$$\,\,\,$$ sec<sup>2</sup>xdx = 2t dt
<br><br>as sec<sup>2</sup>x = 1 + tan<sup>2</sup>x = 1 + t<sup>4</sup>
<br><br>= $$\int {{{{{\sec }^4}x\,dx} \over {2\sqrt {\tan x} }}} $$
<br><br>= $$\int {{{{{\sec }^2}x\left( {{{\sec }^2}x\,dx} \right)} \over {2\sqrt {\tan x} }}} $$
<br><br>= $$\int {{{\left( {1 + {t^4}} \right)2t\,dt} \over {2t}}} $$
<br><br>= $$\int {\left( {1 + {t^4}} \right)} \,dt$$
<br><br>= t + $${{{t^5}} \over 5}$$ + k
<br><br>= $$\sqrt {\tan x} $$ + $${1 \over 5}$$ tan$$^{{5 \over 2}}$$x + k
<br><br>By comparing with the given equation, we get
<br><br>A = $${1 \over 2}$$, B = $${5 \over 2}$$, C = $${1 \over 5}$$
<br><br>$$\therefore\,\,\,$$ A + B + C = $${{16} \over 5}$$
|
mcq
|
jee-main-2016-online-9th-april-morning-slot
|
5rgUmlOIq0Q0L7jm
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
Let $${I_n} = \int {{{\tan }^n}x\,dx} ,\,\left( {n > 1} \right).$$
<br/><br/>If $${I_4} + {I_6}$$ = $$a{\tan ^5}x + b{x^5} + C$$, where C is a constant of integration,
<br/><br/>then the ordered pair $$\left( {a,b} \right)$$ is equal to
|
[{"identifier": "A", "content": "$$\\left( {{1 \\over 5},0} \\right)$$"}, {"identifier": "B", "content": "$$\\left( {{1 \\over 5}, - 1} \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - {1 \\over 5},0} \\right)$$"}, {"identifier": "D", "content": "$$\\left( { - {1 \\over 5},1} \\right)$$"}]
|
["A"]
| null |
Given,
<br><br>In = $$\int {{{\tan }^n}x\,dx,\,\,\,n > 1} $$
<br><br>$$\therefore\,\,\,$$ I<sub>4</sub> = $$\int {{{\tan }^4}x\,dx} $$
<br><br>and I<sub>6</sub> = $$\int {{{\tan }^6}} x\,dx$$
<br><br>$$\therefore\,\,\,$$ I = I<sub>4</sub> + I<sub>6</sub>
<br><br>= $$\int {\left( {{{\tan }^4}x + {{\tan }^6}x} \right)} dx$$
<br><br>= $$\int {{{\tan }^4}} x\left( {1 + {{\tan }^2}x} \right)dx$$
<br><br>= $$\int {{{\tan }^4}} x.{\sec ^2}x\,dx$$
<br><br>Let, tanx = t
<br><br>$$ \Rightarrow $$$$\,\,\,$$ sec<sup>2</sup>x dx = dt
<br><br>$$\therefore\,\,\,$$ I = $$\int {{t^4}\,dt} $$
<br><br>= $${1 \over 5}$$ t<sup>5</sup> + C
<br><br>= $${1 \over 5}$$ tan<sup>5</sup>x + C
<br><br>$$\therefore\,\,\,$$ By comparing with the question, we get
<br><br>A = $${1 \over 5}$$, B = 0
|
mcq
|
jee-main-2017-offline
|
4FTEIQa2Co5vKRg6Ns20r
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}dx = x - {K \over {\sqrt A }}{{\tan }^{ - 1}}} $$ $$\left( {{{K\,\tan x + 1} \over {\sqrt A }}} \right) + C,(C\,\,$$ is a constant of integration) then the ordered pair (K, A) is equal to :
|
[{"identifier": "A", "content": "(2, 1)"}, {"identifier": "B", "content": "($$-$$2, 3)"}, {"identifier": "C", "content": "(2, 3)"}, {"identifier": "D", "content": "($$-$$2, 1)"}]
|
["C"]
| null |
Given,
<br><br>$$\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}} \,\,dx$$
<br><br>Let tanx = t
<br><br>$$ \Rightarrow $$$$\,\,\,$$ sec<sup>2</sup>x dx = dt
<br><br>$$ \Rightarrow $$$$\,\,\,$$ dx = $${{dt} \over {{{\sec }^2}x}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ $$dx = {{dt} \over {1 + {{\tan }^2}x}}$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ dx = $${{dt} \over {1 + {t^2}}}$$
<br><br>$$\therefore\,\,\,$$ $$\int {{{\tan x} \over {1 + \tan x + {{\tan }^2}x}}} \,\,dx$$
<br><br>= $$\int {{t \over {1 + t + {t^2}}}} \times {{dt} \over {1 + {t^2}}}$$
<br><br>= $$\int {{t \over {\left( {1 + t + {t^2}} \right)\left( {1 + {t^2}} \right)}}} \,\,dt$$
<br><br>= $$\int {\left( {{1 \over {1 + {t^2}}} - {1 \over {1 + t + {t^2}}}} \right)} \,\,dt$$
<br><br>= $${\tan ^{ - 1}}\left( t \right) - \int {{1 \over {1 + t + {t^2} + {1 \over 4} - {1 \over 4}}}} \,\,dt$$
<br><br>= $$x - \int {{1 \over {{t^2} + t + {1 \over 4} + 1 - {1 \over 4}}}} \,\,dt$$
<br><br>= $$x - \int {{1 \over {{{\left( {t + {1 \over 2}} \right)}^2} + {{\left( {{{\sqrt 3 } \over 2}} \right)}^2}}}} $$
<br><br>= $$x - {1 \over {{{\sqrt 3 } \over 2}}} + {\tan ^{ - 1}}\left( {{{2t + 1} \over {\sqrt 3 }}} \right) + c$$
<br><br>= $$x - {2 \over {\sqrt 3 }}\,\,{\tan ^{ - 1}}\left( {{{2\tan x + 1} \over {\sqrt 3 }}} \right) + c$$
<br><br>$$\therefore\,\,\,$$ By comparing
<br><br>A = 3 and K = 2
|
mcq
|
jee-main-2018-online-16th-april-morning-slot
|
sWW8X3j3VBEBvcDd
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
The integral
<br/><br/>$$\int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}} dx$$
<br/><br/>is equal to
|
[{"identifier": "A", "content": "$${{ - 1} \\over {1 + {{\\cot }^3}x}} + C$$"}, {"identifier": "B", "content": "$${1 \\over {3\\left( {1 + {{\\tan }^3}x} \\right)}} + C$$"}, {"identifier": "C", "content": "$${{ - 1} \\over {3\\left( {1 + {{\\tan }^3}x} \\right)}} + C$$"}, {"identifier": "D", "content": "$${1 \\over {1 + {{\\cot }^3}x}} + C$$ "}]
|
["C"]
| null |
Given,
<br><br>$$\int {{{{{\sin }^2}x{{\cos }^2}x} \over {{{\left( {{{\sin }^5}x + {{\cos }^3}x{{\sin }^2}x + {{\sin }^3}x{{\cos }^2}x + {{\cos }^5}x} \right)}^2}}}}\,dx $$
<br><br>$$ = \int {{{{{\sin }^2}x\,{{\cos }^2}x} \over {{{\left[ {{{\sin }^3}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right) + {{\cos }^3}x\left( {{{\sin }^2}x + {{\cos }^2}x} \right)} \right]}^2}}}} \,dx$$
<br><br>$$ = \int {{{{{\sin }^2}x\,{{\cos }^2}x} \over {{{\left( {{{\sin }^3}x + {{\cos }^3}x} \right)}^2}}}} dx$$
<br><br>$$ = \int {{{{{\sin }^2}x\,\,{{\cos }^2}x} \over {{{\cos }^6}x{{\left( {1 + {{\tan }^3}x} \right)}^2}}}} \,\,dx$$
<br><br>$$ = \int {{{{{\sin }^2}x} \over {{{\cos }^4}x{{\left( {1 + {{\tan }^3}} \right)}^2}}}} \,\,dx$$
<br><br>$$ = \int {{{{{\tan }^2}x.{{\sec }^2}x} \over {{{\left( {1 + {{\tan }^3}x} \right)}^2}}}} \,dx$$
<br><br>[ Let $$\,\,\,\,\,\,{\tan ^3}x = t$$
<br><br>$$\,\,\,\,\,\,$$ $$ \Rightarrow 3{\tan ^2}x\,{\sec ^2}x\,\,dx = dt$$ ]
<br><br>$$ = {1 \over 3}\int {{{3{{\tan }^2}x\,\,{{\sec }^2}x\,\,dx} \over {{{\left( {1 + {{\tan }^3}x} \right)}^2}}}} $$
<br><br>$$ = {1 \over 3}\int {{{dt} \over {{{\left( {1 + t} \right)}^2}}}} $$
<br><br>$$ = {1 \over 3} \times {{{{\left( {1 + t} \right)}^{ - 2 + 1}}} \over { - 2 + 1}} + c$$
<br><br>$$ = {1 \over 3} \times {{ - 1} \over {\left( {1 + t} \right)}} + c$$
<br><br>$$ = {{ - 1} \over {3\left( {1 + {{\tan }^3}x} \right)}} + c$$
|
mcq
|
jee-main-2018-offline
|
Qj3TsBFFhWo0Z64Jt54cd
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
Let n $$ \ge $$ 2 be a natural number and $$0 < \theta < {\pi \over 2}.$$ Then $$\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta $$ is equal to - (where C is a constant of integration)
|
[{"identifier": "A", "content": "$${n \\over {{n^2} - 1}}{\\left( {1 + {1 \\over {{{\\sin }^{n - 1}}\\theta }}} \\right)^{{{n + 1} \\over n}}} + C$$"}, {"identifier": "B", "content": "$${n \\over {{n^2} - 1}}{\\left( {1 - {1 \\over {{{\\sin }^{n + 1}}\\theta }}} \\right)^{{{n + 1} \\over n}}} + C$$"}, {"identifier": "C", "content": "$${n \\over {{n^2} - 1}}{\\left( {1 - {1 \\over {{{\\sin }^{n - 1}}\\theta }}} \\right)^{{{n + 1} \\over n}}} + C$$"}, {"identifier": "D", "content": "$${n \\over {{n^2} + 1}}{\\left( {1 - {1 \\over {{{\\sin }^{n - 1}}\\theta }}} \\right)^{{{n + 1} \\over n}}} + C$$"}]
|
["C"]
| null |
$$\int {{{{{\left( {{{\sin }^n}\theta - \sin \theta } \right)}^{1/n}}\cos \theta } \over {{{\sin }^{n + 1}}\theta }}} \,d\theta $$
<br><br>$$ = \int {{{\sin \theta {{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)}^{1/n}}} \over {{{\sin }^{n + 1}}\theta }}} \,d\theta $$
<br><br>Put $$1 - {1 \over {{{\sin }^{n - 1}}\theta }} = t$$
<br><br>So $${{\left( {n - 1} \right)} \over {{{\sin }^n}\theta }}\cos \theta d\theta = dt$$
<br><br>Now $${1 \over {n - 1}}\int {{{\left( t \right)}^{{1 \over n} + 1}}dt} $$
<br><br>$$ = {1 \over {\left( {n - 1} \right)}}{{{{\left( t \right)}^{{1 \over n} + 1}}} \over {{1 \over n} + 1}} + C$$
<br><br>$$ = {n \over {\left( {n^2 - 1} \right)}}{\left( {1 - {1 \over {{{\sin }^{n - 1}}\theta }}} \right)^{{1 \over n} + 1}} + C$$
|
mcq
|
jee-main-2019-online-10th-january-morning-slot
|
H8ln4uLN3lq8lQeMmZ3rsa0w2w9jxacphio
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
Let $$a \in \left( {0,{\pi \over 2}} \right)$$ be fixed. If the integral
<br/><br>$$\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx$$ = A(x) cos 2$$\alpha $$ + B(x) sin 2$$\alpha $$ + C, where C is a
<br/><br/>constant of integration, then the functions A(x) and B(x) are respectively : </br>
|
[{"identifier": "A", "content": "$$x - \\alpha $$ and $${\\log _e}\\left| {\\cos \\left( {x - \\alpha } \\right)} \\right|$$"}, {"identifier": "B", "content": "$$x + \\alpha $$ and $${\\log _e}\\left| {\\sin \\left( {x - \\alpha } \\right)} \\right|$$"}, {"identifier": "C", "content": "$$x + \\alpha $$ and $${\\log _e}\\left| {\\sin \\left( {x + \\alpha } \\right)} \\right|$$"}, {"identifier": "D", "content": "$$x - \\alpha $$ and $${\\log _e}\\left| {\\sin \\left( {x - \\alpha } \\right)} \\right|$$"}]
|
["D"]
| null |
<p>To solve the given integral, first we simplify the expression in the integral as follows :</p>
<p>$$\int {{{\tan x + \tan \alpha } \over {\tan x - \tan \alpha }}} dx = \int {{{\sin x \over \cos x} + {\sin \alpha \over \cos \alpha }} \over {{\sin x \over \cos x} - {\sin \alpha \over \cos \alpha }}} dx$$ </p>
<p>Simplifying further, this becomes :</p>
<p>$$\int {{{\sin x\cos \alpha + \sin \alpha \cos x} \over {\sin x\cos \alpha - \sin \alpha \cos x}}} dx$$ </p>
<p>By using the formula for sin(a + b) = sin a cos b + cos a sin b and sin(a - b) = sin a cos b - cos a sin b, we get :</p>
<p>$$\int {{{\sin(x + \alpha)}} \over {\sin(x - \alpha)}} dx$$ </p>
<p>Now, let's use the substitution method. Let t = x - α, therefore x = y + α and dx = dt. Substituting these values into the integral, we get :</p>
<p>= $$\int {{{\sin(t + 2\alpha)}} \over {\sin y}} dy$$ </p>
$$
\begin{aligned}
& =\int \frac{\sin t \cos 2 \alpha+\sin 2 \alpha \cos t}{\sin t} d t \\\\
& =\int\left(\cos 2 \alpha+\sin 2 \alpha \frac{\cos t}{\sin t}\right) d t
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& =t(\cos 2 \alpha)+(\sin 2 \alpha) \log _e|\sin t|+C \\\\
& =(x-\alpha) \cos 2 \alpha+(\sin 2 \alpha) \log _e|\sin (x-\alpha)|+C \\\\
& =A(x) \cos 2 \alpha+B(x) \sin 2 \alpha+C \text { (given) }
\end{aligned}
$$
<br/><br/>Now on comparing, we get
<br/><br/>$$
A(x)=x-\alpha \text { and } B(x)=\log _e|\sin (x-\alpha)|
$$
|
mcq
|
jee-main-2019-online-12th-april-evening-slot
|
2TA2SAVPabICEfCIXG3rsa0w2w9jx2b6x3x
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$\int {{x^5}} {e^{ - {x^2}}}dx = g\left( x \right){e^{ - {x^2}}} + c$$, where c is a constant of integration, then $$g$$(–1) is equal to :
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "- 1"}, {"identifier": "C", "content": "$$ - {5 \\over 2}$$"}, {"identifier": "D", "content": "$$ - {1 \\over 2}$$"}]
|
["C"]
| null |
Let x<sup>2</sup> = t<br><br>
$$ \Rightarrow {1 \over 2}\int {{t^2}{e^{ - t}}dt} $$<br><br>
$$ \Rightarrow {1 \over 2}\left[ { - {t^2}{e^{ - t}} + \int {2t{e^{ - t}}dt} } \right]$$<br><br>
$$ \Rightarrow {{ - {t^2}{e^{ - t}}} \over 2} - t{e^{ - t}} - {e^{ - t}}$$<br><br>
$$ \Rightarrow \left( { - {{{x^4}} \over 2} - {x^2} - 1} \right){e^{ - {x^2}}} + c$$<br><br>
Then $$g(x) = - {{{x^4}} \over 2} - {x^2} - 1$$<br><br>
$$ \Rightarrow g( - 1) = - {1 \over 2} - 1 - 1$$<br><br>
$$ \Rightarrow - {5 \over 2}$$
|
mcq
|
jee-main-2019-online-10th-april-evening-slot
|
NGp83ukLX0F1lubeJ33rsa0w2w9jwxvclk0
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$\int {{{dx} \over {{{\left( {{x^2} - 2x + 10} \right)}^2}}}} = A\left( {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {{f\left( x \right)} \over {{x^2} - 2x + 10}}} \right) + C$$
<br/><br/> where C is a constant of integration then :
|
[{"identifier": "A", "content": "A =$${1 \\over {54}}$$ and f(x) = 9(x\u20131)<sup>2</sup>"}, {"identifier": "B", "content": "A =$${1 \\over {54}}$$ and f(x) = 3(x\u20131)"}, {"identifier": "C", "content": "A =$${1 \\over {81}}$$ and f(x) = 3(x\u20131)"}, {"identifier": "D", "content": "A =$${1 \\over {27}}$$ and f(x) = 9(x\u20131)<sup>2</sup>"}]
|
["B"]
| null |
$$\int {{{dx} \over {{{({x^2} - 2x + 10)}^2}}} = \int {{{dx} \over {{{({{(x - 1)}^2} + 9)}^2}}}} } $$<br><br>
$$Let{\rm{ }}{\left( {x{\rm{ }}-{\rm{ }}1} \right)^2}{\rm{ }} = {\rm{ }}9ta{n^2}\theta \,\,\,\,...\left( i \right)$$<br><br>
$$ \Rightarrow \tan \theta = {{x - 1} \over 3}$$<br><br>
On Differentiating ...(i)<br><br>
2(x – 1)dx = 18tan$$\theta $$ sec<sup>2</sup>$$\theta $$ d$$\theta $$<br><br>
$$ \therefore I = \int {{{18\tan \theta {{\sec }^2}\theta d\theta } \over {2 \times 3\tan \theta \times 81{{\sec }^4}\theta }}} $$<br><br>
$$I = {1 \over {27}}\int {{{\cos }^2}\theta \,d\theta } = {1 \over {27}} \times {1 \over 2}\int {\left( {1 + \cos 2\theta } \right)} d\theta $$<br><br>
$$I = {1 \over {54}}\left\{ {\theta + {{\sin 2\theta } \over 2}} \right\} + c$$<br><br>
$$I = {1 \over {54}}\left\{ {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {1 \over 2} \times {{2\left( {{{x - 1} \over 3}} \right)} \over {1 + {{\left( {{{x - 1} \over 3}} \right)}^2}}}} \right\} + c$$<br><br>
$$I = {1 \over {54}}\left[ {{{\tan }^{ - 1}}\left( {{{x - 1} \over 3}} \right) + {{3(x - 1)} \over {{x^2} - 2x + 10}}} \right] + c$$<br><br>
then $$A = {1 \over {54}}$$<br><br>
f(x) = 3(x – 1)
|
mcq
|
jee-main-2019-online-10th-april-morning-slot
|
QtWndUPN5ICUBfhiUWRfe
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
The integral $$\int {{\rm{se}}{{\rm{c}}^{{\rm{2/ 3}}}}\,{\rm{x }}\,{\rm{cose}}{{\rm{c}}^{{\rm{4 / 3}}}}{\rm{x \,dx}}} $$ is equal to
(Hence C is a constant of integration)
|
[{"identifier": "A", "content": "-3/4 tan <sup>- 4 / 3</sup> x + C"}, {"identifier": "B", "content": "3tan<sup>\u20131/3</sup>x + C"}, {"identifier": "C", "content": "\u20133cot<sup>\u20131/3</sup>x+ C"}, {"identifier": "D", "content": "- 3tan<sup>\u20131/3</sup>x + C"}]
|
["D"]
| null |
$$\int {{{\sec }^{{2 \over 3}}}} x\cos e{c^{{4 \over 3}}}xdx$$
<br><br>= $$\int {{{{{\sec }^{{2 \over 3}}}x} \over {\cos e{c^{{2 \over 3}}}x}}\cos e{c^2}xdx} $$
<br><br>= $$\int {{1 \over {{{\cot }^{{2 \over 3}}}x}}\cos e{c^2}xdx} $$
<br><br>Let cot x = t<sup>3</sup>
<br><br>$$ \Rightarrow $$ - cosec<sup>2</sup>x dx = 3t<sup>2</sup>dt
<br><br>= $$ - 3\int {{{{t^2}dt} \over {{t^2}}}} $$
<br><br>= -3t + C
<br><br>= -3$${{{\cot }^{{1 \over 3}}}x}$$ + C
<br><br>= -3$${{{\tan }^{ - {1 \over 3}}}x}$$ + C
|
mcq
|
jee-main-2019-online-9th-april-morning-slot
|
AQn2qq06q2fjAXy7QknRR
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$\int {{{dx} \over {{x^3}{{(1 + {x^6})}^{2/3}}}} = xf(x){{(1 + {x^6})}^{{1 \over 3}}} + C} $$ <br/>
where C is a constant of integration, then the
function ƒ(x) is equal to
|
[{"identifier": "A", "content": "$${3 \\over {{x^2}}}$$"}, {"identifier": "B", "content": "$$ - {1 \\over {6{x^3}}}$$"}, {"identifier": "C", "content": "$$ - {1 \\over {2{x^3}}}$$"}, {"identifier": "D", "content": "$$ - {1 \\over {2{x^2}}}$$"}]
|
["C"]
| null |
I = $$\int {{{dx} \over {{x^3}{{\left( {1 + {x^6}} \right)}^{{2 \over 3}}}}}} $$
<br><br>= $$\int {{{dx} \over {{x^7}{{\left( {{1 \over {{x^6}}} + 1} \right)}^{{2 \over 3}}}}}} $$
<br><br>Let $${{1 \over {{x^6}}} + 1}$$ = t
<br><br>$$ \Rightarrow $$ $${{ - 6} \over {{x^7}}}dx = dt$$
<br><br>$$ \Rightarrow $$ $${{dx} \over {{x^7}}} = - {{dt} \over 6}$$
<br><br>$$ \therefore $$ I = $$ - {1 \over 6}\int {{{dt} \over {{t^{{2 \over 3}}}}}} $$
<br><br>= $$ - {1 \over 6}\left[ {{{{t^{{1 \over 3}}}} \over {{1 \over 3}}}} \right] + C$$
<br><br>= $$ - {1 \over 2}{\left[ {{1 \over {{x^6}}} + 1} \right]^{{1 \over 3}}} + C$$
<br><br>= $$ - {1 \over 2}{{{{\left( {1 + {x^6}} \right)}^{{1 \over 3}}}} \over {{x^2}}} \times {x \over x} + C$$
<br><br>= $$x.\left( { - {1 \over {2{x^3}}}} \right).{\left( {1 + {x^6}} \right)^{{1 \over 3}}} + C$$
<br><br>$$ \therefore $$ f(x) = $${ - {1 \over {2{x^3}}}}$$
|
mcq
|
jee-main-2019-online-8th-april-evening-slot
|
ye4tbM04B4WyItNxoFvXn
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
The integral $$\int {{{3{x^{13}} + 2{x^{11}}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^4}}}} \,dx$$ is equal to : (where C is a constant of integration)
|
[{"identifier": "A", "content": "$${{{x^{12}}} \\over {6{{\\left( {2{x^4} + 3{x^2} + 1} \\right)}^3}}}$$ + $$C$$"}, {"identifier": "B", "content": "$${{{x^4}} \\over {6{{\\left( {2{x^4} + 3{x^2} + 1} \\right)}^3}}} + C$$"}, {"identifier": "C", "content": "$${{{x^{12}}} \\over {{{\\left( {2{x^4} + 3{x^2} + 1} \\right)}^3}}} + C$$"}, {"identifier": "D", "content": "$${{{x^4}} \\over {{{\\left( {2{x^4} + 3{x^2} + 1} \\right)}^3}}} + C$$"}]
|
["A"]
| null |
$$\int {{{3{x^{13}} + 2{x^{11}}} \over {{{\left( {2{x^4} + 3{x^2} + 1} \right)}^4}}}} dx$$
<br><br>$$\int {{{\left( {{3 \over {{x^3}}} + {2 \over {{x^5}}}} \right)dx} \over {{{\left( {2 + {3 \over {{x^2}}} + {1 \over {{x^4}}}} \right)}^4}}}} $$
<br><br>Let $$\left( {2 + {3 \over {{x^2}}} + {1 \over {{x^4}}}} \right) = t$$
<br><br>$$ - {1 \over 2}\int {{{dt} \over {{t^4}}}} = {1 \over {6{t^3}}} + C$$
<br><br>$$ \Rightarrow {{{x^{12}}} \over {6{{\left( {2{x^4} + 3{x^2} + 1} \right)}^3}}} + C$$
|
mcq
|
jee-main-2019-online-12th-january-evening-slot
|
xn6ue9zHp1nBwlYZ056NP
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$\int {{{x + 1} \over {\sqrt {2x - 1} }}} \,dx$$ = f(x) $$\sqrt {2x - 1} $$ + C, where C is a constant of integration, then f(x) is equal to :
|
[{"identifier": "A", "content": "$${2 \\over 3}$$ (x $$-$$ 4)"}, {"identifier": "B", "content": "$${1 \\over 3}$$ (x + 4)"}, {"identifier": "C", "content": "$${1 \\over 3}$$ (x + 1)"}, {"identifier": "D", "content": "$${2 \\over 3}$$ (x + 2)"}]
|
["B"]
| null |
$$\sqrt {2x - 1} = t \Rightarrow 2x - 1 = {t^2} \Rightarrow 2dx = 2t.dt$$
<br><br>$$\int {{{x + 1} \over {\sqrt {2x - 1} }}dx = \int {{{{{{t^2} + 1} \over 2} + 1} \over t}tdt = \int {{{{t^2} + 3} \over 2}dt} } } $$
<br><br>$$ = {1 \over 2}\left( {{{{t^3}} \over 3} + 3t} \right) = {t \over 6}\left( {{t^2} + 9} \right) + c$$
<br><br>$$ = \sqrt {2x - 1} \left( {{{2x - 1 + 9} \over 6}} \right) + c = \sqrt {2x - 1} \left( {{{x + 4} \over 3}} \right) + c$$
<br><br>$$ \Rightarrow f\left( x \right) = {{x + 4} \over 3}$$
|
mcq
|
jee-main-2019-online-11th-january-evening-slot
|
wH0XiFnWxUutIMcwlOwvx
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} $$ dx = A(x)$${\left( {\sqrt {1 - {x^2}} } \right)^m}$$ + C, for a suitable chosen integer m and a function A(x), where C is a constant of integration, then (A(x))<sup>m</sup> equals :
|
[{"identifier": "A", "content": "$${1 \\over {27{x^6}}}$$"}, {"identifier": "B", "content": "$${{ - 1} \\over {27{x^9}}}$$"}, {"identifier": "C", "content": "$${1 \\over {9{x^4}}}$$"}, {"identifier": "D", "content": "$${1 \\over {3{x^3}}}$$"}]
|
["B"]
| null |
$$\int {{{\sqrt {1 - {x^2}} } \over {{x^4}}}} $$ dx = A(x)$${\left( {\sqrt {1 - {x^2}} } \right)^m}$$ + C
<br><br>$$\int {{{\left| x \right|\sqrt {{1 \over {{x^2}}} - 1} } \over {{x^4}}}} \,dx,$$
<br><br>Put $${1 \over {{x^2}}} - 1 = t \Rightarrow {{dt} \over {dx}} = {{ - 2} \over {{x^3}}}$$
<br><br><b>Case-I</b> $$x \ge 0$$
<br><br>$$ - {1 \over 2}\int {\sqrt t \,dt\, \Rightarrow {{{t^{3/2}}} \over 3}} + C$$
<br><br>$$ \Rightarrow - {1 \over 3}{\left( {{1 \over {{x^2}}} - 1} \right)^{3/2}}$$
<br><br>$$ \Rightarrow {{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^2}}} + C$$
<br><br>$$A(x) = - {1 \over {3{x^3}}}\,\,$$ and $$m = 3$$
<br><br>$${(A(x))^m} = {\left( { - {1 \over {3{x^3}}}} \right)^3} = - {1 \over {27{x^9}}}$$
<br><br><b>Case-II</b> $$x \le 0$$
<br><br>We get $${{{{\left( {\sqrt {1 - {x^2}} } \right)}^3}} \over { - 3{x^3}}} + C$$
<br><br>$$A(x) = {1 \over { - 3{x^3}}},\,\,\,m = 3$$
<br><br>$${(A(x))^m} = {{ - 1} \over {27{x^9}}}$$
|
mcq
|
jee-main-2019-online-11th-january-morning-slot
|
GNwlKDvLfwVDVMJ3J4cOC
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$\int \, $$x<sup>5</sup>.e<sup>$$-$$4x<sup>3</sup></sup> dx = $${1 \over {48}}$$e<sup>$$-$$4x<sup>3</sup></sup> f(x) + C, where C is a constant of inegration, then f(x) is equal to -
|
[{"identifier": "A", "content": "$$-$$2x<sup>3</sup> $$-$$ 1"}, {"identifier": "B", "content": "$$-$$ 2x<sup>3</sup> + 1 "}, {"identifier": "C", "content": "4x<sup>3</sup> + 1"}, {"identifier": "D", "content": "$$-$$4x<sup>3</sup> $$-$$ 1"}]
|
["D"]
| null |
$$\int {{x^5}} .{e^{ - 4{x^3}}}\,dx = {1 \over {48}}{e^{ - 4{x^3}}}f\left( x \right) + c$$
<br><br>Put $${x^3} = t$$
<br><br>$$3{x^2}\,dx = dt$$
<br><br>$$\int {{x^3}.{e^{ - 4{x^3}}}.\,{x^2}} dx$$
<br><br>$${1 \over 3}\int {t.{e^{ - 4t}}dt} $$
<br><br>$${1 \over 3}\left[ {t.{{{e^{ - 4t}}} \over { - 4}} - \int {{{{e^{ - 4t}}} \over { - 4}}dt} } \right]$$
<br><br>$$ - {{{e^{ - 4t}}} \over {48}}\left[ {4t + 1} \right] + c$$
<br><br>$${{ - {e^{ - 4{x^3}}}} \over {48}}\left[ {4{x^3} + 1} \right] + c$$
<br><br>$$ \therefore $$ $$f(x) = - 1 - 4{x^3}$$
|
mcq
|
jee-main-2019-online-10th-january-evening-slot
|
XbsoEUqQUgLdlc6T8Qlx5
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx,\,\left( {x \ge 0} \right),$$
<br/><br/>$$f\left( 0 \right) = 0,$$ then the value of $$f(1)$$ is :
|
[{"identifier": "A", "content": "$$ - $$ $${1 \\over 2}$$"}, {"identifier": "B", "content": "$$ - $$ $${1 \\over 4}$$"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over 4}$$"}]
|
["D"]
| null |
$$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{{\left( {{x^2} + 1 + 2{x^7}} \right)}^2}}}} \,dx$$
<br><br>$$f\left( x \right) = \int {{{5{x^8} + 7{x^6}} \over {{x^{14}}{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx$$
<br><br>$$f\left( x \right) = \int {{{5{x^{ - 6}} + 7{x^{ - 8}}} \over {{{\left( {{x^{ - 5}} + {x^{ - 7}} + 2} \right)}^2}}}} \,dx$$
<br><br>Let $${x^{ - 5}} + {x^{ - 7}} + 2 = t$$
<br><br>$$\left( { - 5{x^{ - 6}} - 7{x^{ - 8}}} \right)dx = dt$$
<br><br>$$\left( {5{x^{ - 6}} + 7{x^{ - 8}}} \right)dx = - dt$$
<br><br>$$f(x) = \int {{{ - dt} \over {{t^2}}}} = {1 \over t} + c$$
<br><br>$$f\left( x \right) = {1 \over {{x^{ - 5}} + {x^{ - 7}} + 2}} + c$$
<br><br>$$f\left( x \right) = {{{x^7}} \over {{x^2} + 1 + 2{x^7}}} + c$$
<br><br>$$f\left( 0 \right) = 0$$
<br><br>$$ \therefore $$ $$c = 0$$
<br><br>$$f\left( x \right) = {{{x^7}} \over {\left( {{x^2} + 1 + 2{x^7}} \right)}}$$
<br><br>$$f(1) = {1 \over {1 + 1 + 2}} = {1 \over 4}$$
|
mcq
|
jee-main-2019-online-9th-january-evening-slot
|
Hj0hWZ3LNJm6lC7Dh5vpM
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
For x<sup>2</sup> $$ \ne $$ n$$\pi $$ + 1, n $$ \in $$ N (the set of natural numbers), the integral <br/><br/>$$\int {x\sqrt {{{2\sin ({x^2} - 1) - \sin 2({x^2} - 1)} \over {2\sin ({x^2} - 1) + \sin 2({x^2} - 1)}}} dx} $$ is equal to :
<br/><br/>(where c is a constant of integration)
|
[{"identifier": "A", "content": "$${\\log _e}\\left| {{1 \\over 2}{{\\sec }^2}\\left( {{x^2} - 1} \\right)} \\right| + c$$"}, {"identifier": "B", "content": "$${1 \\over 2}{\\log _e}\\left| {\\sec \\left( {{x^2} - 1} \\right)} \\right| + c$$"}, {"identifier": "C", "content": "$${1 \\over 2}{\\log _e}\\left| {{{\\sec }^2}\\left( {{{{x^2} - 1} \\over 2}} \\right)} \\right| + c$$"}, {"identifier": "D", "content": "$${\\log _e}\\left| {\\sec \\left( {{{{x^2} - 1} \\over 2}} \\right)} \\right| + c$$"}]
|
["D"]
| null |
$$\int {x\sqrt {{{2\sin \left( {{x^2} - } \right) - \sin 2\left( {{x^2} - 1} \right)} \over {2\sin \left( {{x^2} - 1} \right) + \sin 2\left( {{x^2} - 1} \right)}}} } \,\,dx$$
<br><br>$$ = \int {x\sqrt {{{2\sin \left( {{x^2} - 1} \right) - 2sin\left( {{x^2} - 1} \right)\cos \left( {{x^2} - 1} \right)} \over {2\sin \left( {{x^2} - 1} \right) + 2\sin \left( {{x^2} - 1} \right)\cos \left( {{x^2} - 1} \right)}}} } \,\,dx$$
<br><br>$$ = \int {x\sqrt {{{1 - \cos \left( {{x^2} - 1} \right)} \over {1 + \cos \left( {{x^2} - 1} \right)}}} } \,dx$$
<br><br>$$ = \int {x\sqrt {{{2{{\sin }^2}\left( {{{{x^2} - 1} \over 2}} \right)} \over {2{{\cos }^2}\left( {{{{x^2} - 1} \over 2}} \right)}}} } \,dx$$
<br><br>$$ = \int {x\tan } \left( {{{{x^2} - 1} \over 2}} \right)dx$$
<br><br>put $${{{x^2} - 1} \over 2} = t$$
<br><br>$$ \Rightarrow {x^2} - 1 = 2t$$
<br><br>$$ \Rightarrow 2xdx = 2dt$$
<br><br>$$ \Rightarrow xdx = dt$$
<br><br>$$ \therefore $$ $$\int {\tan t\,dt} $$
<br><br>$$ = \ln \left| {\sec t} \right| + C$$
<br><br>$$ = \ln \left| {\sec \left( {{{{x^2} - 1} \over 2}} \right)} \right| + C$$
|
mcq
|
jee-main-2019-online-9th-january-morning-slot
|
7CoM1UAK7oFjOeAfUm7k9k2k5khl5cw
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}} = \lambda \tan \theta + 2{\log _e}\left| {f\left( \theta \right)} \right| + C$$<br/><br/>
where C is a constant of integration, then the
ordered pair ($$\lambda $$, ƒ($$\theta $$)) is equal to :
|
[{"identifier": "A", "content": "(\u20131, 1 \u2013 tan$$\\theta $$)"}, {"identifier": "B", "content": "(1, 1 + tan$$\\theta $$)"}, {"identifier": "C", "content": "(\u20131, 1 + tan$$\\theta $$)"}, {"identifier": "D", "content": "(1, 1 \u2013 tan$$\\theta $$)"}]
|
["C"]
| null |
I = $$\int {{{d\theta } \over {{{\cos }^2}\theta \left( {\tan 2\theta + \sec 2\theta } \right)}}}$$
<br><br>= $$\int {{{{{\sec }^2}\theta d\theta } \over {{{2\tan \theta } \over {1 - {{\tan }^2}\theta }} + {{1 + {{\tan }^2}\theta } \over {1 - {{\tan }^2}\theta }}}}} $$
<br><br>= $$\int {{{\left( {1 - {{\tan }^2}\theta } \right){{\sec }^2}\theta d\theta } \over {{{\left( {1 + {{\tan }}\theta } \right)}^2}}}} $$
<br><br>tan$$\theta $$ = t $$ \Rightarrow $$ sec<sup>2</sup> $$\theta $$ d$$\theta $$ = dt
<br><br>= $$\int {{{\left( {1 - {t^2}} \right)dt} \over {{{\left( {1 + {t}} \right)}^2}}}} $$
<br><br>= $$\int {{{\left( {1 - t} \right)\left( {1 + t} \right)dt} \over {{{\left( {1 + {t}} \right)}^2}}}} $$
<br><br>= $$\int {{{\left( {1 - t} \right)} \over {\left( {1 + t} \right)}}} dt$$
<br><br>= $$\int {\left( {{1 \over {\left( {1 + t} \right)}} - {t \over {\left( {1 + t} \right)}}} \right)} dt$$
<br><br>= $${\log _e}\left| {1 + t} \right|$$ - $$\int {\left( {{{1 + t} \over {\left( {1 + t} \right)}} - {1 \over {\left( {1 + t} \right)}}} \right)} dt$$
<br><br>= $${\log _e}\left| {1 + t} \right|$$ - t + $${\log _e}\left| {1 + t} \right|$$ + C
<br><br>= 2$${\log _e}\left| {1 + t} \right|$$ - t + C
<br><br>= 2$${\log _e}\left| {1 + \tan \theta } \right|$$ - tan $$\theta $$ + C
<br><br>$$ \therefore $$ $$\lambda $$ = -1 and ƒ($$\theta $$) = 1 + tan$$\theta $$
|
mcq
|
jee-main-2020-online-9th-january-evening-slot
|
H0FHSffmm2l7UtVzAnjgy2xukfqemvyt
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If
<br/>$$\int {{{\cos \theta } \over {5 + 7\sin \theta - 2{{\cos }^2}\theta }}} d\theta $$ = A$${\log _e}\left| {B\left( \theta \right)} \right| + C$$,
<br/><br/>where C is a constant of integration, then $${{{B\left( \theta \right)} \over A}}$$
<br/>can be :
|
[{"identifier": "A", "content": "$${{2\\sin \\theta + 1} \\over {5\\left( {\\sin \\theta + 3} \\right)}}$$"}, {"identifier": "B", "content": "$${{2\\sin \\theta + 1} \\over {\\sin \\theta + 3}}$$"}, {"identifier": "C", "content": "$${{5\\left( {2\\sin \\theta + 1} \\right)} \\over {\\sin \\theta + 3}}$$"}, {"identifier": "D", "content": "$${{5\\left( {\\sin \\theta + 3} \\right)} \\over {2\\sin \\theta + 1}}$$"}]
|
["C"]
| null |
$$\int {{{\cos \theta } \over {5 + 7\sin \theta - 2{{\cos }^2}\theta }}} d\theta $$
<br><br>= $$\int {{{\cos \theta d\theta } \over {5 + 7\sin \theta - 2\left( {1 - {{\sin }^2}\theta } \right)}}} $$
<br><br>= $$\int {{{\cos \theta d\theta } \over {3 + 7\sin \theta + 2{{\sin }^2}\theta }}} $$
<br><br>Let sin $$\theta $$ = t
<br>$$ \Rightarrow $$ cos$$\theta $$d$$\theta $$ = dt
<br><br>= $$\int {{{dt} \over {3 + 7t + 2{t^2}}}} $$
<br><br>= $$\int {{{dt} \over {\left( {2t + 1} \right)\left( {t + 3} \right)}}} $$
<br><br>= $${1 \over 5}\int {\left( {{2 \over {2t + 1}} - {1 \over {t + 3}}} \right)dt} $$
<br><br>= $${1 \over 5}\ln \left| {{{2t + 1} \over {t + 3}}} \right|$$ + C
<br><br>= $${1 \over 5}\ln \left| {{{2\sin \theta + 1} \over {\sin \theta + 3}}} \right|$$ + C
<br><br>$$ \therefore $$ A = $${{1 \over 5}}$$ and B($$\theta $$) = $${{{2\sin \theta + 1} \over {\sin \theta + 3}}}$$
<br><br>$$ \therefore $$ $${{{B\left( \theta \right)} \over A}}$$ = $${{5\left( {2\sin \theta + 1} \right)} \over {\sin \theta + 3}}$$
|
mcq
|
jee-main-2020-online-5th-september-evening-slot
|
aEHNBpLgZlUpf8cadcjgy2xukfjjgzep
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If <br/>$$\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $$ = $$g\left( x \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}} + c$$<br/><br/>
where c is a constant of integration,
then g(0) is equal to :
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "e"}, {"identifier": "D", "content": "e<sup>2</sup>"}]
|
["B"]
| null |
I = $$\int {\left( {{e^{2x}} + 2{e^x} - {e^{ - x}} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx} $$
<br><br>= $$\int {\left( {\left( {{e^{2x}} + {e^x} - 1} \right) + \left( {{e^x} - {e^{ - x}}} \right)} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx$$
<br><br>= $$\int {\left( {{e^{2x}} + {e^x} - 1} \right)} {e^{\left( {{e^x} + {e^{ - x}}} \right)}}dx$$
<br><br>+ $$\int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx} $$
<br><br>= $$\int {{{\left( {{e^{2x}} + {e^x} - 1} \right){e^{\left( {{e^x} + {e^{ - x}}} \right)}}} \over {{e^x}}}.} {e^x}dx$$
<br><br>$$ + \int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx} $$
<br><br>= $$\int {\left( {{e^x} - {e^{ - x}} + 1} \right)} {e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}dx$$
<br><br>+ $$\int {{e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} - {e^{ - x}}} \right)dx} $$
<br><br>Let $${{e^x} + {e^{ - x}} + x}$$ = t $$ \Rightarrow $$ $${\left( {{e^x} - {e^{ - x}} + 1} \right)}$$dx = dt
<br><br>and let $${{e^x} + {e^{ - x}}}$$ = u $$ \Rightarrow $$ $${\left( {{e^x} - {e^{ - x}}} \right)dx}$$ = du
<br><br>$$ \therefore $$ I = $$\int {{e^t}} dt + \int {{e^u}} du$$
<br><br>= $${{e^t}}$$ + $${{e^u}}$$ + C
<br><br>= $${e^{\left( {{e^x} + {e^{ - x}} + x} \right)}}$$ + $${{e^{\left( {{e^x} + {e^{ - x}}} \right)}}}$$ + C
<br><br>= $${e^{\left( {{e^x} + {e^{ - x}}} \right)}}\left( {{e^x} + 1} \right)$$ + C
<br><br>$$ \therefore $$ g(x) = e<sup>x</sup> + 1
<br><br>$$ \Rightarrow $$ g(0) = 2
|
mcq
|
jee-main-2020-online-5th-september-morning-slot
|
E6ZDVN1s3vLFiRj1o5jgy2xukf7g2hr5
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
Let $$f\left( x \right) = \int {{{\sqrt x } \over {{{\left( {1 + x} \right)}^2}}}dx\left( {x \ge 0} \right)} $$. Then f(3) – f(1) is eqaul to :
|
[{"identifier": "A", "content": "$$ - {\\pi \\over {12}} + {1 \\over 2} + {{\\sqrt 3 } \\over 4}$$"}, {"identifier": "B", "content": "$$ {\\pi \\over {12}} + {1 \\over 2} - {{\\sqrt 3 } \\over 4}$$"}, {"identifier": "C", "content": "$$ - {\\pi \\over 6} + {1 \\over 2} + {{\\sqrt 3 } \\over 4}$$"}, {"identifier": "D", "content": "$${\\pi \\over 6} + {1 \\over 2} - {{\\sqrt 3 } \\over 4}$$"}]
|
["B"]
| null |
$$\int {{{\sqrt x } \over {{{(1 + x)}^2}}}} dx(x > 0)$$<br><br>Put x = tan<sup>2</sup>$$\theta $$ $$ \Rightarrow $$ 2xdx = 2tan$$\theta $$sec<sup>2</sup>$$\theta $$d$$\theta $$<br><br>$$I = \int {{{2{{\tan }^2}\theta .{{\sec }^2}\theta } \over 2}} d\theta = \int {2{{\sin }^2}\theta d\theta = \int {(1 - \cos 2\theta )d\theta } } $$<br><br>$$ = \theta - {{\sin 2\theta } \over 2} + c$$<br><br>$$ \Rightarrow f(x) = \theta - {1 \over 2} \times {{2\tan \theta } \over {1 + {{\tan }^2}\theta }} + c$$ $$f(x) = \theta - {{\tan \theta } \over {1 + {{\tan }^2}\theta }} + c = {\tan ^{ - 1}}\sqrt x - {{\sqrt x } \over {1 + x}} + c$$<br><br>Now $$f(3) - f(1) = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) - {{\sqrt 3 } \over {1 + 3}} - {\tan ^{ - 1}}(1) + {1 \over 2}$$<br><br>$$ = {\pi \over 3} - {{\sqrt 3 } \over 4} - {\pi \over 4} + {1 \over 2}$$<br><br>$$ = {\pi \over 12} + {1 \over 2} - {{\sqrt 3 } \over 4}$$
|
mcq
|
jee-main-2020-online-4th-september-morning-slot
|
m2nLubDEJo1qVTMWIO7k9k2k5ith9ii
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
The integral $$\int {{{dx} \over {{{(x + 4)}^{{8 \over 7}}}{{(x - 3)}^{{6 \over 7}}}}}} $$ is equal to :<br/>
(where C is a constant of integration)
|
[{"identifier": "A", "content": "$${1 \\over 2}{\\left( {{{x - 3} \\over {x + 4}}} \\right)^{{3 \\over 7}}} + C$$"}, {"identifier": "B", "content": "$${\\left( {{{x - 3} \\over {x + 4}}} \\right)^{{1 \\over 7}}} + C$$"}, {"identifier": "C", "content": "$$ - {1 \\over {13}}{\\left( {{{x - 3} \\over {x + 4}}} \\right)^{{{13} \\over 7}}} + C$$"}, {"identifier": "D", "content": "-$${\\left( {{{x - 3} \\over {x + 4}}} \\right)^{-{1 \\over 7}}} + C$$"}]
|
["B"]
| null |
$$\int {{{dx} \over {{{(x + 4)}^{{8 \over 7}}}{{(x - 3)}^{{6 \over 7}}}}}} $$
<br><br>= $$\int {{{dx} \over {{{\left( {x + 4} \right)}^2}{{\left( {{{x - 3} \over {x + 4}}} \right)}^{{6 \over 7}}}}}} $$
<br><br>Put $${{{x - 3} \over {x + 4}}}$$ = t
<br><br>$$ \Rightarrow $$ $$\left\{ {{{\left( {x + 4} \right) - \left( {x - 3} \right)} \over {{{\left( {x + 4} \right)}^2}}}} \right\}dx$$ = dt
<br><br>$$ \Rightarrow $$ $${{dx} \over {{{\left( {x + 4} \right)}^2}}} = {{dt} \over 7}$$
<br><br>= $${1 \over 7}\int {{{dt} \over {{{\left( t \right)}^{{6 \over 7}}}}}} $$
<br><br>= $${1 \over 7}\left( {{{{t^{{1 \over 7}}}} \over {{1 \over 7}}}} \right)$$ + C
<br><br>= $${\left( {{{x - 3} \over {x + 4}}} \right)^{{1 \over 7}}}$$ + C
|
mcq
|
jee-main-2020-online-9th-january-morning-slot
|
V06zxcvduCHTTdCKQv7k9k2k5gzjkvq
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$\int {{{\cos xdx} \over {{{\sin }^3}x{{\left( {1 + {{\sin }^6}x} \right)}^{2/3}}}}} = f\left( x \right){\left( {1 + {{\sin }^6}x} \right)^{1/\lambda }} + c$$
<br/><br/>where c is a constant of integration, then $$\lambda f\left( {{\pi \over 3}} \right)$$ is equal to
|
[{"identifier": "A", "content": "$${9 \\over 8}$$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "-2"}, {"identifier": "D", "content": "$$-{9 \\over 8}$$"}]
|
["C"]
| null |
Given I = $$\int {{{\cos xdx} \over {{{\sin }^3}x{{\left( {1 + {{\sin }^6}x} \right)}^{2/3}}}}}$$
<br><br>Let sin x = t
<br><br>$$ \Rightarrow $$ cos xdx = dt
<br><br>$$ \therefore $$ I = $$\int {{{dt} \over {{t^3}{{\left( {1 + {t^6}} \right)}^{2/3}}}}} $$
<br><br>I = $$\int {{{dt} \over {{t^7}{{\left( {{1 \over {{t^6}}} + 1} \right)}^{2/3}}}}} $$
<br><br>Let $${{1 \over {{t^6}}} + 1}$$ = z<sup>3</sup>
<br><br>$$ \Rightarrow $$ $$ - {6 \over {{t^7}}}dt = 3{z^2}dz$$
<br><br>$$ \Rightarrow $$ $${{dt} \over {{t^7}}} = {{{z^2}} \over { - 2}}dz$$
<br><br>So I = $$ - \int {{{{z^2}dz} \over {2{{\left( {{z^3}} \right)}^{2/3}}}}} $$
<br><br>I = $$ - \int {{{dz} \over 2}} $$
<br><br>I = $$ - {z \over 2} + C$$
<br><br>I = $$ - {1 \over 2}{\left( {1 + {1 \over {{t^6}}}} \right)^{{1 \over 3}}} + C$$
<br><br>I = $$ - {1 \over 2}{\left( {1 + {1 \over {{{\sin }^6}x}}} \right)^{{1 \over 3}}} + C$$
<br><br>I = $$ - {1 \over {2{{\sin }^2}x}}{\left( {{{\sin }^6}x + 1} \right)^{{1 \over 3}}} + C$$
<br><br>$$ \therefore $$ $$\lambda $$ = 3 and f(x) = $$ - {1 \over 2}$$cosec<sup>2</sup> x
<br><br>Then $$\lambda f\left( {{\pi \over 3}} \right)$$ = 3 $$ \times $$ $$ - {1 \over 2}$$cosec<sup>2</sup> $${\pi \over 3}$$
<br><br>= 3 $$ \times $$ $$ - {1 \over 2}$$ $$ \times $$ $${4 \over 3}$$ = -2
|
mcq
|
jee-main-2020-online-8th-january-morning-slot
|
jgUAE7YDRC8UIu9j6Gjgy2xukf45aw9x
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$\int {{{\sin }^{ - 1}}\left( {\sqrt {{x \over {1 + x}}} } \right)} dx$$ = A(x)$${\tan ^{ - 1}}\left( {\sqrt x } \right)$$ + B(x) + C,
<br/>where C is a constant of integration, then the
ordered pair (A(x), B(x)) can be :
|
[{"identifier": "A", "content": "(x + 1, -$${\\sqrt x }$$)"}, {"identifier": "B", "content": "(x + 1, $${\\sqrt x }$$)"}, {"identifier": "C", "content": "(x - 1, -$${\\sqrt x }$$)"}, {"identifier": "D", "content": "(x - 1, $${\\sqrt x }$$)"}]
|
["A"]
| null |
Given, I = $$\int {{{\sin }^{ - 1}}\left( {\sqrt {{x \over {1 + x}}} } \right)} dx$$
<br><br>Let $${\sin ^{ - 1}}\left( {{{\sqrt x } \over {\sqrt {1 + x} }}} \right)$$ = $$\theta $$
<br><br>$$ \Rightarrow $$ $${{{\sqrt x } \over {\sqrt {1 + x} }} = \sin \theta }$$
<br><br>$$ \Rightarrow $$ tan $$\theta $$ = $${{{\sqrt x } \over 1}}$$
<br><br>$$ \Rightarrow $$ $$\theta $$ = $${\tan ^{ - 1}}\left( {\sqrt x } \right)$$
<br><br>$$ \therefore $$ I = $$\int {{{\tan }^{ - 1}}\left( {\sqrt x } \right)dx} $$
<br><br>= $$\int {{{\tan }^{ - 1}}\left( {\sqrt x } \right).1dx} $$
<br><br>Applying integration by parts,
<br><br>I = $${\tan ^{ - 1}}\left( {\sqrt x } \right).x - \int {{1 \over {1 + x}}{1 \over {2\sqrt x }}xdx} $$
<br><br>Let $${\sqrt x }$$ = t
<br><br>$$ \Rightarrow $$ x = t<sup>2</sup>
<br><br>$$ \Rightarrow $$ dx = 2tdt
<br><br>$$ \therefore $$ I = $${\tan ^{ - 1}}\left( {\sqrt x } \right).x - \int {{{{t^2}} \over {\left( {1 + {t^2}} \right)\left( {2t} \right)}}2tdt} $$
<br><br>= $${\tan ^{ - 1}}\left( {\sqrt x } \right).x - \int {{{\left( {{t^2} + 1} \right) - 1} \over {\left( {1 + {t^2}} \right)}}dt} $$
<br><br>= $${\tan ^{ - 1}}\left( {\sqrt x } \right).x - t + {\tan ^{-1}}t + c$$
<br><br>= $${\tan ^{ - 1}}\left( {\sqrt x } \right).x - \sqrt x + {\tan ^{ - 1}}\sqrt x + c$$
<br><br>$$ \therefore $$ A(x) = x + 1, B(x) = –$${\sqrt x }$$
|
mcq
|
jee-main-2020-online-3rd-september-evening-slot
|
nGPDGf6uHUWzEnzMjQ1klreafmf
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \sin 2x} }}} dx = a{\sin ^{ - 1}}\left( {{{\sin x + \cos x} \over b}} \right) + c$$, where c is a constant of integration, then
the ordered pair (a, b) is equal to :
|
[{"identifier": "A", "content": "(-1, 3)"}, {"identifier": "B", "content": "(1, 3)"}, {"identifier": "C", "content": "(1, -3)"}, {"identifier": "D", "content": "(3, 1)"}]
|
["B"]
| null |
Given $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \sin 2x} }}} dx$$
<br><br>Write sin2x = 1 + sin2x - 1
<br><br>= $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {1 + \sin 2x - 1} \right]} }}} dx$$
<br><br>= $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\sin }^2}x + {{\cos }^2}x + 2\sin x\cos x - 1} \right]} }}} dx$$
<br><br>= $$\int {{{\cos x - \sin x} \over {\sqrt {8 - \left[ {{{\left( {\sin x + \cos x} \right)}^2} - 1} \right]} }}} dx$$
<br><br>= $$\int {{{\cos x - \sin x} \over {\sqrt {9 - {{\left( {\sin x + \cos x} \right)}^2}} }}} dx$$
<br><br>put sin x + cos x = t
<br>$$ \Rightarrow $$ (cos x – sin x) dx = dt
<br><br>= $$\int {{{dt} \over {\sqrt {9 - {{\left( t \right)}^2}} }}} $$
<br><br>= $${\sin ^{ - 1}}\left( {{t \over 3}} \right)$$ + C
<br><br>= $${\sin ^{ - 1}}\left( {{{\sin x + \cos x} \over 3}} \right) + C$$
<br><br>$$ \therefore $$ a = 1 and b = 3
|
mcq
|
jee-main-2021-online-24th-february-morning-slot
|
rymMYDE5Xap5FUPpvJ1kls58c1v
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
The value of the integral <br/>$$\int {{{\sin \theta .\sin 2\theta ({{\sin }^6}\theta + {{\sin }^4}\theta + {{\sin }^2}\theta )\sqrt {2{{\sin }^4}\theta + 3{{\sin }^2}\theta + 6} } \over {1 - \cos 2\theta }}} \,d\theta $$ is :
|
[{"identifier": "A", "content": "$${1 \\over {18}}{\\left[ {9 - 2{{\\cos }^6}\\theta - 3{{\\cos }^4}\\theta - 6{{\\cos }^2}\\theta } \\right]^{{3 \\over 2}}} + c$$"}, {"identifier": "B", "content": "$${1 \\over {18}}{\\left[ {11 - 18{{\\sin }^2}\\theta + 9{{\\sin }^4}\\theta - 2{{\\sin }^6}\\theta } \\right]^{{3 \\over 2}}} + c$$"}, {"identifier": "C", "content": "$${1 \\over {18}}{\\left[ {11 - 18{{\\cos }^2}\\theta + 9{{\\cos }^4}\\theta - 2{{\\cos }^6}\\theta } \\right]^{{3 \\over 2}}} + c$$"}, {"identifier": "D", "content": "$${1 \\over {18}}{\\left[ {9 - 2{{\\sin }^6}\\theta - 3{{\\sin }^4}\\theta - 6{{\\sin }^2}\\theta } \\right]^{{3 \\over 2}}} + c$$"}]
|
["C"]
| null |
$$\int {{{2{{\sin }^2}\theta \cos \theta ({{\sin }^6}\theta + {{\sin }^4}\theta + {{\sin }^2}\theta )\sqrt {2{{\sin }^4}\theta + 3{{\sin }^2}\theta + 6} } \over {2{{\sin }^2}\theta }}d\theta } $$<br><br>Let sin$$\theta$$ = t, cos$$\theta$$ d$$\theta$$ = dt<br><br>$$ = \int {({t^6} + {t^4} + {t^2})\sqrt {2{t^4} + 3{t^2} + 6} \,dt} $$
<br><br>$$= \int {({t^5} + {t^3} + t)\sqrt {2{t^6} + 3{t^4} + 6{t^2}} dt} $$<br><br>Let $$2{t^6} + 3{t^4} + 6{t^2} = z$$<br><br>$$12({t^5} + {t^3} + t)dt = dz$$<br><br>$$ = {1 \over {12}}\int {\sqrt z dz = {1 \over {18}}{z^{3/2}} + c} $$<br><br>$$ = {1 \over {18}}[{(2{\sin ^6}\theta + 3{\sin ^4}\theta + 6{\sin ^2}\theta )^{3/2}} + C$$<br><br>$$ = {1 \over {18}}{[(1 - {\cos ^2}\theta )(2{(1 - {\cos ^2}\theta )^2} + 3 - 3{\cos ^2}\theta + 6)]^{3/2}} + C$$<br><br>$$ = {1 \over {18}}{[(1 - {\cos ^2}\theta )(2{\cos ^4}\theta - 7{\cos ^2}\theta + 11)]^{3/2}} + C$$<br><br>$$ = {1 \over {18}}{[ - 2{\cos ^6}\theta + 9{\cos ^4}\theta - 18{\cos ^2}\theta + 11]^{3/2}} + C$$
|
mcq
|
jee-main-2021-online-25th-february-morning-slot
|
qaDgVipCZFCuXu1Pxz1klt7lrht
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
The integral $$\int {{{{e^{3{{\log }_e}2x}} + 5{e^{2{{\log }_e}2x}}} \over {{e^{4{{\log }_e}x}} + 5{e^{3{{\log }_e}x}} - 7{e^{2{{\log }_e}x}}}}} dx$$, x > 0, is equal to : (where c is a constant of integration)
|
[{"identifier": "A", "content": "$${\\log _e}\\sqrt {{x^2} + 5x - 7} + c$$"}, {"identifier": "B", "content": "$$4{\\log _e}|{x^2} + 5x - 7| + c$$"}, {"identifier": "C", "content": "$${1 \\over 4}{\\log _e}|{x^2} + 5x - 7| + c$$"}, {"identifier": "D", "content": "$${\\log _e}|{x^2} + 5x - 7| + c$$"}]
|
["B"]
| null |
$$\int {{{{e^{3{{\log }_e}2x}} + 5{e^{2{{\log }_e}2x}}} \over {{e^{4{{\log }_e}x}} + 5{e^{3{{\log }_e}x}} - 7{e^{2{{\log }_e}x}}}}dx} $$<br><br>$$ = \int {{{8{x^3} + 5(4{x^2})} \over {{x^4} + 5{x^3} - 7{x^2}}}} $$<br><br>$$ = \int {{{8{x^3} + 20{x^2}} \over {{x^4} + 5{x^3} - 7{x^2}}}} $$<br><br>$$ = \int {{{8x + 20} \over {{x^2} + 5x - 7}}} $$<br><br>$$ = \int {{{4(2x + 5)} \over {{x^2} + 5x - 7}}} $$
<br><br>Let $$\left\{ \matrix{
{x^2} + 5x - 7 = t \hfill \cr
(2x + 5)dx = dt \hfill \cr} \right\}$$<br><br>$$ = \int {{{4dt} \over t}} $$<br><br>$$ = 4\ln \left| t \right| + C$$<br><br>$$ = 4\ln \left| {({x^2} + 5x - 7)} \right| + C$$
<br><br>Here C is integral constant.
|
mcq
|
jee-main-2021-online-25th-february-evening-slot
|
OlbUl68mvnNzXdDFhr1kmizm1hi
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
For real numbers $$\alpha$$, $$\beta$$, $$\gamma$$ and $$\delta $$, if <br/>$$\int {{{({x^2} - 1) + {{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)} \over {({x^4} + 3{x^2} + 1){{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)}}dx} $$<br/><br/> $$ = \alpha {\log _e}\left( {{{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)} \right) + \beta {\tan ^{ - 1}}\left( {{{\gamma ({x^2} + 1)} \over x}} \right) + \delta {\tan ^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right) + C$$ <br/><br/>where C is an arbitrary constant, then the value of 10($$\alpha$$ + $$\beta$$$$\gamma$$ + $$\delta$$) is equal to ______________.
|
[]
| null |
6
|
$$\int {{{({x^2} - 1) + {{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)} \over {({x^4} + 3{x^2} + 1){{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)}}dx} $$
<br><br>= $$\int {{{{x^2} - 1} \over {({x^4} + 3{x^2} + 1){{\tan }^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right)}}dx + \int {{1 \over {{x^4} + 3{x^2} + 1}}dx} } $$<br><br>Let, $${I_1} = \int {{{1 - {1 \over {{x^2}}}} \over {\left[ {{{\left( {x + {1 \over x}} \right)}^2} + 1} \right]{{\tan }^{ - 1}}\left( {x + {1 \over x}} \right)}}} dx$$
<br><br>and $${I_2} = \int {{{dx} \over {{x^4} + 3x + 1}}} $$
<br><br>$${\tan ^{ - 1}}\left( {x + {1 \over x}} \right) = t$$<br><br>$${I_1} = \int {{{dt} \over t}} $$<br><br>$${I_1} = \ln (t) = \ln \left| {{{\tan }^{ - 1}}\left( {x + {1 \over x}} \right)} \right|$$<br><br>Now<br><br>$${I_2} = \int {{{dx} \over {{x^4} + 3x + 1}}} $$<br><br>$$ = {1 \over 2}\int {{{({x^2} + 1) - ({x^2} - 1)} \over {{x^4} + 3{x^2} + 1}}} dx$$<br><br>$$ = {1 \over 2}\left[ {\int {{{1 + {1 \over {{x^2}}}} \over {{x^2} + 3 + {1 \over {{x^2}}}}}dx - } \int {{{\left( {1 - {1 \over {{x^2}}}} \right)} \over {{x^2} + 3 + {1 \over {{x^2}}}}}dx} } \right]$$
<br><br>$$ = {1 \over 2}\left[ {\int {{{1 + {1 \over {{x^2}}}} \over {\left[ {{{\left( {x - {1 \over x}} \right)}^2} + 5} \right]}}} dx - \int {{{1 - {1 \over {{x^2}}}} \over {\left[ {{{\left( {x + {1 \over x}} \right)}^2} + 1} \right]}}} dx} \right]$$
<br><br>Let $${x - {1 \over x} = u}$$ and $${x + {1 \over x} = v}$$<br><br>$$ = {1 \over 2}\left[ {\int {{{du} \over {{u^2} + {{\left( {\sqrt 5 } \right)}^2}}} - \int {{{dv} \over {{v^2} + 1}}} } } \right]$$<br><br>$${I_2} = {1 \over {2\sqrt 5 }}{\tan ^{ - 1}}\left( {{{x - {1 \over x}} \over {\sqrt 5 }}} \right) - {1 \over 2}{\tan ^{ - 1}}\left( {x + {1 \over x}} \right)$$<br><br>$$I = {I_1} + {I_2} =$$
<br><br>$$ \ln \left| {{{\tan }^{ - 1}}\left( {x + {1 \over x}} \right)} \right| + {1 \over {2\sqrt 5 }}\ln \left( {{{{x^2} - 1} \over {\sqrt 5 x}}} \right) - {1 \over 2}{\tan ^{ - 1}}\left( {{{{x^2} + 1} \over x}} \right) + C$$<br><br>$$\alpha = 1,\beta = {1 \over {2\sqrt 5 }},\lambda = {1 \over {\sqrt 5 }},\delta = - {1 \over 2}$$<br><br>$$ \therefore $$ $$10(\alpha + \beta \lambda + \delta ) = 10\left[ {1 + {1 \over {10}} - {1 \over 2}} \right]$$<br><br>$$ = 10\left( {{1 \over {10}} + {1 \over 2}} \right)$$<br><br>$$ = 1 + 5 = 6$$
|
integer
|
jee-main-2021-online-16th-march-evening-shift
|
0zdXCvMp55v13ujE0o1kmlicnc5
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
The integral $$\int {{{(2x - 1)\cos \sqrt {{{(2x - 1)}^2} + 5} } \over {\sqrt {4{x^2} - 4x + 6} }}} dx$$ is equal to (where c is a constant of integration)
|
[{"identifier": "A", "content": "$${1 \\over 2}\\sin \\sqrt {{{(2x - 1)}^2} + 5} + c$$"}, {"identifier": "B", "content": "$${1 \\over 2}\\cos \\sqrt {{{(2x + 1)}^2} + 5} + c$$"}, {"identifier": "C", "content": "$${1 \\over 2}\\cos \\sqrt {{{(2x - 1)}^2} + 5} + c$$"}, {"identifier": "D", "content": "$${1 \\over 2}\\sin \\sqrt {{{(2x + 1)}^2} + 5} + c$$"}]
|
["A"]
| null |
$$\int {{{(2x - 1)\cos \sqrt {{{(2x - 1)}^2} + 5} } \over {\sqrt {{{(2x - 1)}^2} + 5} }}} dx$$<br><br>$${(2x - 1)^2} + 5 = {t^2}$$<br><br>$$2(2x - 1)2dx = 2t\,dt$$<br><br>$$2\sqrt {{t^2} - 5} dx = t\,dt$$<br><br>So, $$\int {{{\sqrt {{t^2} - 5} \cos t} \over {2\sqrt {{t^2} - 5} }}dt = {1 \over 2}\sin t + c} $$<br><br>$$ = {1 \over 2}\sin \sqrt {{{(2x - 1)}^2} + 5} + c$$
|
mcq
|
jee-main-2021-online-18th-march-morning-shift
|
2PEg8x92SALhxOO63W1kmlm3mz5
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$f(x) = \int {{{5{x^8} + 7{x^6}} \over {{{({x^2} + 1 + 2{x^7})}^2}}}dx,(x \ge 0),f(0) = 0} $$ and $$f(1) = {1 \over K}$$, then the value of K is
|
[]
| null |
4
|
$$\int {{{5{x^8} + 7{x^6}} \over {{{(2{x^7} + {x^2} + 1)}^2}}}dx = \int {{{5{x^8} + 7{x^6}} \over {{x^{14}}{{\left( {2 + {1 \over {{x^5}}} + {1 \over {{x^7}}}} \right)}^2}}}dx} } $$<br><br>$$\int {{{{5 \over {{x^6}}} + {7 \over {{x^8}}}} \over {{{\left( {2 + {1 \over {{x^5}}} + {1 \over {{x^7}}}} \right)}^2}}}dx} $$<br><br>put $$2 + {1 \over {{x^5}}} + {1 \over {{x^7}}} = t$$<br><br>$$ \Rightarrow - \left( {{5 \over {{x^6}}} + {7 \over {{x^8}}}} \right)dx = dt$$<br><br>$$\int {{{ - dt} \over {{t^2}}} = {1 \over t} + c} $$<br><br>$$ \Rightarrow f(x) = {1 \over {2 + {1 \over {{x^5}}} + {1 \over {{x^7}}}}} + C = {{{x^7}} \over {2{x^7} + 1 + {x^2}}} + C$$<br><br>$$f(0) = 0 \Rightarrow C = 0$$<br><br>$$f(x) = {1 \over 4} = {1 \over k}$$<br><br>$$ \Rightarrow k = 4$$
|
integer
|
jee-main-2021-online-18th-march-morning-shift
|
1ktepadjh
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $$\int {{{dx} \over {{{({x^2} + x + 1)}^2}}} = a{{\tan }^{ - 1}}\left( {{{2x + 1} \over {\sqrt 3 }}} \right) + b\left( {{{2x + 1} \over {{x^2} + x + 1}}} \right) + C} $$, x > 0 where C is the constant of integration, then the value of $$9\left( {\sqrt 3 a + b} \right)$$ is equal to _____________.
|
[]
| null |
15
|
$\int \frac{d x}{\left(x^2+x+1\right)^2}$
<br/><br/>$=\int \frac{d x}{\left[\left(x+\frac{1}{2}\right)^2+\left(\frac{\sqrt{3}}{2}\right)^2\right]^2}$
<br/><br/>Let $x+\frac{1}{2}=\frac{\sqrt{3}}{2} \tan \theta$
<br/><br/>$$
\Rightarrow d x=\frac{\sqrt{3}}{2} \sec ^2 \theta d \theta
$$
<br/><br/>$\therefore \int \frac{\frac{\sqrt{3}}{2} \sec ^2 \theta d \theta}{\frac{9}{16}\left(\tan ^2 \theta+1\right)^2}
$
<br/><br/>$ =\frac{8}{3 \sqrt{3}} \int \frac{\sec ^2 \theta d \theta}{\sec ^4 \theta} $
<br/><br/>$ =\frac{8}{3 \sqrt{3}} \int \cos ^2 \theta d \theta=\frac{8}{3 \sqrt{3}} \int \frac{1+\cos 2 \theta}{2} d \theta $
<br/><br/>$ =\frac{4}{3 \sqrt{3}}\left(\theta+\frac{\sin 2 \theta}{2}\right)+C $
<br/><br/>= $ \frac{4}{2 \sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)+\frac{4}{3 \sqrt{3}} \frac{\frac{2 x+1}{\sqrt{3}}}{1+\left(\frac{2 x+1}{\sqrt{3}}\right)^2}+C$
<br/><br/>$=\frac{4}{3 \sqrt{3}} \tan ^{-1}\left(\frac{2 x+1}{3}\right)+\frac{1}{3} \frac{2 x+1}{\left(x^2+x+1\right)}+C$
<br/><br/>$\therefore a=\frac{4}{3 \sqrt{3}}, b=\frac{1}{3}$
<br/><br/>Hence, $9(\sqrt{3} a+b)=9\left(\frac{4}{3}+\frac{1}{3}\right)=15$
|
integer
|
jee-main-2021-online-27th-august-morning-shift
|
1ktiptozs
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
The integral $$\int {{1 \over {\root 4 \of {{{(x - 1)}^3}{{(x + 2)}^5}} }}} \,dx$$ is equal to : (where C is a constant of integration)
|
[{"identifier": "A", "content": "$${3 \\over 4}{\\left( {{{x + 2} \\over {x - 1}}} \\right)^{{1 \\over 4}}} + C$$"}, {"identifier": "B", "content": "$${3 \\over 4}{\\left( {{{x + 2} \\over {x - 1}}} \\right)^{{5 \\over 4}}} + C$$"}, {"identifier": "C", "content": "$${4 \\over 3}{\\left( {{{x - 1} \\over {x + 2}}} \\right)^{{1 \\over 4}}} + C$$"}, {"identifier": "D", "content": "$${4 \\over 3}{\\left( {{{x - 1} \\over {x + 2}}} \\right)^{{5 \\over 4}}} + C$$"}]
|
["C"]
| null |
$$\int {{{dx} \over {{{(x - 1)}^{3/4}}{{(x + 2)}^{5/4}}}}} $$<br><br>$$ = \int {{{dx} \over {{{\left( {{{x + 2} \over {x - 1}}} \right)}^{5/4}}.\,{{(x - 1)}^2}}}} $$<br><br>put $${{x + 2} \over {x - 1}} = t$$<br><br>$$ = - {1 \over 3}\int {{{dt} \over {{t^{5/4}}}}} $$<br><br>$$ = {4 \over 3}.{1 \over {{t^{1/4}}}} + C$$<br><br>$$ = {4 \over 3}{\left( {{{x - 1} \over {x + 2}}} \right)^{1/4}} + C$$
|
mcq
|
jee-main-2021-online-31st-august-morning-shift
|
1l58fbnfg
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
<p>If $$\int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx = g(x) + c} $$, $$g(1) = 0$$, then $$g\left( {{1 \over 2}} \right)$$ is equal to :</p>
|
[{"identifier": "A", "content": "$${\\log _e}\\left( {{{\\sqrt 3 - 1} \\over {\\sqrt 3 + 1}}} \\right) + {\\pi \\over 3}$$"}, {"identifier": "B", "content": "$${\\log _e}\\left( {{{\\sqrt 3 + 1} \\over {\\sqrt 3 - 1}}} \\right) + {\\pi \\over 3}$$"}, {"identifier": "C", "content": "$${\\log _e}\\left( {{{\\sqrt 3 + 1} \\over {\\sqrt 3 - 1}}} \\right) - {\\pi \\over 3}$$"}, {"identifier": "D", "content": "$${1 \\over 2}{\\log _e}\\left( {{{\\sqrt 3 - 1} \\over {\\sqrt 3 + 1}}} \\right) - {\\pi \\over 6}$$"}]
|
["A"]
| null |
Given, $$\int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx = g(x) + c} $$, $$g(1) = 0$$
<br/><br/>Let I = $\int {{1 \over x}\sqrt {{{1 - x} \over {1 + x}}} dx}$
<br/><br/>= $\int {{1 \over x}\sqrt {{{\left( {1 - x} \right)\left( {1 + x} \right)} \over {\left( {1 + x} \right)\left( {1 + x} \right)}}} } dx$
<br/><br/>$=\int \frac{1-x}{x \sqrt{1-x^2}} d x$
<br/><br/>$=\int \frac{1}{x \sqrt{1-x^2}} d x-\int \frac{1}{\sqrt{1-x^2}} d x$
<br/><br/>Put $x=\frac{1}{t}$
<br/><br/>$d x=-\frac{1}{t^2}$
<br/><br/>$=\int \frac{\frac{-1}{t^2}}{\frac{1}{t} \sqrt{1-\frac{1}{t^2}}} d t-\sin ^{-1}(x)+C_1$
<br/><br/>$=\int \frac{-d t}{\sqrt{t^2-1}}-\sin ^{-1}(x)+C_1$
<br/><br/>$=-\ln \left|t+\sqrt{t^2-1}\right|-\sin ^{-1}(x)+C_1$
<br/><br/>$=-\ln \left|\frac{1}{\mathrm{x}}+\sqrt{\frac{1}{\mathrm{x}^2}-1}\right|-\sin ^{-1}(\mathrm{x})+\mathrm{C}_1$ [Putting values of t]
<br/><br/>$=-\ln \left|\frac{1}{\mathrm{x}}+\sqrt{\frac{1}{\mathrm{x}^2}-1}\right|-\left(\frac{\pi}{2}-\cos ^{-1} \mathrm{x}\right)+\mathrm{C}_1$
<br/><br/>$=-\ln \left|\frac{1}{\mathrm{x}}+\sqrt{\frac{1}{\mathrm{x}^2}-1}\right|+\cos ^{-1}(\mathrm{x})-\frac{\pi}{2}+\mathrm{C}_1$
<br/><br/>$$ \therefore $$ $g(x)=\cos ^{-1}(x)-\ell n\left|\frac{1}{x}+\sqrt{\frac{1}{x^2}-1}\right|$
<br/><br/>So, $g(1)=\cos ^{-1}(1)-\ell n|1|=0$
<br/><br/>$g\left(\frac{1}{2}\right)=\cos ^{-1}\left(\frac{1}{2}\right)-\ell n|2+\sqrt{3}|$
<br/><br/>$ = {\pi \over 3} + \ln \left( {{1 \over {2 + \sqrt 3 }}} \right)$
<br/><br/>$=\frac{\pi}{3}+\ln \left(\frac{\sqrt{3}-1}{\sqrt{3}+1}\right)$
|
mcq
|
jee-main-2022-online-26th-june-evening-shift
|
ldr0pnrl
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
If $\int \sqrt{\sec 2 x-1} d x=\alpha \log _e\left|\cos 2 x+\beta+\sqrt{\cos 2 x\left(1+\cos \frac{1}{\beta} x\right)}\right|+$ constant, then $\beta-\alpha$ is equal to ____________.
|
[]
| null |
1
|
<p>$$I = \int {\sqrt {\sec 2x - 1} dx=\int {\sqrt {{{1 - \cos 2x} \over {\cos 2x}}} dx} } $$</p>
<p>$$ = \int {\sqrt {{{2{{\sin }^2}x} \over {2{{\cos }^2}x - 1}}} dx} $$</p>
<p>Let $$\sqrt {2\cos } x = t - \sqrt 2 \sin xdx = dt$$</p>
<p>$$I = \int { - {{dt} \over {\sqrt {{t^2} - 1} }} = - \ln \left| {t + \sqrt {{t^2} - {a^2}} } \right| + c} $$</p>
<p>$$ = - \ln \left| {\sqrt 2 \cos x + \sqrt {2{{\cos }^2}} x - 1} \right| + c$$</p>
<p>$$ = - {1 \over 2}\ln \left| {2{{\cos }^2}x + \cos 2x + 2\sqrt 2 \sqrt {\cos 2x.{{\cos }^2}x} } \right| + c$$</p>
<p>$$ = - {1 \over 2}\ln \left| {2\cos 2x + 1 + 2\sqrt {\cos 2x(1 + \cos 2x)} } \right| + c$$</p>
<p>$$ = - {1 \over 2}\ln \left| {\cos 2x + {1 \over 2} + \sqrt {\cos 2x(1 + \cos 2x)} } \right| + c$$</p>
<p>$$\therefore$$ $$\alpha = {{ - 1} \over 2},\beta = {1 \over 2}$$</p>
<p>$$\therefore$$ $$\beta - \alpha = 1$$</p>
|
integer
|
jee-main-2023-online-30th-january-evening-shift
|
1ldv1n42c
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
<p>Let $$f(x) = \int {{{2x} \over {({x^2} + 1)({x^2} + 3)}}dx} $$. If $$f(3) = {1 \over 2}({\log _e}5 - {\log _e}6)$$, then $$f(4)$$ is equal to</p>
|
[{"identifier": "A", "content": "$${\\log _e}19 - {\\log _e}20$$"}, {"identifier": "B", "content": "$${\\log _e}17 - {\\log _e}18$$"}, {"identifier": "C", "content": "$${1 \\over 2}({\\log _e}19 - {\\log _e}17)$$"}, {"identifier": "D", "content": "$${1 \\over 2}({\\log _e}17 - {\\log _e}19)$$"}]
|
["D"]
| null |
$f(x)=\int \frac{2 x}{\left(x^{2}+1\right)\left(x^{2}+3\right)} d x$
<br/><br/>
$$
\begin{aligned}
& \text { Put } x^{2}=t \Rightarrow 2 x d x=d t \\\\
& f(x)=\int \frac{d t}{(t+1)(t+3)}=\int \frac{d t}{(t+2)^{2}-1} \\\\
& =\frac{1}{2} \log _{e}\left|\frac{t+1}{t+3}\right|+C \\\\
& f(x)=\frac{1}{2} \log _{e}\left(\frac{x^{2}+1}{x^{2}+3}\right)+C \Rightarrow \\\\
& f(3)=\frac{1}{2} \log _{e}\left(\frac{10}{12}\right)+C \\\\
& \because f(3)+\frac{1}{2}\left(\log _{e} 5-\log _{e} 6\right) \Rightarrow C=0 \\\\
& f(x)=\frac{1}{2} \log _{e}\left(\frac{x^{2}+1}{x^{2}+3}\right) \Rightarrow \\\\
& f(4)=\frac{1}{2}\left(\log _{e} 17-\log _{e} 19\right)
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-25th-january-morning-shift
|
lgnz42oa
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
Let $f(x)=\int \frac{d x}{\left(3+4 x^{2}\right) \sqrt{4-3 x^{2}}},|x|<\frac{2}{\sqrt{3}}$. If $f(0)=0$<br/><br/> and $f(1)=\frac{1}{\alpha \beta} \tan ^{-1}\left(\frac{\alpha}{\beta}\right)$,
$\alpha, \beta>0$, then $\alpha^{2}+\beta^{2}$ is equal to ____________.
|
[]
| null |
28
|
$$
\begin{aligned}
& f(x)=\int \frac{d x}{\left(3+4 x^2\right) \sqrt{4-3 x^2}} \\\\
& x=\frac{1}{t} \\\\
& =\int \frac{\frac{-1}{t^2} d t}{\frac{\left(3 t^2+4\right)}{t^2} \frac{\sqrt{4 t^2-3}}{t}} \\\\
& =\int \frac{-d t \cdot t}{\left(3 t^2+4\right) \sqrt{4 t^2-3}}: \text { Put } 4 t^2-3=z^2 \\\\
& =-\frac{1}{4} \int \frac{z d z}{\left(3\left(\frac{z^2+3}{4}\right)+4\right) z}
\end{aligned}
$$<br/><br/>
$$
\begin{aligned}
& =\int \frac{-d z}{3 z^2+25}=-\frac{1}{3} \int \frac{\mathrm{dz}}{\mathrm{z}^2+\left(\frac{5}{\sqrt{3}}\right)^2} \\\\
& =-\frac{1}{3} \frac{\sqrt{3}}{5} \tan ^{-1}\left(\frac{\sqrt{3} z}{5}\right)+C \\\\
& =-\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{5} \sqrt{4 t^2-3}\right)+C \\\\
& f(x)=-\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{5} \sqrt{\frac{4-3 \mathrm{x}^2}{x^2}}\right)+C \\\\
& \because \mathrm{f}(0)=0 \because \mathrm{c}=\frac{\pi}{10 \sqrt{3}}
\end{aligned}
$$<br/><br/>$$
\begin{aligned}
& f(1)=-\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{\sqrt{3}}{5}\right)+\frac{\pi}{10 \sqrt{3}} \\\\
& f(1)=\frac{1}{5 \sqrt{3}} \cot ^{-1}\left(\frac{\sqrt{3}}{5}\right)=\frac{1}{5 \sqrt{3}} \tan ^{-1}\left(\frac{5}{\sqrt{3}}\right) \\\\
& \alpha=5: \beta=\sqrt{3} \therefore \alpha^2+\beta^2=28
\end{aligned}
$$
|
integer
|
jee-main-2023-online-15th-april-morning-shift
|
1lgrgryfu
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
<p>Let $$I(x)=\int \sqrt{\frac{x+7}{x}} \mathrm{~d} x$$ and $$I(9)=12+7 \log _{e} 7$$. If $$I(1)=\alpha+7 \log _{e}(1+2 \sqrt{2})$$, then $$\alpha^{4}$$ is equal to _________.</p>
|
[]
| null |
64
|
Given integral: $$\int \sqrt{\frac{x+7}{x}} \, dx$$
<br/><br/>Let's make the substitution $$x = t^2$$. Then, $$dx = 2t \, dt$$.
<br/><br/>Substituting these values, the integral becomes :
<br/><br/>$$\int 2 \sqrt{t^2 + 7} \, dt$$
<br/><br/>Now, let's evaluate this integral :
<br/><br/>$$I(t) = 2\left(\frac{t}{2} \sqrt{t^2+7} + \frac{7}{2} \ln\left|t + \sqrt{t^2+7}\right|\right) + C$$
<br/><br/>Substituting back $$t = \sqrt{x}$$, we have :
<br/><br/>$$I(x) = 2\left(\frac{\sqrt{x}}{2} \sqrt{x+7} + \frac{7}{2} \ln\left|\sqrt{x} + \sqrt{x+7}\right|\right) + C$$
<br/><br/>Simplifying further :
<br/><br/>$$I(x) = \sqrt{x} \sqrt{x+7} + 7 \ln\left|\sqrt{x} + \sqrt{x+7}\right| + C$$
<p>We are given that $$I(9) = 12+7 \ln 7$$.</p>
<p>Let's substitute $$x = 9$$ and solve for the constant $$C$$:</p>
<p>$$12+7 \ln 7 = \sqrt{9} \sqrt{9+7}+7 \ln |\sqrt{9}+\sqrt{9+7}|+C$$
<br/><br/>$$ \Rightarrow $$ $$12+7 \ln 7 = 3 \sqrt{16}+7 \ln |\sqrt{9}+\sqrt{16}|+C$$
<br/><br/>$$ \Rightarrow $$ $$12+7 \ln 7 = 3 \cdot 4+7 \ln (3+\sqrt{16})+C$$
<br/><br/>$$ \Rightarrow $$ $$12+7 \ln 7 = 12+7 \ln (3+4)+C$$
<br/><br/>$$ \Rightarrow $$ $$12+7 \ln 7 = 12+7 \ln 7+C$$</p>
<p>From this equation, we can see that $$C = 0$$.</p>
<p>Now, we need to calculate $$I(1)$$ :</p>
<p>$$I(1) = \sqrt{1} \sqrt{1+7}+7 \ln |\sqrt{1}+\sqrt{1+7}|$$
<br/><br/>$$ \Rightarrow $$ $$I(1) = 1 \sqrt{8}+7 \ln (1+\sqrt{8})$$
<br/><br/>$$ \Rightarrow $$ $$I(1) = \sqrt{8}+7 \ln (1+2 \sqrt{2})$$</p>
<p>Therefore, $$\alpha = \sqrt{8}$$.</p>
<p>Finally, to find $$\alpha^4$$:</p>
<p>$$\alpha^4 = \left(\sqrt{8}\right)^4$$
<br/><br/>$$ \Rightarrow $$ $$\alpha^4 = 8^2$$
<br/><br/>$$ \Rightarrow $$ $$\alpha^4 = 64$$</p>
<p>Hence, $$\alpha^4$$ is equal to 64.</p>
|
integer
|
jee-main-2023-online-12th-april-morning-shift
|
1lgzzmzjh
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
<p>Let $$I(x)=\int \frac{(x+1)}{x\left(1+x e^{x}\right)^{2}} d x, x > 0$$. If $$\lim_\limits{x \rightarrow \infty} I(x)=0$$, then $$I(1)$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$\\frac{e+1}{e+2}-\\log _{e}(e+1)$$"}, {"identifier": "B", "content": "$$\\frac{e+1}{e+2}+\\log _{e}(e+1)$$"}, {"identifier": "C", "content": "$$\\frac{e+2}{e+1}-\\log _{e}(e+1)$$"}, {"identifier": "D", "content": "$$\\frac{e+2}{e+1}+\\log _{e}(e+1)$$"}]
|
["C"]
| null |
$$
\begin{aligned}
& \mathrm{I}=\int \frac{x+1}{x\left(1+x e^x\right)^2} d x \\\\
& \text { Put } 1+x e^x=t \Rightarrow x e^x=t-1 \\\\
& \Rightarrow\left(x e^x+e^x\right) d x=d t \\\\
& \Rightarrow e^x(x+1) d x=d t
\end{aligned}
$$
<br/><br/>$$
\therefore I=\int \frac{d t}{e^x \cdot x t^2}=\int \frac{d t}{(t-1) t^2}
$$
<br/><br/>Let $\frac{1}{t^2(t-1)}=\frac{\mathrm{A}}{(t-1)}+\frac{\mathrm{B} t+\mathrm{C}}{t^2}$
<br/><br/>$$
\Rightarrow 1=\mathrm{A} t^2+(\mathrm{B} t+\mathrm{C})(t-1)
$$
<br/><br/>Comparing coefficients of $t^2, t$ and constant terms, we get
<br/><br/>$$
A+B=0, C-B=0,-C=1
$$
<br/><br/>On solving above equations, we get
<br/><br/>$$
\begin{aligned}
& \mathrm{C}=-1,=\mathrm{B}, \mathrm{A}=1 \\\\
& \therefore \mathrm{I}=\int \frac{1}{t-1} d t+\int \frac{-t-1}{t^2} d t \\\\
& =\int \frac{1}{t-1} d t-\int \frac{1}{t} d t-\int \frac{1}{t^2} d t \\\\
& =\log |t-1|-\log |t|+\frac{1}{t}+\mathrm{C} \\\\
& \Rightarrow \mathrm{I}=\log \left|x e^x\right|-\log \left|1+x e^x\right|+\frac{1}{1+x e^x}+c
\end{aligned}
$$
<br/><br/>$$
=\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x}+C
$$
<br/><br/>Now, $\lim _{x \rightarrow \infty} \mathrm{I}(x)=0$
<br/><br/>$$
\begin{aligned}
& \Rightarrow \lim _{x \rightarrow \infty}\left\{\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x}+\mathrm{C}\right\}=0 \\\\
& \Rightarrow \lim _{x \rightarrow \infty}\left\{\log \left|\frac{e^x}{\frac{1}{x}+e^x}\right|+\frac{\frac{1}{x}}{\frac{1}{x}+e^x}+\mathrm{C}\right\} \\\\
& \Rightarrow 0+0+\mathrm{C}=0 \Rightarrow \mathrm{C}=0 \\\\
& \therefore \mathrm{I}(x)=\log \left|\frac{x e^x}{1+x e^x}\right|+\frac{1}{1+x e^x} \\\\
& \Rightarrow \mathrm{I}(1)=\log \left|\frac{e}{1+e}\right|+\frac{1}{1+e}=1-\log (1+e)+\frac{1}{1+e} \\\\
& =\frac{2+e}{1+e}-\log |1+e|
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-8th-april-morning-shift
|
jaoe38c1lscnrzev
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
<p>$$\text { The integral } \int \frac{\left(x^8-x^2\right) \mathrm{d} x}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} \text { is equal to : }$$</p>
|
[{"identifier": "A", "content": "$$\\log _{\\mathrm{e}}\\left(\\left|\\tan ^{-1}\\left(x^3+\\frac{1}{x^3}\\right)\\right|\\right)^{1 / 3}+\\mathrm{C}$$\n"}, {"identifier": "B", "content": "$$\\log _{\\mathrm{e}}\\left(\\left|\\tan ^{-1}\\left(x^3+\\frac{1}{x^3}\\right)\\right|\\right)+\\mathrm{C}$$\n"}, {"identifier": "C", "content": "$$\\log _{\\mathrm{e}}\\left(\\left|\\tan ^{-1}\\left(x^3+\\frac{1}{x^3}\\right)\\right|\\right)^{1 / 2}+\\mathrm{C}$$\n"}, {"identifier": "D", "content": "$$\\log _{\\mathrm{e}}\\left(\\left|\\tan ^{-1}\\left(x^3+\\frac{1}{x^3}\\right)\\right|\\right)^3+\\mathrm{C}$$"}]
|
["A"]
| null |
<p>$$I=\int \frac{x^8-x^2}{\left(x^{12}+3 x^6+1\right) \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)} d x$$</p>
<p>$$\begin{aligned}
& \text { Let } \tan ^{-1}\left(x^3+\frac{1}{x^3}\right)=t \\
& \Rightarrow \frac{1}{1+\left(x^3+\frac{1}{x^3}\right)^2} \cdot\left(3 x^2-\frac{3}{x^4}\right) d x=d t \\
& \Rightarrow \frac{x^6}{x^{12}+3 x^6+1} \cdot \frac{3 x^6-3}{x^4} d x=d t \\
& I=\frac{1}{3} \int \frac{d t}{t}=\frac{1}{3} \ln |t|+C \\
& I=\frac{1}{3} \ln \left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|+C \\
& I=\ln \left|\tan ^{-1}\left(x^3+\frac{1}{x^3}\right)\right|^{1 / 3}+C
\end{aligned}$$</p>
<p>Hence option (1) is correct.</p>
|
mcq
|
jee-main-2024-online-27th-january-evening-shift
|
jaoe38c1lseyv2z0
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
<p>For $$x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$$, if $$y(x)=\int \frac{\operatorname{cosec} x+\sin x}{\operatorname{cosec} x \sec x+\tan x \sin ^2 x} d x$$, and $$\lim _\limits{x \rightarrow\left(\frac{\pi}{2}\right)^{-}} y(x)=0$$ then $$y\left(\frac{\pi}{4}\right)$$ is equal to</p>
|
[{"identifier": "A", "content": "$$-\\frac{1}{\\sqrt{2}} \\tan ^{-1}\\left(\\frac{1}{\\sqrt{2}}\\right)$$\n"}, {"identifier": "B", "content": "$$\\tan ^{-1}\\left(\\frac{1}{\\sqrt{2}}\\right)$$\n"}, {"identifier": "C", "content": "$$\\frac{1}{2} \\tan ^{-1}\\left(\\frac{1}{\\sqrt{2}}\\right)$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{\\sqrt{2}} \\tan ^{-1}\\left(-\\frac{1}{2}\\right)$$"}]
|
["D"]
| null |
<p>$$\begin{aligned}
& y(x)=\int \frac{\left(1+\sin ^2 x\right) \cos x}{1+\sin ^4 x} d x \\
& \text { Put } \sin x=t \\
& =\int \frac{1+t^2}{t^4+1} d t=\frac{1}{\sqrt{2}} \tan ^{-1} \frac{\left(t-\frac{1}{t}\right)}{\sqrt{2}}+C \\
& x=\frac{\pi}{2}, t=1 \quad \therefore C=0 \\
& y\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}} \tan ^{-1}\left(-\frac{1}{2}\right)
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-29th-january-morning-shift
|
jaoe38c1lsfkcmax
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
<p>If $$\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-\theta)}} d x=A \sqrt{\cos \theta \tan x-\sin \theta}+B \sqrt{\cos \theta-\sin \theta \cot x}+C$$, where $$C$$ is the integration constant, then $$A B$$ is equal to</p>
|
[{"identifier": "A", "content": "$$2 \\sec \\theta$$\n"}, {"identifier": "B", "content": "$$8 \\operatorname{cosec}(2 \\theta)$$\n"}, {"identifier": "C", "content": "$$4 \\operatorname{cosec}(2 \\theta)$$\n"}, {"identifier": "D", "content": "$$4 \\sec \\theta$$"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& \text {} \int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x \sin (x-\theta)}} d x \\
& I=\int \frac{\sin ^{\frac{3}{2}} x+\cos ^{\frac{3}{2}} x}{\sqrt{\sin ^3 x \cos ^3 x(\sin x \cos \theta-\cos x \sin \theta)}} d x \\
& =\int \frac{\sin ^{\frac{3}{2}} x}{\sin ^{\frac{3}{2}} x \cos ^2 x \sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\cos ^{\frac{3}{2}} x}{\sin ^2 x \cos ^{\frac{3}{2}} x \sqrt{\cos \theta-\cot x \sin \theta}} d x= \\
& \int \frac{\sec ^2 x}{\sqrt{\tan x \cos \theta-\sin \theta}} d x+\int \frac{\operatorname{cosec}^2 x}{\sqrt{\cos \theta-\cot x \sin \theta}} d x \\
&
\end{aligned}$$</p>
<p>$$\mathrm{I}=\mathrm{I}_1+\mathrm{I}_2$$ ...... {Let}</p>
<p>For $$\mathrm{I}_1$$, let $$\tan \mathrm{x} \cos \theta-\sin \theta=\mathrm{t}^2$$</p>
<p>$$\sec ^2 x d x=\frac{2 t d t}{\cos \theta}$$</p>
<p>For $$\mathrm{I}_2$$, let $$\cos \theta-\cot \mathrm{x} \sin \theta=\mathrm{z}^2$$</p>
<p>$$\operatorname{cosec}^2 x d x=\frac{2 z d z}{\sin \theta}$$</p>
<p>$$\begin{aligned}
& I=I_1+I_2 \\
& =\int \frac{2 t d t}{\cos \theta t}+\int \frac{2 z d z}{\sin \theta z} \\
& =\frac{2 t}{\cos \theta}+\frac{2 z}{\sin \theta} \\
& =2 \sec \theta \sqrt{\tan x \cos \theta-\sin \theta}+2 \operatorname{cosec} \theta \sqrt{\cos \theta-\cot x \sin \theta} \\
& \text { Comparing } \\
& \quad A B=8 \operatorname{cosec} 2 \theta
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-29th-january-evening-shift
|
luy6z4yq
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
<p>Let $$\int \frac{2-\tan x}{3+\tan x} \mathrm{~d} x=\frac{1}{2}\left(\alpha x+\log _e|\beta \sin x+\gamma \cos x|\right)+C$$, where $$C$$ is the constant of integration. Then $$\alpha+\frac{\gamma}{\beta}$$ is equal to :</p>
|
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "4"}]
|
["D"]
| null |
<p>$$I=\int \frac{2-\tan x}{3+\tan x} d x=\int \frac{2 \cos x-\sin x}{3 \cos x+\sin x} d x$$</p>
<p>Put, $$2 \cos x-\sin x=a(-3 \sin x+\cos x)+ b(3 \cos x+\sin x)$$</p>
<p>$$\begin{aligned}
& a+3 b=2 \quad \text{.... (i)}\\
& -3 a+b=-1 \quad \text{.... (ii)}
\end{aligned}$$</p>
<p>From equation (i) and (ii) $$a=b=\frac{1}{2}$$</p>
<p>$$I=\frac{1}{2} \int \frac{-3 \sin x+\cos x}{3 \cos x+\sin x} d x+\frac{1}{2} \int \frac{3 \cos x+\sin x}{3 \cos x+\sin x} d x$$</p>
<p>$$I=\frac{1}{2} \ln |3 \cos x+\sin x|+\frac{1}{2} x= \frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|)$$</p>
<p>On comparing, we get</p>
<p>$$I=\frac{1}{2} \ln |3 \cos x+\sin x|+\frac{1}{2} x=\frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|)$$</p>
<p>On comparing, we get</p>
<p>$$\begin{aligned}
& \Rightarrow \frac{1}{2}(x+\log (|3 \cos x+\sin x|))= \\
& \qquad \frac{1}{2}(d x+\log |\beta \sin x+\gamma \cos x|)
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \alpha=1, \beta=1, \gamma=3 \\
& \alpha+\frac{\gamma}{\beta}=1+\frac{3}{1}=4
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-9th-april-morning-shift
|
lv3vefpq
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
<p>If $$\int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} \mathrm{~d} x=\mathrm{A}\left(\frac{\alpha x-1}{\beta x+3}\right)^B+\mathrm{C}$$, where $$\mathrm{C}$$ is the constant of integration, then the value of $$\alpha+\beta+20 \mathrm{AB}$$ is _________.</p>
|
[]
| null |
7
|
<p>$$\begin{aligned}
& I=\int \frac{1}{\sqrt[5]{(x-1)^4(x+3)^6}} d x \\
& \frac{x+3}{x-1}=t \Rightarrow d x=\frac{-4}{(t-1)^2} d t \Rightarrow x=\left(\frac{3+t}{t-1}\right) \\
& \Rightarrow(x-1)^4(x+3)^6=(x-1)^5(x+3)^5\left(\frac{x+3}{x-1}\right) \\
& I=\int \frac{\frac{-4}{(t-1)^2} d t}{t^{1 / 5}\left(\frac{3+t}{t-1}-1\right)\left(\frac{3+t}{t-3}+3\right)} \\
& I=\int \frac{-4 d t}{t^{1 / 5}(16 t)}=\frac{5}{4}\left(\frac{x-1}{x+3}\right)^{1 / 5}+c
\end{aligned}$$</p>
<p>Comparing,</p>
<p>$$\begin{aligned}
& \Rightarrow A=\frac{5}{4}, B=\frac{1}{5}, \alpha=1, \beta=1 \\
& \Rightarrow \alpha+\beta+20 A B=1+1+20 \times \frac{5}{4} \times \frac{1}{5}=7
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-8th-april-evening-shift
|
lv5gs8hd
|
maths
|
indefinite-integrals
|
integration-by-substitution
|
<p>Let $$I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x$$. If $$I(0)=3$$, then $$I\left(\frac{\pi}{12}\right)$$ is equal to</p>
|
[{"identifier": "A", "content": "$$\\sqrt3$$"}, {"identifier": "B", "content": "$$2\\sqrt3$$"}, {"identifier": "C", "content": "$$6\\sqrt3$$"}, {"identifier": "D", "content": "$$3\\sqrt3$$"}]
|
["D"]
| null |
<p>$$\begin{aligned}
& I(x)=\int \frac{6}{\sin ^2 x(1-\cot x)^2} d x \\
& I(x)=\int \frac{6}{(\sin x-\cos x)^2} d x \\
& =\int \frac{6 \sec ^2 x}{(\tan x-1)^2} d x
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \text { Let } \tan x=t \Rightarrow \sec ^2 x d x=d t \\
& =\int \frac{6 d t}{(t-1)^2} \\
& =-\frac{6}{(t-1)}+c \\
& =\frac{-6}{(\tan x-1)}+c \\
& I(x)=\frac{6}{1-\tan x}+c \\
& I(0)=3 \\
& \frac{6}{1-\tan 0}+c=3 \\
& c=-3
\end{aligned}$$</p>
<p>$$\begin{aligned}
I(x) & =\frac{6}{1-\tan x}-3 \\
I\left(\frac{\pi}{12}\right) & =\frac{6}{1-\tan \left(\frac{\pi}{12}\right)}-3 \\
& =\frac{6}{1-(2-\sqrt{3})}-3 \\
& =\frac{6}{\sqrt{3}-1}-3 \\
& =\frac{6-3 \sqrt{3}+3}{\sqrt{3}-1} \\
& =\frac{9-3 \sqrt{3}}{\sqrt{3}-1} \\
& =\frac{3 \sqrt{3}(\sqrt{3}-1)}{\sqrt{3}-1} \\
& =3 \sqrt{3}
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-8th-april-morning-shift
|
i7dl4gKHTGzDLWWS
|
maths
|
indefinite-integrals
|
standard-integral
|
$$\int {{{dx} \over {\cos x - \sin x}}} $$ is equal to
|
[{"identifier": "A", "content": "$${1 \\over {\\sqrt 2 }}\\log \\left| {\\tan \\left( {{x \\over 2} + {{3\\pi } \\over 8}} \\right)} \\right| + C$$ "}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 2 }}\\log \\left| {\\cot \\left( {{x \\over 2}} \\right)} \\right| + C$$ "}, {"identifier": "C", "content": "$${1 \\over {\\sqrt 2 }}\\log \\left| {\\tan \\left( {{x \\over 2} - {{3\\pi } \\over 8}} \\right)} \\right| + C$$ "}, {"identifier": "D", "content": "$$\\,{1 \\over {\\sqrt 2 }}\\log \\left| {\\tan \\left( {{x \\over 2} - {\\pi \\over 8}} \\right)} \\right| + C$$ "}]
|
["A"]
| null |
$$\int {{{dx} \over {\cos x - \sin x}}} $$
<br><br>$$ = \int {{{dx} \over {\sqrt 2 \cos \left( {x + {\pi \over 4}} \right)}}} $$
<br><br>$$ = {1 \over {\sqrt 2 }}\int {\sec \left( {x + {\pi \over 4}} \right)dx} $$
<br><br>$$ = {1 \over {\sqrt 2 }}\log \left| {\tan \left( {{\pi \over 4} + {x \over 2} + {\pi \over 8}} \right)} \right| + C$$
<br><br>$$\left[ \, \right.$$ As $$\int {\sec x\,dx} = \log \left| {\tan \left( {{\pi \over 4} + {x \over 2}} \right)} \right|$$ $$\left. \, \right]$$
<br><br>$$ = {1 \over {\sqrt 2 }}\log \left| {\tan \left( {{x \over 2} + {{3\pi } \over 8}} \right)} \right| + C$$
|
mcq
|
aieee-2004
|
MCOERemigEhmIwRq
|
maths
|
indefinite-integrals
|
standard-integral
|
If $$\int {{{\sin x} \over {\sin \left( {x - \alpha } \right)}}dx = Ax + B\log \sin \left( {x - \alpha } \right), + C,} $$ then value of
<br/>$$(A, B)$$ is
|
[{"identifier": "A", "content": "$$\\left( { - \\cos \\alpha ,\\sin \\alpha } \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { \\cos \\alpha ,\\sin \\alpha } \\right)$$"}, {"identifier": "C", "content": "$$\\left( { - \\sin \\alpha ,\\cos \\alpha } \\right)$$ "}, {"identifier": "D", "content": "$$\\left( { \\sin \\alpha ,\\cos \\alpha } \\right)$$"}]
|
["B"]
| null |
$$\int {{{\sin x} \over {\sin \left( {x - \alpha } \right)}}} dx$$
<br><br>$$ = \int {{{\sin \left( {x - \alpha + \alpha } \right)} \over {\sin \left( {x - \alpha } \right)}}} dx$$
<br><br>$$ = \int {{{\sin \left( {x - \alpha } \right)\cos \alpha + \cos \left( {x - \alpha } \right)\sin \alpha } \over {\sin \left( {x - \alpha } \right)}}} $$
<br><br>$$ = \int {\left\{ {\cos \alpha + \sin \alpha \,\cot \left. {\left( {x - \alpha } \right)} \right\}} \right.} dx$$
<br><br>$$ = \left( {\cos \alpha } \right)x + \left( {\sin \alpha } \right)\log \,\sin \left( {x - \alpha } \right) + C$$
<br><br>$$\therefore$$ $$A = \cos \alpha ,$$ $$B = \sin \alpha $$
|
mcq
|
aieee-2004
|
0TSlPppNziu0Gv6m
|
maths
|
indefinite-integrals
|
standard-integral
|
$$\int {{{dx} \over {\cos x + \sqrt 3 \sin x}}} $$ equals
|
[{"identifier": "A", "content": "$$\\log \\,\\tan \\,\\left( {{x \\over 2} + {\\pi \\over {12}}} \\right) + C$$ "}, {"identifier": "B", "content": "$$\\log \\,\\tan \\,\\left( {{x \\over 2} - {\\pi \\over {12}}} \\right) + C$$"}, {"identifier": "C", "content": "$$\\,{1 \\over 2}\\,\\log \\,\\tan \\,\\left( {{x \\over 2} + {\\pi \\over {12}}} \\right) + C$$ "}, {"identifier": "D", "content": "$$\\,{1 \\over 2}\\,\\log \\,\\tan \\,\\left( {{x \\over 2} - {\\pi \\over {12}}} \\right) + C$$ "}]
|
["C"]
| null |
$$I = \int {{{dx} \over {\cos x + \sqrt 3 \sin x}}} $$
<br><br>$$ \Rightarrow I = \int {{{dx} \over {2\left[ {{1 \over 2}\cos x + {{\sqrt 3 } \over 2}\sin x} \right]}}} $$
<br><br>$$ = {1 \over 2}\int {{{dx} \over {\left[ {\sin {\pi \over 6}\cos x + \cos {\pi \over 6}\sin x} \right]}}} $$
<br><br>$$ = {1 \over 2}.\int {{{dx} \over {\sin \left( {x + {\pi \over 6}} \right)}}} $$
<br><br>$$ \Rightarrow I = {1 \over 2}.\int {\cos ec\left( {x + {\pi \over 6}} \right)dx} $$
<br><br>But we know that
<br><br>$$\int {\cos ec\,x\,dx} = \log \left| {\left( {\tan x/2} \right)} \right| + C$$
<br><br>$$\therefore$$ $$I = {1 \over 2}.\log \,\tan \left( {{x \over 2} + {\pi \over {12}}} \right) + C$$
|
mcq
|
aieee-2007
|
iKxtxDJH7ZA06wso35mnL
|
maths
|
indefinite-integrals
|
standard-integral
|
The integral
<br/><br/>$$\int {\sqrt {1 + 2\cot x(\cos ecx + \cot x)\,} \,\,dx} $$
<br/><br/>$$\left( {0 < x < {\pi \over 2}} \right)$$ is equal to :
<br/><br/>(where C is a constant of integration)
|
[{"identifier": "A", "content": "4 log(sin $${x \\over 2}$$ ) + C"}, {"identifier": "B", "content": "2 log(sin $${x \\over 2}$$ ) + C"}, {"identifier": "C", "content": "2 log(cos $${x \\over 2}$$ ) + C"}, {"identifier": "D", "content": "4 log(cos $${x \\over 2}$$) + C"}]
|
["B"]
| null |
Let, I = $$\int {\sqrt {1 + 2\cot x\cos ec + 2{{\cot }^2}x} .dx} $$<br><br>
$$ \Rightarrow $$ I = $$\int {\sqrt {{{{{\sin }^2}x + 2\cos x + 2{{\cos }^2}x} \over {{{\sin }^2}x}}} .dx} $$<br><br>
$$ \Rightarrow $$ I = $$\int {\sqrt {{{1 + 2\cos x + {{\cos }^2}x} \over {\sin x}}} .dx} $$<br><br>
$$ \Rightarrow $$ I = $$\int {\left| {{{1 + \cos x} \over {\sin x}}} \right|dx} $$<br><br>
$$ \Rightarrow $$ I = $$\int {\left| {\cos ec\,x + \cot x} \right|.dx} $$<br><br>
$$ \Rightarrow $$ I = $$\log \left| {\cos ec\,x - \cot x} \right| + \log \left| {\sin x} \right| + C$$<br><br>
$$ \Rightarrow $$ I = $$\log \left| {1 - \cos x} \right| + C$$<br><br>
$$ \Rightarrow $$ I = $$\log \left| {2{{\sin }^2}{x \over 2}} \right| + C$$<br><br>
$$ \Rightarrow $$ I = $$\log \left| {{{\sin }^2}{x \over 2}} \right| + \log 2+ C$$<br><br>
$$ \Rightarrow $$ I = 2$$\log \left| {{{\sin }}{x \over 2}} \right| + C_1$$<br>
|
mcq
|
jee-main-2017-online-8th-april-morning-slot
|
fBGcqpp56sjJxOFIirviI
|
maths
|
indefinite-integrals
|
standard-integral
|
If $$\,\,\,$$ f$$\left( {{{3x - 4} \over {3x + 4}}} \right)$$ = x + 2, x $$ \ne $$ $$-$$ $${4 \over 3}$$, and
<br/><br/>$$\int {} $$f(x) dx = A log$$\left| {} \right.$$1 $$-$$ x $$\left| {} \right.$$ + Bx + C,
<br/><br/>then the ordered pair (A, B) is equal to :
<br/><br/>(where C is a constant of integration)
|
[{"identifier": "A", "content": "$$\\left( {{8 \\over 3},{2 \\over 3}} \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { - {8 \\over 3},{2 \\over 3}} \\right)$$ "}, {"identifier": "C", "content": "$$\\left( { - {8 \\over 3}, - {2 \\over 3}} \\right)$$ "}, {"identifier": "D", "content": "$$\\left( { {8 \\over 3}, - {2 \\over 3}} \\right)$$ "}]
|
["B"]
| null |
Given,
<br><br>f$$\left( {{{3x - 4} \over {3x + 4}}} \right)$$ = x + 2, x $$ \ne $$ $$-$$ $${4 \over 3}$$
<br><br>Let, $${{3x - 4} \over {3x + 4}}$$ = t
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 3x $$-$$ 4 = 3tx + 4t
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 3x $$-$$ 3tx = 4t + 4
<br><br>$$ \Rightarrow $$$$\,\,\,$$ x = $${{4t + 4} \over {3 - 3t}}$$
<br><br>So, f(t) = $${{4t + 4} \over {3 - 3t}}$$ + 2 = $${{10 - 2t} \over {3 - 3t}}$$
<br><br>$$\therefore\,\,\,$$ f (x) = $${{10 - 2x} \over {3 - 3x}}$$
<br><br>$$\therefore\,\,\,$$ $$\int {f(x)\,dx} $$
<br><br>= $$\int {{{2x - 10} \over {3x - 3}}} \,dx$$
<br><br>= $$\int {{{2x} \over {3x - 3}} - 10\int {{{dx} \over {3x - 3}}} } $$
<br><br>= $${2 \over 3}\int {{{x - 1} \over {x + 1}}} dx + {2 \over 3}\int {{{dx} \over {x - 1}} - {{10} \over 3}} \int {{{dx} \over {x - 1}}} $$
<br><br>= $${2 \over 3}$$ x + $${2 \over 3}$$ log $$\left| {x - 1} \right|$$ $$-$$ $${{10} \over 3}$$ log $$\left| {x - 1} \right|$$ + C
<br><br>= $${2 \over 3}$$ x $$-$$ $${8 \over 3}$$ log $$\left| {x - 1} \right|$$ + C
<br><br>$$\therefore\,\,\,$$ A = $$-$$ $${8 \over 3}$$ and B = $${2 \over 3}$$
|
mcq
|
jee-main-2017-online-9th-april-morning-slot
|
u7bfYkLqQ1Ri6xiLX6tgK
|
maths
|
indefinite-integrals
|
standard-integral
|
If $$f\left( {{{x - 4} \over {x + 2}}} \right) = 2x + 1,$$ (x $$ \in $$ <b>R</b> $$-$${1, $$-$$ 2}), then $$\int f \left( x \right)dx$$ is equal to :
<br/>(where C is a constant of integration)
|
[{"identifier": "A", "content": "12 log<sub>e</sub> | 1 $$-$$ x | + 3x + C"}, {"identifier": "B", "content": "$$-$$ 12 log<sub>e</sub> | 1 $$-$$ x | $$-$$ 3x + C"}, {"identifier": "C", "content": "12 log<sub>e</sub> | 1 $$-$$ x | $$-$$ 3x + C"}, {"identifier": "D", "content": "$$-$$ 12 log<sub>e</sub> | 1 $$-$$ x | + 3x + C"}]
|
["B"]
| null |
Let, $${{{x - 4} \over {x + 2}}}$$ = t<br><br>
$$ \Rightarrow $$ x - 4 = t(x+2)<br><br>
$$ \Rightarrow $$ x (1 -t) = 2(t+2)<br><br>
$$ \Rightarrow $$ x = $${{2(t + 2)} \over {1 - t}}$$<br><br>
$$ \therefore $$ f(t) = 2($${{2(t + 2)} \over {1 - t}}$$) + 1<br><br>
= $${{4t + 8} \over {1 - t}} + 1$$<br><br>
= $${{3t + 9} \over {1 - t}}$$<br><br>
= $${{3(t + 3)} \over {1 - t}}$$<br><br>
= $${{3(t - 1 + 4)} \over {1 - t}}$$<br><br>
= - 3 + $$12 \over 1 - t$$<br><br>
$$ \therefore $$ f(x) = - 3 + $$12 \over 1 - x$$<br><br>
$$ \therefore $$ $$\int {f(x)dx} $$<br><br>
= $$\int {\left( { - 3 + {{12} \over {1 - x}}} \right)dx} $$<br><br>
= $$-12{\log _e}|1 - x|$$ - 3x + C
|
mcq
|
jee-main-2018-online-15th-april-morning-slot
|
1YrtLQqQIm1HgxI1ExfLd
|
maths
|
indefinite-integrals
|
standard-integral
|
If $$\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} \,\,dx = A\sqrt {7 - 6x - {x^2}} + B{\sin ^{ - 1}}\left( {{{x + 3} \over 4}} \right) + C$$
<br/>(where C is a constant of integration), then the ordered pair (A, B) is equal to :
|
[{"identifier": "A", "content": "(2, 1)"}, {"identifier": "B", "content": "($$-$$ 2, $$-$$1)"}, {"identifier": "C", "content": "($$-$$ 2, 1)"}, {"identifier": "D", "content": "(2, $$-$$1)"}]
|
["B"]
| null |
We can write,
<br><br>7 - 6x - x<sup>2</sup> = 16 - (x + 3)<sup>2</sup>
<br><br>and $${d \over {dx}}\left( {7 - 6x - {x^2}} \right) = - (2x + 6)$$
<br><br>So, $$\int {{{2x + 5} \over {\sqrt {7 - 6x - {x^2}} }}} dx$$
<br><br>= $$\int {{{2x + 6 - 1} \over {\sqrt {7 - 6x - {x^2}} }}} dx$$
<br><br>= $$\int {{{2x + 6} \over {\sqrt {7 - 6x - {x^2}} }}} dx$$ -
<br> $$\int {{1 \over {\sqrt {16 - {{(x + 3)}^2}} }}} dx$$
<br><br>= $$\int {{{2x + 6} \over {\sqrt {7 - 6x - {x^2}} }}} dx$$ -
<br> $$\int {{1 \over {\sqrt {{{\left( 4 \right)}^2} - {{(x + 3)}^2}} }}} dx$$
<br><br>= $$ - 2\sqrt {7 - 6x - {x^2}} $$ - $${\sin ^{ - 1}}\left( {{{x + 3} \over 4}} \right)$$ + C
<br><br>By comparing with the given equation in the question we get,
<br><br>A = - 2 and B = - 1
|
mcq
|
jee-main-2018-online-15th-april-evening-slot
|
PYnBJilIrz8XFwr7gR18hoxe66ijvwvro3u
|
maths
|
indefinite-integrals
|
standard-integral
|
$$\int {{e^{\sec x}}}$$ $$(\sec x\tan xf(x) + \sec x\tan x + se{x^2}x)dx$$<br/>
= e<sup>secx</sup>f(x) + C then a possible choice of f(x) is :-
|
[{"identifier": "A", "content": "x sec x + tan x + 1/2"}, {"identifier": "B", "content": "sec x + xtan x - 1/2"}, {"identifier": "C", "content": "sec x - tan x - 1/2"}, {"identifier": "D", "content": "sec x + tan x + 1/2"}]
|
["D"]
| null |
Given
<br><br>$$\int {{e^{\sec x}}\left( {\sec x\tan x\,f(x) + (sec\,x\,tan\,x\, + se{c^2}x)} \right)} $$<br><br>
$$ = {e^{\sec x}}f(x) + C$$
<br><br>Differentiating both sides with respect to x,
<br><br>$${e^{\sec x}}.\sec x\tan x\,f\left( x \right)$$ + $${e^{\sec x}}\left( {\sec x\tan x + {{\sec }^2}x} \right)$$
<br><br>= $${e^{\sec x}}$$. f'(x) + f(x)$${e^{\sec x}}$$.$$\sec x\tan x$$
<br><br>$$ \Rightarrow $$ $${e^{\sec x}}\left( {\sec x\tan x + {{\sec }^2}x} \right)$$ = $${e^{\sec x}}$$. f'(x)
<br><br>$$ \Rightarrow $$ f'(x) = $${\sec x\tan x + {{\sec }^2}x}$$
<br><br>$$ \therefore $$ $$f(x) = \int {\left( {\left( {\sec x\tan x} \right) + {{\sec }^2}x} \right)dx} $$<br><br>
$$ \therefore $$ $$f(x) = \sec x + \tan x + C$$
<br><br>From question we can not find the value of C. So we have to choose any random value of C where ($$\sec x + \tan x$$) present.
<br><br>Only in option D, ($$\sec x + \tan x$$) term present. So only possible option is D.
|
mcq
|
jee-main-2019-online-9th-april-evening-slot
|
dsxm8kHFIFngSyvhIT3rsa0w2w9jx5brt6e
|
maths
|
indefinite-integrals
|
standard-integral
|
The integral $$\int {{{2{x^3} - 1} \over {{x^4} + x}}} dx$$ is equal to :
<br/>(Here C is a constant of integration)
|
[{"identifier": "A", "content": "$${\\log _e}{{\\left| {{x^3} + 1} \\right|} \\over {{x^2}}} + C$$"}, {"identifier": "B", "content": "$${1 \\over 2}{\\log _e}{{\\left| {{x^3} + 1} \\right|} \\over {{x^2}}} + C$$"}, {"identifier": "C", "content": "$${\\log _e}\\left| {{{{x^3} + 1} \\over x}} \\right| + C$$"}, {"identifier": "D", "content": "$${1 \\over 2}{\\log _e}{{{{\\left( {{x^3} + 1} \\right)}^2}} \\over {\\left| {{x^3}} \\right|}} + C$$"}]
|
["C"]
| null |
$$\int {{{2{x^3} - 1} \over {{x^4} + x}}dx = \int {{{2x - {x^{ - 2}}} \over {{x^2} + {x^{ - 1}}}}dx = \ln ({x^2} + {x^{ - 1}}} } ) + c$$<br><br>
$$ \Rightarrow \ln ({x^3} + 1) - \ln x + c$$
|
mcq
|
jee-main-2019-online-12th-april-morning-slot
|
p6chnBVIYFmsOS0wrDfOu
|
maths
|
indefinite-integrals
|
standard-integral
|
$$\int {{{\sin {{5x} \over 2}} \over {\sin {x \over 2}}}dx} $$ is equal to<br/>
(where c is a constant of integration)
|
[{"identifier": "A", "content": "2x + sinx + 2sin2x + c"}, {"identifier": "B", "content": "x + 2sinx + sin2x + c"}, {"identifier": "C", "content": "x + 2sinx + 2sin2x + c"}, {"identifier": "D", "content": "2x + sinx + sin2x + c"}]
|
["B"]
| null |
$$\int {{{\sin {{5x} \over 2}} \over {\sin {x \over 2}}}dx} $$
<br><br>= $$\int {{{2\cos {x \over 2}.\sin {{5x} \over 2}} \over {2\cos {x \over 2}.\sin {x \over 2}}}} dx $$
<br><br>= $$\int {{{\sin \left( {{{5x} \over 2} + {x \over 2}} \right) + \sin \left( {{{5x} \over 2} - {x \over 2}} \right)} \over {\sin x}}} dx$$
<br><br>= $$\int {{{\sin 3x + \sin 2x} \over {\sin x}}} dx$$
<br><br>Use sin2x = 2sinxcosx and sin3x = 3sinx
– 4sin<sup>3</sup>x
<br><br>= $$\int {{{3\sin x - 4{{\sin }^3}x + 2\sin x\cos x} \over {\sin x}}} dx$$
<br><br>= $$\int {\left( {3 - 4{{\sin }^2}x + 2\cos x} \right)} dx$$
<br><br>= $$\int {\left( {3 - 2\left( {1 - \cos 2x} \right) + 2\cos x} \right)} dx$$
<br><br>= $$\int {\left( {1 + 2\cos 2x + 2\cos x} \right)} dx$$
<br><br>= x + sin2x + 2sinx + C
|
mcq
|
jee-main-2019-online-8th-april-morning-slot
|
RognVI7rIVMeBEOJCg7k9k2k5iqewxy
|
maths
|
indefinite-integrals
|
standard-integral
|
If ƒ'(x) = tan<sup>–1</sup>(secx + tanx), $$ - {\pi \over 2} < x < {\pi \over 2}$$,
<br/>and
ƒ(0) = 0, then ƒ(1) is equal to :
|
[{"identifier": "A", "content": "$${1 \\over 4}$$"}, {"identifier": "B", "content": "$${{\\pi - 1} \\over 4}$$"}, {"identifier": "C", "content": "$${{\\pi + 1} \\over 4}$$"}, {"identifier": "D", "content": "$${{\\pi + 2} \\over 4}$$"}]
|
["C"]
| null |
ƒ'(x) = tan<sup>–1</sup>(secx + tanx)
<br><br>$$ \Rightarrow $$ ƒ'(x) = $${\tan ^{ - 1}}\left( {{{1 + \sin x} \over {\cos x}}} \right)$$ = $${\tan ^{ - 1}}\left( {{{1 + \tan {x \over 2}} \over {1 - \tan {x \over 2}}}} \right)$$
<br><br>= $${\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} + {x \over 2}} \right)} \right)$$
<br><br>As $$ - {\pi \over 2} < x < {\pi \over 2}$$
<br><br>$$ \Rightarrow $$ $$0 < {\pi \over 4} + {x \over 2} < {\pi \over 2}$$
<br><br>$$ \therefore $$ ƒ'(x) = $${\pi \over 4} + {x \over 2}$$
<br><br>$$ \Rightarrow $$ f(x) = $${\pi \over 4}x + {{{x^2}} \over 4} + c$$
<br><br>$$ \because $$ ƒ(0) = 0 $$ \Rightarrow $$ c = 0
<br><br>$$ \therefore $$ f(x) = $${\pi \over 4}x + {{{x^2}} \over 4}$$
<br><br>$$ \Rightarrow $$ f(1) = $${{{\pi + 1} \over 4}}$$
|
mcq
|
jee-main-2020-online-9th-january-morning-slot
|
GbIwcD7jCGMKEpGbhmjgy2xukf7gh1qh
|
maths
|
indefinite-integrals
|
standard-integral
|
The integral $$\int {{{\left( {{x \over {x\sin x + \cos x}}} \right)}^2}dx} $$ is equal to <br/>
(where C is a constant of integration):
|
[{"identifier": "A", "content": "$$\\sec x - {{x\\tan x} \\over {x\\sin x + \\cos x}} + C$$"}, {"identifier": "B", "content": "$$\\sec x + {{x\\tan x} \\over {x\\sin x + \\cos x}} + C$$"}, {"identifier": "C", "content": "$$\\tan x - {{x\\sec x} \\over {x\\sin x + \\cos x}} + C$$"}, {"identifier": "D", "content": "$$\\tan x + {{x\\sec x} \\over {x\\sin x + \\cos x}} + C$$"}]
|
["C"]
| null |
$${\int {\left( {{x \over {x\sin x + \cos x}}} \right)} ^2}dx $$
<br><br>$$= \int {\left( {{x \over {\cos x}}} \right).{{x\cos x\,dx} \over {{{(x\sin x + \cos x)}^2}}}} $$<br><br>= $${x \over {\cos x}}\left( { - {1 \over {x\sin x + \cos x}}} \right) + \int {\left( {{{\cos x + x\sin x} \over {{{\cos }^2}x}}} \right)} \left( {{1 \over {x\sin x + \cos x}}} \right)dx$$<br><br>$$ = - {{x\sec x} \over {x\sin x + \cos x}} + \int {{{\sec }^2}x\,dx} $$<br><br>$$ = - {{x\sec x} \over {x\sin x + \cos x}} + \tan x + C$$
|
mcq
|
jee-main-2020-online-4th-september-morning-slot
|
1l57o9ptf
|
maths
|
indefinite-integrals
|
standard-integral
|
If $$\int {{{({x^2} + 1){e^x}} \over {{{(x + 1)}^2}}}dx = f(x){e^x} + C} $$, where C is a constant, then $${{{d^3}f} \over {d{x^3}}}$$ at x = 1 is equal to :
|
[{"identifier": "A", "content": "$$ - {3 \\over 4}$$"}, {"identifier": "B", "content": "$${3 \\over 4}$$"}, {"identifier": "C", "content": "$$ - {3 \\over 2}$$"}, {"identifier": "D", "content": "$${3 \\over 2}$$"}]
|
["B"]
| null |
<p>$$I = \int {{{{e^x}({x^2} + 1)} \over {{{(x + 1)}^2}}}dx = f(x){e^x} + c} $$</p>
<p>$$ = \int {{{{e^x}({x^2} - 1 + 1 + 1)} \over {{{(x + 1)}^2}}}dx} $$</p>
<p>$$ = \int {{e^x}\left[ {{{x - 1} \over {x + 1}} + {2 \over {{{(x + 1)}^2}}}} \right]dx} $$</p>
<p>$$ = {e^x}\left( {{{x - 1} \over {x + 1}}} \right) + c$$</p>
<p>$$\therefore$$ $$f(x) = {{x - 1} \over {x + 1}}$$</p>
<p>$$f(x) = 1 - {2 \over {x + 1}}$$</p>
<p>$$f'(x) = 2{\left( {{1 \over {x + 1}}} \right)^2}$$</p>
<p>$$f''(x) = - 4{\left( {{1 \over {x + 1}}} \right)^3}$$</p>
<p>$$f'''(x) = {{12} \over {{{(x + 1)}^4}}}$$</p>
<p>for $$x = 1$$</p>
<p>$$f'''(1) = {{12} \over {{2^4}}} = {{12} \over {16}} = {3 \over 4}$$</p>
|
mcq
|
jee-main-2022-online-27th-june-morning-shift
|
1l6hzevej
|
maths
|
indefinite-integrals
|
standard-integral
|
<p>$$
\text { The integral } \int \frac{\left(1-\frac{1}{\sqrt{3}}\right)(\cos x-\sin x)}{\left(1+\frac{2}{\sqrt{3}} \sin 2 x\right)} d x \text { is equal to }
$$</p>
|
[{"identifier": "A", "content": "$$\\frac{1}{2} \\log _{e}\\left|\\frac{\\tan \\left(\\frac{x}{2}+\\frac{\\pi}{12}\\right)}{\\tan \\left(\\frac{x}{2}+\\frac{\\pi}{6}\\right)}\\right|+C$$"}, {"identifier": "B", "content": "$$\\frac{1}{2} \\log _{e}\\left|\\frac{\\tan \\left(\\frac{x}{2}+\\frac{\\pi}{6}\\right)}{\\tan \\left(\\frac{x}{2}+\\frac{\\pi}{3}\\right)}\\right|+C$$"}, {"identifier": "C", "content": "$$\n\\log _{e}\\left|\\frac{\\tan \\left(\\frac{x}{2}+\\frac{\\pi}{6}\\right)}{\\tan \\left(\\frac{x}{2}+\\frac{\\pi}{12}\\right)}\\right|+C$$"}, {"identifier": "D", "content": "$$\\frac{1}{2} \\log _{e}\\left|\\frac{\\tan \\left(\\frac{x}{2}-\\frac{\\pi}{12}\\right)}{\\tan \\left(\\frac{x}{2}-\\frac{\\pi}{6}\\right)}\\right|+C\n$$"}]
|
["A"]
| null |
<p>$$ = \int {{{\left( {1 - {1 \over {\sqrt 3 }}} \right)(\cos x - \sin x)} \over {\left( {1 + {2 \over {\sqrt 3 }}\sin 2x} \right)}}dx} $$</p>
<p>$$ = \int {{{\left( {{{\sqrt 3 - 1} \over {\sqrt 3 }}} \right)\sqrt 2 \sin \left( {{\pi \over 4} - x} \right)} \over {\left( {{2 \over {\sqrt 3 }}} \right)\left( {\sin {\pi \over 3} + \sin 2x} \right)}}dx} $$</p>
<p>$$ = \int {{{{{(\sqrt 3 - 1)} \over {\sqrt 2 }}\sin \left( {{\pi \over 4} - x} \right)} \over {\left( {\sin {\pi \over 3} + \sin 2x} \right)}}dx} $$</p>
<p>$$ = \int {{{{{\sqrt 3 - 1} \over {2\sqrt 2 }}\sin \left( {{\pi \over 4} - x} \right)} \over {\sin \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right)}}dx} $$</p>
<p>$$ = {1 \over 2}\int {{{2\sin {\pi \over {12}}\sin \left( {{\pi \over 4} - x} \right)} \over {\sin \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right)}}dx} $$</p>
<p>$$ = {1 \over 2}\int {{{\cos \left( {{\pi \over 6} - x} \right) - \cos \left( {{\pi \over 3} - x} \right)} \over {\sin \left( {{\pi \over 6} + x} \right)\cos \left( {{\pi \over 6} - x} \right)}}dx} $$</p>
<p>$$ = {1 \over 2}\left[ {\int {{\mathop{\rm cosec}\nolimits} \left( {{\pi \over 6} + x} \right)dx - \int {\sec \left( {{\pi \over 6} - x} \right)dx} } } \right]$$</p>
<p>$$ = {1 \over 2}\left[ {\ln \left| {\tan \left( {{\pi \over {12}} + {x \over 2}} \right)} \right| - \int {\cos ec\left( {{\pi \over 3} - x} \right)dx} } \right]$$</p>
<p>$$ = {1 \over 2}\left[ {\ln \left| {\tan \left( {{\pi \over {12}} + {x \over 2}} \right)} \right| - \ln \left| {{\pi \over 6} + {x \over 2}} \right|} \right] + C$$</p>
<p>$$ = {1 \over 2}\ln \left| {{{\tan \left( {{\pi \over 2} + {x \over 2}} \right)} \over {\tan \left( {{\pi \over 6} + {x \over 2}} \right)}}} \right| + C$$</p>
|
mcq
|
jee-main-2022-online-26th-july-evening-shift
|
1lgvpqqkm
|
maths
|
indefinite-integrals
|
standard-integral
|
<p>For $$\alpha, \beta, \gamma, \delta \in \mathbb{N}$$, if $$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$$
, where $$e=\sum_\limits{n=0}^{\infty} \frac{1}{n !}$$ and $$\mathrm{C}$$ is constant of integration, then $$\alpha+2 \beta+3 \gamma-4 \delta$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$-8$$"}, {"identifier": "B", "content": "$$-4$$"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "4<br/><br/>"}]
|
["D"]
| null |
We have,
<br/><br/>$$\int\left(\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right) \log _{e} x d x=\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$$
<br/><br/>$$
\begin{aligned}
\left(\frac{x}{e}\right)^{2 x} & =\left(\frac{e^{\log _e x}}{e}\right)^{2 x} \left[\because x=e^{\log _e x}\right]
\\\\
& =e^{2\left(x \log _e x-x\right)} .........(i)
\end{aligned}
$$
<br/><br/>$$
\text { Also, }\left(\frac{e}{x}\right)^{2 x}=\left(\frac{x}{e}\right)^{-2 x}=e^{-2\left(x \log _e x-x\right)}
$$ .........(ii)
<br/><br/>Let,
<br/><br/>$$
\begin{aligned}
I & =\int\left[\left(\frac{x}{e}\right)^{2 x}+\left(\frac{e}{x}\right)^{2 x}\right] \log _e x d x \\\\
& =\int\left[e^{2\left(x \log _e x-x\right)}+e^{-2\left(x \log _e x-x\right)}\right] \log _e x d x ~~~~\text{ [From Eqs. (i) and (ii)] }
\end{aligned}
$$
<br/><br/>Let $x \log _e x-x=t$
<br/><br/>$$
\log _e x d x=d t
$$
<br/><br/>$$
\begin{aligned}
I & =\int\left(e^{2 t}+e^{-2 t}\right) d t=\frac{e^{2 t}}{2}-\frac{e^{-2 t}}{2}+C \\\\
& =\frac{1}{2}\left(\frac{x}{e}\right)^{2 x}-\frac{1}{2}\left(\frac{e}{x}\right)^{2 x}+C
\end{aligned}
$$
<br/><br/>On comparing I with $\frac{1}{\alpha}\left(\frac{x}{e}\right)^{\beta x}-\frac{1}{\gamma}\left(\frac{e}{x}\right)^{\delta x}+C$
<br/><br/>$$
\begin{aligned}
& \alpha=2, \beta=2, \gamma=2, \delta=2 \\\\
& \therefore \alpha+2 \beta+3 \gamma-4 \delta=2+2 \times 2+3 \times 2-4 \times 2=4
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-10th-april-evening-shift
|
1lgyl5enm
|
maths
|
indefinite-integrals
|
standard-integral
|
<p>The integral $$
\int\left[\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right] \ln \left(\frac{e x}{2}\right) d x
$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$\\left(\\frac{x}{2}\\right)^{x}+\\left(\\frac{2}{x}\\right)^{x}+C$$"}, {"identifier": "B", "content": "$$\\left(\\frac{x}{2}\\right)^{x}-\\left(\\frac{2}{x}\\right)^{x}+C$$"}, {"identifier": "C", "content": "$$\\left(\\frac{x}{2}\\right)^{x} \\log _{2}\\left(\\frac{2}{x}\\right)+C$$"}, {"identifier": "D", "content": "None"}]
|
["B"]
| null |
<p>To solve the integral:</p>
<p>$ I = \int\left[\left(\frac{x}{2}\right)^x+\left(\frac{2}{x}\right)^x\right] \ln \left(\frac{e x}{2}\right) \, dx $</p>
<p>we start by simplifying the logarithmic term:</p>
<p>$ \ln\left(\frac{e x}{2}\right) = \ln(e) + \ln\left(\frac{x}{2}\right) = 1 + \ln\left(\frac{x}{2}\right) $</p>
<p>So the integral becomes:</p>
<p>$ I = \int \left[ \left( \frac{x}{2} \right)^x + \left( \frac{2}{x} \right)^x \right] \left[ 1 + \ln\left( \frac{x}{2} \right) \right] \, dx $</p>
<p>Let's denote:</p>
<p><p>$ A = \left( \dfrac{x}{2} \right)^x $</p></p>
<p><p>$ B = \left( \dfrac{2}{x} \right)^x $</p></p>
<p>Then, the integral can be split:</p>
<p>$ I = \int \left[ A(1 + \ln\left( \frac{x}{2} \right)) + B(1 + \ln\left( \frac{x}{2} \right)) \right] \, dx $</p>
<h3>Calculating $ \int A(1 + \ln\left( \frac{x}{2} \right)) \, dx $:</h3>
<p>First, find the derivative of $ A $:</p>
<p>$ \frac{dA}{dx} = A \left[ \ln\left( \frac{x}{2} \right) + 1 \right] $</p>
<p>This means:</p>
<p>$ A \left[ 1 + \ln\left( \frac{x}{2} \right) \right] = \frac{dA}{dx} $</p>
<p>Therefore:</p>
<p>$ \int A \left[ 1 + \ln\left( \frac{x}{2} \right) \right] \, dx = \int \frac{dA}{dx} \, dx = A + C_1 = \left( \frac{x}{2} \right)^x + C_1 $</p>
<h3>Calculating $ \int B(1 + \ln\left( \frac{x}{2} \right)) \, dx $:</h3>
<p>Note that:</p>
<p>$ \ln\left( \frac{x}{2} \right) = \ln x - \ln 2 $</p>
<p>$ \ln\left( \frac{2}{x} \right) = \ln 2 - \ln x $</p>
<p>The derivative of $ B $ is:</p>
<p>$ \frac{dB}{dx} = B \left[ \ln\left( \frac{2}{x} \right) + 1 \right] = B \left[ \ln 2 - \ln x + 1 \right] $</p>
<p>But in our integrand, we have $ B \left[ 1 + \ln\left( \frac{x}{2} \right) \right] = B \left[ 1 + \ln x - \ln 2 \right] $. Notice that:</p>
<p>$ 1 + \ln x - \ln 2 = - \left( \ln 2 - \ln x + 1 \right) $</p>
<p>Thus:</p>
<p>$ B \left[ 1 + \ln\left( \frac{x}{2} \right) \right] = - \frac{dB}{dx} $</p>
<p>Therefore:</p>
<p>$ \int B \left[ 1 + \ln\left( \frac{x}{2} \right) \right] \, dx = -\int \frac{dB}{dx} \, dx = -B + C_2 = -\left( \frac{2}{x} \right)^x + C_2 $</p>
<h3>Combining the Results:</h3>
<p>Adding both integrals:</p>
<p>$ I = \left( \frac{x}{2} \right)^x - \left( \frac{2}{x} \right)^x + C $</p>
<p><strong>Answer: Option B</strong></p>
<p>$ \left( \frac{x}{2} \right)^x - \left( \frac{2}{x} \right)^x + C $</p>
|
mcq
|
jee-main-2023-online-8th-april-evening-shift
|
lvb294vs
|
maths
|
indefinite-integrals
|
standard-integral
|
<p>If $$\int \frac{1}{\mathrm{a}^2 \sin ^2 x+\mathrm{b}^2 \cos ^2 x} \mathrm{~d} x=\frac{1}{12} \tan ^{-1}(3 \tan x)+$$ constant, then the maximum value of $$\mathrm{a} \sin x+\mathrm{b} \cos x$$, is :</p>
|
[{"identifier": "A", "content": "$$\\sqrt{41}$$\n"}, {"identifier": "B", "content": "$$\\sqrt{39}$$\n"}, {"identifier": "C", "content": "$$\\sqrt{40}$$\n"}, {"identifier": "D", "content": "$$\\sqrt{42}$$"}]
|
["C"]
| null |
<p>$$\begin{aligned}
& \int \frac{1}{a^2 \sin ^2 x+b^2 \cos ^2 x} d x=\frac{1}{12} \tan ^{-1}(3 \tan x)+c \\
& I=\int \frac{\sec ^2 x}{b^2+a^2 \tan ^2 x} d x \\
& \tan x=t \\
& \Rightarrow \sec ^2 x d x=d t \\
& I=\int \frac{d t}{b^2+a^2 t^2} \\
& =\frac{1}{b a} \tan ^{-1}\left(\frac{a t}{b}\right)+c \\
& I=\frac{1}{a b} \tan ^{-1}\left(\frac{a}{b} \tan x\right)+c \\
& \Rightarrow a b=12 \text { and } \frac{a}{b}=3\\
& \Rightarrow a^2=36 \text { and } b^2=4\\
& \Rightarrow \text { Maximum value of } a \sin x+b \cos x \text { is } \sqrt{a^2+b^2}
\end{aligned}$$</p>
<p>$$\quad=\sqrt{36+4}$$</p>
<p>$$\quad=\sqrt{40}$$</p>
|
mcq
|
jee-main-2024-online-6th-april-evening-shift
|
uudnlgNnuCb2c1ku
|
maths
|
inverse-trigonometric-functions
|
domain-and-range-of-inverse-trigonometric-functions
|
The trigonometric equation $${\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a$$ has a solution for :
|
[{"identifier": "A", "content": "$$\\left| a \\right| \\ge {1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "B", "content": "$${1 \\over 2} < \\left| a \\right| < {1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "C", "content": "all real values of $$a$$ "}, {"identifier": "D", "content": "$$\\left| a \\right| \\le {1 \\over {\\sqrt 2 }}$$ "}]
|
["D"]
| null |
Given that,
<br><br>$${\sin ^{ - 1}}x = 2{\sin ^{ - 1}}a$$
<br><br>We know,
<br><br>$$ - {\pi \over 2} \le {\sin ^{ - 1}}x \le {\pi \over 2}$$
<br><br>$$\therefore$$ $$\,\,\,$$ $$ - {\pi \over 2} \le 2{\sin ^{ - 1}}a \le {\pi \over 2}$$
<br><br>$$ \Rightarrow - {\pi \over 4} \le {\sin ^{ - 1}}a \le {\pi \over 4}$$
<br><br>$$ \Rightarrow \sin \left( { - {\pi \over 4}} \right) \le a \le \sin \left( {{\pi \over 4}} \right)$$
<br><br>$$ \Rightarrow - {1 \over {\sqrt 2 }} \le a \le {1 \over {\sqrt 2 }}$$
<br><br>$$\therefore$$ $$\,\,\,\,\,\,\,\,\left| a \right| \le {1 \over {\sqrt 2 }}$$
|
mcq
|
aieee-2003
|
nLj5VndTCR1FFqA1CBjgy2xukewsg59s
|
maths
|
inverse-trigonometric-functions
|
domain-and-range-of-inverse-trigonometric-functions
|
The domain of the function
<br/>f(x) = $${\sin ^{ - 1}}\left( {{{\left| x \right| + 5} \over {{x^2} + 1}}} \right)$$ is (–
$$\infty $$, -a]$$ \cup $$[a, $$\infty $$). Then a is equal to :
|
[{"identifier": "A", "content": "$${{\\sqrt {17} - 1} \\over 2}$$"}, {"identifier": "B", "content": "$${{1 + \\sqrt {17} } \\over 2}$$"}, {"identifier": "C", "content": "$${{\\sqrt {17} } \\over 2} + 1$$"}, {"identifier": "D", "content": "$${{\\sqrt {17} } \\over 2}$$"}]
|
["B"]
| null |
f(x) = $${\sin ^{ - 1}}\left( {{{\left| x \right| + 5} \over {{x^2} + 1}}} \right)$$
<br><br>$$ \therefore $$ $$ - 1 \le {{\left| x \right| + 5} \over {{x^2} + 1}} \le 1$$
<br><br>Since |x| + 5 & x<sup>2</sup>
+ 1 is always positive
<br><br>So $${{\left| x \right| + 5} \over {{x^2} + 1}} \ge 0$$
<br><br>That means this inequality $$ - 1 \le {{\left| x \right| + 5} \over {{x^2} + 1}}$$ always right. So we can ignore it.
<br><br>So for domain :
<br><br>$${{\left| x \right| + 5} \over {{x^2} + 1}} \le 1$$
<br><br>$$ \Rightarrow $$ $${{x^2} - \left| x \right| - 4 \ge 0}$$
<br><br>$$ \Rightarrow $$ $$\left( {\left| x \right| - {{1 - \sqrt {17} } \over 2}} \right)\left( {\left| x \right| - {{1 + \sqrt {17} } \over 2}} \right) \ge 0$$
<br><br>$$ \Rightarrow $$ |x| $$ \ge $$ $${{{1 + \sqrt {17} } \over 2}}$$ or |x| $$ \le $$ $${{{1 - \sqrt {17} } \over 2}}$$
<br><br>As $${{{1 - \sqrt {17} } \over 2}}$$ is < 0 and |x| always $$ \ge $$ 0. So <br>|x| $$ \le $$ $${{{1 - \sqrt {17} } \over 2}}$$ not possible.
<br><br>$$ \therefore $$ |x| $$ \ge $$ $${{{1 + \sqrt {17} } \over 2}}$$
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265699/exam_images/wxftevmnlm0p8wyfumnh.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265486/exam_images/v91g6oghwytiqyntvyy4.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 2nd September Morning Slot Mathematics - Inverse Trigonometric Functions Question 59 English Explanation"></picture>
<br><br>x $$ \in $$ $$\left( { - \infty , - {{1 + \sqrt {17} } \over 2}} \right) \cup \left( {{{1 + \sqrt {17} } \over 2},\infty } \right)$$
<br><br>So, a = $${{{1 + \sqrt {17} } \over 2}}$$
|
mcq
|
jee-main-2020-online-2nd-september-morning-slot
|
y6Fc34VqoMktCyXZqP1kmkm3flo
|
maths
|
inverse-trigonometric-functions
|
domain-and-range-of-inverse-trigonometric-functions
|
The number of solutions of the equation <br/><br/>$${\sin ^{ - 1}}\left[ {{x^2} + {1 \over 3}} \right] + {\cos ^{ - 1}}\left[ {{x^2} - {2 \over 3}} \right] = {x^2}$$, for x$$\in$$[$$-$$1, 1], and [x] denotes the greatest integer less than or equal to x, is :
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "Infinite"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "4"}]
|
["A"]
| null |
There are three cases possible for $$x \in [ - 1,1]$$<br><br>Case I : $$x \in \left[ { - 1, - \sqrt {{2 \over 3}} } \right)$$<br><br>$$ \therefore $$ $${\sin ^{ - 1}}(1) + {\cos ^{ - 1}}(0) = {x^2}$$<br><br>$$ \Rightarrow {x^2} = {\pi \over 2} + {\pi \over 2} = \pi $$<br><br>$$ \Rightarrow x = \pm \sqrt \pi $$ $$ \to $$ (Reject)<br><br>Case II : $$x \in \left( { - \sqrt {{2 \over 3}} ,\sqrt {{2 \over 3}} } \right)$$<br><br>$$ \therefore $$ $${\sin ^{ - 1}}(0) + {\cos ^{ - 1}}( - 1) = {x^2}$$<br><br>$$ \Rightarrow 0 + \pi = {x^2}$$<br><br>$$ \Rightarrow x = \pm \sqrt x $$ $$ \to $$ (Reject)<br><br>Case III : $$x \in \left( {\sqrt {{2 \over 3}} ,1} \right)$$<br><br>$$ \therefore $$ $${\sin ^{ - 1}}(0) + {\cos ^{ - 1}}(0) = {x^2}$$<br><br>$$ \Rightarrow {x^2} - \pi \Rightarrow x - \pm \sqrt x $$ (Reject)<br><br>$$ \therefore $$ No solution. There, the correct answer is (1).
|
mcq
|
jee-main-2021-online-17th-march-evening-shift
|
1krpwpsl7
|
maths
|
inverse-trigonometric-functions
|
domain-and-range-of-inverse-trigonometric-functions
|
The number of real roots of the equation $${\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = {\pi \over 4}$$ is :
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "0"}]
|
["D"]
| null |
$${\tan ^{ - 1}}\sqrt {x(x + 1)} + {\sin ^{ - 1}}\sqrt {{x^2} + x + 1} = {\pi \over 4}$$<br><br>For equation to be defined,<br><br>x<sup>2</sup> + x $$\ge$$ 0<br><br>$$\Rightarrow$$ x<sup>2</sup> + x + 1 $$\ge$$ 1<br><br>$$\therefore$$ Only possibility that the equation is defined <br><br>x<sup>2</sup> + x = 0 $$\Rightarrow$$ x = 0; x = $$-$$1<br><br>None of these values satisfy<br><br>$$\therefore$$ No of roots = 0
|
mcq
|
jee-main-2021-online-20th-july-morning-shift
|
1kru9uhcb
|
maths
|
inverse-trigonometric-functions
|
domain-and-range-of-inverse-trigonometric-functions
|
If the domain of the function $$f(x) = {{{{\cos }^{ - 1}}\sqrt {{x^2} - x + 1} } \over {\sqrt {{{\sin }^{ - 1}}\left( {{{2x - 1} \over 2}} \right)} }}$$ is the interval ($$\alpha$$, $$\beta$$], then $$\alpha$$ + $$\beta$$ is equal to :
|
[{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$${1 \\over 2}$$"}, {"identifier": "D", "content": "1"}]
|
["A"]
| null |
$$O \le {x^2} - x + 1 \le 1$$<br><br>$$ \Rightarrow {x^2} - x \le 0$$<br><br>$$ \Rightarrow x \in [0,1]$$<br><br>Also, $$0 < {\sin ^{ - 1}}\left( {{{2x - 1} \over 2}} \right) \le {\pi \over 2}$$<br><br>$$ \Rightarrow 0 < {{2x - 1} \over 2} \le 1$$<br><br>$$ \Rightarrow 0 < 2x - 1 \le 2$$<br><br>$$1 < 2x \le 3$$<br><br>$${1 \over 2} < x \le {3 \over 2}$$<br><br>Taking intersection <br><br>$$x \in \left( {{1 \over 2},1} \right]$$<br><br>$$ \Rightarrow \alpha = {1 \over 2},\beta = 1$$<br><br>$$ \Rightarrow \alpha + \beta = {3 \over 2}$$
|
mcq
|
jee-main-2021-online-22th-july-evening-shift
|
1ktczv71w
|
maths
|
inverse-trigonometric-functions
|
domain-and-range-of-inverse-trigonometric-functions
|
The domain of the function $${{\mathop{\rm cosec}\nolimits} ^{ - 1}}\left( {{{1 + x} \over x}} \right)$$ is :
|
[{"identifier": "A", "content": "$$\\left( { - 1, - {1 \\over 2}} \\right] \\cup (0,\\infty )$$"}, {"identifier": "B", "content": "$$\\left[ { - {1 \\over 2},0} \\right) \\cup [1,\\infty )$$"}, {"identifier": "C", "content": "$$\\left( { - {1 \\over 2},\\infty } \\right) - \\{ 0\\} $$"}, {"identifier": "D", "content": "$$\\left[ { - {1 \\over 2},\\infty } \\right) - \\{ 0\\} $$"}]
|
["D"]
| null |
$${{1 + x} \over x} \in ( - \infty , - 1] \cup [1,\infty )$$<br><br>$${1 \over x} \in ( - \infty , - 2] \cup [0,\infty )$$<br><br>$$x \in \left[ { - {1 \over 2},0} \right) \cup (0,\infty )$$<br><br>$$x \in \left[ { - {1 \over 2},0} \right) \cup \{ 0\} $$
|
mcq
|
jee-main-2021-online-26th-august-evening-shift
|
1ktk4rt98
|
maths
|
inverse-trigonometric-functions
|
domain-and-range-of-inverse-trigonometric-functions
|
The domain of the function<br/><br/>$$f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right)$$ is :
|
[{"identifier": "A", "content": "$$\\left[ {0,{1 \\over 4}} \\right]$$"}, {"identifier": "B", "content": "$$[ - 2,0] \\cup \\left[ {{1 \\over 4},{1 \\over 2}} \\right]$$"}, {"identifier": "C", "content": "$$\\left[ {{1 \\over 4},{1 \\over 2}} \\right] \\cup \\{ 0\\} $$"}, {"identifier": "D", "content": "$$\\left[ {0,{1 \\over 2}} \\right]$$"}]
|
["C"]
| null |
$$f(x) = {\sin ^{ - 1}}\left( {{{3{x^2} + x - 1} \over {{{(x - 1)}^2}}}} \right) + {\cos ^{ - 1}}\left( {{{x - 1} \over {x + 1}}} \right)$$<br><br>$$ - 1 \le {{x - 1} \over {x + 1}} \le 1 \Rightarrow 0 \le x < \infty $$ .... (1)<br><br>$$ - 1 \le {{3{x^2} + x - 1} \over {{{(x - 1)}^2}}} \le 1 \Rightarrow x \in \left[ {{{ - 1} \over 4},{1 \over 2}} \right] \cup \{ 0\} $$ .... (2)<br><br>(1) & (2)<br><br>$$\Rightarrow$$ Domain = $$\left[ {{1 \over 4},{1 \over 2}} \right] \cup \{ 0\} $$
|
mcq
|
jee-main-2021-online-31st-august-evening-shift
|
1l545i5vo
|
maths
|
inverse-trigonometric-functions
|
domain-and-range-of-inverse-trigonometric-functions
|
<p>The domain of the function $${\cos ^{ - 1}}\left( {{{2{{\sin }^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right)} \over \pi }} \right)$$ is :</p>
|
[{"identifier": "A", "content": "$$R - \\left\\{ { - {1 \\over 2},{1 \\over 2}} \\right\\}$$"}, {"identifier": "B", "content": "$$( - \\infty , - 1] \\cup [1,\\infty ) \\cup \\{ 0\\} $$"}, {"identifier": "C", "content": "$$\\left( { - \\infty ,{{ - 1} \\over 2}} \\right) \\cup \\left( {{1 \\over 2},\\infty } \\right) \\cup \\{ 0\\} $$"}, {"identifier": "D", "content": "$$\\left( { - \\infty ,{{ - 1} \\over {\\sqrt 2 }}} \\right] \\cup \\left[ {{1 \\over {\\sqrt 2 }},\\infty } \\right) \\cup \\{ 0\\} $$"}]
|
["D"]
| null |
<p>$$ - 1 \le {{2{{\sin }^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right)} \over \pi } \le 1$$</p>
<p>$$ \Rightarrow - {\pi \over 2} \le {\sin ^{ - 1}}\left( {{1 \over {4{x^2} - 1}}} \right) \le {\pi \over 2}$$</p>
<p>$$ \Rightarrow - 1 \le {1 \over {4{x^2} - 1}} \le 1$$</p>
<p>$$\therefore$$ $${1 \over {4{x^2} - 1}} + 1 \ge 0$$</p>
<p>$$ \Rightarrow {{1 + 4{x^2} - 1} \over {4{x^2} - 1}} \ge 0$$</p>
<p>$$ \Rightarrow {{4{x^2}} \over {4{x^2} - 1}} \ge 0$$ </p>
<p>$$ \Rightarrow $$ $${{4{x^2}} \over {\left( {2x + 1} \right)\left( {2x - 1} \right)}} \ge 0$$ ...... (1)</p>
<p>$$\therefore$$ $$x \in \left( { - \alpha , - {1 \over 2}} \right) \cup \{ 0\} \cup \left( {{1 \over 2},\alpha } \right)$$ .....(2)</p>
<p>And $${1 \over {4{x^2} - 1}} - 1 \le 0$$</p>
<p>$$ \Rightarrow {{1 - 4{x^2} + 1} \over {4{x^2} - 1}} \le 0$$</p>
<p>$$ \Rightarrow {{2 - 4{x^2}} \over {4{x^2} - 1}} \le 0$$</p>
<p>$$ \Rightarrow {{2{x^2} - 1} \over {4{x^2} - 1}} \ge 0$$</p>
<p>$$ \Rightarrow $$ $${{\left( {\sqrt 2 x + 1} \right)\left( {\sqrt 2 x - 1} \right)} \over {\left( {2x + 1} \right)\left( {2x - 1} \right)}} \ge 0$$ ...... (3)</p>
<p>$$x \in \left( { - \alpha , - {1 \over {\sqrt 2 }}} \right) \cup \left( { - {1 \over 2},{1 \over 2}} \right) \cup \left( {{1 \over {\sqrt 2 }},\alpha } \right)$$ .....(4)</p>
<p>From (3) and (4), we get</p>
<p>$$\therefore$$ $$x \in \left[ { - \alpha , - {1 \over {\sqrt 2 }}} \right) \cup \left[ {{1 \over {\sqrt 2 }},\alpha } \right) \cup \{ 0\} $$</p>
|
mcq
|
jee-main-2022-online-29th-june-morning-shift
|
1l5c1qzrq
|
maths
|
inverse-trigonometric-functions
|
domain-and-range-of-inverse-trigonometric-functions
|
<p>The domain of the function <br/><br/>$$f(x) = {{{{\cos }^{ - 1}}\left( {{{{x^2} - 5x + 6} \over {{x^2} - 9}}} \right)} \over {{{\log }_e}({x^2} - 3x + 2)}}$$ is :</p>
|
[{"identifier": "A", "content": "$$( - \\infty ,1) \\cup (2,\\infty )$$"}, {"identifier": "B", "content": "$$(2,\\infty )$$"}, {"identifier": "C", "content": "$$\\left[ { - {1 \\over 2},1} \\right) \\cup (2,\\infty )$$"}, {"identifier": "D", "content": "$$\\left[ { - {1 \\over 2},1} \\right) \\cup (2,\\infty ) - \\left\\{ 3,{{{3 + \\sqrt 5 } \\over 2},{{3 - \\sqrt 5 } \\over 2}} \\right\\}$$"}]
|
["D"]
| null |
$-1 \leq \frac{x^{2}-5 x+6}{x^{2}-9} \leq 1$ and $x^{2}-3 x+2>0, \neq 1$
<br/><br/>
$$
\frac{(x-3)(2 x+1)}{x^{2}-9} \geq 0 \mid \frac{5(x-3)}{x^{2}-9} \geq 0
$$
<br/><br/>
The solution to this inequality is
<br/><br/>
$x \in\left[\frac{-1}{2}, \infty\right)-\{3\}$
<br/><br/>
for $x^{2}-3 x+2>0$ and $\neq 1$
<br/><br/>
$$
x \in(-\infty, 1) \cup(2, \infty)-\left\{\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right\}
$$
<br/><br/>
Combining the two solution sets (taking intersection)
<br/><br/>
$$
x \in\left[-\frac{1}{2}, 1\right) \cup(2, \infty)-\left\{\frac{3-\sqrt{5}}{2}, \frac{3+\sqrt{5}}{2}\right\}
$$
|
mcq
|
jee-main-2022-online-24th-june-morning-shift
|
1l6kib0jf
|
maths
|
inverse-trigonometric-functions
|
domain-and-range-of-inverse-trigonometric-functions
|
<p>The domain of the function $$f(x)=\sin ^{-1}\left[2 x^{2}-3\right]+\log _{2}\left(\log _{\frac{1}{2}}\left(x^{2}-5 x+5\right)\right)$$, where [t] is the greatest integer function, is :</p>
|
[{"identifier": "A", "content": "$$\n\\left(-\\sqrt{\\frac{5}{2}}, \\frac{5-\\sqrt{5}}{2}\\right)\n$$"}, {"identifier": "B", "content": "$$\n\\left(\\frac{5-\\sqrt{5}}{2}, \\frac{5+\\sqrt{5}}{2}\\right)\n$$"}, {"identifier": "C", "content": "$$\n\\left(1, \\frac{5-\\sqrt{5}}{2}\\right)\n$$"}, {"identifier": "D", "content": "$$\n\\left[1, \\frac{5+\\sqrt{5}}{2}\\right)\n$$"}]
|
["C"]
| null |
<p>$$ - 1 \le 2{x^2} - 3 < 2$$</p>
<p>or $$2 \le 2{x^2} < 5$$</p>
<p>or $$1 \le {x^2} < {5 \over 2}$$</p>
<p>$$x \in \left( { - \sqrt {{5 \over 2}} , - 1} \right] \cup \left[ {1,\sqrt {{5 \over 2}} } \right)$$</p>
<p>$${\log _{{1 \over 2}}}({x^2} - 5x + 5) > 0$$</p>
<p>$$0 < {x^2} - 5x + 5 < 1$$</p>
<p>$${x^2} - 5x + 5 > 0$$ & $${x^2} - 5x + 4 < 0$$</p>
<p>$$x \in \left( { - \infty ,{{5 - \sqrt 5 } \over 2}} \right) \cup \left( {{{5 + \sqrt 5 } \over 2},\infty } \right)$$</p>
<p>& $$x \in ( - \infty ,1) \cup (4,\infty )$$</p>
<p>Taking intersection</p>
<p>$$x \in \left( {1,{{5 - \sqrt 5 } \over 2}} \right)$$</p>
|
mcq
|
jee-main-2022-online-27th-july-evening-shift
|
1l6rfcn42
|
maths
|
inverse-trigonometric-functions
|
domain-and-range-of-inverse-trigonometric-functions
|
<p>The domain of the function $$f(x)=\sin ^{-1}\left(\frac{x^{2}-3 x+2}{x^{2}+2 x+7}\right)$$ is :</p>
|
[{"identifier": "A", "content": "$$[1, \\infty)$$"}, {"identifier": "B", "content": "$$[-1,2]$$"}, {"identifier": "C", "content": "$$[-1, \\infty)$$"}, {"identifier": "D", "content": "$$(-\\infty, 2]$$"}]
|
["C"]
| null |
$f(x)=\sin ^{-1}\left(\frac{x^{2}-3 x+2}{x^{2}+2 x+7}\right)$
<br/><br/>$$
-1 \leq \frac{x^{2}-3 x+2}{x^{2}+2 x+7} \leq 1
$$
<br/><br/>$$
\begin{aligned}
& \frac{x^{2}-3 x+2}{x^{2}+2 x+7} \leq 1 \\\\
& x^{2}-3 x+2 \leq x^{2}+2 x+7 \\\\
& 5 x \geq-5 \\\\
& x \geq-1
\end{aligned}
$$
<br/><br/>And $$
\frac{x^{2}-3 x+2}{x^{2}+2 x+7} \geq-1
$$
<br/><br/>$$
x^{2}-3 x+2 \geq-x^{2}-2 x-7
$$
<br/><br/>$2 x^{2}-x+9 \geq 0$
<br/><br/>$$
x \in R
$$
<br/><br/>(i) $\cap$ (ii)
<br/><br/>Domain $\in[-1, \infty)$
|
mcq
|
jee-main-2022-online-29th-july-evening-shift
|
lgnwv1sg
|
maths
|
inverse-trigonometric-functions
|
domain-and-range-of-inverse-trigonometric-functions
|
If the domain of the function
<br/><br/>$f(x)=\log _{e}\left(4 x^{2}+11 x+6\right)+\sin ^{-1}(4 x+3)+\cos ^{-1}\left(\frac{10 x+6}{3}\right)$ is $(\alpha, \beta]$, then
<br/><br/>$36|\alpha+\beta|$ is equal to :
|
[{"identifier": "A", "content": "72"}, {"identifier": "B", "content": "54"}, {"identifier": "C", "content": "45"}, {"identifier": "D", "content": "63"}]
|
["C"]
| null |
<p>To find the domain of the function, we need to consider the individual functions and their respective domains. We have:</p>
<p><ol>
<li>$f_1(x) = \ln(4x^2 + 11x + 6)$</li>
<li>$f_2(x) = \sin^{-1}(4x + 3)$</li>
<li>$f_3(x) = \cos^{-1}\left(\frac{10x + 6}{3}\right)$</li>
</ol></p>
<p><ol>
<li>For $f_1(x)$:</li>
</ol></p>
<p>$$
4x^2 + 11x + 6 > 0
$$</p>
<p>Factoring the quadratic expression:</p>
<p>$$
(4x + 3)(x + 2) > 0
$$</p>
<p>From this inequality, we have:</p>
<p>$$
x \in (-\infty, -2) \cup \left(-\frac{3}{4}, \infty\right)
$$</p>
<ol>
<li>For $f_2(x)$:</li>
</ol>
<p>$$
-1 \le 4x + 3 \le 1
$$</p>
<p>From these inequalities, we get:</p>
<p>$$
x \in \left[-1, -\frac{1}{2}\right]
$$</p>
<ol>
<li>For $f_3(x)$:</li>
</ol>
<p>$$
-1 \le \frac{10x + 6}{3} \le 1
$$</p>
<p>From these inequalities, we get:</p>
<p>$$
x \in \left[-\frac{9}{10}, -\frac{3}{10}\right]
$$</p>
<p>Now, we need to find the intersection of the domains of the three functions:</p>
<p>$$
\left(-\infty, -2\right) \cup \left(-\frac{3}{4}, \infty\right) \cap \left[-1, -\frac{1}{2}\right] \cap \left[-\frac{9}{10}, -\frac{3}{10}\right]
$$</p>
<p>To find the intersection, let's analyze the intervals:</p>
<ul>
<li>The interval $(-\infty, -2) \cup \left(-\frac{3}{4}, \infty\right)$ contains all $x$ values less than $-2$ and greater than $-\frac{3}{4}$.</li>
<li>The interval $\left[-1, -\frac{1}{2}\right]$ contains all $x$ values between $-1$ and $-\frac{1}{2}$.</li>
<li>The interval $\left[-\frac{9}{10}, -\frac{3}{10}\right]$ contains all $x$ values between $-\frac{9}{10}$ and $-\frac{3}{10}$.</li>
</ul>
<p>Looking at the intervals, we can see that the intersection is:</p>
<p>$$
x \in \left(-\frac{3}{4}, -\frac{1}{2}\right]
$$</p>
<p>Thus, the domain of the function is $(\alpha, \beta] = \left(-\frac{3}{4}, -\frac{1}{2}\right]$. Now, we need to find the value of $36|\alpha + \beta|$:</p>
<p>$$
36\left|-\frac{3}{4} - \frac{1}{2}\right| = 36\left|-\frac{5}{4}\right| =45
$$</p>
|
mcq
|
jee-main-2023-online-15th-april-morning-shift
|
1lgvqqyhz
|
maths
|
inverse-trigonometric-functions
|
domain-and-range-of-inverse-trigonometric-functions
|
<p>If the domain of the function $$f(x)=\sec ^{-1}\left(\frac{2 x}{5 x+3}\right)$$ is $$[\alpha, \beta) \mathrm{U}(\gamma, \delta]$$, then $$|3 \alpha+10(\beta+\gamma)+21 \delta|$$ is equal to _________.</p>
|
[]
| null |
24
|
Given that $f(x)=\sec ^{-1}\left(\frac{2 x}{5 x+3}\right)$
<br/><br/>Since, the domain for $\sec ^{-1} x$ is $|x| \geq 1$
<br/><br/>Therefore $\left|\frac{2 x}{5 x+3}\right| \geq 1$
<br/><br/>$$
\begin{aligned}
& \frac{2 x}{5 x+3} \leq-1 \text { or } \frac{2 x}{5 x+3} \geq 1 \\\\
& \frac{2 x+5 x+3}{5 x+3} \leq 0 \text { or } \frac{2 x-5 x-3}{5 x+3} \geq 0 \\\\
& \frac{7 x+3}{5 x+3} \leq 0 \text { or } \frac{-3(x+1)}{5 x+3} \geq 0
\end{aligned}
$$
<br/><br/><b>Case I :</b> $7 x+3 \leq 0$ and $5 x+3>0$
<br/><br/>$$
\begin{array}{rlrl}
& x \leq-\frac{3}{7} \text { or } x > -\frac{3}{5} \\\\
& \Rightarrow -\frac{3}{5} < x \leq-\frac{3}{7}
\end{array}
$$
<br/><br/><b>Case II :</b> $7 x+3 \geq 0$ and $5 x+3<0$
<br/><br/>$$
x \geq-\frac{3}{7} \text { and } x<-\frac{3}{5}
$$
<br/><br/>Which is not possible
<br/><br/><b>Case III :</b> $x+1 \geq 0$ and $5 x+3<0$
<br/><br/>$$
\begin{aligned}
& x \geq-1 \text { and } x<-\frac{3}{5} \\\\
& \Rightarrow -1 \leq x<-\frac{3}{5}
\end{aligned}
$$
<br/><br/><b>Case IV :</b> $x+1 \leq 0$ and $5 x+3 \geq 0$
<br/><br/>$$
x \leq-1 \text { and } x \geq-\frac{3}{5}
$$
<br/><br/>Which is not possible
<br/><br/>$\therefore$ Domain is $\left[-1,-\frac{3}{5}\right) \cup\left(-\frac{3}{5},-\frac{3}{7}\right]$
<br/><br/>$$
\therefore \alpha=-1, \beta=-\frac{3}{5}, \gamma=-\frac{3}{5}, \delta=-\frac{3}{7}
$$
<br/><br/>$$ \therefore $$ $$
|3 \alpha+10(\beta+\gamma)+21 \delta|
=\left|-3+10\left(-\frac{3}{5}-\frac{3}{5}\right)+21\left(-\frac{3}{7}\right)\right|=24
$$
|
integer
|
jee-main-2023-online-10th-april-evening-shift
|
lv0vxckz
|
maths
|
inverse-trigonometric-functions
|
domain-and-range-of-inverse-trigonometric-functions
|
<p>If the domain of the function $$\sin ^{-1}\left(\frac{3 x-22}{2 x-19}\right)+\log _{\mathrm{e}}\left(\frac{3 x^2-8 x+5}{x^2-3 x-10}\right)$$ is $$(\alpha, \beta]$$, then $$3 \alpha+10 \beta$$ is equal to:</p>
|
[{"identifier": "A", "content": "95"}, {"identifier": "B", "content": "100"}, {"identifier": "C", "content": "97"}, {"identifier": "D", "content": "98"}]
|
["C"]
| null |
<p>$$\begin{aligned}
& \sin ^{-1}\left(\frac{3 x-22}{2 x-19}\right)+\log _e\left(\frac{3 x^2-8 x+5}{x^2-3 x-10}\right) \\
& -1 \leq \frac{3 x-22}{2 x-19} \leq 1 \\
& \frac{3 x-22}{2 x-19}+1 \geq 0 \text { and } \frac{3 x-22}{2 x-19}-1 \leq 0 \\
& \frac{3 x-22+2 x-19}{2 x-19} \geq 0 \text { and } \frac{3 x-22-2 x+19}{2 x-19} \leq 0 \\
& \Rightarrow \frac{5 x-41}{2 x-19} \geq 0 \text { and } \frac{x-3}{2 x-19} \leq 0 \\
& x \in\left(-\infty, \frac{41}{5}\right] \cup\left(\frac{19}{2}, \infty\right) \text { and } x \in\left[3, \frac{19}{2}\right)
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \quad x \in\left[3, \frac{41}{5}\right] \quad \text{.... (1)}\\
& \text { and, } \frac{3 x^2-8 x+5}{x^2-3 x-10}>0 \\
& \frac{(3 x-5)(x-1)}{(x-5)(x-2)}>0 \\
& \Rightarrow \quad x \in(-\infty,-2) \cup\left[1, \frac{5}{3}\right] \cup(5, \infty) \ldots
\end{aligned}$$</p>
<p>Taking intersection of individual domains</p>
<p>$$x \in\left(5, \frac{41}{5}\right]$$</p>
<p>$$\begin{aligned}
& \Rightarrow \quad \alpha=5 \text { and } \beta=\frac{41}{5} \\
& \Rightarrow 3 \alpha+10 \beta=15+82 \\
& =97
\end{aligned}$$</p>
<p>$$\therefore \quad$$ Option (4) is correct</p>
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mcq
|
jee-main-2024-online-4th-april-morning-shift
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59dUaVXasYJWZEiRdguhw
|
maths
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inverse-trigonometric-functions
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principal-value-of-inverse-trigonometric-functions
|
Considering only the principal values of inverse functions, the set
<br/>A = { x $$ \ge $$ 0: tan<sup>$$-$$1</sup>(2x) + tan<sup>$$-$$1</sup>(3x) = $${\pi \over 4}$$}
|
[{"identifier": "A", "content": "contains two elements "}, {"identifier": "B", "content": "contains more than two elements "}, {"identifier": "C", "content": "is an empty set "}, {"identifier": "D", "content": "is a singleton "}]
|
["D"]
| null |
tan<sup>$$-$$1</sup>(2x) + tan<sup>$$-$$1</sup>(3x) = $$\pi $$/4
<br><br>$$ \Rightarrow \,\,{{5x} \over {1 - 6{x^2}}}$$ = 1
<br><br>$$ \Rightarrow $$ 6x<sup>2</sup> + 5x $$-$$ 1 = 0
<br><br>x = $$-$$1 or x = $${1 \over 6}$$
<br><br>x = $${1 \over 6}$$
<br><br>$$ \because $$ x > 0
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mcq
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jee-main-2019-online-12th-january-morning-slot
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r1lgCyi2I37gkOBHCY1kmix729h
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maths
|
inverse-trigonometric-functions
|
principal-value-of-inverse-trigonometric-functions
|
Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy <br/><br/>$${\sin ^{ - 1}}\left( {{{3x} \over 5}} \right) + {\sin ^{ - 1}}\left( {{{4x} \over 5}} \right) = {\sin ^{ - 1}}x$$ is equal to :
|
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "1"}]
|
["C"]
| null |
$${\sin ^{ - 1}}{{3x} \over 5} + {\sin ^{ - 1}}{{4x} \over 5} = {\sin ^{ - 1}}x$$<br><br>$${\sin ^{ - 1}}\left( {{{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\sqrt {1 - {{9{x^2}} \over {25}}} } \right) = {\sin ^{ - 1}}x$$<br><br>$${{3x} \over 5}\sqrt {1 - {{16{x^2}} \over {25}}} + {{4x} \over 5}\sqrt {1 - {{9{x^2}} \over {25}}} = x$$<br><br>$$x = 0 $$ or $$3\sqrt {25 - 16{x^2}} + 4\sqrt {25 - 9{x^2}} = 25$$<br><br>$$4\sqrt {25 - 9{x^2}} = 25 - 3\sqrt {25 - 16{x^2}} $$
<br><br>Squaring we get<br><br>$$16(25 - 9{x^2}) = 625 - 9(25 - 16{x^2}) - 150\sqrt {25 - 16{x^2}} $$<br><br>$$400 = 625 + 225 - 150\sqrt {25 - 16{x^2}} $$<br><br>$$\sqrt {25 - 16{x^2}} = 3 \Rightarrow 25 - 16{x^2} = 9$$<br><br>$$ \Rightarrow {x^2} = 1$$<br><br>Put x = 0, 1, $$-$$1 in the original equation<br><br>We see that all values satisfy the original equation.<br><br>Number of solution = 3
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mcq
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jee-main-2021-online-16th-march-evening-shift
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1ktnznemb
|
maths
|
inverse-trigonometric-functions
|
principal-value-of-inverse-trigonometric-functions
|
$${\cos ^{ - 1}}(\cos ( - 5)) + {\sin ^{ - 1}}(\sin (6)) - {\tan ^{ - 1}}(\tan (12))$$ is equal to :<br/><br/>(The inverse trigonometric functions take the principal values)
|
[{"identifier": "A", "content": "3$$\\pi$$ $$-$$ 11"}, {"identifier": "B", "content": "4$$\\pi$$ $$-$$ 9"}, {"identifier": "C", "content": "4$$\\pi$$ $$-$$ 11"}, {"identifier": "D", "content": "3$$\\pi$$ + 1"}]
|
["C"]
| null |
$${\cos ^{ - 1}}(\cos ( - 5)) + {\sin ^{ - 1}}(\sin (6)) - {\tan ^{ - 1}}(\tan (12))$$<br><br>$$ = (2\pi - 5) + (6 - 2\pi ) - (12 - 4\pi )$$<br><br>$$ = 4\pi - 11$$.
|
mcq
|
jee-main-2021-online-1st-september-evening-shift
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1l57ou1of
|
maths
|
inverse-trigonometric-functions
|
principal-value-of-inverse-trigonometric-functions
|
<p>$${\sin ^1}\left( {\sin {{2\pi } \over 3}} \right) + {\cos ^{ - 1}}\left( {\cos {{7\pi } \over 6}} \right) + {\tan ^{ - 1}}\left( {\tan {{3\pi } \over 4}} \right)$$ is equal to :</p>
|
[{"identifier": "A", "content": "$${{11\\pi } \\over {12}}$$"}, {"identifier": "B", "content": "$${{17\\pi } \\over {12}}$$"}, {"identifier": "C", "content": "$${{31\\pi } \\over {12}}$$"}, {"identifier": "D", "content": "$$-$$$${{3\\pi } \\over {4}}$$"}]
|
["A"]
| null |
<p>$${\sin ^{ - 1}}\left( {{{\sqrt 3 } \over 2}} \right) + {\cos ^{ - 1}}\left( {{{ - \sqrt 3 } \over 2}} \right) + {\tan ^{ - 1}}\left( { - 1} \right)$$</p>
<p>$$ = {\pi \over 3} + {{5\pi } \over 6} - {\pi \over 4}$$</p>
<p>$$ = {{4\pi + 10\pi - 3\pi } \over {12}} = {{11\pi } \over {12}}$$</p>
|
mcq
|
jee-main-2022-online-27th-june-morning-shift
|
1l58gom2l
|
maths
|
inverse-trigonometric-functions
|
principal-value-of-inverse-trigonometric-functions
|
<p>If the inverse trigonometric functions take principal values then <br/><br/>$${\cos ^{ - 1}}\left( {{3 \over {10}}\cos \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right) + {2 \over 5}\sin \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right)} \right)$$ is equal to :</p>
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "$${\\pi \\over 4}$$"}, {"identifier": "C", "content": "$${\\pi \\over 3}$$"}, {"identifier": "D", "content": "$${\\pi \\over 6}$$"}]
|
["C"]
| null |
<p>$${\cos ^{ - 1}}\left( {{3 \over {10}}\cos \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right) + {2 \over 5}\sin \left( {{{\tan }^{ - 1}}\left( {{4 \over 3}} \right)} \right)} \right)$$</p>
<p>$$ = {\cos ^{ - 1}}\left( {{3 \over {10}}\,.\,{3 \over 5} + {2 \over 5}\,.\,{4 \over 5}} \right)$$</p>
<p>$$ = {\cos ^{ - 1}}\left( {{1 \over 2}} \right) = {\pi \over 3}$$</p>
|
mcq
|
jee-main-2022-online-26th-june-evening-shift
|
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