question_id
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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
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stringclasses 293
values | explanation
stringlengths 13
9.38k
| question_type
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|---|---|---|---|---|---|---|---|---|---|---|
1l5c2cwqz
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$\mathop {Max}\limits_{0\, \le x\, \le 2} \left\{ {{{9 - {x^2}} \over {5 - x}}} \right\} = \alpha $$ and $$\mathop {Min}\limits_{0\, \le x\, \le 2} \left\{ {{{9 - {x^2}} \over {5 - x}}} \right\} = \beta $$.</p>
<p>If $$\int\limits_{\beta - {8 \over 3}}^{2\alpha - 1} {Max\left\{ {{{9 - {x^2}} \over {5 - x}},x} \right\}dx = {\alpha _1} + {\alpha _2}{{\log }_e}\left( {{8 \over {15}}} \right)} $$ then $${\alpha _1} + {\alpha _2}$$ is equal to _____________.</p>
|
[]
| null |
34
|
Let $f(x)=\frac{x^{2}-9}{x-5} \Rightarrow f^{\prime}(x)=\frac{(x-1)(x-9)}{(x-5)^{2}}$
<br/><br/>
So, $\alpha=f(1)=2$ and $\beta=\min (f(0), f(2))=\frac{5}{3}$
<br/><br/>
Now, $\int_{-1}^{3} \max \left\{\frac{x^{2}-9}{x-5}, x\right\} d x=\int_{-1}^{9 / 5} \frac{x^{2}-9}{x-5} d x+\int_{9 / 5}^{3} x d x$
<br/><br/>
$$
=\int_{-1}^{9 / 5}\left(x+5+\frac{16}{x-5}\right) d x+\left.\frac{x^{2}}{2}\right|_{9 / 5} ^{3}
$$
<br/><br/>
$$
=\frac{28}{25}+14+16 \ln \left(\frac{8}{15}\right)+\frac{72}{25}=18+16 \ln \left(\frac{8}{15}\right)
$$
<br/><br/>
Clearly $\alpha_{1}=18$ and $\alpha_{2}=16$, so $\alpha_{1}+\alpha_{2}=34$.
|
integer
|
jee-main-2022-online-24th-june-morning-shift
|
1l5w1ciod
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$f(t) = \int\limits_0^t {{e^{{x^3}}}\left( {{{{x^8}} \over {{{({x^6} + 2{x^3} + 2)}^2}}}} \right)dx} $$. If $$f(1) + f'(1) = \alpha e - {1 \over 6}$$, then the value of 150$$\alpha$$ is equal to ___________.</p>
|
[]
| null |
16
|
<p>Given,</p>
<p>$$f(t) = \int\limits_0^t {{e^{{x^3}}}\left( {{{{x^8}} \over {{{({x^6} + 2{x^3} + 2)}^2}}}} \right)dx} $$</p>
<p>$$f'(t) = {e^{{t^3}}}\left( {{{{t^8}} \over {{{({t^6} + 2{t^3} + 2)}^2}}}} \right)$$</p>
<p>$$\therefore$$ $$f'(1) = {e^1}\left( {{1 \over {{{(1 + 2 + 2)}^2}}}} \right)$$</p>
<p>$$ = {e \over {{5^2}}}$$</p>
<p>Now, $$f(t) = \int\limits_0^t {{e^{{x^3}}}\left( {{{{x^8}} \over {{{({x^6} + 2{x^2} + 2)}^2}}}} \right)dx} $$</p>
<p>Let $${x^3} = z \Rightarrow 3{x^2}dx = dz$$</p>
<p>$$ = \int\limits_0^{{t^3}} {{e^z}\left( {{{{x^6}\,.\,{x^2}dx} \over {{{({x^6} + 2{x^3} + 2)}^2}}}} \right)} $$</p>
<p>$$ = {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{{{z^2}dz} \over {{{({z^2} + 2z + 2)}^2}}}} \right)} $$</p>
<p>$$ = {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{{{z^2} + 2z + 2 - 2z - 2} \over {{{({z^2} + 2z + 2)}^2}}}} \right)} dz$$</p>
<p>$$ = {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{{{z^2} + 2z + 2} \over {{{({z^2} + 2z + 2)}^2}}} - {{2z + 2} \over {{{({z^2} + 2z + 2)}^2}}}} \right)dz} $$</p>
<p>$$ = {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {{1 \over {{z^2} + 2z + 2}} - {{2z + 2} \over {{{({z^2} + 2z + 2)}^2}}}} \right)dz} $$</p>
<p>$$ = {1 \over 3}\int\limits_0^{{t^3}} {{e^z}\left( {f(z) + f'(z)} \right)dz} $$</p>
<p>$$ = {1 \over 3}\left[ {{e^z}f(z)} \right]_0^{{t^3}}$$</p>
<p>$$ = {1 \over 3}\left[ {{e^z} \times {1 \over {{z^2} + 2z + 2}}} \right]_0^{{t^3}}$$</p>
<p>$$ = {1 \over 3}\left[ {{e^{{t^3}}} \times {1 \over {{t^6} + 2{t^3} + 2}} - {1 \over 2}} \right]$$</p>
<p>$$\therefore$$ $$f(t) = {1 \over 3}\left[ {{{{e^{{t^3}}}} \over {{t^6} + 2{t^3} + 2}} - {1 \over 2}} \right]$$</p>
<p>So, $$f(1) = {1 \over 3}\left[ {{e \over {1 + 2 + 2}} - {1 \over 2}} \right]$$</p>
<p>$$ = {1 \over 3}\left[ {{e \over 5} - {1 \over 2}} \right]$$</p>
<p>Given, $$f(1) + f'(1) = \alpha e - {1 \over 6}$$</p>
<p>$$ \Rightarrow {1 \over 3}\left[ {{e \over 5} - {1 \over 2}} \right] + {e \over {25}} = \alpha e - {1 \over 6}$$</p>
<p>$$ \Rightarrow {e \over {15}} - {1 \over 6} + {e \over {25}} = \alpha e - {1 \over 6}$$</p>
<p>$$ \Rightarrow {e \over {15}} + {e \over {25}} = \alpha e$$</p>
<p>$$ \Rightarrow {{10e + 6e} \over {150}} = \alpha e$$</p>
<p>$$ \Rightarrow {{16e} \over {150}} = \alpha e$$</p>
<p>$$ \Rightarrow \alpha = {{16} \over {150}}$$</p>
<p>$$\therefore$$ $$150\alpha = 150 \times {{16} \over {150}} = 16$$</p>
|
integer
|
jee-main-2022-online-30th-june-morning-shift
|
1l6dvfwkf
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>For any real number $$x$$, let $$[x]$$ denote the largest integer less than equal to $$x$$. Let $$f$$ be a real valued function defined on the interval $$[-10,10]$$ by $$f(x)=\left\{\begin{array}{l}x-[x], \text { if }[x] \text { is odd } \\ 1+[x]-x, \text { if }[x] \text { is even } .\end{array}\right.$$
Then the value of $$\frac{\pi^{2}}{10} \int_{-10}^{10} f(x) \cos \pi x \,d x$$ is :</p>
|
[{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "0"}]
|
["A"]
| null |
<p>Case 1 :</p>
<p>Let $$0 \le x < 1$$</p>
<p>then $$\left[ x \right] = 0$$, which is even</p>
<p>$$\therefore$$ $$f(x) = 1 + \left[ x \right] - x$$</p>
<p>$$ = 1 + 0 - x$$</p>
<p>$$ = 1 - x$$</p>
<p>Case 2 :</p>
<p>Let $$1 \le x < 2$$</p>
<p>then $$\left[ x \right] = 1$$, which is odd</p>
<p>$$\therefore$$ $$f(x) = x - \left[ x \right]$$</p>
<p>$$ = x - 1$$</p>
<p>Case 3 :</p>
<p>Let $$2 \le x < 3$$</p>
<p>then $$\left[ x \right] = 2$$, which is even</p>
<p>$$\therefore$$ $$f(x) = 1 + \left[ x \right] - x$$</p>
<p>$$ = 1 + 2 - x$$</p>
<p>$$ = 3 - x$$</p>
<p>Case 4 :</p>
<p>Let $$3 \le x < 4$$</p>
<p>then $$\left[ x \right] = 3$$, which is odd</p>
<p>$$\therefore$$ $$f(x) = x - \left[ x \right]$$</p>
<p>$$ = x - 3$$</p>
<p>$$\therefore$$ $$f(x) = \left\{ {\matrix{
{1 - x} & ; & {0 \le x < 1} \cr
{x - 1} & ; & {1 \le x < 2} \cr
{3 - x} & ; & {2 \le x < 3} \cr
{x - 3} & ; & {3 \le x < 4} \cr
} } \right.$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l79fc2ri/bc49e861-7a20-4168-b847-928ce9f0304f/283cb1e0-24aa-11ed-8d2e-5f0df5271c2d/file-1l79fc2rj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l79fc2ri/bc49e861-7a20-4168-b847-928ce9f0304f/283cb1e0-24aa-11ed-8d2e-5f0df5271c2d/file-1l79fc2rj.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 25th July Morning Shift Mathematics - Definite Integration Question 106 English Explanation"></p>
<p>$$\therefore$$ $$f(x)$$ is periodic and period of $$f(x) = 2$$</p>
<p>And period of $$\cos \pi x = {{2\pi } \over \pi } = 2$$</p>
<p>$$\therefore$$ Period of $$f(x)\cos \pi x = 2$$</p>
<p>Now,</p>
<p>$$I = {{{\pi ^2}} \over {10}}\int_{ - 10}^{10} {f(x)\cos \pi x\,dx} $$</p>
<p>$$ = {{{\pi ^2}} \over {10}}\int_{ - 10}^{ - 10 + 10 \times 2} {f(x)\cos \pi x\,dx} $$</p>
<p>$$ = {{{\pi ^2}} \over {10}}\int_0^{10 \times 2} {f(x)\cos \pi x\,dx} $$</p>
<p>$$ = {{{\pi ^2}} \over {10}} \times 10\int_0^2 {f(x)\cos \pi x\,dx} $$</p>
<p>$$ = {\pi ^2}\int_0^2 {f(x)\cos \pi x\,dx} $$</p>
<p>$$\therefore$$ $$I = {\pi ^2}\left[ {\int_0^1 {f(x)\cos \pi x\,dx + \int_1^2 {f(x)\cos \pi x\,dx} } } \right]$$</p>
<p>$$ = {\pi ^2}\left[ {\int_0^1 {(1 - x)\cos \pi x\,dx + \int_1^2 {(x - 1)\cos \pi x\,dx} } } \right]$$</p>
<p>$$ = {\pi ^2}\left[ {\int_0^1 {\cos \pi x\,dx - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - \int_1^2 {\cos \pi x\,dx} } } } } \right]$$</p>
<p>$$ = {\pi ^2}\left[ {{1 \over \pi }\left[ {\sin \pi x} \right]_0^1 - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - {1 \over \pi }\left[ {\sin \pi x} \right]_1^2} } } \right]$$</p>
<p>$$ = {\pi ^2}\left[ {0 - \int_0^1 {x\cos \pi x\,dx + \int_1^2 {x\cos \pi x\,dx - 0} } } \right]$$</p>
<p>$$ = {\pi ^2}\left[ { - \left[ {x{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x} \right]_0^1 + \left[ {x{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x} \right]_1^2} \right]$$</p>
<p>$$\left[ {\mathrm{As}\,\int {x\cos \pi x\,dx = x\,.\,\int {\cos \pi x - \int {\left( {1\,.\,{{\sin \pi x} \over \pi }} \right)dx = x\,.\,{{\sin \pi x} \over \pi } + {1 \over {{\pi ^2}}}\cos \pi x + c} } } } \right]$$</p>
<p>$$ = {\pi ^2}\left[ { - \left[ {\left( {1\,.\,{{\sin \pi } \over \pi } + {1 \over {{\pi ^2}}}\,.\,\cos \pi } \right) - \left( {0 + {1 \over {{\pi ^2}}}\,.\,\cos 0} \right)} \right] + \left[ {\left( {2\,.\,{{\sin 2\pi } \over \pi } + {1 \over {{\pi ^2}}}\cos 2\pi } \right) - \left( {1\,.\,{{\sin \pi } \over \pi } + {1 \over {{\pi ^2}}}\cos \pi } \right)} \right]} \right]$$</p>
<p>$$ = {\pi ^2}\left[ { - \left\{ {\left( { - {1 \over {{\pi ^2}}}} \right) - \left( {{1 \over {{\pi ^2}}}} \right)} \right\} + \left( {\left( { + {1 \over {{\pi ^2}}}} \right) - \left( { - {1 \over {{\pi ^2}}}} \right)} \right.} \right]$$</p>
<p>$$ = {\pi ^2}\left[ { - \left( { - {2 \over {{\pi ^2}}}} \right) + {2 \over {{\pi ^2}}}} \right]$$</p>
<p>$$ = {\pi ^2}\left[ {{2 \over {{\pi ^2}}} + {2 \over {{\pi ^2}}}} \right]$$</p>
<p>$$ = {\pi ^2} \times {4 \over {{\pi ^2}}}$$</p>
<p>$$ = 4$$</p>
|
mcq
|
jee-main-2022-online-25th-july-morning-shift
|
1l6f157cx
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$[t]$$ denote the greatest integer less than or equal to $$t$$. Then the value of the integral $$\int_{-3}^{101}\left([\sin (\pi x)]+e^{[\cos (2 \pi x)]}\right) d x$$ is equal to</p>
|
[{"identifier": "A", "content": "$$\\frac{52(1-e)}{e}$$"}, {"identifier": "B", "content": "$$\\frac{52}{e}$$"}, {"identifier": "C", "content": "$$\\frac{52(2+e)}{e}$$"}, {"identifier": "D", "content": "$$\\frac{104}{e}$$"}]
|
["B"]
| null |
<p>$$I = \int_{ - 3}^{101} {\left( {\left[ {\sin (\pi x)} \right] + {e^{[\cos (2\pi x)]}}} \right)dx} $$</p>
$$[\sin \pi x]$$ is periodic with period 2 and $${{e^{[\cos (2\pi x)]}}}$$ is periodic with period 1.</p>
<p>So,</p>
<p>$$I = 52\int_0^2 {\left( {\left[ {\sin \pi x} \right] + {e^{[\cos 2\pi x]}}} \right)dx} $$</p>
<p>$$ = 52\left\{ {\int_1^2 { - 1} \,dx + \int_{{1 \over 4}}^{{3 \over 4}} {{e^{ - 1}}\,dx + \int_{{5 \over 4}}^{{7 \over 4}} {{e^{ - 1}}\,dx + \int_0^{{1 \over 4}} {{e^0}\,dx + \int_{{3 \over 4}}^{{5 \over 4}} {{e^0}\,dx + \int_{{7 \over 4}}^2 {{e^0}\,dx} } } } } } \right\}$$</p>
<p>$$ = {{52} \over e}$$</p>
|
mcq
|
jee-main-2022-online-25th-july-evening-shift
|
1l6f3l2tc
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $${a_n} = \int\limits_{ - 1}^n {\left( {1 + {x \over 2} + {{{x^2}} \over 3} + \,\,.....\,\, + \,\,{{{x^{n - 1}}} \over n}} \right)dx} $$ for every n $$\in$$ N. Then the sum of all the elements of the set {n $$\in$$ N : a<sub>n</sub> $$\in$$ (2, 30)} is ____________.</p>
|
[]
| null |
5
|
<p>$$\because$$ $${a_n} = \int\limits_{ - 1}^n {\left( {1 + {x \over 2} + {{{x^2}} \over 3}\, + \,....\, + \,{{{x^{n - 1}}} \over n}} \right)dx} $$</p>
<p>$$ = \left[ {x + {{{x^2}} \over {{2^2}}} + {{{x^3}} \over {{3^2}}}\, + \,......\, + \,{{{x^n}} \over {{n^2}}}} \right]_{ - 1}^n$$</p>
<p>$${a_n} = {{n + 1} \over {{1^2}}} + {{{n^2} - 1} \over {{2^2}}} + {{{n^3} + 1} \over {{3^2}}} + {{{n^4} - 1} \over {{4^2}}}\, + \,...\, + \,{{{n^n} + {{( - 1)}^{n + 1}}} \over {{n^2}}}$$</p>
<p>Here, $${a_1} = 2,\,{a_2} = {{2 + 1} \over 1} + {{{2^2} - 1} \over 2} = 3 + {3 \over 2} = {9 \over 2}$$</p>
<p>$${a_3} = 4 + 2 + {{28} \over 9} = {{100} \over 9}$$</p>
<p>$${a_4} = 5 + {{15} \over 4} + {{65} \over 9} + {{255} \over {16}} > 31$$.</p>
<p>$$\therefore$$ The required set is $$\{ 2,3\} $$. $$\because$$ $${a_n} \in (2,30)$$</p>
<p>$$\therefore$$ Sum of elements = 5.</p>
|
integer
|
jee-main-2022-online-25th-july-evening-shift
|
1l6gjh7xh
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If $$\mathrm{n}(2 \mathrm{n}+1) \int_{0}^{1}\left(1-x^{\mathrm{n}}\right)^{2 \mathrm{n}} \mathrm{d} x=1177 \int_{0}^{1}\left(1-x^{\mathrm{n}}\right)^{2 \mathrm{n}+1} \mathrm{~d} x$$, then $$\mathrm{n} \in \mathbf{N}$$ is equal to ______________.</p>
|
[]
| null |
24
|
<p>$$\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = \int_0^1 {1\,.\,{{(1 - {x^n})}^{2n + 1}}dx} } $$</p>
<p>$$ = \left[ {{{(1 - {x^n})}^{2n + 1}}\,.\,x} \right]_0^1 - \int_0^1 {x\,.\,(2n + 1){{(1 - {x^n})}^{2n}}\,.\, - n{x^{n - 1}}dx} $$</p>
<p>$$ = n(2n + 1)\int_0^1 {(1 - (1 - {x^n})){{(1 - {x^n})}^{2n}}dx} $$</p>
<p>$$ = n(2n + 1)\int_0^1 {{{(1 - {x^n})}^{2n}}dx - n(2n + 1)\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx} } $$</p>
<p>$$(1 + n(2n + 1))\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = n(2n + 1)\int_0^1 {{{(1 - {x^n})}^{2n}}dx} } $$</p>
<p>$$(2{n^2} + n + 1)\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx = 1177\int_0^1 {{{(1 - {x^n})}^{2n + 1}}dx} } $$</p>
<p>$$\therefore$$ $$2{n^2} + n + 1 = 1177$$</p>
<p>$$2{n^2} + n - 1176 = 0$$</p>
<p>$$\therefore$$ $$n = 24$$ or $$ - {{49} \over 2}$$</p>
<p>$$\therefore$$ $$n = 24$$</p>
|
integer
|
jee-main-2022-online-26th-july-morning-shift
|
1l6hydgfm
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>$$
\int\limits_{0}^{20 \pi}(|\sin x|+|\cos x|)^{2} d x \text { is equal to }
$$</p>
|
[{"identifier": "A", "content": "$$10(\\pi+4)$$"}, {"identifier": "B", "content": "$$10(\\pi+2)$$"}, {"identifier": "C", "content": "$$20(\\pi-2)$$"}, {"identifier": "D", "content": "$$20(\\pi+2)$$"}]
|
["D"]
| null |
<p>$$I = \int\limits_0^{20\pi } {{{\left( {|\sin x| + |\cos x|} \right)}^2}\,dx} $$</p>
<p>$$ = 20\int\limits_0^\pi {\left( {1 + |\sin 2x|} \right)\,dx} $$</p>
<p>$$ = 40\int\limits_0^{{\pi \over 2}} {(1 + \sin 2x)\,dx} $$</p>
<p>$$ = \left. {40\left( {x - {{\cos 2x} \over 2}} \right)} \right|_0^{{\pi \over 2}}$$</p>
<p>$$ = 40\left( {{\pi \over 2} + {1 \over 2} + {1 \over 2}} \right) = 20(\pi + 2)$$</p>
|
mcq
|
jee-main-2022-online-26th-july-evening-shift
|
1l6jbku4w
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a function defined as</p>
<p>$$f(x)=a \sin \left(\frac{\pi[x]}{2}\right)+[2-x], a \in \mathbb{R}$$ where $$[t]$$ is the greatest integer less than or equal to $$t$$. If $$\mathop {\lim }\limits_{x \to -1 } f(x)$$ exists, then the value of $$\int\limits_{0}^{4} f(x) d x$$ is equal to</p>
|
[{"identifier": "A", "content": "$$-$$1"}, {"identifier": "B", "content": "$$-$$2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}]
|
["B"]
| null |
<p>$$f(x) = a\sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]\,a \in R$$</p>
<p>Now,</p>
<p>$$\because$$ $$\mathop {\lim }\limits_{x \to - 1} f(x)$$ exist</p>
<p>$$\therefore$$ $$\mathop {\lim }\limits_{x \to - {1^ - }} f(x) = \mathop {\lim }\limits_{x \to - {1^ + }} f(x)$$</p>
<p>$$ \Rightarrow a\sin \left( {{{ - 2\pi } \over 2}} \right) + 3 = a\sin \left( {{{ - \pi } \over 2}} \right) + 2$$</p>
<p>$$ \Rightarrow - a = 1 \Rightarrow a = - 1$$</p>
<p>Now, $$\int_0^4 {f(x)dx = \int_0^4 {\left( { - \sin \left( {{{\pi [x]} \over 2}} \right) + [2 - x]} \right)dx} } $$</p>
<p>$$ = \int_0^1 {1dx + \int_1^2 { - 1dx + \int_2^3 { - 1dx + \int_3^4 {(1 - 2)dx} } } } $$</p>
<p>$$ = 1 - 1 - 1 - 1 = - 2$$</p>
|
mcq
|
jee-main-2022-online-27th-july-morning-shift
|
1l6jbqfar
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$
I=\int_{\pi / 4}^{\pi / 3}\left(\frac{8 \sin x-\sin 2 x}{x}\right) d x
$$. Then</p>
|
[{"identifier": "A", "content": "$${\\pi \\over 2} < I < {{3\\pi } \\over 4}$$"}, {"identifier": "B", "content": "$${\\pi \\over 5} < I < {{5\\pi } \\over {12}}$$"}, {"identifier": "C", "content": "$${{5\\pi } \\over {12}} < I < {{\\sqrt 2 } \\over 3}\\pi $$"}, {"identifier": "D", "content": "$${{3\\pi } \\over 4} < I < \\pi $$"}]
|
["C"]
| null |
<p>I comes out around 1.536 which is not satisfied by any given options.</p>
<p>$$\int\limits_{\pi /4}^{\pi /3} {{{8x - 2x} \over x}dx > I > \int\limits_{\pi /4}^{\pi /3} {{{8\sin x - 2x} \over x}dx} } $$</p>
<p>$${\pi \over 2} > I > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x} \over x} - 2} \right)dx} $$</p>
<p>$${{\sin x} \over x}$$ is decreasing in $$\left( {{\pi \over 4},{\pi \over 3}} \right)$$ so it attains maximum at $$x = {\pi \over 4}$$</p>
<p>$$I > \int\limits_{\pi /4}^{\pi /3} {\left( {{{8\sin x/3} \over {x/3}} - 2} \right)dx} $$</p>
<p>$$I > \sqrt 3 - {\pi \over 6}$$</p>
|
mcq
|
jee-main-2022-online-27th-july-morning-shift
|
1l6jdqp3s
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let a function $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be defined as :</p>
<p>$$f(x)= \begin{cases}\int\limits_{0}^{x}(5-|t-3|) d t, & x>4 \\ x^{2}+b x & , x \leq 4\end{cases}$$</p>
<p>where $$\mathrm{b} \in \mathbb{R}$$. If $$f$$ is continuous at $$x=4$$, then which of the following statements is NOT true?</p>
|
[{"identifier": "A", "content": "$$f$$ is not differentiable at $$x=4$$"}, {"identifier": "B", "content": "$$f^{\\prime}(3)+f^{\\prime}(5)=\\frac{35}{4}$$"}, {"identifier": "C", "content": "$$f$$ is increasing in $$\\left(-\\infty, \\frac{1}{8}\\right) \\cup(8, \\infty)$$"}, {"identifier": "D", "content": "$$f$$ has a local minima at $$x=\\frac{1}{8}$$"}]
|
["C"]
| null |
<p>$$\because$$ f(x) is continuous at x = 4</p>
<p>$$ \Rightarrow f({4^ - }) = f({4^ + })$$</p>
<p>$$ \Rightarrow 16 + 4b = \int\limits_0^4 {(5 - |t - 3|)dt} $$</p>
<p>$$ = \int\limits_0^3 {(2 + t)dt + \int\limits_3^4 {(8 - t)dt} } $$</p>
<p>$$ = \left. {2t + {{{t^2}} \over 2}} \right)_0^3 + \left. {8t - {{{t^2}} \over 3}} \right]_3^4$$</p>
<p>$$ = 6 + {9 \over 2} - 0 + (32 - 8) - \left( {24 - {9 \over 2}} \right)$$</p>
<p>$$16 + 4b = 15$$</p>
<p>$$ \Rightarrow b = {{ - 1} \over 4}$$</p>
<p>$$ \Rightarrow f(x) = \left\{ {\matrix{
{\int\limits_0^x {5 - |t - 3|\,dt} } & {x > 4} \cr
{{x^2} - {x \over 4}} & {x \le 4} \cr
} } \right.$$</p>
<p>$$ \Rightarrow f'(x) = \left\{ {\matrix{
{5 - |x - 3|} & {x > 4} \cr
{2x - {1 \over 4}} & {x \le 4} \cr
} } \right.$$</p>
<p>$$ \Rightarrow f'(x) = \left\{ {\matrix{
{8 - x} & {x > 4} \cr
{2x - {1 \over 4}} & {x \le 4} \cr
} } \right.$$</p>
<p>$$f'(x) < 0 = x \in \left( { - \infty ,{1 \over 8}} \right) \cup (8,\infty )$$</p>
<p>$$f'(3) + f'(5) = 6 - {1 \over 4} = {{35} \over 4}$$</p>
<p>$$f'(x) = 0 \Rightarrow x = {1 \over 8}$$ have local minima</p>
<p>$$\therefore$$ (C) is only incorrect option.</p>
|
mcq
|
jee-main-2022-online-27th-july-morning-shift
|
1l6kiq196
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$f(x)=2+|x|-|x-1|+|x+1|, x \in \mathbf{R}$$.</p>
<p>Consider</p>
<p>$$(\mathrm{S} 1): f^{\prime}\left(-\frac{3}{2}\right)+f^{\prime}\left(-\frac{1}{2}\right)+f^{\prime}\left(\frac{1}{2}\right)+f^{\prime}\left(\frac{3}{2}\right)=2$$</p>
<p>$$(\mathrm{S} 2): \int\limits_{-2}^{2} f(x) \mathrm{d} x=12$$</p>
<p>Then,</p>
|
[{"identifier": "A", "content": "both (S1) and (S2) are correct"}, {"identifier": "B", "content": "both (S1) and (S2) are wrong"}, {"identifier": "C", "content": "only (S1) is correct"}, {"identifier": "D", "content": "only (S2) is correct"}]
|
["D"]
| null |
<p>$$f(x) = 2 + |x| - |x - 1| + |x + 1|,\,x \in R$$</p>
<p>$$\therefore$$ $$f(x) = \left\{ {\matrix{
{ - x} & , & {x < - 1} \cr
{x + 2} & , & { - 1 \le x < 0} \cr
{3x + 2} & , & {0 \le x < 1} \cr
{x + 4} & , & {x \ge 1} \cr
} } \right.$$</p>
<p>$$\therefore$$ $$f'\left( { - {3 \over 2}} \right) + f'\left( { - {1 \over 2}} \right) + f'\left( {{1 \over 2}} \right) + f'\left( {{3 \over 2}} \right) = - 1 + 1 + 3 + 1 = 4$$</p>
<p>and $$\int\limits_{ - 2}^2 {f(x)dx = \int\limits_{ - 2}^{ - 1} {f(x)dx + \int\limits_{ - 1}^0 {f(x)dx + \int\limits_0^1 {f(x)dx + \int\limits_1^2 {f(x)dx} } } } } $$</p>
<p>$$ = \left[ { - {{{x^2}} \over 2}} \right]_2^{ - 1} + \left[ {{{{{(x + 2)}^2}} \over 2}} \right]_{ - 1}^0 + \left[ {{{{{(3x + 2)}^2}} \over 6}} \right]_0^1 + \left[ {{{{{(x + 4)}^2}} \over 2}} \right]_1^2$$</p>
<p>$$ = {3 \over 2} + {3 \over 2} + {7 \over 2} + {{11} \over 2} = {{24} \over 2} = 12$$</p>
<p>$$\therefore$$ Only (S2) is correct</p>
|
mcq
|
jee-main-2022-online-27th-july-evening-shift
|
1l6kjxlll
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>$$\int\limits_{0}^{2}\left(\left|2 x^{2}-3 x\right|+\left[x-\frac{1}{2}\right]\right) \mathrm{d} x$$, where [t] is the greatest integer function, is equal to :</p>
|
[{"identifier": "A", "content": "$$\\frac{7}{6}$$"}, {"identifier": "B", "content": "$$\\frac{19}{12}$$"}, {"identifier": "C", "content": "$$\\frac{31}{12}$$"}, {"identifier": "D", "content": "$$\\frac{3}{2}$$"}]
|
["B"]
| null |
<p>$$\int\limits_0^2 {|2{x^2} - 3x|dx + \int\limits_0^2 {\left[ {x - {1 \over 2}} \right]dx} } $$</p>
<p>$$ = \int\limits_0^{3/2} {(3x - 2{x^2})dx + \int\limits_{3/2}^2 {(2{x^2} - 3x)dx + \int\limits_0^{1/2} { - 1dx + \int\limits_{1/2}^{3/2} {0\,dx + \int\limits_{3/2}^2 {1dx} } } } } $$</p>
<p>$$ = \left. {\left( {{{3{x^2}} \over 2} - {{2{x^3}} \over 3}} \right)} \right|_0^{3/2} + \left. {\left( {{{2{x^3}} \over 3} - {{3{x^2}} \over 2}} \right)} \right|_{3/2}^2 - {1 \over 2} + {1 \over 2}$$</p>
<p>$$ = \left( {{{27} \over 8} - {{27} \over {12}}} \right) + \left( {{{16} \over 3} - 6 - {{27} \over {12}} + {{27} \over 8}} \right)$$</p>
<p>$$ = {{19} \over {12}}$$</p>
|
mcq
|
jee-main-2022-online-27th-july-evening-shift
|
1l6klps4r
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$f(x)=\min \{[x-1],[x-2], \ldots,[x-10]\}$$ where [t] denotes the greatest integer $$\leq \mathrm{t}$$. Then $$\int\limits_{0}^{10} f(x) \mathrm{d} x+\int\limits_{0}^{10}(f(x))^{2} \mathrm{~d} x+\int\limits_{0}^{10}|f(x)| \mathrm{d} x$$ is equal to ________________.</p>
|
[]
| null |
385
|
<p>$$\because$$ $$f(x) = \min \,\{ [x - 1],[x - 2],\,......,\,[x - 10]\} = [x - 10]$$</p>
<p>Also $$|f(x)| = \left\{ {\matrix{
{ - f(x),} & {if\,x \le 10} \cr
{f(x),} & {if\,x \ge 10} \cr
} } \right.$$</p>
<p>$$\therefore$$ $$\int\limits_0^{10} {f(x)dx + \int\limits_0^{10} {{{(f(x))}^2}dx + \int\limits_0^{10} {( - f(x))dx} } } $$</p>
<p>$$ = \int\limits_0^{10} {{{(f(x))}^2}dx} $$</p>
<p>$$ = {10^2} + {9^2} + {8^2}\, + \,.....\, + \,{1^2}$$</p>
<p>$$ = {{10 \times 11 \times 21} \over 6} = 385$$</p>
|
integer
|
jee-main-2022-online-27th-july-evening-shift
|
1l6m6kw3s
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If $$\int\limits_{0}^{\sqrt{3}} \frac{15 x^{3}}{\sqrt{1+x^{2}+\sqrt{\left(1+x^{2}\right)^{3}}}} \mathrm{~d} x=\alpha \sqrt{2}+\beta \sqrt{3}$$, where $$\alpha, \beta$$ are integers, then $$\alpha+\beta$$ is equal to __________.</p>
|
[]
| null |
10
|
<p>Put $$x = \tan \theta \Rightarrow dx = {\sec ^2}\theta \,d\theta $$</p>
<p>$$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15{{\tan }^3}\theta \,.\,{{\sec }^2}\theta \,d\theta } \over {\sqrt {1 + {{\tan }^2}\theta + \sqrt {{{\sec }^6}\theta } } }}} $$</p>
<p>$$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15{{\tan }^2}\theta {{\sec }^2}\theta \,d\theta } \over {\sec \theta \sqrt {1 + \sec \theta } }}} $$</p>
<p>$$ \Rightarrow I = \int\limits_0^{{\pi \over 3}} {{{15({{\sec }^2}\theta - 1)\sec \theta \tan \theta \,d\theta } \over {\left( {\sqrt {1 + \sec \theta } } \right)}}} $$</p>
<p>Now put $$1 + \sec \theta = {t^2}$$</p>
<p>$$ \Rightarrow \sec \theta \tan \theta \,d\theta = 2tdt$$</p>
<p>$$ \Rightarrow I = \int\limits_{\sqrt 2 }^{\sqrt 3 } {{{15\left( {{{({t^2} - 1)}^2} - 1} \right)2t\,dt} \over t}} $$</p>
<p>$$ \Rightarrow I = 30\int\limits_{\sqrt 2 }^{\sqrt 3 } {({t^4} - 2{t^2} + 1 - 1)\,dt} $$</p>
<p>$$ \Rightarrow I = 30\int\limits_{\sqrt 2 }^{\sqrt 3 } {({t^4} - 2{t^2})\,dt} $$</p>
<p>$$ \Rightarrow I = \left. {30\left( {{{{t^5}} \over 5} - {{2{t^3}} \over 3}} \right)} \right|_{\sqrt 2 }^{\sqrt 3 }$$</p>
<p>$$ = 30\left[ {\left( {{9 \over 5}\sqrt 3 - 2\sqrt 3 } \right) - \left( {{{4\sqrt 2 } \over 5} - {{4\sqrt 2 } \over 3}} \right)} \right]$$</p>
<p>$$ = \left( {54\sqrt 3 - 60\sqrt 3 } \right) - \left( {24\sqrt 2 - 40\sqrt 2 } \right)$$</p>
<p>$$ = 16\sqrt 2 - 6\sqrt 3 $$</p>
<p>$$\therefore$$ $$\alpha = 16$$ and $$\beta = - 6$$</p>
<p>$$\alpha + \beta = 10.$$</p>
|
integer
|
jee-main-2022-online-28th-july-morning-shift
|
1l6nm7c4a
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$I_{n}(x)=\int_{0}^{x} \frac{1}{\left(t^{2}+5\right)^{n}} d t, n=1,2,3, \ldots .$$ Then :</p>
|
[{"identifier": "A", "content": "$$50 I_{6}-9 I_{5}=x I_{5}^{\\prime}$$"}, {"identifier": "B", "content": "$$50 I_{6}-11 I_{5}=x I_{5}^{\\prime}$$"}, {"identifier": "C", "content": "$$50 I_{6}-9 I_{5}=I_{5}^{\\prime}$$"}, {"identifier": "D", "content": "$$50 I_{6}-11 I_{5}=I_{5}^{\\prime}$$"}]
|
["A"]
| null |
<p>$${I_n}(x) = \int\limits_0^x {{1 \over {{{({t^2} + 5)}^n}}}dt} $$</p>
<p>$$ = \int\limits_0^x {{1 \over {\underbrace {{{({t^2} + 5)}^n}}_I}} \times \mathop I\limits_{II} \,dt} $$</p>
<p>$$ = \left. {{t \over {{{({t^2} + 5)}^n}}}} \right|_0^x - \int\limits_0^x {{{ - 2nt} \over {{{({t^2} + 5)}^{n + 1}}}} \times t\,dt} $$</p>
<p>$$ = {x \over {{{({x^2} + 5)}^n}}} + \int\limits_0^x {2n\left( {{{{t^2} + 5 - 5} \over {{{({t^2} + 5)}^{n + 1}}}}} \right)dt} $$</p>
<p>$${I_n}(x) = {x \over {{{({x^2} + 5)}^n}}} + 2n\,{I_n}(x) - 10n\,{I_{n + 1}}(x)$$</p>
<p>$$10n\,{I_{n + 1}}(x) - (2n - 1)\,{I_n}(x) = xI{'_n}(x)$$</p>
<p>For $$n = 5$$</p>
<p>$$50{I_6}(x) - 9{I_5}(x) = xI{'_5}(x)$$</p>
|
mcq
|
jee-main-2022-online-28th-july-evening-shift
|
1l6npjdpk
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The value of the integral $$\int\limits_{0}^{\frac{\pi}{2}} 60 \frac{\sin (6 x)}{\sin x} d x$$ is equal to _________.</p>
|
[]
| null |
104
|
<p>$$I = \int\limits_0^{{\pi \over 2}} {60\,.\,{{\sin 6x} \over {\sin x}}dx} $$</p>
<p>$$ = 60\,.\,2\int\limits_0^{{\pi \over 2}} {(3 - 4{{\sin }^2}x)(4{{\cos }^2}x - 3)\cos x\,dx} $$</p>
<p>$$ = 120\int\limits_0^{{\pi \over 2}} {(3 - 4{{\sin }^2}x)(1 - 4{{\sin }^2}x)\cos x\,dx} $$</p>
<p>Let $$\sin x = t \Rightarrow \cos xdx = dt$$</p>
<p>$$ = 120\int\limits_0^1 {(3 - 4{t^2})(1 - 4{t^2})dt} $$</p>
<p>$$ = 120\int\limits_0^1 {(3 - 16{t^2} + 16{t^4})dt} $$</p>
<p>$$ = 120\left[ {3t - {{16{t^3}} \over 3} + {{16{t^5}} \over 5}} \right]_0^1$$</p>
<p>$$ = 104$$</p>
|
integer
|
jee-main-2022-online-28th-july-evening-shift
|
1l6p1b4ew
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The integral $$\int\limits_{0}^{\frac{\pi}{2}} \frac{1}{3+2 \sin x+\cos x} \mathrm{~d} x$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$\\tan ^{-1}(2)$$"}, {"identifier": "B", "content": "$$\\tan ^{-1}(2)-\\frac{\\pi}{4}$$"}, {"identifier": "C", "content": "$$\\frac{1}{2} \\tan ^{-1}(2)-\\frac{\\pi}{8}$$"}, {"identifier": "D", "content": "$$\\frac{1}{2}$$"}]
|
["B"]
| null |
<p>$$I = \int\limits_0^{\pi /2} {{1 \over {3 + 2\sin x + \cos x}}dx} $$</p>
<p>$$ = \int\limits_0^{\pi /2} {{{(1 + {{\tan }^2}x/2)dx} \over {3(1 + {{\tan }^2}x/2) + 2(2\tan x/2) + (1 - {{\tan }^2}x/2)}}} $$</p>
<p>Let $$\tan x/2 = t \Rightarrow {\sec ^2}x/2dx = 2dt$$</p>
<p>$$I = \int\limits_0^1 {{{2dt} \over {4 + 2{t^2} + 4t}}} $$</p>
<p>$$ = \int\limits_0^1 {{{dt} \over {{t^2} + 2t + 2}} = \int\limits_0^1 {{{dt} \over {{{(t + 1)}^2} + 1}}} } $$</p>
<p>$$ = \left. {{{\tan }^{ - 1}}(t + 1)} \right|_0^1 = {\tan ^{ - 1}}2 - {\pi \over 4}$$</p>
|
mcq
|
jee-main-2022-online-29th-july-morning-shift
|
1l6p2ihmv
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If $$f(\alpha)=\int\limits_{1}^{\alpha} \frac{\log _{10} \mathrm{t}}{1+\mathrm{t}} \mathrm{dt}, \alpha>0$$, then $$f\left(\mathrm{e}^{3}\right)+f\left(\mathrm{e}^{-3}\right)$$ is equal to :</p>
|
[{"identifier": "A", "content": "9"}, {"identifier": "B", "content": "$$\\frac{9}{2}$$"}, {"identifier": "C", "content": "$$\\frac{9}{\\log _{e}(10)}$$"}, {"identifier": "D", "content": "$$\\frac{9}{2 \\log _{e}(10)}$$"}]
|
["D"]
| null |
<p>$$f(\alpha ) = \int_1^\alpha {{{{{\log }_{10}}t} \over {1 + t}}dt} $$ ...... (i)</p>
<p>$$f\left( {{1 \over \alpha }} \right) = \int_1^{{1 \over \alpha }} {{{{{\log }_{10}}t} \over {1 + t}}dt} $$</p>
<p>Substituting $$t \to {1 \over p}$$</p>
<p>$$f\left( {{1 \over \alpha }} \right) = \int_1^\alpha {{{{{\log }_{10}}\left( {{1 \over p}} \right)} \over {1 + {1 \over p}}}\left( {{{ - 1} \over {{p^2}}}} \right)dp} $$</p>
<p>$$ = \int_1^\alpha {{{{{\log }_{10}}p} \over {p(p + 1)}}dp = \int_1^\alpha {\left( {{{{{\log }_{10}}t} \over t} - {{{{\log }_{10}}t} \over {t + 1}}} \right)dt} } $$ ....... (ii)</p>
<p>By (i) + (ii)</p>
<p>$$f(\alpha ) + f\left( {{1 \over \alpha }} \right) = \int_1^\alpha {{{{{\log }_{10}}t} \over t}dt = \int_1^\alpha {{{\ln t} \over t}\,.\,{{\log }_{10}}e\,dt} } $$</p>
<p>$$ = {{{{(\ln \alpha )}^2}} \over {2{{\log }_e}10}}$$</p>
<p>$$\alpha = {e^3} \Rightarrow f({e^3}) + f({e^{ - 3}}) = {9 \over {2{{\log }_e}10}}$$</p>
|
mcq
|
jee-main-2022-online-29th-july-morning-shift
|
1l6re2c2a
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If $$[t]$$ denotes the greatest integer $$\leq t$$, then the value of $$\int_{0}^{1}\left[2 x-\left|3 x^{2}-5 x+2\right|+1\right] \mathrm{d} x$$ is :</p>
|
[{"identifier": "A", "content": "$$\\frac{\\sqrt{37}+\\sqrt{13}-4}{6}$$"}, {"identifier": "B", "content": "$$\\frac{\\sqrt{37}-\\sqrt{13}-4}{6}$$"}, {"identifier": "C", "content": "$$\\frac{-\\sqrt{37}-\\sqrt{13}+4}{6}$$"}, {"identifier": "D", "content": "$$\\frac{-\\sqrt{37}+\\sqrt{13}+4}{6}$$"}]
|
["A"]
| null |
<p>$$\int_0^1 {\left[ {2x - |3{x^2} - 5x + 2| + 1} \right]dx} $$</p>
<p>$$3{x^2} - 5x + 2 = 0$$</p>
<p>$$ \Rightarrow 3{x^2} - 3x - 2x + 2 = 0$$</p>
<p>$$ \Rightarrow 3x(x - 1) - 2(x - 1) = 0$$</p>
<p>$$ \Rightarrow (x - 1)(3x - 2) = 0$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7e9s1mi/e836d5da-b39f-4776-82f3-0bfd2186d40f/58c699a0-2754-11ed-a077-1f1e3989e798/file-1l7e9s1mj.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7e9s1mi/e836d5da-b39f-4776-82f3-0bfd2186d40f/58c699a0-2754-11ed-a077-1f1e3989e798/file-1l7e9s1mj.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th July Evening Shift Mathematics - Definite Integration Question 84 English Explanation 1"></p>
<p>$$\therefore$$ In 0 to $${2 \over 3},\,|3{x^2} - 5x + 2| = 3{x^2} - 5x + 2$$</p>
<p>And in $${2 \over 3}$$ to 1, $$|3{x^2} - 5x + 2| = - (3{x^2} - 5x + 2)$$</p>
<p>$$\therefore$$ $$\int_0^{{2 \over 3}} {\left[ {2x - 3{x^2} + 5x - 2 + 1} \right]dx + \int_{{2 \over 3}}^1 {\left[ {2x + 3{x^2} - 5x + 2 + 1} \right]dx} } $$</p>
<p>$$ = \int_0^{{2 \over 3}} {\left[ { - 3{x^2} + 7x - 1} \right]dx + \int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} } $$</p>
<p>Now, let $$f(x) = - 3{x^2} + 7x - 1$$</p>
<p>$$ \Rightarrow f'(x) = - 6x + 7$$</p>
<p>$$ = - 6\left( {x - {7 \over 6}} \right)$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7e9ullv/94ab05fd-e1dc-4b97-860b-aead321779c9/9fd49a40-2754-11ed-a077-1f1e3989e798/file-1l7e9ullw.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7e9ullv/94ab05fd-e1dc-4b97-860b-aead321779c9/9fd49a40-2754-11ed-a077-1f1e3989e798/file-1l7e9ullw.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th July Evening Shift Mathematics - Definite Integration Question 84 English Explanation 2"></p>
<p>$$f(0) = - 1$$</p>
<p>$$f\left( {{2 \over 3}} \right) = - 3 \times {4 \over 9} + {{14} \over 3} - 1$$</p>
<p>$$ = {7 \over 3} = 2.33$$</p>
<p>$$\therefore$$ Range of $$f(x) = \left[ { - 1,{7 \over 3}} \right]$$</p>
<p>Integer value between $$ - 1$$ to $${7 \over 3} = - 1,0,1,2,{7 \over 3}$$</p>
<p>When $$f(x) = - 1$$</p>
<p>$$ \Rightarrow - 3{x^2} + 7x - 1 = - 1$$</p>
<p>$$ \Rightarrow 3{x^2} - 7x = 0$$</p>
<p>$$ \Rightarrow x(3x - 7) = 0$$</p>
<p>$$ \Rightarrow x = 0,{7 \over 3}$$ (not possible as $${7 \over 3} \notin (0,2)$$)</p>
<p>When $$f(x) = 0$$</p>
<p>$$ \Rightarrow - 3{x^2} + 7x - 1 = 0$$</p>
<p>$$ \Rightarrow 3{x^2} - 7x + 1 = 0$$</p>
<p>$$ \Rightarrow x = {{7\, \pm \,\sqrt {49 - 12} } \over 6}$$</p>
<p>$$ = {{7\, \pm \sqrt {37} } \over 6}$$</p>
<p>$$\therefore$$ $$x = {{7 - \sqrt {37} } \over 6}$$</p>
<p>When $$f(x) = 1$$</p>
<p>$$ \Rightarrow - 3{x^2} + 7x - 1 = 1$$</p>
<p>$$ \Rightarrow 3{x^2} - 7x + 2 = 0$$</p>
<p>$$ \Rightarrow x = {{7\, \pm \,\sqrt {49 - 24} } \over 6}$$</p>
<p>$$ = {{7\, \pm \,\sqrt {25} } \over 6}$$</p>
<p>$$ = {{7\, \pm \,5} \over 6}$$</p>
<p>$$ \Rightarrow x = 2,\,{1 \over 3}$$</p>
<p>When $$f(x) = 2$$</p>
<p>$$ \Rightarrow - 3{x^2} + 7x - 1 = 2$$</p>
<p>$$ \Rightarrow 3{x^2} - 7x + 3 = 0$$</p>
<p>$$ \Rightarrow x = {{7\, \pm \,\sqrt {49 - 36} } \over 6}$$</p>
<p>$$ = {{7\, \pm \,\sqrt {13} } \over 6}$$</p>
<p>$$\therefore$$ $$x = {{7\, - \sqrt {13} } \over 6}$$</p>
<p>$$\therefore$$ Possible values of $$x = 0,\,{{7\, - \,\sqrt {37} } \over 6},{1 \over 3},{{7\, - \,\sqrt {13} } \over 6},{2 \over 3}$$</p>
<p>$$\therefore$$ $$\int_0^{{2 \over 3}} {\left[ { - 3{x^2} + 7x - 1} \right]dx} $$</p>
<p>$$ = \int_0^{{{7\, - \,\sqrt {37} } \over 6}} {( - 1)dx + \int_{{{7\, - \,\sqrt {37} } \over 6}}^{{1 \over 3}} {0\,dx + \int_{{1 \over 3}}^{{{7\, - \,\sqrt {13} } \over 6}} {(1)\,dx + \int_{{{7\, - \,\sqrt {13} } \over 6}}^{{2 \over 3}} {2\,dx} } } } $$</p>
<p>$$ = {{ - 7\, + \,\sqrt {37} } \over 6} + 0 + {{7\, - \,\sqrt {13} } \over 6} - {1 \over 3} + 2\left[ {{2 \over 3} - {{7\, - \,\sqrt {13} } \over 6}} \right]$$</p>
<p>$$ = {{ - 7 + \sqrt {37} + 7 - \sqrt {13} - 2 + 8 - 14 + 2\sqrt {13} } \over 6}$$</p>
<p>$$ = {{\sqrt {37} + \sqrt {13} - 8} \over 6}$$</p>
<p>Now, $$\int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} $$</p>
<p>Let $$f(x) = 3{x^2} - 3x + 3$$</p>
<p>$$f'(x) = 6x - 3 = 3(2x - 1)$$</p>
<p>$$f'(x) = 0 \Rightarrow 2x - 1 = 0 \Rightarrow x = {1 \over 2}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7e9vv1e/90dc8535-185a-41da-a8c1-52842238a21e/c2ecac20-2754-11ed-a077-1f1e3989e798/file-1l7e9vv1f.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7e9vv1e/90dc8535-185a-41da-a8c1-52842238a21e/c2ecac20-2754-11ed-a077-1f1e3989e798/file-1l7e9vv1f.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 29th July Evening Shift Mathematics - Definite Integration Question 84 English Explanation 3"></p>
<p>$$f\left( {{2 \over 3}} \right) = 3\left( {{4 \over 9} - {2 \over 3} + 1} \right)$$</p>
<p>$$ = {7 \over 3} = 2.3$$</p>
<p>$$f(1) = 3(1 - 1 + 1)$$</p>
<p>$$ = 3$$</p>
<p>No integer values present in between $${7 \over 3}$$ and 3.</p>
<p>$$\therefore$$ Possible value of $$x = {2 \over 3},\,1$$</p>
<p>So, integration don't break anywhere.</p>
<p>$$\therefore$$ $$\int_{{2 \over 3}}^1 {\left[ {3{x^2} - 3x + 3} \right]dx} $$</p>
<p>$$ = \int_{{2 \over 3}}^1 {\left[ {2.33} \right]\,dx} $$</p>
<p>$$ = \int_{{2 \over 3}}^1 {2\,dx = 2\left( {1 - {2 \over 3}} \right) = {2 \over 3}} $$</p>
<p>$$\therefore$$ Value of Integration</p>
<p>$$ = {{\sqrt {37} + \sqrt {13} - 1} \over 6} + {2 \over 3}$$</p>
<p>$$ = {{\sqrt {37} + \sqrt {13} - 8 + 4} \over 6}$$</p>
<p>$$ = {{\sqrt {37} + \sqrt {13} - 4} \over 6}$$</p>
|
mcq
|
jee-main-2022-online-29th-july-evening-shift
|
1ldo61zd1
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The value of the integral <br/><br/>$$\int\limits_{ - {\pi \over 4}}^{{\pi \over 4}} {{{x + {\pi \over 4}} \over {2 - \cos 2x}}dx} $$ is :</p>
|
[{"identifier": "A", "content": "$${{{\\pi ^2}} \\over {6\\sqrt 3 }}$$"}, {"identifier": "B", "content": "$${{{\\pi ^2}} \\over 6}$$"}, {"identifier": "C", "content": "$${{{\\pi ^2}} \\over {3\\sqrt 3 }}$$"}, {"identifier": "D", "content": "$${{{\\pi ^2}} \\over {12\\sqrt 3 }}$$"}]
|
["A"]
| null |
$$
I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{x+\frac{\pi}{4}}{(2-\cos 2 x)} d x
$$
<br/><br/>Using $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$
<br/><br/>$$
I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}}\left(\frac{-x+\frac{\pi}{4}}{2-\cos 2 x}\right) d x
$$
<br/><br/>$\begin{aligned} & \therefore 2 I=\int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \frac{\pi d x}{2(2-\cos 2 x)} \\\\ & \Rightarrow I=\frac{2 \pi}{4} \int_0^{\frac{\pi}{4}}\left(\frac{d x}{\left.\frac{2-1-\tan ^2 x}{1+\tan ^2 x}\right)}\right. \\\\ & \Rightarrow I=\frac{\pi}{2} \int_0^{\frac{\pi}{4}}\left(\frac{1+\tan ^2 x}{1+3 \tan ^2 x}\right) d x\end{aligned}$
<br/><br/>Now,
<br/><br/>$$
\begin{aligned}
& \tan x=t \\\\
& =\frac{\pi}{2} \int_0^1 \frac{d t}{1+3 t^2} \\\\
& =\frac{\pi}{2}\left[\frac{\tan ^{-1}(\sqrt{3} t)}{\sqrt{3}}\right]_0^1 \\\\
& =\frac{\pi}{2 \sqrt{3}}\left(\frac{\pi}{3}\right)=\frac{\pi^2}{6 \sqrt{3}}
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-1st-february-evening-shift
|
1ldo73jao
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If $$\int\limits_0^\pi {{{{5^{\cos x}}(1 + \cos x\cos 3x + {{\cos }^2}x + {{\cos }^3}x\cos 3x)dx} \over {1 + {5^{\cos x}}}} = {{k\pi } \over {16}}} $$, then k is equal to _____________.</p>
|
[]
| null |
13
|
$$
\begin{aligned}
& \mathrm{I}=\int_0^\pi \frac{5^{\cos x}\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right)}{1+5^{\cos x}} d x \\\\
& I=\int_0^\pi \frac{5^{-\cos x}\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right)}{1+5^{-\cos x}} d x \\\\
& 2 \mathrm{I}=\int_0^\pi\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right) d x \\\\
& {2 I}={2} \int_0^{\frac{\pi}{2}}\left(1+\cos x \cos 3 x+\cos ^2 x+\cos ^3 x \cos 3 x\right) d x
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& I=\int_0^{\frac{\pi}{2}}\left(1+\sin x(-\sin 3 x)+\sin ^2 x-\sin ^3 x \sin 3 x\right) d x \\\\
& 2 I=\int_0^{\frac{\pi}{2}}\left(3+\cos 4 x+\cos ^3 x \cos 3 x-\sin ^3 x \sin 3 x\right) d x \\\\
& 2 I=\int_0^{\frac{\pi}{2}} 3+\cos 4 x+\left(\frac{\cos 3 x+3 \cos x}{4}\right) \cos 3 x-\sin 3 x\left(\frac{3 \sin x-\sin 3 x}{4}\right) d x
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& 2 \mathrm{I}=\int_0^{\frac{\pi}{2}}\left(3+\cos 4 \mathrm{x}+\frac{1}{4}+\frac{3}{4} \cos 4 \mathrm{x}\right) \mathrm{dx} \\\\
& 2 \mathrm{I}=\frac{13}{4} \times \frac{\pi}{2}+\frac{7}{4}\left(\frac{\sin 4 \mathrm{x}}{4}\right)_0^{\frac{\pi}{2}} \Rightarrow \mathrm{I}=\frac{13 \pi}{16}
\end{aligned}
$$
|
integer
|
jee-main-2023-online-1st-february-evening-shift
|
ldo82cg6
|
maths
|
definite-integration
|
properties-of-definite-integration
|
Let $\alpha>0$. If $\int\limits_0^\alpha \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} \mathrm{~d} x=\frac{16+20 \sqrt{2}}{15}$, then $\alpha$ is equal to :
|
[{"identifier": "A", "content": "4\n"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "$2 \\sqrt{2}$"}, {"identifier": "D", "content": "$\\sqrt{2}$"}]
|
["B"]
| null |
$I=\int\limits_{0}^{\alpha} \frac{x}{\sqrt{x+\alpha}-\sqrt{x}} d x, \alpha>0$
<br/><br/>$$
\begin{aligned}
& =\frac{1}{\alpha} \int\limits_{0}^{\alpha} x(\sqrt{x+\alpha}+\sqrt{x}) d x
\end{aligned}
$$
<br/><br/>$$\eqalign{
& = {1 \over \alpha }\int\limits_0^\alpha {\left( {x\sqrt {x + \alpha } + x\sqrt x } \right)} dx \cr
& = {1 \over \alpha }\int\limits_0^\alpha {\left[ {\left( {x + \alpha - \alpha } \right)\sqrt {x + \alpha } + {x^{{3 \over 2}}}} \right]} dx \cr} $$
<br/><br/>$$
\begin{aligned}
& =\frac{1}{\alpha}\int\limits_0^\alpha \left[(x+\alpha)^{3 / 2}-\alpha(x+\alpha)^{1 / 2}+x^{3 / 2}\right] d x \\\\
& =\frac{1}{\alpha}\left[\frac{2}{5}(x+\alpha)^{5 / 2}-\alpha \frac{2}{3}(x+\alpha)^{3 / 2}+\frac{2}{5} x^{5 / 2}\right]_0^\alpha \\\\
& =\frac{1}{\alpha}\left(\frac{2}{5}(2 \alpha)^{5 / 2}-\frac{2 \alpha}{3}(2 \alpha)^{3 / 2}+\frac{2}{5} \alpha^{5 / 2}-\frac{2}{5} \alpha^{5 / 2}+\frac{2}{3} \alpha^{5 / 2}\right) \\\\
& =\frac{1}{\alpha}\left(\frac{2^{7 / 2} \alpha^{5 / 2}}{5}-\frac{2^{5 / 2} \alpha^{5 / 2}}{3}+\frac{2}{3} \alpha^{5 / 2}\right)=\alpha^{3 / 2}\left(\frac{2^{7 / 2}}{5}-\frac{2^{5 / 2}}{3}+\frac{2}{3}\right) \\\\
& =\frac{\alpha^{3 / 2}}{15}(24 \sqrt{2}-20 \sqrt{2}+10)=\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10)
\end{aligned}
$$
<br/><br/>Now, $\frac{\alpha^{3 / 2}}{15}(4 \sqrt{2}+10)=\frac{16+20 \sqrt{2}}{15}=\frac{2 \sqrt{2}(4 \sqrt{2}+10)}{15}$
<br/><br/>$$
\Rightarrow \alpha^{3 / 2}=2 \sqrt{2}=2^{3 / 2} $$
<br/><br/>$$\Rightarrow \alpha=2
$$
|
mcq
|
jee-main-2023-online-31st-january-evening-shift
|
1ldonnoj6
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If $$\int_\limits{0}^{1}\left(x^{21}+x^{14}+x^{7}\right)\left(2 x^{14}+3 x^{7}+6\right)^{1 / 7} d x=\frac{1}{l}(11)^{m / n}$$ where $$l, m, n \in \mathbb{N}, m$$ and $$n$$ are coprime then $$l+m+n$$ is equal to ____________.</p>
|
[]
| null |
63
|
$I=\int_{0}^{1}\left(x^{21}+x^{14}+x^{7}\right)\left(2 x^{14}+3 x^{7}+6\right)^{1 / 7} d x$ <br/><br/>$I=\int_{0}^{1}\left(x^{20}+x^{13}+x^{6}\right)\left(2 x^{21}+3 x^{14}+6 x^{7}\right)^{1 / 7} d x$
<br/><br/>Let $2 x^{21}+3 x^{14}+6 x^{7}=t$
<br/><br/>$\Rightarrow 42\left(x^{20}+x^{13}+x^{6}\right) d x=d t$
<br/><br/>$I=\frac{1}{42} \int_{0}^{11} t^{1 / 7} d t=\frac{1}{42} \frac{7}{8}\left[t^{8 / 7}\right]_{0}^{11}$
<br/><br/>$=\frac{1}{48} (11)^{8/7}$
<br/><br/>$\therefore \quad I=48, m=8, n=7$
<br/><br/>$\therefore \quad l+m+n=63$
|
integer
|
jee-main-2023-online-1st-february-morning-shift
|
1ldoo6cqc
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a differentiable function such that $$f^{\prime}(x)+f(x)=\int_\limits{0}^{2} f(t) d t$$. If $$f(0)=e^{-2}$$, then $$2 f(0)-f(2)$$ is equal to ____________.</p>
|
[]
| null |
1
|
$f^{\prime}(x)+f(x)=k$
<br/><br/>$$
\begin{aligned}
& \Rightarrow e^{x} f(x)=k e^{x}+c \\\\
& f(x)=k+c e^{-x} \\\\
& k=\int_{0}^{2}\left(k+c e^{-t}\right) d t \\\\
& k=2 k+\left.c \cdot \frac{e^{-t}}{-1}\right|_{0} ^{2} \\\\
& k=2 k+c\left(\frac{e^{-2}}{-1}+1\right) \\\\
& -k=c\left(1-\frac{1}{e^{2}}\right) \\\\
& f(x)=c e^{-x}-c\left(1-\frac{1}{e^{2}}\right) \\\\
& f(0)=c-c+\frac{c}{e^{2}}=\frac{1}{e^{2}} \Rightarrow c=1 \\\\
& f(2)=e^{-2}-r\left(1-e^{-2}\right) \\\\
& =2 e^{-2}-1 \\\\
& 2f(0)-f(2)=1
\end{aligned}
$$
|
integer
|
jee-main-2023-online-1st-february-morning-shift
|
1ldpsjlsl
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$\alpha \in (0,1)$$ and $$\beta = {\log _e}(1 - \alpha )$$. Let $${P_n}(x) = x + {{{x^2}} \over 2} + {{{x^3}} \over 3}\, + \,...\, + \,{{{x^n}} \over n},x \in (0,1)$$. Then the integral $$\int\limits_0^\alpha {{{{t^{50}}} \over {1 - t}}dt} $$ is equal to</p>
|
[{"identifier": "A", "content": "$$ - \\left( {\\beta + {P_{50}}\\left( \\alpha \\right)} \\right)$$"}, {"identifier": "B", "content": "$$\\beta - {P_{50}}(\\alpha )$$"}, {"identifier": "C", "content": "$${P_{50}}(\\alpha ) - \\beta $$"}, {"identifier": "D", "content": "$$\\beta + {P_{50}} - (\\alpha )$$"}]
|
["A"]
| null |
$\int_{0}^{\alpha} \frac{t^{50}}{1-t} d t$
<br/><br/>$$
\begin{aligned}
& = \int_{0}^{\alpha} \frac{t^{50}-1+1}{1-t}\\\\
& =-\int_{0}^{\alpha}\left(\frac{1-t^{50}}{1-t}-\frac{1}{1-t}\right) d t\\\\
& =-\int_{0}^{\alpha}\left(1+t+\ldots . .+t^{49}\right)+\int_{0}^{\alpha} \frac{1}{1-t} d t
\end{aligned}
$$
<br/><br/>$=-\left(\frac{\alpha^{50}}{50}+\frac{\alpha^{49}}{49}+\ldots . .+\frac{\alpha^{1}}{1}\right)+\left(\frac{\ln (1-\mathrm{f})}{-1}\right)_{0}^{\alpha}$
<br/><br/>$=-\mathrm{P}_{50}(\alpha)-\ln (1-\alpha)$
<br/><br/>$=-\mathrm{P}_{50}(\alpha)-\beta$
|
mcq
|
jee-main-2023-online-31st-january-morning-shift
|
1ldpt0own
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The value of $$\int_\limits{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$\\frac{10}{3}-\\sqrt{3}+\\log _{e} \\sqrt{3}$$"}, {"identifier": "B", "content": "$$\\frac{7}{2}-\\sqrt{3}-\\log _{e} \\sqrt{3}$$"}, {"identifier": "C", "content": "$$\\frac{10}{3}-\\sqrt{3}-\\log _{e} \\sqrt{3}$$"}, {"identifier": "D", "content": "$$-2+3\\sqrt{3}+\\log _{e} \\sqrt{3}$$"}]
|
["A"]
| null |
Let I = $$\int_\limits{\frac{\pi}{3}}^{\frac{\pi}{2}} \frac{(2+3 \sin x)}{\sin x(1+\cos x)} d x$$
<br/><br/>$$
=\int\limits_{\pi / 3}^{\pi / 2} \frac{2}{\sin x(1+\cos x)} d x+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{1+\cos x} d x
$$
<br/><br/>$$
\begin{aligned}
=\int\limits_{\pi / 3}^{\pi / 2} \frac{2(1-\cos x)}{\sin x\left(1-\cos ^2 x\right)} & d x +\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{1+\cos x} d x
\end{aligned}
$$
<br/><br/>$$
=\int\limits_{\pi / 3}^{\pi / 2} \frac{2(1-\cos x)}{\sin ^3 x} d x+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{2 \cos ^2 \frac{x}{2}} d x
$$
<br/><br/>$$
\begin{array}{r}
=\int\limits_{\pi / 3}^{\pi / 2} \frac{2}{\sin ^3 x} d x-\int\limits_{\pi / 3}^{\pi / 2} 2 \cot x \operatorname{cosec}^2 x d x
+\int\limits_{\pi / 3}^{\pi / 2} \frac{3}{2} \sec ^2 \frac{x}{2} d x
\end{array}
$$
<br/><br/>$$
=2 I_1-2 I_2+\frac{3}{2} I_3
$$
<br/><br/>Now,
<br/><br/>$$
\begin{aligned}
I_1 & =\int\limits_{\pi / 3}^{\pi / 2} \operatorname{cosec}^3 x d x \\\\
& =\int\limits_{\pi / 3}^{\pi / 2} \sqrt{1+\cot ^2 x} \operatorname{cosec}^2 x d x
\end{aligned}
$$
<br/><br/>Put $\cot x=t \Rightarrow-\operatorname{cosec}^2 x d x=d t$
<br/><br/>When, $x=\frac{\pi}{3} \Rightarrow t=\frac{1}{\sqrt{3}} \text { and } x=\frac{\pi}{2}
\Rightarrow t=0$
<br/><br/>$$
\begin{aligned}
\therefore & I_1=-\int\limits_{\frac{1}{\sqrt{3}}}^0 \sqrt{1+t^2} d t \\\\
= & \int\limits_0^{\frac{1}{\sqrt{3}}} \sqrt{1+t^2} d t \\\\
= & {\left[\frac{t}{2} \sqrt{1+t^2}+\frac{1}{2} \log \left[t+\sqrt{1+t^2}\right]\right]_0^{\frac{1}{\sqrt{3}}} } \\\\
= & \frac{1}{3}+\frac{1}{2} \log \sqrt{3}
\end{aligned}
$$
<br/><br/>$$
I_2=\int\limits_{\pi / 3}^{\pi / 2} \cot x \operatorname{cosec}^2 x d x
$$
<br/><br/>Put $\cot x=t \Rightarrow \operatorname{cosec}^2 x d x=-d t$
<br/><br/>When, $x=\frac{\pi}{3}$
$\Rightarrow t=\frac{1}{\sqrt{3}}$ and $x=\frac{\pi}{2} \Rightarrow t=0$
<br/><br/>$$
\therefore \begin{aligned}
I_2 & =-\int\limits_{\frac{1}{\sqrt{3}}}^0 t d t \\\\
& =\int\limits_0^{\frac{1}{\sqrt{3}} t} t d t=\left[\frac{t^2}{2}\right]_0^{\frac{1}{\sqrt{3}}}=\frac{1}{6}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& I_3=\int\limits_{\frac{\pi}{3}}^{\frac{\pi}{2}} \sec ^2 \frac{x}{2} d x \\\\
= & 2\left(\tan \frac{\pi}{4}-\tan \frac{\pi}{6}\right)=2\left(1-\frac{1}{\sqrt{3}}\right)
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& \therefore \quad I=2 I_1-2 I_2+\frac{3}{2} I_3 \\\\
& =2\left(\frac{1}{3}+\frac{1}{2} \log \sqrt{3}\right)-2\left(\frac{1}{6}\right) +\frac{3}{2}\left(2\left(1-\frac{1}{\sqrt{3}}\right)\right)
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& =\frac{2}{3}+\log \sqrt{3}-\frac{1}{3}+3-\sqrt{3} \\\\
& =\frac{10}{3}-\sqrt{3}+\log _e \sqrt{3}
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-31st-january-morning-shift
|
1ldr6vx1p
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If [t] denotes the greatest integer $$\le \mathrm{t}$$, then the value of $${{3(e - 1)} \over e}\int\limits_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx} $$ is :</p>
|
[{"identifier": "A", "content": "$$\\mathrm{e^8-e}$$"}, {"identifier": "B", "content": "$$\\mathrm{e^7-1}$$"}, {"identifier": "C", "content": "$$\\mathrm{e^9-e}$$"}, {"identifier": "D", "content": "$$\\mathrm{e^8-1}$$"}]
|
["A"]
| null |
<p>$$I = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{[x] + [{x^3}]}}dx} $$</p>
<p>$$ = {{3(e - 1)} \over e}\int_1^2 {{x^2}{e^{1 + [{x^3}]}}dx} $$ ($$\because$$ $$[x] = 1$$ when $$x \in (12)$$)</p>
<p>$$ = 3(e - 1)\int_1^2 {{x^2}{e^{[{x^3}]}}dx} $$</p>
<p>Let $${x^3} = t$$</p>
<p>$$I = (e - 1)\int_1^8 {{e^{[t]}}dt} $$</p>
<p>$$ = ({e^{ - 1}})({e^1} + {e^2} + {e^3}\, + \,...\, + \,{e^7})$$</p>
<p>$$ = (e - 1)e{{({e^7} - 1)} \over {e - 1}}$$</p>
<p>$$ = {e^8} - e$$</p>
|
mcq
|
jee-main-2023-online-30th-january-morning-shift
|
1ldsfe8t4
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The value of the integral $$\int_1^2 {\left( {{{{t^4} + 1} \over {{t^6} + 1}}} \right)dt} $$ is</p>
|
[{"identifier": "A", "content": "$${\\tan ^{ - 1}}{1 \\over 2} - {1 \\over 3}{\\tan ^{ - 1}}8 + {\\pi \\over 3}$$"}, {"identifier": "B", "content": "$${\\tan ^{ - 1}}2 - {1 \\over 3}{\\tan ^{ - 1}}8 + {\\pi \\over 3}$$"}, {"identifier": "C", "content": "$${\\tan ^{ - 1}}2 + {1 \\over 3}{\\tan ^{ - 1}}8 - {\\pi \\over 3}$$"}, {"identifier": "D", "content": "$${\\tan ^{ - 1}}{1 \\over 2} + {1 \\over 3}{\\tan ^{ - 1}}8 - {\\pi \\over 3}$$"}]
|
["C"]
| null |
<p>$$\int_1^2 {{{{t^4} + 1} \over {{t^6} + 1}}dt} $$</p>
<p>$$ = \int_1^2 {{{{{({t^2} + 1)}^2}} \over {{t^6} + 1}}dt - 2\int_1^2 {{{{t^2}} \over {{t^6} + 1}}dt} } $$</p>
<p>$$ = \int_1^2 {{{{t^2} + 1} \over {{t^4} - {t^2} + 1}}dt - 2\int_1^2 {{{{t^2}} \over {{{({t^3})}^2} + 1}}dt} } $$</p>
<p>$$ = \left. {{{\tan }^{ - 1}}(2t + \sqrt 3 ) + {{\tan }^{ - 1}}(2t - \sqrt 3 )} \right|_1^2 - \left. {{2 \over 3}{{\tan }^{ - 1}}({t^3})} \right|_1^2$$</p>
<p>$$ = {\tan ^{ - 1}}(4 + \sqrt 3 ) + {\tan ^{ - 1}}(4 - \sqrt 3 ) - {\tan ^{ - 1}}(2 + \sqrt 3 ) - {\tan ^{ - 1}}(2 + \sqrt 3 ) - {\tan ^{ - 1}}(2\sqrt 3 ) - {2 \over 3}({\tan ^{ - 1}}8 - {\tan ^{ - 1}}1)$$</p>
<p>$$ = {\tan ^{ - 1}}2 + {1 \over 3}{\tan ^{ - 1}}8 - {\pi \over 3}$$</p>
|
mcq
|
jee-main-2023-online-29th-january-evening-shift
|
1ldsfpdr8
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The value of the integral $$\int\limits_{1/2}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} $$ is equal to :</p>
|
[{"identifier": "A", "content": "$${\\pi \\over 2}{\\log _e}2$$"}, {"identifier": "B", "content": "$${\\pi \\over 4}{\\log _e}2$$"}, {"identifier": "C", "content": "$${1 \\over 2}{\\log _e}2$$"}, {"identifier": "D", "content": "$$\\pi {\\log _e}2$$"}]
|
["A"]
| null |
<p>$$I = \int\limits_{{1 \over 2}}^2 {{{{{\tan }^{ - 1}}x} \over x}dx} $$ ..... (i)</p>
<p>$$x \to {1 \over x}$$</p>
<p>$$I = \int\limits_{{1 \over 2}}^2 {{1 \over x}{{\tan }^{ - 1}}{1 \over x}dx} $$ ..... (ii)</p>
<p>$$2I = \int\limits_{{1 \over 2}}^2 {{1 \over x}\,.\,{\pi \over 2}dx} $$</p>
<p>$$ = \left. {{\pi \over 2}\ln x} \right|_{{1 \over 2}}^2 = \pi \ln 2$$</p>
<p>$$ \Rightarrow I = {\pi \over 2}\ln 2$$</p>
|
mcq
|
jee-main-2023-online-29th-january-evening-shift
|
1ldsuma7b
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$f(x) = x + {a \over {{\pi ^2} - 4}}\sin x + {b \over {{\pi ^2} - 4}}\cos x,x \in R$$ be a function which<br/><br/> satisfies $$f(x) = x + \int\limits_0^{\pi /2} {\sin (x + y)f(y)dy} $$. then $$(a+b)$$ is equal to</p>
|
[{"identifier": "A", "content": "$$ - 2\\pi (\\pi + 2)$$"}, {"identifier": "B", "content": "$$ - \\pi (\\pi - 2)$$"}, {"identifier": "C", "content": "$$ - \\pi (\\pi + 2)$$"}, {"identifier": "D", "content": "$$ - 2\\pi (\\pi - 2)$$"}]
|
["A"]
| null |
$f(x)=x+\int\limits_{0}^{\pi / 2}(\sin x \cos y+\cos x \sin y) f(y) d y$
<br/><br/>
$f(x)=x+\int\limits_{0}^{\pi / 2}((\cos y f(y) d y) \sin x+(\sin y f(y) d y) \cos x)\quad....(1)$
<br/><br/>
On comparing with
<br/><br/>
$f(x)=x+\frac{a}{\pi^{2}-4} \sin x+\frac{b}{\pi^{2}-4} \cos x, x \in \mathbb{R}$ then
<br/><br/>
$\Rightarrow \frac{a}{\pi^{2}-4}=\int_{0}^{\pi / 2} \cos y f(y) d y \quad....(2)$
<br/><br/>
$\Rightarrow \frac{b}{\pi^{2}-4}=\int_{0}^{\pi / 2} \sin y f(y) d y \quad....(3)$
<br/><br/>
<b>Add (2) and (3)</b>
<br/><br/>
$\frac{a+b}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y) f(y) d y \quad....(4)$
<br/><br/>
$\frac{a+b}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y) f\left(\frac{\pi}{2}-y\right) d y\quad....(5)$
<br/><br/>
<b>Add (4) and (5)</b>
<br/><br/>
$\frac{2(a+b)}{\pi^{2}-4}=\int_{0}^{\pi / 2}(\sin y+\cos y)\left(\frac{\pi}{2}+\frac{(a+b)}{\pi^{2}-4}(\sin y+\cos y)\right) d y$
<br/><br/>
$=\pi+\frac{a+b}{\pi^{2}-4}\left(\frac{\pi}{2}+1\right)$
<br/><br/>
$(a+b)=-2 \pi(\pi+2)$
|
mcq
|
jee-main-2023-online-29th-january-morning-shift
|
1ldu5exe2
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The integral $$16\int\limits_1^2 {{{dx} \over {{x^3}{{\left( {{x^2} + 2} \right)}^2}}}} $$ is equal to</p>
|
[{"identifier": "A", "content": "$${{11} \\over {12}} + {\\log _e}4$$"}, {"identifier": "B", "content": "$${{11} \\over 6} + {\\log _e}4$$"}, {"identifier": "C", "content": "$${{11} \\over {12}} - {\\log _e}4$$"}, {"identifier": "D", "content": "$${{11} \\over 6} - {\\log _e}4$$"}]
|
["D"]
| null |
$I=\int \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}$ <br/><br/>$=\frac{1}{4} \int \frac{x}{x^{2}+2} d x+\frac{1}{4} \int \frac{x}{\left(x^{2}+2\right)^{2}}-\frac{1}{4} \int \frac{d x}{x}+\frac{1}{4} \int \frac{d x}{x^{3}}$<br/><br/> $=\frac{1}{8} \ln \left(x^{2}+2\right)-\frac{\ln x}{4}-\frac{1}{8\left(x^{2}+2\right)}-\frac{1}{8 x^{3}}$
<br/><br/>
Now, $16 \int_{1}^{2} \frac{d x}{x^{3}\left(x^{2}+2\right)^{2}}=2 \ln 6-2 \ln 3-4 \ln 2+\frac{11}{6}$ <br/><br/>$=\frac{11}{6}-\ln 4$
|
mcq
|
jee-main-2023-online-25th-january-evening-shift
|
1ldu6241m
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If $$\int\limits_{{1 \over 3}}^3 {|{{\log }_e}x|dx = {m \over n}{{\log }_e}\left( {{{{n^2}} \over e}} \right)} $$, where m and n are coprime natural numbers, then $${m^2} + {n^2} - 5$$ is equal to _____________.</p>
|
[]
| null |
20
|
$I=\int\limits_{\frac{1}{3}}^{3}|\ln x| d x=-\int\limits_{\frac{1}{3}}^{1} \ln x d x+\int\limits_{1}^{3} \ln x d x$
<br/><br/>
$$
\begin{aligned}
& \left.=-[x \ln x-x]_{\frac{1}{3}}^{1}+x \ln x-x\right]_{1}^{3} \\\\
& =-\left[(0-1)-\left(\frac{1}{3} \ln 3-\frac{1}{3}\right)\right]+[(3 \ln 3-3)-(0-1)] \\\\
& =\frac{2}{3}-\frac{1}{3} \ln 3+3 \ln 3-2
\end{aligned}
$$<br/><br/>
$$
\begin{aligned}
& =\frac{8}{3} \ln 3-\frac{4}{3} \\\\
& =\frac{4}{3}(2 \ln 3-\ln e) \\\\
& =\frac{4}{3} \ln \left(\frac{3^{2}}{e}\right) \\\\
& m=4, m^{m}=3 \\\\
& m^{2}+n^{2}-5=20
\end{aligned}
$$
|
integer
|
jee-main-2023-online-25th-january-evening-shift
|
1ldv16kbb
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The minimum value of the function $$f(x) = \int\limits_0^2 {{e^{|x - t|}}dt} $$ is :</p>
|
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "$$2(e-1)$$"}, {"identifier": "C", "content": "$$e(e-1)$$"}, {"identifier": "D", "content": "$$2e-1$$"}]
|
["B"]
| null |
$$
f(x)=\int_0^2 e^{|x-t|} d t
$$<br/><br/>
For $x>2$<br/><br/>
$$
f(x)=\int_0^2 e^{x-t} d t=e^x\left(1-e^{-2}\right)
$$<br/><br/>
For $x<0$<br/><br/>
$$
f(x)=\int_0^2 e^{t-x} d t=e^{-x}\left(e^2-1\right)
$$<br/><br/>
For $x \in[0,2]$<br/><br/>
$$
\begin{aligned}
& f(x)=\int_0^x e^{x-t} d t + \int_x^2 e^{t-x} d t \\\\
& =e^{2-x}+e^x-2
\end{aligned}
$$<br/><br/>
For $x>2$<br/><br/>
$$
\left.f(x)\right|_{\min =e^2-1}
$$<br/><br/>
For $\mathrm{x}<0$<br/><br/>
$$
\left.f(x)\right|_{\min =e^2-1}
$$<br/><br/>
For $x \in[0,2]$<br/><br/>
$$
\left.f(x)\right|_{\min }=2(e-1)
$$
|
mcq
|
jee-main-2023-online-25th-january-morning-shift
|
1ldwx9a09
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>$$\int\limits_{{{3\sqrt 2 } \over 4}}^{{{3\sqrt 3 } \over 4}} {{{48} \over {\sqrt {9 - 4{x^2}} }}dx} $$ is equal to :</p>
|
[{"identifier": "A", "content": "$${\\pi \\over 2}$$"}, {"identifier": "B", "content": "$${\\pi \\over 3}$$"}, {"identifier": "C", "content": "$${\\pi \\over 6}$$"}, {"identifier": "D", "content": "$$2\\pi $$"}]
|
["D"]
| null |
$$
\int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x
$$<br/><br/>
We have $\int \frac{d x}{\sqrt{a^2-x^2}}=\sin ^{-1} \frac{x}{a}+C$<br/><br/>
Hence $\int_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}} \frac{48}{\sqrt{9-4 x^2}} d x=\frac{48}{2} \times\left[\sin ^{-1} \frac{2 x}{3}\right]_{\frac{3 \sqrt{2}}{4}}^{\frac{3 \sqrt{3}}{4}}$<br/><br/>
$=24 \times\left[\sin ^{-1}\left(\frac{2}{3} \times \frac{3 \sqrt{3}}{4}\right)-\sin ^{-1}\left(\frac{2}{3} \times \frac{3 \sqrt{2}}{4}\right)\right]$
$=24 \times\left[\sin ^{-1} \frac{\sqrt{3}}{2}-\sin ^{-1} \frac{1}{\sqrt{2}}\right]$<br/><br/>
$=24 \times\left(\frac{\pi}{3}-\frac{\pi}{4}\right)$<br/><br/>
$=24 \times \frac{\pi}{12}=2 \pi$
|
mcq
|
jee-main-2023-online-24th-january-evening-shift
|
1ldyc3v3b
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The value of $$12\int\limits_0^3 {\left| {{x^2} - 3x + 2} \right|dx} $$ is ____________</p>
|
[]
| null |
22
|
<p>Let I = $$\int\limits_0^3 {|{x^2} - 3x + 2|dx} $$</p>
<p>$$ = \int\limits_0^3 {|(x - 1)(x - 2)|dx} $$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1le2gufhw/2197edc8-523a-4ea7-bfd4-68c608ff54fb/16df8b40-ab6c-11ed-b566-111c81fc645a/file-1le2gufhx.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1le2gufhw/2197edc8-523a-4ea7-bfd4-68c608ff54fb/16df8b40-ab6c-11ed-b566-111c81fc645a/file-1le2gufhx.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 24th January Morning Shift Mathematics - Definite Integration Question 63 English Explanation"></p>
<p>$$ = \int\limits_0^1 { + ({x^2} - 3x + 2)dx + \int\limits_1^2 { - ({x^2} - 3x + 2)dx + \int\limits_2^3 {({x^2} - 3x + 2)dx} } } $$</p>
<p>$$ = \left[ {{{{x^3}} \over 3} - {{3{x^2}} \over 2} + 2x} \right]_0^1 - \left[ {{{{x^3}} \over 3} - {{3{x^2}} \over 2} + 2x} \right]_1^2 + \left[ {{{{x^3}} \over 3} - {{3{x^2}} \over 2} + 2x} \right]_2^3$$</p>
<p>$$ = \left( {{1 \over 3} - {3 \over 2} + 2} \right) - \left[ {\left( {{8 \over 3} - 6 + 4} \right) - \left( {{1 \over 3} - {3 \over 2} + 2} \right)} \right] + \left[ {\left( {{{27} \over 3} - {{27} \over 2} + 6} \right) - \left( {{8 \over 3} - 6 + 4} \right)} \right]$$</p>
<p>$$ = {5 \over 6} - \left( {{2 \over 3} - {5 \over 6}} \right) + \left( {{3 \over 2} - {2 \over 3}} \right)$$</p>
<p>$$ = {5 \over 6} + {5 \over 6} - {2 \over 3} - {2 \over 3} + {3 \over 2}$$</p>
<p>$$ = {{10} \over 6} - {4 \over 3} + {3 \over 2}$$</p>
<p>$$ = {{10 - 8 + 9} \over 6} = {{11} \over 6}$$</p>
<p>$$ \therefore $$ 12I = $$12 \times {{11} \over 6}$$ = 22</p>
|
integer
|
jee-main-2023-online-24th-january-morning-shift
|
1ldyc80xq
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The value of $${8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{(\cos x)}^{2023}}} \over {{{(\sin x)}^{2023}} + {{(\cos x)}^{2023}}}}dx} $$ is ___________</p>
|
[]
| null |
2
|
<p>Let, $$I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{(\cos x)}^{2023}}} \over {{{(\sin x)}^{2023}} + {{(\cos x)}^{2023}}}}dx} $$ ..... (1)</p>
<p>Using formula,</p>
<p>$$\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} } $$</p>
<p>$$I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{\left[ {\cos \left( {{\pi \over 2} - x} \right)} \right]}^{2023}}} \over {{{\left[ {\sin \left( {{\pi \over 2} - x} \right)} \right]}^{2023}} + {{\left[ {\cos \left( {{\pi \over 2} - x} \right)} \right]}^{2023}}}}dx} $$</p>
<p>$$ = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{(\sin x)}^{2023}}} \over {{{(\cos x)}^{2023}} + {{(\sin x)}^{2023}}}}dx} $$ ..... (2)</p>
<p>Adding equation (1) and (2), we get</p>
<p>$$2I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {{{{{(\sin x)}^{2023}} + {{(\cos x)}^{2023}}} \over {{{(\sin x)}^{2023}} + {{(\cos x)}^{203}}}}dx} $$</p>
<p>$$ \Rightarrow 2I = {8 \over \pi }\int\limits_0^{{\pi \over 2}} {dx} $$</p>
<p>$$ \Rightarrow 2I = {8 \over \pi }\left[ x \right]_0^{{\pi \over 2}}$$</p>
<p>$$ \Rightarrow 2I = {8 \over \pi } \times {\pi \over 2}$$</p>
<p>$$ \Rightarrow 2I = 4$$</p>
<p>$$ \Rightarrow I = 2$$</p>
|
integer
|
jee-main-2023-online-24th-january-morning-shift
|
lgnwfxh4
|
maths
|
definite-integration
|
properties-of-definite-integration
|
If $\int\limits_{0}^{1} \frac{1}{\left(5+2 x-2 x^{2}\right)\left(1+e^{(2-4 x)}\right)} d x=\frac{1}{\alpha} \log _{e}\left(\frac{\alpha+1}{\beta}\right), \alpha, \beta>0$, then $\alpha^{4}-\beta^{4}$
is equal to :
|
[{"identifier": "A", "content": "-21"}, {"identifier": "B", "content": "21"}, {"identifier": "C", "content": "19"}, {"identifier": "D", "content": "0"}]
|
["B"]
| null |
<ol>
<li>The given integral is:</li>
</ol>
<p>$$
I=\int_0^1 \frac{dx}{\left(5+2x-2x^2\right)\left(1+e^{2-4x}\right)}
$$ .............(i)</p>
<ol>
<li>Perform a substitution, $x \rightarrow 1-x$. This gives:</li>
</ol>
<p>$$
I=\int_0^1 \frac{dx}{\left(5+2(1-x)-2(1-x)^2\right)\left(1+e^{2-4(1-x)}\right)}
$$</p>
<ol>
<li>Simplify the expression inside the integral:</li>
</ol>
<p>$$
I=\int_0^1 \frac{dx}{\left(5+2-2x-2(1-2x+x^2)\right)\left(1+e^{4x -2}\right)}
$$ ............(ii)</p>
<ol>
<li>Add the original integral (i) and the integral after substitution (ii):</li>
</ol>
<p>$$
2I=\int_0^1 \frac{dx}{5+2x-2x^2}
$$</p>
<ol>
<li>Factor the quadratic expression in the denominator:</li>
</ol>
<p>$$
2I=\int_0^1 \frac{dx}{2\left(\frac{11}{4}-\left(x-\frac{1}{2}\right)^2\right)}
$$</p>
<ol>
<li>To solve this integral, we can perform a change of variables using the substitution <br/><br/>$x-\frac{1}{2}= \frac{\sqrt{11}}{2}\tan{u}$, then $dx=\frac{\sqrt{11}}{2}\sec^2{u} du$:</li>
</ol>
<p>$$
I=\frac{1}{\sqrt{11}}\int \frac{\sec^2{u}}{1+\tan^2{u}}du
$$</p>
<ol>
<li>Using the identity $\sec^2{u}=1+\tan^2{u}$:</li>
</ol>
<p>$$
I=\frac{1}{\sqrt{11}}\int du = \frac{1}{\sqrt{11}}(u+C)
$$</p>
<ol>
<li>Now we need to find the limits of the integral after the substitution. If $x=0$, then $u=\tan^{-1}\left(-\frac{1}{\sqrt{11}}\right)$. If $x=1$, then $u=\tan^{-1}\left(\frac{1}{\sqrt{11}}\right)$. So, the integral becomes:</li>
</ol>
<p>$$
I=\frac{1}{\sqrt{11}}\left[\tan^{-1}\left(\frac{1}{\sqrt{11}}\right)-\tan^{-1}\left(-\frac{1}{\sqrt{11}}\right)\right]
$$</p>
<ol>
<li>Using the properties of the arctangent function, we can rewrite the integral as:</li>
</ol>
<p>$$
I=\frac{1}{\sqrt{11}} \ln\left(\frac{\sqrt{11}+1}{\sqrt{10}}\right)
$$</p>
<ol>
<li>From this result, we have $\alpha=\sqrt{11}$ and $\beta=\sqrt{10}$. Now, we can find $\alpha^4 - \beta^4$:</li>
</ol>
<p>$$
\alpha^4 - \beta^4 = (11)^2 - (10)^2 = 121 - 100 = 21
$$</p>
<p>Thus, $\alpha^4 - \beta^4 = 21$.</p>
|
mcq
|
jee-main-2023-online-15th-april-morning-shift
|
1lgoxpnhf
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The value of $${{{e^{ - {\pi \over 4}}} + \int\limits_0^{{\pi \over 4}} {{e^{ - x}}{{\tan }^{50}}xdx} } \over {\int\limits_0^{{\pi \over 4}} {{e^{ - x}}({{\tan }^{49}}x + {{\tan }^{51}}x)dx} }}$$ is</p>
|
[{"identifier": "A", "content": "51"}, {"identifier": "B", "content": "50"}, {"identifier": "C", "content": "25"}, {"identifier": "D", "content": "49"}]
|
["B"]
| null |
<p>We're given the expression:</p>
<p>$$\frac{e^{-\pi/4} + \int_0^{\pi / 4} e^{-x} \tan ^{50} x dx}{\int_0^{\pi / 4} e^{-x}(\tan x)^{49} dx + \int_0^{\pi / 4} e^{-x}(\tan x)^{51} dx}$$</p>
<p>Notice that the integrals in the numerator and denominator have the same form. They both involve an integral of $e^{-x} \tan^n x$ from 0 to $\pi / 4$, where $n$ is an integer. Let's denote this integral as $I(n)$:</p>
<p>$$I(n) = \int_0^{\pi / 4} e^{-x} \tan^n x dx$$</p>
<p>We can then rewrite the original expression in terms of $I(n)$:</p>
<p>$$\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)}$$</p>
<p>Now, we'll apply the method of integration by parts, which states that for two functions $u(x)$ and $v(x)$:</p>
<p>$$\int u dv = uv - \int v du$$</p>
<p>We'll choose:</p>
<p>$$u = \tan^n x, \quad dv = e^{-x} dx$$</p>
<p>Then we get:</p>
<p>$$du = n \tan^{n-1} x \sec^2 x dx, \quad v = -e^{-x}$$</p>
<p>Applying integration by parts, we have:</p>
<p>$$I(n) = -e^{-x} \tan^n x \Bigg|_0^{\pi / 4} + n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x \sec^2 x dx$$</p>
<p>Since $\tan(\pi / 4) = 1$, the first term evaluates to:</p>
<p>$$-e^{-\pi / 4}$$</p>
<p>The second term becomes:</p>
<p>$$n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x (1 + \tan^2 x) dx = n \int_0^{\pi / 4} e^{-x} \tan^{n-1} x dx + n \int_0^{\pi / 4} e^{-x} \tan^{n+1} x dx$$</p>
<p>This is equal to:</p>
<p>$$n(I(n-1) + I(n+1))$$</p>
<p>So we have:</p>
<p>$$I(n) = -e^{-\pi / 4} + n(I(n-1) + I(n+1))$$</p>
<p>Now we can substitute $n = 50$ into this equation:</p>
<p>$$I(50) = -e^{-\pi / 4} + 50(I(49) + I(51))$$</p>
<p>So the original expression becomes:</p>
<p>$$\frac{e^{-\pi/4} + I(50)}{I(49) + I(51)} = \frac{e^{-\pi/4} - e^{-\pi / 4} + 50(I(49) + I(51))}{I(49) + I(51)} = 50$$</p>
|
mcq
|
jee-main-2023-online-13th-april-evening-shift
|
1lgoy1rhw
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$f_{n}=\int_\limits{0}^{\frac{\pi}{2}}\left(\sum_\limits{k=1}^{n} \sin ^{k-1} x\right)\left(\sum_\limits{k=1}^{n}(2 k-1) \sin ^{k-1} x\right) \cos x d x, n \in \mathbb{N}$$. Then $$f_{21}-f_{20}$$ is equal to _________</p>
|
[]
| null |
41
|
Given, $$f_{n}=\int_\limits{0}^{\frac{\pi}{2}}\left(\sum_\limits{k=1}^{n} \sin ^{k-1} x\right)\left(\sum_\limits{k=1}^{n}(2 k-1) \sin ^{k-1} x\right) \cos x d x$$
<br/><br/>$$
\begin{aligned}
& \text { Put } \sin x=t \\\\
& \cos x d x=d t \\\\
& f_n=\int_0^1\left(\sum_{k=1}^n(t)^{k-1}\right)\left(\sum_{k=1}^n(2 k-1)(t)^{k-1}\right) d t
\end{aligned}
$$
<br/><br/>$$ \therefore $$ $$
\begin{aligned}
f_{21}=\int_0^1\left(\sum_{k=1}^{21}(t)^{20}\right)\left(\sum_{k=1}^{21}(2 k-1)(t)^{20}\right) d t
\end{aligned}
$$
<br/><br/>= $$\int\limits_0^1 {\left( {{t^0} + {t^1} + {t^2} + ...... + {t^{20}}} \right)\left( {1{t^0} + 3{t^1} + 5{t^2} + ...... + 41{t^{20}}} \right)} dt$$
<br/><br/>$$ \therefore $$ $$
\begin{aligned}
f_{20}=\int_0^1\left(\sum_{k=1}^{20}(t)^{19}\right)\left(\sum_{k=1}^{20}(2 k-1)(t)^{19}\right) d t
\end{aligned}
$$
<br/><br/>= $$\int\limits_0^1 {\left( {{t^0} + {t^1} + {t^2} + ...... + {t^{19}}} \right)\left( {1{t^0} + 3{t^1} + 5{t^2} + ...... + 39{t^{19}}} \right)} dt$$
<br/><br/>Now, $$
f_{21}-f_{20}
$$
<br/><br/>$$
\begin{aligned}
=\int_0^1\left(1+t+t^2+\ldots .\right. & \left.+t^{19}\right)(41) t^{20} \\
& +\left(1+3 t+5 t^2+\ldots . .+41 t^{20}\right) t^{20} d t
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& =\left(\frac{1}{21}+\frac{1}{22}+\ldots+\frac{1}{40}\right) 41+\left(\frac{1}{21}+\frac{3}{22}+\ldots . .+\frac{39}{40}+\frac{41}{41}\right) \\\\
& =\left[\frac{42}{21}+\frac{44}{22}+\frac{46}{23}+\ldots .+\frac{80}{40}+\frac{41}{41}\right] \\\\
& =40+1=41
\end{aligned}
$$
|
integer
|
jee-main-2023-online-13th-april-evening-shift
|
1lgpy0x97
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>$$\int_\limits{0}^{\infty} \frac{6}{e^{3 x}+6 e^{2 x}+11 e^{x}+6} d x=$$</p>
|
[{"identifier": "A", "content": "$$\\log _{e}\\left(\\frac{256}{81}\\right)$$"}, {"identifier": "B", "content": "$$\\log _{e}\\left(\\frac{64}{27}\\right)$$"}, {"identifier": "C", "content": "$$\\log _{e}\\left(\\frac{32}{27}\\right)$$"}, {"identifier": "D", "content": "$$\\log _{e}\\left(\\frac{512}{81}\\right)$$"}]
|
["C"]
| null |
$$
\begin{aligned}
& \mathrm{l}=\int_0^{\infty} \frac{6}{\left(\mathrm{e}^{\mathrm{x}}+1\right)\left(\mathrm{e}^{\mathrm{x}}+2\right)\left(\mathrm{e}^{\mathrm{x}}+3\right)} \mathrm{dx} \\\\
& =6 \int_0^{\infty}\left(\frac{\frac{1}{2}}{\mathrm{e}^{\mathrm{x}}+1}+\frac{-1}{\mathrm{e}^{\mathrm{x}}+2}+\frac{\frac{1}{2}}{\mathrm{e}^{\mathrm{x}}+3}\right) \mathrm{dx} \\\\
& =3 \int_0^{\infty} \frac{\mathrm{e}^{-\mathrm{x}}}{1+\mathrm{e}^{-\mathrm{x}}} \mathrm{dx}-6 \int_0^{\infty} \frac{\mathrm{e}^{-\mathrm{x}} \mathrm{dx}}{1+2 \mathrm{e}^{-\mathrm{x}}}+3 \int_0^{\infty} \frac{\mathrm{e}^{-\mathrm{x}}}{1+3 \mathrm{e}^{-\mathrm{x}}} \mathrm{dx} \\\\
& =3\left[-\ln \left(1+\mathrm{e}^{-\mathrm{x}}\right)\right]_0^{\infty}+6 \frac{1}{2}\left[\ln \left(1+2 \mathrm{e}^{-\mathrm{x}}\right)\right]_0^{\infty} -\frac{3}{3}\left[\ln \left(1+3 \mathrm{e}^{-\mathrm{x}}\right)\right]_0^{\infty} \\\\
& =3 \ln 2-3 \ln 3+\ln 4 \\\\
& =3 \ln \frac{2}{3}+\ln 4 \\\\
& =\ln \frac{32}{27}
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-13th-april-morning-shift
|
1lgq0ygkp
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let for $$x \in \mathbb{R}, S_{0}(x)=x, S_{k}(x)=C_{k} x+k \int_{0}^{x} S_{k-1}(t) d t$$, where
<br/><br/>$$C_{0}=1, C_{k}=1-\int_{0}^{1} S_{k-1}(x) d x, k=1,2,3, \ldots$$ Then $$S_{2}(3)+6 C_{3}$$ is equal to ____________.</p>
|
[]
| null |
18
|
Given,
<br/><br/>$$
S_k(x)=C_k x+k \int_0^x S_{k-1}(t) d t
$$
<br/><br/>Put $\mathrm{k}=2$ and $\mathrm{x}=3$
<br/><br/>$$
\mathrm{S}_2(3)=\mathrm{C}_2(3)+2 \int_0^3 \mathrm{~S}_1(\mathrm{t}) \mathrm{dt}
$$ .........(i)
<br/><br/>Also,
<br/><br/>$$
\begin{aligned}
& S_1(x)=C_1(x)+\int_0^x S_0(t) d t \\\\
& =C_1 x+\frac{x^2}{2} \\\\
& S_2(3)=3 C_2+2 \int_0^3\left(C_1 t+\frac{t^2}{2}\right) d t \\\\
& =3 C_2+9 C_1+9
\end{aligned}
$$
<br/><br/>Also,
<br/><br/>$$
\begin{aligned}
& \mathrm{C}_1=1-\int_0^1 \mathrm{~S}_0(\mathrm{x}) \mathrm{dx}=\frac{1}{2} \\\\
& \mathrm{C}_2=1-\int_0^1 \mathrm{~S}_1(\mathrm{x}) \mathrm{dx}=0 \\\\
& \mathrm{C}_3=1-\int_0^1 \mathrm{~S}_2(\mathrm{x}) \mathrm{dx} \\\\
& =1-\int_0^1\left(\mathrm{C}_2 \mathrm{x}+\mathrm{C}_1 \mathrm{x}^2+\frac{\mathrm{x}^3}{3}\right) \mathrm{dx} \\\\
& =\frac{3}{4}
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& S_2(x)=C_2 x+2 \int_0^x S_1(t) d t \\\\
&=C_2 x+C_1 x^2+\frac{x^3}{3} \\\\
& = S_2(3)+6 C_3=6 C_3+3 C_2+9 C_1+9 \\\\
&=18
\end{aligned}
$$
|
integer
|
jee-main-2023-online-13th-april-morning-shift
|
1lgrgllfd
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If $$\int_\limits{-0.15}^{0.15}\left|100 x^{2}-1\right| d x=\frac{k}{3000}$$, then $$k$$ is equal to ___________.</p>
|
[]
| null |
575
|
$$
\int\limits_{-0.15}^{0.15}\left|100 x^2-1\right| d x=2 \int\limits_0^{0.15}\left|100 x^2-1\right| \mathrm{dx}
$$
<br/><br/>$$
\text { Now } 100 x^2-1=0 \Rightarrow x^2=\frac{1}{100} \Rightarrow x=0.1
$$
<br/><br/>$$
I=2\left[\int_0^{0.1}\left(1-100 x^2\right) d x+\int_{0.1}^{0.15}\left(100 x^2-1\right) d x\right]
$$
<br/><br/>$$
\begin{aligned}
I & =2\left[x-\frac{100}{3} x^3\right]_0^{0.1}+2\left[\frac{100 x^3}{3}-x\right]_{0.1}^{0.15} \\\\
& =2\left[0.1-\frac{0.1}{3}\right]+2\left[\frac{0.3375}{3}-0.15-\frac{0.1}{3}+0.1\right] \\\\
& =2\left[0.2-\frac{0.2}{3}+0.1125-0.15\right] \\\\
& =2\left[\frac{5}{100}-\frac{2}{30}+\frac{1125}{10000}\right]=2\left(\frac{1500-2000+3375}{30000}\right) \\\\
& =\frac{575}{3000} \Rightarrow \mathrm{k}=575
\end{aligned}
$$
|
integer
|
jee-main-2023-online-12th-april-morning-shift
|
1lgsvib75
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a continuous function satisfying $$\int_\limits{0}^{\frac{\pi}{2}} f(\sin 2 x) \sin x d x+\alpha \int_\limits{0}^{\frac{\pi}{4}} f(\cos 2 x) \cos x d x=0$$, then the value of $$\alpha$$ is :</p>
|
[{"identifier": "A", "content": "$$-\\sqrt{3}$$"}, {"identifier": "B", "content": "$$\\sqrt{2}$$"}, {"identifier": "C", "content": "$$-\\sqrt{2}$$"}, {"identifier": "D", "content": "$$\\sqrt{3}$$"}]
|
["C"]
| null |
The integral equation is given by :
<br/><br/>$$\int\limits_0^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0$$
<br/><br/>Step 1 : Break the first integral into two parts :
<br/><br/>$$I = \int\limits_0^{\frac{\pi}{4}} f(\sin 2x) \sin x \, dx + \int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}} f(\sin 2x) \sin x \, dx + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx$$
<br/><br/>3. Apply the King's property, $\int\limits_a^b f(x) dx = \int\limits_a^b f(a+b-x) dx$, to the first integral and substitute $x - \frac{\pi}{4} = t$ in the second integral.
<br/><br/>This gives :
<br/><br/>$$
\int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \sin(\frac{\pi}{4}-x) \, dx + \int\limits_0^{\frac{\pi}{4}} f(\cos 2t) \sin(\frac{\pi}{4}+t) \, dt + \alpha \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0
$$
<br/><br/>$$
\int\limits_0^{\frac{\pi}{4}} f(\cos 2x) [2 \sin \frac{\pi}{4} \cos x + \alpha \cos x] \, dx = 0
$$
<br/><br/>Then, noticing that $2 \sin \frac{\pi}{4} = \sqrt{2}$, you can factor out the term $\cos x$:
<br/><br/>$$
= (\alpha + \sqrt{2}) \int\limits_0^{\frac{\pi}{4}} f(\cos 2x) \cos x \, dx = 0
$$
<br/><br/>In order for this equation to hold true, either the integral of the function is zero, or the term outside the integral is zero. Since we have no reason to assume that the integral of the function is zero, we set the term outside the integral to zero, yielding the solution:
<br/><br/>$$
\alpha + \sqrt{2} = 0 \Rightarrow \alpha = -\sqrt{2}
$$
<br/><br/>So, the correct answer to the original problem is $\alpha = -\sqrt{2}$, which corresponds to Option C.
|
mcq
|
jee-main-2023-online-11th-april-evening-shift
|
1lgsvsu3q
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let the function $$f:[0,2] \rightarrow \mathbb{R}$$ be defined as</p>
<p>$$f(x)= \begin{cases}e^{\min \left\{x^{2}, x-[x]\right\},} & x \in[0,1) \\ e^{\left[x-\log _{e} x\right]}, & x \in[1,2]\end{cases}$$</p>
<p>where $$[t]$$ denotes the greatest integer less than or equal to $$t$$. Then the value of the integral $$\int_\limits{0}^{2} x f(x) d x$$ is :</p>
|
[{"identifier": "A", "content": "$$2 e-1$$"}, {"identifier": "B", "content": "$$2 e-\\frac{1}{2}$$"}, {"identifier": "C", "content": "$$1+\\frac{3 e}{2}$$"}, {"identifier": "D", "content": "$$(e-1)\\left(e^{2}+\\frac{1}{2}\\right)$$"}]
|
["B"]
| null |
$$
\begin{aligned}
\operatorname{Minimum}\left\{\mathrm{x}^2,\{\mathrm{x}\}\right\} & =\mathrm{x}^2 ; \mathrm{x} \in[0,1) \\\\
{\left[\mathrm{x}-\log _{\mathrm{e}} \mathrm{x}\right] } & =1 ; \mathrm{x} \in[1,2)
\end{aligned}
$$
<br/><br/>$$
\therefore \mathrm{f}(\mathrm{x})=\left\{\begin{array}{l}
\mathrm{e}^{\mathrm{x}^2} ; \mathrm{x} \in[0,1) \\\\
\mathrm{e} ; \mathrm{x} \in[1,2)
\end{array}\right.
$$
<br/><br/>$$
\begin{aligned}
& \int\limits_0^2 x f(x)=\int\limits_0^1 x \cdot e^{x^2} d x+\int\limits_1^2 x \cdot e d x \\\\
& x^2=t \Rightarrow 2 x d x=d t
\end{aligned}
$$
<br/><br/>$$
=\frac{1}{2} \int\limits_0^1 e^t d t+\left.e \frac{x^2}{2}\right|_1 ^2
$$
<br/><br/>$$
\begin{aligned}
& =\frac{1}{2}(\mathrm{e}-1)+\frac{1}{2}(4-1) \mathrm{e} \\\\
& =2 \mathrm{e}-\frac{1}{2}
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-11th-april-evening-shift
|
1lguu4z56
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The value of the integral $$\int_\limits{-\log _{e} 2}^{\log _{e} 2} e^{x}\left(\log _{e}\left(e^{x}+\sqrt{1+e^{2 x}}\right)\right) d x$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$\\log _{e}\\left(\\frac{(2+\\sqrt{5})^{2}}{\\sqrt{1+\\sqrt{5}}}\\right)+\\frac{\\sqrt{5}}{2}$$"}, {"identifier": "B", "content": "$$\\log _{e}\\left(\\frac{\\sqrt{2}(2+\\sqrt{5})^{2}}{\\sqrt{1+\\sqrt{5}}}\\right)-\\frac{\\sqrt{5}}{2}$$"}, {"identifier": "C", "content": "$$\\log _{e}\\left(\\frac{2(2+\\sqrt{5})}{\\sqrt{1+\\sqrt{5}}}\\right)-\\frac{\\sqrt{5}}{2}$$"}, {"identifier": "D", "content": "$$\\log _{e}\\left(\\frac{\\sqrt{2}(3-\\sqrt{5})^{2}}{\\sqrt{1+\\sqrt{5}}}\\right)+\\frac{\\sqrt{5}}{2}$$"}]
|
["B"]
| null |
$$\int_\limits{-\log _{e} 2}^{\log _{e} 2} e^{x}\left(\log _{e}\left(e^{x}+\sqrt{1+e^{2 x}}\right)\right) d x$$
<br/><br/>Let $e^x=t \Rightarrow e^x d x=d t$
<br/><br/>When, $x \rightarrow-\log _e 2$, then $t \rightarrow \frac{1}{2}$
<br/><br/>When, $x \rightarrow \log _e 2$, then $t \rightarrow 2$
<br/><br/>$$
I=\int_\limits{\frac{1}{2}}^2\left[\log _e\left(t+\sqrt{1+t^2}\right)\right] d t
$$ ...........(i)
<br/><br/>On applying integration by part method in Eq. (i), we get
<br/><br/>$$
\begin{aligned}
& I=\left[t \log _e\left(t+\sqrt{1+t^2}\right)\right]_{1 / 2}^2-\int_{1 / 2}^2 \frac{t}{t+\sqrt{1+t^2}}\left(1+\frac{2 t}{2 \sqrt{1+t^2}}\right) d t \\\\
& =2 \log _e(2+\sqrt{5})-\frac{1}{2} \log _e\left(\frac{1+\sqrt{5}}{2}\right)-\int_{1 / 2}^2 \frac{t}{\sqrt{1+t^2}} d t
\end{aligned}
$$
<br/><br/>$$
=\log _e\left(\frac{(2+\sqrt{5})^2}{\left(\frac{1+\sqrt{5}}{2}\right)^{1 / 2}}\right)-\frac{1}{2} \int_{1 / 2}^2 \frac{2 t}{\sqrt{1+t^2}} d t
$$ .............(ii)
<br/><br/>Let $\quad I_1=\int_{1 / 2}^2 \frac{2 t}{\sqrt{1+t^2}} d t$
<br/><br/>Let $1+t^2=w$
<br/><br/>$2 t d t=d w$
<br/><br/>When, $t \rightarrow \frac{1}{2}$, then $w=\frac{5}{4}$
<br/><br/>When, $t \rightarrow 2$, then $w=5$
<br/><br/>$$
\begin{aligned}
I_1 & =\int_{5 / 4}^5 \frac{1}{\sqrt{w}} d w \\\\
& =[2 \sqrt{w}]_{5 / 4}^5 \\\\
& =2\left[\sqrt{5}-\frac{\sqrt{5}}{2}\right]=\sqrt{5}
\end{aligned}
$$
<br/><br/>On substitute value of $I_1$ in Eq. (ii), we get
<br/><br/>$$
I=\log _e\left(\frac{\sqrt{2}(2+\sqrt{5})^2}{\sqrt{1+\sqrt{5}}}\right)-\frac{\sqrt{5}}{2}
$$
|
mcq
|
jee-main-2023-online-11th-april-morning-shift
|
1lguwr95k
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>For $$m, n > 0$$, let $$\alpha(m, n)=\int_\limits{0}^{2} t^{m}(1+3 t)^{n} d t$$. If $$11 \alpha(10,6)+18 \alpha(11,5)=p(14)^{6}$$, then $$p$$ is equal to ___________.</p>
|
[]
| null |
32
|
We have, $\alpha(m, n)=\int\limits_0^2 t^m(1+3 t)^n d t$
<br/><br/>Also, $11 \alpha(10,6)+18 \alpha(11,5)=P(14)^6$
<br/><br/>$\Rightarrow 11 \int\limits_0^2 t^{10}(1+3 t)^6 d t+18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6$
<br/><br/>Using integration by part for expression $t^{10}(1+3 t)^6$
<br/><br/>$$
\begin{array}{r}
\left.\Rightarrow 11\left[(1+3 t)^6 \times \frac{t^{11}}{11}\right]_0^2-\int\limits_0^2 6(1+3 t)^5 \times 3 \times \frac{t^{11}}{11} d t\right] \\\\
+18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6
\end{array}
$$
<br/><br/>$$
\begin{aligned}
& \Rightarrow 7^6 \times 2^{11}-18 \int\limits_0^2 t^{11}(1+3 t)^5 d t+18 \int\limits_0^2 t^{11}(1+3 t)^5 d t=P(14)^6 \\\\
& \Rightarrow 7^6 \times 2^{11}=P(14)^6 \Rightarrow(7 \times 2)^6 \times 2^5=P(14)^6 \\\\
& \Rightarrow P=2^5=32
\end{aligned}
$$
|
integer
|
jee-main-2023-online-11th-april-morning-shift
|
1lgyoktep
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$[t]$$ denote the greatest integer function. If $$\int_\limits{0}^{2.4}\left[x^{2}\right] d x=\alpha+\beta \sqrt{2}+\gamma \sqrt{3}+\delta \sqrt{5}$$, then $$\alpha+\beta+\gamma+\delta$$ is equal to __________.</p>
|
[]
| null |
6
|
$$
\begin{aligned}
\int\limits_0^{2.4}\left[x^2\right] d x & =\int\limits_0^1\left[x^2\right] d x+\int\limits_1^{\sqrt{2}}\left[x^2\right] d x \\\\
& +\int\limits_{\sqrt{2}}^{\sqrt{3}}\left[x^2\right] d x+\int\limits_{\sqrt{3}}^2\left[x^2\right] d x+\int\limits_2^{\sqrt{5}}\left[x^2\right] d x+\int\limits_{\sqrt{5}}^{2.4}\left[x^2\right] d x
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& =\int_0 0 . d x+\int_1^{\sqrt{2}} 1 \cdot d x+\int_{\sqrt{2}}^{\sqrt{3}} 2 d x+\int_{\sqrt{3}}^2 3 d x+\int_2^{\sqrt{5}} 4 d x+\int_{\sqrt{5}}^{2.4} 5 d x \\\\
& = 0+[x]_1^{\sqrt{2}}+2[x]_{\sqrt{2}}^{\sqrt{3}}+3[x]_{\sqrt{3}}^2+4[x]_2^{\sqrt{5}}+5[x]_{\sqrt{5}}^{2.4} \\\\
& =\sqrt{2}-1+2 \sqrt{3}-2 \sqrt{2}+6-3 \sqrt{3}+4 \sqrt{5}-8+12-5 \sqrt{5} \\\\
& =-\sqrt{2}-\sqrt{3}-\sqrt{5}+9 \\\\
& \therefore \alpha=9, \beta=-1, \gamma=-1, \delta=-1 \\\\
& \text { So, } \alpha+\beta+\gamma+\delta=9-1-1-1=6
\end{aligned}
$$
|
integer
|
jee-main-2023-online-8th-april-evening-shift
|
1lh0010vg
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$[t]$$ denote the greatest integer $$\leq t$$. Then $$\frac{2}{\pi} \int_\limits{\pi / 6}^{5 \pi / 6}(8[\operatorname{cosec} x]-5[\cot x]) d x$$ is equal to __________.</p>
|
[]
| null |
14
|
$$
\begin{aligned}
& \text { Let } \mathrm{I}=\frac{2}{\pi} \int\limits_{\frac{\pi}{6}}^{ \frac{5\pi}{6}}\{8[\operatorname{cosec} x]-5[\cot x]\} d x \\\\
& =\frac{2}{\pi}\left[8 \int\limits_{\frac{\pi}{6}}^{\frac{5 \pi}{6}}[\operatorname{cosec} x] d x-5 \int\limits_{\frac{\pi}{6}}^{\frac{5 \pi}{6}}[\cot x] d x\right]
\end{aligned}
$$
<br/><br/>$$
\left.\begin{array}{r}
=\frac{2}{\pi}\left[8 \int\limits_{\pi / 6}^{5 \pi / 6} d x-5\left\{\int\limits_{\pi / 6}^{\pi / 4} d x+\int\limits_{\pi / 4}^{\pi / 2} 0 . d x+\int\limits_{\pi / 2}^{3 \pi / 4}(-1) d x+\right.\right.
\left.\left.+\int\limits_{3 \pi / 4}^{5 \pi / 6}(-2) d x\right\}\right]
\end{array}\right]
$$
<br/><br/>$$
\begin{aligned}
& =\frac{2}{\pi}\left[8 \times\left(\frac{5 \pi}{6} - \frac{\pi}{6}\right)-5\left\{\left(\frac{\pi}{4}-\frac{\pi}{6}\right)-\left(\frac{3 \pi}{4}-\frac{\pi}{2}\right)\right\}\right.\left.-2\left(\frac{5 \pi}{6}-\frac{3 \pi}{4}\right)\right] \\\\
& =\frac{2}{\pi}\left[\frac{16 \pi}{3}+\frac{5 \pi}{3}\right]=14
\end{aligned}
$$
|
integer
|
jee-main-2023-online-8th-april-morning-shift
|
1lh21dq8w
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x > 0$$. Then $$18 \int_\limits{1}^{2} f(x) d x$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$10 \\log _{\\mathrm{e}} 2+6$$"}, {"identifier": "B", "content": "$$5 \\log _{e} 2-3$$"}, {"identifier": "C", "content": "$$10 \\log _{\\mathrm{e}} 2-6$$"}, {"identifier": "D", "content": "$$5 \\log _{\\mathrm{e}} 2+3$$"}]
|
["C"]
| null |
We have,
<br/><br/>$$
5 f(x)+4 f\left(\frac{1}{x}\right)=\frac{1}{x}+3, x>0
$$ ..........(i)
<br/><br/>On replacing $x$ by $\frac{1}{x}$ in (i), we get
<br/><br/>$$
5 f\left(\frac{1}{x}\right)+4 f(x)=x+3
$$ ..........(ii)
<br/><br/>Now, using Eq. (i) $\times 5-$ (ii) $\times 4$, we get
<br/><br/>$$
\begin{aligned}
& 25 f(x)-16 f(x) =\left(\frac{5}{x}+15\right)-(4 x+12) \\\\
&\Rightarrow 9 f(x) =\frac{5}{x}-4 x+3 \\\\
&\Rightarrow f(x) =\frac{1}{9}\left(\frac{5}{x}-4 x+3\right) ...........(iii)
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
\therefore \quad & 18 \int_1^2 f(x) d x=18 \int_1^2 \frac{1}{9}\left(\frac{5}{x}-4 x+3\right) d x \text { [Using Eq. (iii)] }\\\\
= & 2 \int_1^2\left(\frac{5}{x}-4 x+3\right) d x
\end{aligned}
$$
<br/><br/>$$
\begin{aligned}
& =2\left[5 \log _e x-4\left(\frac{x^2}{2}\right)+3 x\right]_1^2 \\\\
& =2\left[\left(5 \log _e 2-2(2)^2+3(2)\right)-\left(5 \log 1-2(1)^2+3(1)\right)\right] \\\\
& =2\left[5 \log _e 2-8+6+2-3\right] [\because \log 1=0]\\\\
& =10 \log _e 2-6
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-6th-april-morning-shift
|
1lh2y8vvh
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$f(x)$$ be a function satisfying $$f(x)+f(\pi-x)=\pi^{2}, \forall x \in \mathbb{R}$$. Then $$\int_\limits{0}^{\pi} f(x) \sin x d x$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$\\pi^{2}$$"}, {"identifier": "B", "content": "$$\\frac{\\pi^{2}}{2}$$"}, {"identifier": "C", "content": "$$2 \\pi^{2}$$"}, {"identifier": "D", "content": "$$\\frac{\\pi^{2}}{4}$$"}]
|
["A"]
| null |
Let $I=\int\limits_0^\pi f(x) \sin x d x$ ..........(i)
<br/><br/>$$
\begin{aligned}
& =\int\limits_0^\pi f(\pi-x) \sin (\pi-x) d x \\\\
& =\int\limits_0^\pi f(\pi-x) \sin x d x ........(ii)
\end{aligned}
$$
<br/><br/>On adding Equations (i) and (ii), we get
<br/><br/>$$
\begin{aligned}
& 2 I=\int\limits_0^\pi[f(x)+f(\pi-x)] \sin x d x \\\\
& \Rightarrow 2 I=\int\limits_0^\pi \pi^2 \sin x d x=\pi^2 \int\limits_0^\pi \sin x d x \\\\
& =\pi^2(-\cos x)_0^\pi=-\pi^2(-1-1)=2 \pi^2 \\\\
& \Rightarrow I=\pi^2
\end{aligned}
$$
|
mcq
|
jee-main-2023-online-6th-april-evening-shift
|
lsam93zl
|
maths
|
definite-integration
|
properties-of-definite-integration
|
If $\int\limits_0^{\frac{\pi}{3}} \cos ^4 x \mathrm{~d} x=\mathrm{a} \pi+\mathrm{b} \sqrt{3}$, where $\mathrm{a}$ and $\mathrm{b}$ are rational numbers, then $9 \mathrm{a}+8 \mathrm{b}$ is equal to :
|
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "$\\frac{3}{2}$"}]
|
["A"]
| null |
<p>To solve the given integral $\int\limits_0^{\frac{\pi}{3}} \cos ^4 x \mathrm{~d} x$, we'll apply a known power-reduction formula that allows us to express even powers of sine and cosine functions in terms of cosine of multiple angles. Specifically for $\cos^4 x$, we can write it in terms of double angles as:</p>
<p>$$
\cos^4 x = \left(\frac{1 + \cos(2x)}{2}\right)^2
$$</p>
<p>We can then expand and simplify the integral using this formula. Let's proceed with this:</p>
<p>$$
\int\limits_0^{\frac{\pi}{3}} \cos^4 x \mathrm{~d}x = \int\limits_0^{\frac{\pi}{3}} \left(\frac{1 + \cos(2x)}{2}\right)^2 \mathrm{~d}x
$$</p>
<p>Now, let's expand the integrand and then integrate term by term:</p>
<p>$$
= \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \cos^2(2x)\right)\mathrm{~d}x
$$</p>
<p>For the $\cos^2(2x)$ term, we again use the power reduction formula:</p>
<p>$$
\cos^2(2x) = \frac{1 + \cos(4x)}{2}
$$</p>
<p>Let's substitute this into the integral and continue:</p>
<p>$$
= \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{4} \left(\frac{1 + \cos(4x)}{2}\right)\right)\mathrm{~d}x
$$</p>
<p>Simplify and integrate:</p>
<p>$$
= \int\limits_0^{\frac{\pi}{3}} \left(\frac{1}{4} + \frac{1}{2} \cos(2x) + \frac{1}{8} + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x
$$</p>
<p>$$
= \int\limits_0^{\frac{\pi}{3}} \left(\frac{3}{8} + \frac{1}{2} \cos(2x) + \frac{1}{8} \cos(4x)\right)\mathrm{~d}x
$$</p>
<p>$$
= \left. \left(\frac{3}{8} x + \frac{1}{4} \sin(2x) + \frac{1}{32} \sin(4x)\right) \right|_0^{\frac{\pi}{3}}
$$</p>
<p>Evaluating this from $0$ to $\frac{\pi}{3}$:</p>
<p>$$
= \left(\frac{3}{8} \cdot \frac{\pi}{3} + \frac{1}{4} \sin\left(2 \cdot \frac{\pi}{3}\right) + \frac{1}{32} \sin\left(4 \cdot \frac{\pi}{3}\right)\right) - \left(\frac{3}{8} \cdot 0 + \frac{1}{4} \sin(0) + \frac{1}{32} \sin(0)\right)
$$</p>
<p>$$
= \frac{\pi}{8} + \frac{1}{4} \sin\left(\frac{2\pi}{3}\right) + \frac{1}{32} \sin\left(\frac{4\pi}{3}\right)
$$</p>
<p>$\sin\left(\frac{2\pi}{3}\right)$ is $\frac{\sqrt{3}}{2}$ and $\sin\left(\frac{4\pi}{3}\right)$ is $-\frac{\sqrt{3}}{2}$:</p>
<p>$$
= \frac{\pi}{8} + \frac{1}{4} \cdot \frac{\sqrt{3}}{2} - \frac{1}{32} \cdot \frac{\sqrt{3}}{2}
$$</p>
<p>$$
= \frac{\pi}{8} + \frac{\sqrt{3}}{8} - \frac{\sqrt{3}}{64}
$$</p>
<p>Now, combining terms we get the final result:</p>
<p>$$
= \frac{\pi}{8} + \frac{8\sqrt{3} - \sqrt{3}}{64}
$$</p>
<p>$$
= \frac{\pi}{8} + \frac{7\sqrt{3}}{64}
$$</p>
<p>Now, let's match this result to the form $\mathrm{a}\pi + \mathrm{b}\sqrt{3}$ and find $a$ and $b$:</p>
<p>$$
a = \frac{1}{8}, \quad b = \frac{7}{64}
$$</p>
<p>Now we find $9a + 8b$:</p>
<p>$$
9a + 8b = 9 \cdot \frac{1}{8} + 8 \cdot \frac{7}{64}
$$</p>
<p>$$
= \frac{9}{8} + \frac{7}{8}
$$</p>
<p>$$
= \frac{16}{8}
$$</p>
<p>$$
= 2
$$</p>
<p>Therefore, the value of $9a + 8b$ is 2, which correspond to Option A.</p>
|
mcq
|
jee-main-2024-online-1st-february-evening-shift
|
lsamq77b
|
maths
|
definite-integration
|
properties-of-definite-integration
|
The value of $\int\limits_0^1\left(2 x^3-3 x^2-x+1\right)^{\frac{1}{3}} \mathrm{~d} x$ is equal to :
|
[{"identifier": "A", "content": "-1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "1"}]
|
["C"]
| null |
$\begin{aligned} & I=\int_0^1\left(2 x^3-3 x^2-x+1\right)^{1 / 3} d x \\\\ & I=\int_0^1\left((2 x-1)\left(x^2-x-1\right)\right)^{1 / 3} d x \\\\ & I=\int_0^1\left[(2(1-x)-1)\left((1-x)^2-(1-x)-1\right)\right]^{1 / 3} d x \\\\ & I=\int_0^1\left((1-2 x)\left(x^2-x-1\right)\right)^{1 / 3} d x\end{aligned}$
<br/><br/>$\begin{aligned} & I=-\int_0^1\left((2 x-1)\left(x^2-x-1\right)\right)^{1 / 3} d x \\\\ & I=-I \\\\ & 2 I=0 \\\\ & I=0\end{aligned}$
|
mcq
|
jee-main-2024-online-1st-february-evening-shift
|
lsao9n4x
|
maths
|
definite-integration
|
properties-of-definite-integration
|
The value of the integral $\int\limits_0^{\pi / 4} \frac{x \mathrm{~d} x}{\sin ^4(2 x)+\cos ^4(2 x)}$ equals :
|
[{"identifier": "A", "content": "$\\frac{\\sqrt{2} \\pi^2}{8}$"}, {"identifier": "B", "content": "$\\frac{\\sqrt{2} \\pi^2}{16}$"}, {"identifier": "C", "content": "$\\frac{\\sqrt{2} \\pi^2}{32}$"}, {"identifier": "D", "content": "$\\frac{\\sqrt{2} \\pi^2}{64}$"}]
|
["C"]
| null |
Take $I=\int\limits_0^{\pi / 4} \frac{x d x}{\sin ^4(2 x)+\cos ^4(2 x)}$
<br/><br/>Let $2 x=t$
<br/><br/>$2 d x=d t$
<br/><br/>$d x=\frac{d t}{2}$
<br/><br/>$\begin{aligned} & I=\int\limits_0^{\pi / 2} \frac{t / 2 \cdot 1 / 2 d t}{\sin ^4 t+\cos ^4 t} \\\\ & I=\frac{1}{4} \int\limits_0^{\pi / 2} \frac{t d t}{\sin ^4 t+\cos ^4 t} \\\\ & =\frac{1}{4} \int\limits_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-t\right) d t}{\sin ^4(\pi / 2-t)+\cos ^4(\pi / 2-t)} \\\\ & =\frac{1}{4} \int\limits_0^{\pi / 2} \frac{\left(\frac{\pi}{2}-t\right)}{\sin ^4 t+\cos ^4 t}\end{aligned}$
<br/><br/>$\begin{aligned} & 2 I=\frac{1}{4} \int\limits_0^{\pi / 2} \frac{\frac{\pi}{2}}{\sin ^4 t+\cos ^4 t} d t \\\\ & 2 I=\frac{\pi}{8} \int\limits_0^{\pi / 2} \frac{d t}{\sin ^4 t+\cos ^4 t} \\\\ & 2 I=\frac{\pi}{8} \int\limits_0^{\pi / 2} \frac{\sec ^4 t}{1+\tan ^4 t} d t\end{aligned}$
<br/><br/>$\begin{aligned} & \text { Put tant }=y \\\\ & \sec ^2 t d t=d y \\\\ & 2 I=\frac{\pi}{8} \int\limits_0^{\infty} \frac{\left(1+y^2\right) d y}{1+y^4} \\\\ & =\frac{\pi}{8} \int\limits_0^{\infty} \frac{1+\frac{1}{y^2}}{y^2+\frac{1}{y^2}-2+2} d y \\\\ & = \frac{\pi}{8} \int\limits_0^{\infty} \frac{\left(1+\frac{1}{y^2}\right) d y}{2+\left(y-\frac{1}{y}\right)^2}\end{aligned}$
<br/><br/>$\begin{aligned} & \text { Put, } y-\frac{1}{y}=4 \\\\ & 2 I=\frac{\pi}{8} \int\limits_{-\infty}^{\infty} \frac{d u}{2+u^2} \\\\ & =\frac{\pi}{8 \sqrt{2}}\left[\tan ^{-1} \frac{y}{\sqrt{2}}\right]_{-\infty}^{\infty} \\\\ & =\frac{\sqrt{2} \pi^2}{32}\end{aligned}$
|
mcq
|
jee-main-2024-online-1st-february-morning-shift
|
lsaq6l6n
|
maths
|
definite-integration
|
properties-of-definite-integration
|
If $\int\limits_{-\pi / 2}^{\pi / 2} \frac{8 \sqrt{2} \cos x \mathrm{~d} x}{\left(1+\mathrm{e}^{\sin x}\right)\left(1+\sin ^4 x\right)}=\alpha \pi+\beta \log _{\mathrm{e}}(3+2 \sqrt{2})$, where $\alpha, \beta$ are integers, then $\alpha^2+\beta^2$ equals :
|
[]
| null |
8
|
$$
I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{\left(1+e^{\sin x}\right)\left(1+\sin ^4 x\right)} d x
$$ ............(1)
<br/><br/>Apply king's rule
<br/><br/>$$
I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x\left(e^{\sin x}\right)}{\left(1+e^{\sin x}\right)\left(1+\sin ^4 x\right)} d x
$$ ..........(2)
<br/><br/>Adding (1) and (2), we get
<br/><br/>$$
\begin{aligned}
& 2 I=\int\limits_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} d x \\\\
& I=\int\limits_0^{\frac{\pi}{2}} \frac{8 \sqrt{2} \cos x}{1+\sin ^4 x} d x,
\end{aligned}
$$
<br/><br/>Let $\sin x=t$
<br/><br/>$$
\begin{aligned}
& I=8 \sqrt{2} \int\limits_0^1 \frac{d t}{1+t^4} \\\\
& =4 \sqrt{2} \int\limits_0^1 \frac{\left(1+\frac{1}{t^2}\right)-\left(1-\frac{1}{t^2}\right)}{t^2+\frac{1}{t^2}} d t
\end{aligned}
$$
<br/><br/>$\begin{aligned} & =4 \sqrt{2} \int\limits_0^1 \frac{\left(1+\frac{1}{t^2}\right) d t}{\left(t-\frac{1}{t^2}\right)^2+2}-4 \sqrt{2} \int\limits_0^1 \frac{\left(1-\frac{1}{t^2}\right) d t}{\left(t+\frac{1}{t^2}\right)^2-2} \\\\ & =4 \sqrt{2} \cdot \frac{1}{\sqrt{2}}\left(\left.\tan ^{-1} \frac{t-\frac{1}{t}}{\sqrt{2}}\right|_0 ^1-4 \sqrt{2} \cdot \frac{1}{2 \sqrt{2}}\left[\log \left|\frac{t+\frac{1}{t}-\sqrt{2}}{t+\frac{1}{t}+\sqrt{2}}\right|\right]_0^1\right. \\\\ & =2 \pi-2 \log \left|\frac{2-\sqrt{2}}{2+\sqrt{2}}\right|\end{aligned}$
<br/><br/>$\begin{aligned} & =2 \pi+2 \log (3+2 \sqrt{2})=\alpha \pi+\beta \log _e(3+2 \sqrt{2}) \\\\ & \Rightarrow \alpha=2, \beta=2 \\\\ & \Rightarrow \alpha^2+\beta^2=8\end{aligned}$
|
integer
|
jee-main-2024-online-1st-february-morning-shift
|
lsbke98l
|
maths
|
definite-integration
|
properties-of-definite-integration
|
If $\int\limits_0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} \mathrm{~d} x=\mathrm{a}+\mathrm{b} \sqrt{2}+\mathrm{c} \sqrt{3}$, where $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are rational numbers, then $2 \mathrm{a}+3 \mathrm{~b}-4 \mathrm{c}$ is equal to :
|
[{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "7"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "8"}]
|
["D"]
| null |
<p>$$\begin{aligned}
& \int_\limits0^1 \frac{1}{\sqrt{3+x}+\sqrt{1+x}} d x=\int_\limits0^1 \frac{\sqrt{3+x}-\sqrt{1+x}}{(3+x)-(1+x)} d x \\
& \frac{1}{2}\left[\int_\limits0^1 \sqrt{3+x} d x-\int_\limits0^1(\sqrt{1+x}) d x\right]
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \frac{1}{2}\left[2 \frac{(3+x)^{\frac{3}{2}}}{3}-\frac{2(1+x)^{\frac{3}{2}}}{3}\right]_0^1 \\
& \frac{1}{2}\left[\frac{2}{3}(8-3 \sqrt{3})-\frac{2}{3}\left(2^{\frac{3}{2}}-1\right)\right] \\
& \frac{1}{3}[8-3 \sqrt{3}-2 \sqrt{2}+1] \\
& =3-\sqrt{3}-\frac{2}{3} \sqrt{2}=a+b \sqrt{2}+c \sqrt{3} \\
& a=3, b=-\frac{2}{3}, c=-1 \\
& 2 a+3 b-4 c=6-2+4=8
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-27th-january-morning-shift
|
lsbkjqqv
|
maths
|
definite-integration
|
properties-of-definite-integration
|
If $(a, b)$ be the orthocentre of the triangle whose vertices are $(1,2),(2,3)$ and $(3,1)$, and $\mathrm{I}_1=\int\limits_{\mathrm{a}}^{\mathrm{b}} x \sin \left(4 x-x^2\right) \mathrm{d} x, \mathrm{I}_2=\int\limits_{\mathrm{a}}^{\mathrm{b}} \sin \left(4 x-x^2\right) \mathrm{d} x$, then $36 \frac{\mathrm{I}_1}{\mathrm{I}_2}$ is equal to :
|
[{"identifier": "A", "content": "80"}, {"identifier": "B", "content": "72"}, {"identifier": "C", "content": "66"}, {"identifier": "D", "content": "88"}]
|
["B"]
| null |
<p>Equation of CE</p>
<p>$$\begin{aligned}
& y-1=-(x-3) \\
& x+y=4
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1cekv6/bbd7771e-59aa-40fe-80f6-e5c8fe8d22a7/2f978020-d3c5-11ee-a50b-bb659a2e1d74/file-1lt1cekv7.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1cekv6/bbd7771e-59aa-40fe-80f6-e5c8fe8d22a7/2f978020-d3c5-11ee-a50b-bb659a2e1d74/file-1lt1cekv7.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) 27th January Morning Shift Mathematics - Definite Integration Question 37 English Explanation"></p>
<p>orthocentre lies on the line $$x+y=4$$</p>
<p>so, $$a+b=4$$</p>
<p>$$I_1=\int_\limits a^b x \sin (x(4-x)) d x\quad$$ ..... (i)</p>
<p>Using king rule</p>
<p>$$I_1=\int_\limits a^b(4-x) \sin (x(4-x)) d x\quad$$ .... (ii)</p>
<p>$$\begin{aligned}
& \text { (i) }+ \text { (ii) } \\
& 2 \mathrm{I}_1=\int_\limits{\mathrm{a}}^{\mathrm{b}} 4 \sin (\mathrm{x}(4-\mathrm{x})) \mathrm{dx} \\
& 2 \mathrm{I}_1=4 \mathrm{I}_2 \\
& \mathrm{I}_1=2 \mathrm{I}_2 \\
& \frac{\mathrm{I}_1}{\mathrm{I}_2}=2 \\
& \frac{36 \mathrm{I}_1}{\mathrm{I}_2}=72
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-27th-january-morning-shift
|
jaoe38c1lscnv0ux
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>For $$0 < \mathrm{a} < 1$$, the value of the integral $$\int_\limits0^\pi \frac{\mathrm{d} x}{1-2 \mathrm{a} \cos x+\mathrm{a}^2}$$ is :</p>
|
[{"identifier": "A", "content": "$$\\frac{\\pi^2}{\\pi+a^2}$$\n"}, {"identifier": "B", "content": "$$\\frac{\\pi^2}{\\pi-a^2}$$\n"}, {"identifier": "C", "content": "$$\\frac{\\pi}{1-\\mathrm{a}^2}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\pi}{1+\\mathrm{a}^2}$$"}]
|
["C"]
| null |
<p>$$\begin{aligned}
& I=\int_\limits0^\pi \frac{d x}{1-2 a \cos x+a^2} ; 0< a<1 \\
& I=\int_\limits0^\pi \frac{d x}{1+2 a \cos x+a^2} \\
& 2 I=2 \int_\limits0^{\pi / 2} \frac{2\left(1+a^2\right)}{\left(1+a^2\right)^2-4 a^2 \cos ^2 x} d x \\
& \Rightarrow I=\int_\limits0^{\pi / 2} \frac{2\left(1+a^2\right) \cdot \sec ^2 x}{\left(1+a^2\right)^2 \cdot \sec ^2 x-4 a^2} d x \\
& \Rightarrow I=\int_\limits0^{\pi / 2} \frac{2 \cdot\left(1+a^2\right) \cdot \sec ^2 x}{\left(1+a^2\right)^2 \cdot \tan ^2 x+\left(1-a^2\right)^2} d x
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \mathrm{I}=\int_\limits0^{\pi / 2} \frac{\frac{2 \cdot \sec ^2 \mathrm{x}}{1+\mathrm{a}^2} \cdot \mathrm{dx}}{\tan ^2 \mathrm{x}+\left(\frac{1-\mathrm{a}^2}{1+\mathrm{a}^2}\right)^2} \\
& \Rightarrow \mathrm{I}=\frac{2}{\left(1-\mathrm{a}^2\right)}\left[\frac{\pi}{2}-0\right] \\
& \mathrm{I}=\frac{\pi}{1-\mathrm{a}^2}
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-27th-january-evening-shift
|
jaoe38c1lscoi3t2
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$f(x)=\int_\limits0^x g(t) \log _{\mathrm{e}}\left(\frac{1-\mathrm{t}}{1+\mathrm{t}}\right) \mathrm{dt}$$, where $$g$$ is a continuous odd function.
If $$\int_{-\pi / 2}^{\pi / 2}\left(f(x)+\frac{x^2 \cos x}{1+\mathrm{e}^x}\right) \mathrm{d} x=\left(\frac{\pi}{\alpha}\right)^2-\alpha$$, then $$\alpha$$ is equal to _________.</p>
|
[]
| null |
2
|
<p>$$f(x)=\int_\limits0^x g(t) \ln \left(\frac{1-t}{1+t}\right) d t$$</p>
<p>$$f(-x)=\int_\limits0^{-x} g(t) \ln \left(\frac{1-t}{1+t}\right) d t$$</p>
<p>$$f(-x)=-\int_\limits0^x g(-y) \ln \left(\frac{1+y}{1-y}\right) d y$$</p>
<p>$$=-\int_\limits0^x g(y) \ln \left(\frac{1-y}{1+y}\right) d y$$ (g is odd)</p>
<p>$$f(-x)=-f(x) \Rightarrow f$$ is also odd</p>
<p>Now,</p>
<p>$$\begin{aligned}
& I=\int_\limits{-\pi / 2}^{\pi / 2}\left(f(x)+\frac{x^2 \cos x}{1+e^x}\right) d x \quad \text{... (1)}\\
& I=\int_\limits{-\pi / 2}^{\pi / 2}\left(f(-x)+\frac{x^2 e^x \cos x}{1+e^x}\right) d x \quad \text{... (2)}\\
& 2 I=\int_\limits{-\pi / 2}^{\pi / 2} x^2 \cos x d x=2 \int_0^{\pi / 2} x^2 \cos x d x
\end{aligned}$$</p>
<p>$$\begin{aligned}
& I=\left(x^2 \sin x\right)_0^{\pi / 2}-\int_\limits0^{\pi / 2} 2 x \sin x d x \\
& =\frac{\pi^2}{4}-2\left(-x \cos x+\int \cos x d x\right)_0^{\pi / 2} \\
& =\frac{\pi^2}{4}-2(0+1)=\frac{\pi^2}{4}-2 \Rightarrow\left(\frac{\pi}{2}\right)^2-2 \\
& \therefore \alpha=2
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-27th-january-evening-shift
|
jaoe38c1lsd4i9ml
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$f, g:(0, \infty) \rightarrow \mathbb{R}$$ be two functions defined by $$f(x)=\int\limits_{-x}^x\left(|t|-t^2\right) e^{-t^2} d t$$ and $$g(x)=\int\limits_0^{x^2} t^{1 / 2} e^{-t} d t$$. Then, the value of $$9\left(f\left(\sqrt{\log _e 9}\right)+g\left(\sqrt{\log _e 9}\right)\right)$$ is equal to :</p>
|
[{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "9"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "6"}]
|
["C"]
| null |
<p>$$\begin{aligned}
& \mathrm{f}(\mathrm{x})=\int_\limits{-\mathrm{x}}^{\mathrm{x}}\left(|\mathrm{t}|-\mathrm{t}^2\right) \mathrm{e}^{-\mathrm{t}^2} \mathrm{dt} \\
& \Rightarrow \mathrm{f}^{\prime}(\mathrm{x})=2 \cdot\left(|\mathrm{x}|-\mathrm{x}^2\right) \mathrm{e}^{-\mathrm{x}^2} \ldots \ldots \ldots . .(1) \\
& \mathrm{g}(\mathrm{x})=\int_\limits0^{\mathrm{x}^2} \mathrm{t}^{\frac{1}{2}} \mathrm{e}^{-t} \mathrm{dt} \\
& \mathrm{g}^{\prime}(\mathrm{x})=\mathrm{xe}^{-\mathrm{x}^2}(2 \mathrm{x})-0 \\
& \mathrm{f}^{\prime}(\mathrm{x})+\mathrm{g}^{\prime}(\mathrm{x})=2 \mathrm{xe}^{-\mathrm{x}^2}-2 \mathrm{x}^2 \mathrm{e}^{-\mathrm{x}^2}+2 \mathrm{x}^2 \mathrm{e}^{-\mathrm{x}^2}
\end{aligned}$$</p>
<p>Integrating both sides w.r.t. $$\mathrm{x}$$</p>
<p>$$\begin{aligned}
& \mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x})=\int_\limits0^\alpha 2 \mathrm{xe}^{-\mathrm{x}^2} \mathrm{dx} \\
& \mathrm{x}^2=\mathrm{t} \\
& \Rightarrow \int_0^{\sqrt{\alpha}} \mathrm{e}^{-\mathrm{t}} \mathrm{dt}=\left[-\mathrm{e}^{-\mathrm{t}}\right]_0^{\sqrt{\alpha}} \\
& =-\mathrm{e}^{\left(\log _c(9)^{-1}\right)+1} \\
& \Rightarrow 9(\mathrm{f}(\mathrm{x})+\mathrm{g}(\mathrm{x}))=\left(1-\frac{1}{9}\right) 9=8
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-31st-january-evening-shift
|
jaoe38c1lsd4ywg3
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>$$\left|\frac{120}{\pi^3} \int_\limits0^\pi \frac{x^2 \sin x \cos x}{\sin ^4 x+\cos ^4 x} d x\right| \text { is equal to }$$ ________.</p>
|
[]
| null |
15
|
<p>$$\begin{aligned}
& \int_\limits0^\pi \frac{x^2 \sin x \cdot \cos x}{\sin ^4 x+\cos ^4 x} d x \\
& =\int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cdot \cos x}{\sin ^4 x+\cos ^4 x}\left(x^2-(\pi-x)^2\right) d x \\
& =\int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cdot \cos x\left(2 \pi x-\pi^2\right)}{\sin ^4 x+\cos ^4 x} \\
& =2 \pi \int_\limits0^{\frac{\pi}{2}} \frac{x \sin x \cos x}{\sin ^4 x+\cos 4 x} d x-\pi^2 \int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin ^4 x+\cos 4 x} d x \\
& =2 \pi \cdot \frac{\pi}{4} \int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cos ^4 x}{\sin ^4 x+\cos ^4 x} d x-\pi^2 \int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cos ^4 x}{\sin ^4 x+\cos ^4 x} d x
\end{aligned}$$</p>
<p>$$\begin{aligned}
& =-\frac{\pi^2}{2} \int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cos x}{\sin ^4 x+\cos ^4 x} d x \\
& =-\frac{\pi^2}{2} \int_\limits0^{\frac{\pi}{2}} \frac{\sin x \cos x d x}{1-2 \sin ^2 x \times \cos ^2 x} \\
& =-\frac{\pi^2}{2} \int_\limits0^{\frac{\pi}{2}} \frac{\sin 2 x}{2-\sin ^2 2 x} d x \\
& =-\frac{\pi^2}{2} \int_\limits0^{\frac{\pi}{2}} \frac{\sin 2 x}{1+\cos ^2 2 x} d x
\end{aligned}$$</p>
<p>Let $$\cos 2 \mathrm{x}=\mathrm{t}$$</p>
|
integer
|
jee-main-2024-online-31st-january-evening-shift
|
jaoe38c1lse5tymd
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If the integral $$525 \int_\limits0^{\frac{\pi}{2}} \sin 2 x \cos ^{\frac{11}{2}} x\left(1+\operatorname{Cos}^{\frac{5}{2}} x\right)^{\frac{1}{2}} d x$$ is equal to $$(n \sqrt{2}-64)$$, then $$n$$ is equal to _________.</p>
|
[]
| null |
176
|
<p>$$I=\int_\limits0^{\frac{\pi}{2}} \sin 2 x \cdot(\cos x)^{\frac{11}{2}}\left(1+(\cos x)^{\frac{5}{2}}\right)^{\frac{1}{2}} d x$$</p>
<p>Put $$\cos x=t^2 \Rightarrow \sin x d x=-2 t d t$$</p>
<p>$$\begin{aligned}
& \therefore \mathrm{I}=4 \int_\limits0^1 \mathrm{t}^2 \cdot \mathrm{t}^{11} \sqrt{\left(1+\mathrm{t}^5\right)}(\mathrm{t}) \mathrm{dt} \\
& \mathrm{I}=4 \int_\limits0^1 \mathrm{t}^{14} \sqrt{1+\mathrm{t}^5} \mathrm{dt}
\end{aligned}$$</p>
<p>Put $$1+\mathrm{t}^5=\mathrm{k}^2$$</p>
<p>$$\Rightarrow 5 \mathrm{t}^4 \mathrm{dt}=2 \mathrm{k} \mathrm{dk}$$</p>
<p>$$\therefore \mathrm{I}=4 \cdot \int_\limits1^{\sqrt{2}}\left(\mathrm{k}^2-1\right)^2 \cdot \mathrm{k} \frac{2 \mathrm{k}}{5} \mathrm{dk}$$</p>
<p>$$\mathrm{I}=\frac{8}{5} \int_\limits1^{\sqrt{2}} \mathrm{k}^6-2 \mathrm{k}^4+\mathrm{k}^2 \mathrm{dk}$$</p>
<p>$$\mathrm{I}=\frac{8}{5}\left[\frac{\mathrm{k}^7}{7}-\frac{2 \mathrm{k}^5}{5}+\frac{\mathrm{k}^3}{3}\right]_1^{\sqrt{2}}$$</p>
<p>$$\mathrm{I}=\frac{8}{5}\left[\frac{8 \sqrt{2}}{7}-\frac{8 \sqrt{2}}{5}+\frac{2 \sqrt{2}}{3}-\frac{1}{7}+\frac{2}{5}-\frac{1}{3}\right]$$</p>
<p>$$\begin{aligned}
&\mathrm{I}=\frac{8}{5}\left[\frac{22 \sqrt{2}}{105}-\frac{8}{105}\right]\\
&\therefore 525 \cdot \mathrm{I}=176 \sqrt{2}-64
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-31st-january-morning-shift
|
jaoe38c1lse60n68
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a function defined by $$f(x)=\frac{4^x}{4^x+2}$$ and $$M=\int_\limits{f(a)}^{f(1-a)} x \sin ^4(x(1-x)) d x, N=\int_\limits{f(a)}^{f(1-a)} \sin ^4(x(1-x)) d x ; a \neq \frac{1}{2}$$. If $$\alpha M=\beta N, \alpha, \beta \in \mathbb{N}$$, then the least value of $$\alpha^2+\beta^2$$ is equal to __________.</p>
|
[]
| null |
5
|
<p>$$\mathrm{f}(\mathrm{a})+\mathrm{f}(1-\mathrm{a})=1$$</p>
<p>$$M=\int_\limits{f(a)}^{f(1-a)}(1-x) \cdot \sin ^4 x(1-x) d x$$</p>
<p>$$\mathrm{M}=\mathrm{N}-\mathrm{M} \qquad 2 \mathrm{M}=\mathrm{N}$$</p>
<p>$$\alpha=2 ; \beta=1 \text {; }$$</p>
<p>Ans. 5</p>
|
integer
|
jee-main-2024-online-31st-january-morning-shift
|
jaoe38c1lsf0gf9u
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If the value of the integral $$\int_\limits{-\frac{\pi}{2}}^{\frac{\pi}{2}}\left(\frac{x^2 \cos x}{1+\pi^x}+\frac{1+\sin ^2 x}{1+e^{\sin x^{2123}}}\right) d x=\frac{\pi}{4}(\pi+a)-2$$, then the value of $$a$$ is</p>
|
[{"identifier": "A", "content": "$$-\\frac{3}{2}$$\n"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "$$\\frac{3}{2}$$\n"}, {"identifier": "D", "content": "2"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& I=\int_\limits{-\pi / 2}^{\pi / 2}\left(\frac{x^2 \cos x}{1+\pi^x}+\frac{1+\sin ^2 x}{1+e^{\sin x^{2023}}}\right) d x \\
& I=\int_\limits{-\pi / 2}^{\pi / 2}\left(\frac{x^2 \cos x}{1+\pi^{-x}}+\frac{1+\sin ^2 x}{1+e^{\sin (-x)^{2023}}}\right) d x
\end{aligned}$$</p>
<p>On Adding, we get</p>
<p>$$2 I=\int_\limits{-\pi / 2}^{\pi / 2}\left(x^2 \cos x+1+\sin ^2 x\right) d x$$</p>
<p>On solving</p>
<p>$$\begin{aligned}
& \mathrm{I}=\frac{\pi^2}{4}+\frac{3 \pi}{4}-2 \\
& \mathrm{a}=3
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-29th-january-morning-shift
|
jaoe38c1lsflco0v
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If $$\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin 2 x} d x=\alpha+\beta \sqrt{2}+\gamma \sqrt{3}$$, where $$\alpha, \beta$$ and $$\gamma$$ are rational numbers, then
$$3 \alpha+4 \beta-\gamma$$ is equal to _________.</p>
|
[]
| null |
6
|
<p>$$\begin{aligned}
& =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}} \sqrt{1-\sin 2 x} d x \\
& =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{3}}|\sin x-\cos x| d x \\
& =\int_\limits{\frac{\pi}{6}}^{\frac{\pi}{4}}(\cos x-\sin x) d x+\int_\limits{\frac{\pi}{4}}^{\frac{\pi}{3}}(\sin x-\cos x) d x \\
& =-1+2 \sqrt{2}-\sqrt{3} \\
& =\alpha+\beta \sqrt{2}+\gamma \sqrt{3} \\
& \alpha=-1, \beta=2, \gamma=-1 \\
& 3 \alpha+4 \beta-\gamma=6
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-29th-january-evening-shift
|
1lsg3nvgv
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$f: \mathbb{R} \rightarrow \mathbb{R}$$ be a function defined by $$f(x)=\frac{x}{\left(1+x^4\right)^{1 / 4}}$$, and $$g(x)=f(f(f(f(x))))$$. Then, $$18 \int_0^{\sqrt{2 \sqrt{5}}} x^2 g(x) d x$$ is equal to</p>
|
[{"identifier": "A", "content": "36"}, {"identifier": "B", "content": "33"}, {"identifier": "C", "content": "39"}, {"identifier": "D", "content": "42"}]
|
["C"]
| null |
<p>$$\begin{aligned}
&f(x)=\frac{x}{\left(1+x^4\right)^{1 / 4}}\\
&f \circ f(x)=\frac{f(x)}{\left(1+f(x)^4\right)^{1 / 4}}=\frac{\frac{x}{\left(1+x^4\right)^{1 / 4}}}{\left(1+\frac{x^4}{1+x^4}\right)^{1 / 4}}=\frac{x}{\left(1+2 x^4\right)^{1 / 4}}\\
&f(f(f(f(x))))=\frac{x}{\left(1+4 x^4\right)^{1 / 4}}
\end{aligned}$$</p>
<p>$$18 \int_\limits0^{\sqrt{2 \sqrt{5}}} \frac{x^3}{\left(1+4 x^4\right)^{1 / 4}} d x$$</p>
<p>$$\begin{aligned}
& \text { Let } 1+4 \mathrm{x}^4=\mathrm{t}^4 \\
& 16 \mathrm{x}^3 \mathrm{dx}=4 \mathrm{t}^3 \mathrm{dt} \\
& \frac{18}{4} \int_\limits1^3 \frac{\mathrm{t}^3 \mathrm{dt}}{\mathrm{t}} \\
& =\frac{9}{2}\left(\frac{\mathrm{t}^3}{3}\right)_1^3 \\
& =\frac{3}{2}[26]=39
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-30th-january-evening-shift
|
1lsg3stoy
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$y=f(x)$$ be a thrice differentiable function in $$(-5,5)$$. Let the tangents to the curve $$y=f(x)$$ at $$(1, f(1))$$ and $$(3, f(3))$$ make angles $$\pi / 6$$ and $$\pi / 4$$, respectively with positive $$x$$-axis. If $$27 \int_\limits1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3}$$ where $$\alpha, \beta$$ are integers, then the value of $$\alpha+\beta$$ equals</p>
|
[{"identifier": "A", "content": "26"}, {"identifier": "B", "content": "$$-$$16"}, {"identifier": "C", "content": "36"}, {"identifier": "D", "content": "$$-$$14"}]
|
["A"]
| null |
<p>$$\begin{aligned}
& y=f(x) \Rightarrow \frac{d y}{d x}=f^{\prime}(x) \\
& \left.\frac{d y}{d x}\right)_{(1, f(1))}=f^{\prime}(1)=\tan \frac{\pi}{6}=\frac{1}{\sqrt{3}} \Rightarrow f^{\prime}(1)=\frac{1}{\sqrt{3}} \\
& \left.\frac{d y}{d x}\right)_{(3, f(3))}=f^{\prime}(3)=\tan \frac{\pi}{4}=1 \Rightarrow f^{\prime}(3)=1 \\
& 27 \int_\limits1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t=\alpha+\beta \sqrt{3} \\
& I=\int_\limits1^3\left(\left(f^{\prime}(t)\right)^2+1\right) f^{\prime \prime}(t) d t \\
& f^{\prime}(t)=z \Rightarrow f^{\prime \prime}(t) d t=d z \\
& z=f^{\prime}(3)=1 \\
& z=f^{\prime}(1)=\frac{1}{\sqrt{3}}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& I=\int_\limits{1 / \sqrt{3}}^1\left(z^2+1\right) d z=\left(\frac{z^3}{3}+z\right)_{1 / \sqrt{3}}^1 \\
& =\left(\frac{1}{3}+1\right)-\left(\frac{1}{3} \cdot \frac{1}{3 \sqrt{3}}+\frac{1}{\sqrt{3}}\right) \\
& =\frac{4}{3}-\frac{10}{9 \sqrt{3}}=\frac{4}{3}-\frac{10}{27} \sqrt{3} \\
& \alpha+\beta \sqrt{3}=27\left(\frac{4}{3}-\frac{10}{27} \sqrt{3}\right)=36-10 \sqrt{3} \\
& \alpha=36, \beta=-10 \\
& \alpha+\beta=36-10=26
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-30th-january-evening-shift
|
1lsg420oq
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$a$$ and $$b$$ be real constants such that the function $$f$$ defined by $$f(x)=\left\{\begin{array}{ll}x^2+3 x+a & , x \leq 1 \\ b x+2 & , x>1\end{array}\right.$$ be differentiable on $$\mathbb{R}$$. Then, the value of $$\int_\limits{-2}^2 f(x) d x$$ equals</p>
|
[{"identifier": "A", "content": "21"}, {"identifier": "B", "content": "19/6"}, {"identifier": "C", "content": "17"}, {"identifier": "D", "content": "15/6"}]
|
["C"]
| null |
<p>To determine the integral of the piecewise function <span class="math-container">$$f$$</span> over the interval <span class="math-container">$$[-2, 2]$$</span>, we first ensure that <span class="math-container">$$f$$</span> is differentiable on <span class="math-container">$$\mathbb{R}$$</span>, as given in the problem statement. Differentiability implies continuity, so <span class="math-container">$$f$$</span> must also be continuous at <span class="math-container">$$x=1$$</span>.</p>
<p>The condition for continuity at <span class="math-container">$$x=1$$</span> is:</p>
<p><span class="math-container">$$x^2 + 3x + a = bx + 2$$</span> at <span class="math-container">$$x=1$$</span>.</p>
<p>This simplifies to:</p>
<p><span class="math-container">$$1 + 3 + a = b(1) + 2$$</span></p>
<p><span class="math-container">$$\Rightarrow a = b - 2$$</span></p>
<p>The first derivative of <span class="math-container">$$f$$</span> gives us two different expressions depending on the value of <span class="math-container">$$x$$</span>:</p>
<p>For <span class="math-container">$$x \leq 1$$</span>, <span class="math-container">$$f'(x) = 2x + 3$$</span>;
and for <span class="math-container">$$x > 1$$</span>, <span class="math-container">$$f'(x) = b$$</span>.</p>
<p>For <span class="math-container">$$f$$</span> to be differentiable at <span class="math-container">$$x=1$$</span>, these derivatives must be equal at that point. Setting <span class="math-container">$$f'(1)$$</span> from both expressions equal to each other gives <span class="math-container">$$2(1) + 3 = b$$</span>, thus <span class="math-container">$$b = 5$$</span>. And from <span class="math-container">$$a = b - 2$$</span>, we have <span class="math-container">$$a = 3$$</span>.</p>
<p>Now, we can calculate the integral of <span class="math-container">$$f$$</span> over the specified interval:</p>
<p><span class="math-container">$$\int\limits_{-2}^1 (x^2 + 3x + 3) \, dx + \int\limits_{1}^2 (5x + 2) \, dx$$</span></p>
<p>$$\begin{aligned}
& =\left[\frac{x^3}{3}+\frac{3 x^2}{2}+3 x\right]_{-2}^1+\left[\frac{5 x^2}{2}+2 x\right]_1^2 \\\\
& =\left(\frac{1}{3}+\frac{3}{2}+3\right)-\left(\frac{-8}{3}+6-6\right)+\left(10+4-\frac{5}{2}-2\right) \\\\
& =6+\frac{3}{2}+12-\frac{5}{2}=17
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-30th-january-evening-shift
|
1lsga9myw
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$f:\left[-\frac{\pi}{2}, \frac{\pi}{2}\right] \rightarrow \mathbf{R}$$ be a differentiable function such that $$f(0)=\frac{1}{2}$$. If the $$\lim _\limits{x \rightarrow 0} \frac{x \int_0^x f(\mathrm{t}) \mathrm{dt}}{\mathrm{e}^{x^2}-1}=\alpha$$, then $$8 \alpha^2$$ is equal to :</p>
|
[{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "16"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& \lim _{x \rightarrow 0} \frac{x \int_0^x f(t) d t}{\left(\frac{e^{x^2}-1}{x^2}\right) \times x^2} \\\\
& \lim _{x \rightarrow 0} \frac{\int_0^x f(t) d t}{x} \quad\left(\lim _{x \rightarrow 0} \frac{e^{x^2}-1}{x^2}=1\right) \\\\
& =\lim _{x \rightarrow 0} \frac{f(x)}{1} \text { (using L Hospital) } \\\\
& f(0)=\frac{1}{2} \\\\
& \alpha=\frac{1}{2} \\\\
& 8 \alpha^2=2
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-30th-january-morning-shift
|
1lsgchdrl
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The value of $$9 \int_\limits0^9\left[\sqrt{\frac{10 x}{x+1}}\right] \mathrm{d} x$$, where $$[t]$$ denotes the greatest integer less than or equal to $$t$$, is</p>
|
[]
| null |
155
|
<p>$$\begin{array}{ll}
\frac{10 x}{x+1}=1 & \Rightarrow x=\frac{1}{9} \\
\frac{10 x}{x+1}=4 & \Rightarrow x=\frac{2}{3} \\
\frac{10 x}{x+1}=9 & \Rightarrow x=9
\end{array}$$</p>
<p>$$\begin{aligned}
& \mathrm{I}=9\left(\int_\limits0^{1 / 9} 0 \mathrm{dx}+\int_\limits{1 / 9}^{2 / 3} 1\mathrm{d} x+\int_\limits{2 / 3}^9 2 \mathrm{dx}\right) \\
& =155
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-30th-january-morning-shift
|
luxwehsa
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The integral $$\int_\limits{1 / 4}^{3 / 4} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x$$ is equal to</p>
|
[{"identifier": "A", "content": "$$-1/2$$"}, {"identifier": "B", "content": "$$-1/4$$"}, {"identifier": "C", "content": "1/4"}, {"identifier": "D", "content": "1/2"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& \int_\limits{\frac{1}{4}}^{\frac{3}{4}} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x \\
& x=\cos 2 \theta \\
& \Rightarrow d x=(-2 \sin 2 \theta \mathrm{d} \theta)
\end{aligned}$$</p>
<p>Take limit as $$\alpha$$ and $$\beta$$</p>
<p>$$\begin{aligned}
& -2 \int_\limits\alpha^\beta \cos 2 \theta \cdot \sin 2 \theta d \theta \\
& =\int_\limits\alpha^\beta \sin 4 \theta d \theta \\
& =\left.\frac{-\cos 4 \theta}{4}\right|_\alpha ^\beta \\
& =-\left.\frac{1}{4}\left(2 \cdot\left(x^2\right)-1\right)\right|_{1 / 4} ^{3 / 4} \\
& =-\left.\frac{1}{4}\left(2 x^2-1\right)\right|_{1 / 4} ^{3 / 4} \\
& =-\frac{1}{4}\left(\frac{18}{16}-1-\frac{2}{16}+1\right) \\
& =-\frac{1}{4}
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-9th-april-evening-shift
|
luxwdx2t
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The value of the integral $$\int_\limits{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x$$ is</p>
|
[{"identifier": "A", "content": "$$\\sqrt{5}-\\sqrt{2}+\\log _e\\left(\\frac{7+4 \\sqrt{5}}{1+\\sqrt{2}}\\right)$$\n"}, {"identifier": "B", "content": "$$\\sqrt{2}-\\sqrt{5}+\\log _e\\left(\\frac{7+4 \\sqrt{5}}{1+\\sqrt{2}}\\right)$$\n"}, {"identifier": "C", "content": "$$\\sqrt{5}-\\sqrt{2}+\\log _e\\left(\\frac{9+4 \\sqrt{5}}{1+\\sqrt{2}}\\right)$$\n"}, {"identifier": "D", "content": "$$\\sqrt{2}-\\sqrt{5}+\\log _e\\left(\\frac{9+4 \\sqrt{5}}{1+\\sqrt{2}}\\right)$$"}]
|
["D"]
| null |
<p>$$\begin{aligned}
& \int_\limits{-1}^2 \log _e\left(x+\sqrt{x^2+1}\right) d x \\
& =\left[x \log _e\left(x+\sqrt{x^2+1}\right)\right]_{-1}^2-\int_\limits{-1}^2 \frac{x}{\left(x+\sqrt{x^2+1}\right)}\left(1+\frac{x}{\sqrt{x^2+1}}\right) d x \\
& =2 \log _2(2+\sqrt{5})+\log _e(\sqrt{2}-1)-\int_\limits{-1}^2 \frac{x}{\sqrt{x^2+1}} d x \\
& =\log _e\left[(2+\sqrt{5})^2(\sqrt{2}-1)\right]-\left[\sqrt{x^2+1}\right]_{-1}^2 \\
& =\log _e\left[\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right]-\sqrt{5}+\sqrt{2} \\
& =\sqrt{2}-\sqrt{5}+\log _e\left[\frac{9+4 \sqrt{5}}{1+\sqrt{2}}\right]
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-9th-april-evening-shift
|
lv0vxbpa
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>$$\text { Let } f(x)=\left\{\begin{array}{lr}
-2, & -2 \leq x \leq 0 \\
x-2, & 0< x \leq 2
\end{array} \text { and } \mathrm{h}(x)=f(|x|)+|f(x)| \text {. Then } \int_\limits{-2}^2 \mathrm{~h}(x) \mathrm{d} x\right. \text { is equal to: }$$</p>
|
[{"identifier": "A", "content": "2"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "1"}]
|
["A"]
| null |
<p>$$f(x)=\left\{\begin{array}{cc}
-2 & -2 \leq x \leq 0 \\
x-2 & 0< x \leq 2
\end{array} h(x)=f|x|+|f(x)|\right.$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwk8ygqe/f2db7258-aa84-4e80-8605-2f6185c78c16/3e8a8d60-198f-11ef-b65b-abc5d1349d93/file-1lwk8ygqf.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwk8ygqe/f2db7258-aa84-4e80-8605-2f6185c78c16/3e8a8d60-198f-11ef-b65b-abc5d1349d93/file-1lwk8ygqf.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 Morning Shift Mathematics - Definite Integration Question 14 English Explanation"></p>
<p>$$\begin{aligned}
& h(x)=\left\{\begin{array}{cc}
-x-2+2=-x & -2 \leq x \leq 0 \\
0 & 0< x \leq 2
\end{array}\right. \\
& \therefore \int_\limits{-2}^2 h(x) d x=\int_\limits{-2}^0-x d x+\int_\limits0^2 0 d x \\
& \left.\frac{x^2}{2}\right|_{-2} ^0=\frac{4}{2}=2
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-4th-april-morning-shift
|
lv0vxdgv
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If the shortest distance between the lines $$\frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$$ and $$\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$$ is $$\frac{38}{3 \sqrt{5}} \mathrm{k}$$, and $$\int_\limits 0^{\mathrm{k}}\left[x^2\right] \mathrm{d} x=\alpha-\sqrt{\alpha}$$, where $$[x]$$ denotes the greatest integer function, then $$6 \alpha^3$$ is equal to _________.</p>
|
[]
| null |
48
|
<p>$$L_1: \frac{x+2}{2}=\frac{y+3}{3}=\frac{z-5}{4}$$</p>
<p>$$\begin{aligned}
& \vec{b}_1=2 \hat{i}+3 \hat{j}+4 \hat{k} \\
& \vec{a}_1=-2 \hat{i}-3 \hat{j}+5 \hat{k}
\end{aligned}$$</p>
<p>$$L_2=\frac{x-3}{1}=\frac{y-2}{-3}=\frac{z+4}{2}$$</p>
<p>$$\begin{aligned}
& \vec{a}_2=3 \hat{i}+2 \hat{j}-4 \hat{k} \\
& \vec{b}_2=1 \hat{i}-3 \hat{j}+2 \hat{k}
\end{aligned}$$</p>
<p>$$d=\left|\frac{\left(\vec{a}_2-\vec{a}_1\right) \cdot\left(\vec{b}_1 \times \vec{b}_2\right)}{\left|\vec{b}_1 \times \vec{b}_2\right|}\right|$$</p>
<p>$$\begin{gathered}
d=\left|\frac{(5 \hat{i}+5 \hat{j}-9 \hat{k}) \cdot(18 \hat{i}-9 \hat{k})}{\sqrt{324+81}}\right| \\
\left|\vec{b}_1 \times \vec{b}\right|\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{k} \\
2 & 3 & 4 \\
1 & -3 & 2
\end{array}\right| \\
\Rightarrow \hat{i}(6+12)-\hat{j}(4-4)+\hat{k}(-6-3) \\
\Rightarrow(18 \hat{i}-9 \hat{k})
\end{gathered}$$</p>
<p>$$\begin{aligned}
& d=\left|\frac{90+81}{9 \sqrt{5}}\right| \\
& d=\frac{171}{9 \sqrt{5}} \\
& \frac{38}{3 \sqrt{5}} k=\frac{171}{9 \sqrt{5}} \\
& \frac{38}{3 \sqrt{5}} k=\frac{57}{3 \sqrt{5}}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& {k=\frac{57}{38}}=\frac{3}{2} \\
& \int_\limits0^{\frac{3}{2}}\left[x^2\right] d x \\
& \int_\limits0^1 0 d x+\int_\limits0^{\sqrt{ }} 1 d x+\int_\limits{\sqrt{2}}^{\frac{3}{2}} 2 d x \\
& 0+(\sqrt{2}-1)+2\left(\frac{3}{2}-\sqrt{2}\right) \\
& \sqrt{2}-1+3-2 \sqrt{2} \\
& 2-\sqrt{2}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \alpha=2 \\
& 6 \alpha^3=6(2)^3=48
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-4th-april-morning-shift
|
lv0vxdop
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If $$\int_0^{\frac{\pi}{4}} \frac{\sin ^2 x}{1+\sin x \cos x} \mathrm{~d} x=\frac{1}{\mathrm{a}} \log _{\mathrm{e}}\left(\frac{\mathrm{a}}{3}\right)+\frac{\pi}{\mathrm{b} \sqrt{3}}$$, where $$\mathrm{a}, \mathrm{b} \in \mathrm{N}$$, then $$\mathrm{a}+\mathrm{b}$$ is equal to _________.</p>
|
[]
| null |
8
|
<p>$$I=\int_\limits0^{\frac{\pi}{4}} \frac{\sin ^2 x}{1+\sin x \cos x} d x=\int_\limits0^{\frac{\pi}{4}} \frac{\sin ^2 x}{\sin ^2 x+\cos ^2 x+\sin x \cos x} d x$$
<p>$$I=\int_\limits0^{\frac{\pi}{4}} \frac{\tan ^2 x}{1+\tan x+\tan ^2 x} d x$$</p>
<p>$$=\int_\limits0^{\frac{\pi}{4}} \frac{\tan x \cdot \sec ^2 x d x}{\left(1+\tan ^2 x\right)\left(1+\tan x+\tan ^2 x\right)}$$</p>
<p>Let $$\tan x=t$$</p>
<p>$$\begin{aligned}
I & =\int_\limits0^1 \frac{t^2}{\left(1+t^2\right)\left(1+t+t^2\right)} d t \\
& =\int_\limits0^1\left(\frac{x}{1+x^2}-\frac{x}{1+x+x^2}\right) d x
\end{aligned}$$</p>
<p>$${1 \over 2}\int\limits_0^1 {{{2x} \over {1 + {x^2}}}dx - \int\limits_0^{} {{{{1 \over 2}(2x + 1) - {1 \over 2}} \over {1 + x + {x^2}}}dx} } $$</p>
<p>$$\begin{aligned}
& =\frac{1}{2} \ln 2-\frac{1}{2} \ln 3 \frac{1}{2} \int_\limits0^1 \frac{d x}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}} \\
& =\frac{1}{2} \ln \frac{2}{3}+\frac{1}{2} \cdot \frac{2}{\sqrt{3}}\left[\tan ^{-1} \frac{2 x+1}{\sqrt{3}}\right]_0^1 \\
& =\frac{1}{2} \ln \frac{2}{3}+\frac{1}{\sqrt{3}}\left(\frac{\pi}{3}-\frac{\pi}{6}\right) \\
& =\frac{1}{2} \ln \frac{2}{3}+\frac{1}{\sqrt{3}} \cdot \frac{\pi}{6} \\
& \therefore \quad a=2, b=6 \\
& \therefore \quad a+b=8
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-4th-april-morning-shift
|
lv2ers9h
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If the value of the integral $$\int\limits_{-1}^1 \frac{\cos \alpha x}{1+3^x} d x$$ is $$\frac{2}{\pi}$$.Then, a value of $$\alpha$$ is</p>
|
[{"identifier": "A", "content": "$$\\frac{\\pi}{2}$$\n"}, {"identifier": "B", "content": "$$\\frac{\\pi}{4}$$\n"}, {"identifier": "C", "content": "$$\\frac{\\pi}{3}$$\n"}, {"identifier": "D", "content": "$$\\frac{\\pi}{6}$$"}]
|
["A"]
| null |
<p>$$\begin{aligned}
& \text { Given, } \int_\limits{-1}^1 \frac{\cos \alpha x}{1+3^x} d x=\frac{2}{\pi} \\
& \begin{aligned}
I & =\int_\limits{-1}^1 \frac{\cos \alpha x}{1+3^x} d x \\
\Rightarrow I & =\int_\limits0^1\left(\frac{\cos \alpha x}{1+3^x}+\frac{\cos \alpha x}{1+3^{-x}}\right) d x \\
& =\int_\limits0^1 \cos \alpha x d x \\
& =\left(\frac{\sin \alpha x}{\alpha}\right)_0^1 \\
& =\frac{\sin \alpha}{\alpha} \\
\Rightarrow & \frac{\sin \alpha}{\alpha}=\frac{2}{\pi} \\
\Rightarrow & \alpha=\frac{\pi}{2}
\end{aligned}
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-4th-april-evening-shift
|
lv3vefcu
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$\int_\limits\alpha^{\log _e 4} \frac{\mathrm{d} x}{\sqrt{\mathrm{e}^x-1}}=\frac{\pi}{6}$$. Then $$\mathrm{e}^\alpha$$ and $$\mathrm{e}^{-\alpha}$$ are the roots of the equation :</p>
|
[{"identifier": "A", "content": "$$2 x^2-5 x+2=0$$\n"}, {"identifier": "B", "content": "$$x^2-2 x-8=0$$\n"}, {"identifier": "C", "content": "$$2 x^2-5 x-2=0$$\n"}, {"identifier": "D", "content": "$$x^2+2 x-8=0$$"}]
|
["A"]
| null |
<p>$$\begin{aligned}
& \text { Let } \sqrt{e^x-1}=t \\
& e^x-1=t^2 \\
& e^x=1+t^2 \\
& e^x=0+2 t-\frac{d t}{d x} \\
& \frac{d t}{d x}=\frac{e^x}{2 t}=\frac{t^2+1}{2 t} \\
& I=\int \frac{2 t}{t\left(1+t^2\right)} d t=2 \tan ^{-1} t \\
& \Rightarrow \quad I=\int_\limits\alpha^{\log ^4} \frac{d x}{\sqrt{e^x-1}} \\
& I=\left.2 \tan ^{-1} \sqrt{e^x-1}\right|_\alpha ^{\log 4} \\
& =2\left(\tan ^{-1} \sqrt{3}-\tan ^{-1} \sqrt{e^\alpha-1}\right)=\frac{\pi}{6} \\
& \Rightarrow \frac{\pi}{3}-\tan ^{-1}\left(\sqrt{e^\alpha-1}\right)=\frac{\pi}{12} \\
& \tan ^{-1}\left(\sqrt{e^\alpha-1}\right)=\frac{\pi}{4} \\
& \Rightarrow e^\alpha-1=1 \\
& e^\alpha=2 \Rightarrow e^{-\alpha}=\frac{1}{2}
\end{aligned}$$</p>
<p>$$\therefore \quad$$ Quadratic equation whose roots are $$e^a$$ & $$e^{-\alpha}$$ is</p>
<p>$$\begin{aligned}
& x^2-\left(e^\alpha+e^{-\alpha}\right) x+e^\alpha \times e^{-\alpha}=0 \\
& x^2-\left(2+\frac{1}{2}\right) x+1=0 \\
& 2 x^2-5 x+2=0
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-8th-april-evening-shift
|
lv5grw2g
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The value of $$k \in \mathbb{N}$$ for which the integral $$I_n=\int_0^1\left(1-x^k\right)^n d x, n \in \mathbb{N}$$, satisfies $$147 I_{20}=148 I_{21}$$ is</p>
|
[{"identifier": "A", "content": "8"}, {"identifier": "B", "content": "14"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "10"}]
|
["C"]
| null |
<p>$$\begin{aligned}
& I(21)=\int_\limits0^1\left(1-x^k\right)^{21} d x \\
& =\int_\limits0^1\left(1-x^k\right)\left(1-x^k\right)^{20} d x \\
& =\int_\limits0^1\left(1-x^k\right)^{20} d x-\int_0 x^k\left(1-x^k\right)^{20} d x \\
& I(21)=I(20)-\int_\limits0^1 x^k\left(1-x^k\right)^{20} d x
\end{aligned}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lw8swbwa/de26032d-c065-4f0a-b1b3-032767344a76/d1fa19a0-1343-11ef-9cb4-095599b9956f/file-1lw8swbwb.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lw8swbwa/de26032d-c065-4f0a-b1b3-032767344a76/d1fa19a0-1343-11ef-9cb4-095599b9956f/file-1lw8swbwb.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) 8th April Morning Shift Mathematics - Definite Integration Question 8 English Explanation"></p>
<p>$$I(21)=I(20)-\left\lfloor\frac{\left(1-x^k\right)^{21}}{-21 k} x-\int_\limits0^1 \frac{(1-x^k)^{21}}{-21 k} d x\right\rfloor$$</p>
<p>$$\begin{aligned}
& I(21)=I(20)-\frac{1}{21 k} I(20) \\
& \Rightarrow[I(21)](21 k+1)=21 K I(20) \\
& \Rightarrow 21 K=147 \Rightarrow K=7
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-8th-april-morning-shift
|
lv7v4o3r
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The integral $$\int_\limits0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} \mathrm{~d} x$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$3 \\pi-50 \\log _e 2+20 \\log _e 5$$\n"}, {"identifier": "B", "content": "$$3 \\pi-25 \\log _e 2+10 \\log _e 5$$\n"}, {"identifier": "C", "content": "$$3 \\pi-10 \\log _{\\mathrm{e}}(2 \\sqrt{2})+10 \\log _{\\mathrm{e}} 5$$\n"}, {"identifier": "D", "content": "$$3 \\pi-30 \\log _e 2+20 \\log _e 5$$"}]
|
["A"]
| null |
<p>$$\int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x$$</p>
<p>$$\begin{aligned}
& \sin x=A(3 \sin x+5 \cos x)+B(3 \cos x-5 \sin x) \\
& \begin{array}{l}
3 A-5 B=1 \\
5 A+3 B=0
\end{array}>A=\frac{3}{34} \quad B=\frac{-5}{34}
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \int_0^{\pi / 4} \frac{136\left[\frac{3}{34}(3 \sin x+5 \cos x)-\frac{5}{34}(3 \cos x-5 \sin x)\right]}{3 \sin x+5 \cos x} d x \\
& \int_0^{\pi / 4} 12 d x-20 \int_0^{\pi / 4} \frac{3 \cos x-5 \sin x}{3 \sin x+5 \cos x} d x \\
& 12 \times \frac{\pi}{4}-20\left[\ln \left|\frac{3}{\sqrt{2}}+\frac{5}{\sqrt{2}}\right|-\ln 5\right] \\
& 3 \pi-20 \ln 2^{5 / 2}+20 \ln 5 \\
& \Rightarrow 3 \pi-50 \ln 2+20 \ln 5
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-5th-april-morning-shift
|
lv7v3ka9
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>The value of $$\int_\limits{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y$$ is :</p>
|
[{"identifier": "A", "content": "$$\\frac{\\pi}{2}$$\n"}, {"identifier": "B", "content": "$$\\pi^2$$\n"}, {"identifier": "C", "content": "$$\\frac{\\pi^2}{2}$$\n"}, {"identifier": "D", "content": "$$2 \\pi^2$$"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& I=\int_\limits{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y \\
& =\int_\limits0^\pi\left(\frac{2 y(1+\sin y)}{1+\cos ^2 y}+\frac{-2 y(1-\sin y)}{1+\cos ^2 y}\right) d y \\
& =\int_\limits0^\pi\left(\frac{2 y+2 y \sin y-2 y+2 y \sin y}{1+\cos ^2 y}\right) d y
\end{aligned}$$</p>
<p>$$I=4 \int_0^\pi\left(\frac{y \sin }{1+\cos ^2 y}\right) d y \quad \text{... (1)}$$</p>
<p>$$\begin{aligned}
& I=4 \int_\limits0^\pi\left(\frac{(\pi-y) \sin (\pi-)}{1+\cos ^2(\pi-y)}\right) d y \\
& I=4\left\lfloor\int_\limits0\left(\frac{\pi \sin y}{1+\cos ^2 y}\right) d y-\int_\limits0^\pi \frac{y \sin y}{1+\cos ^2 y} d y\right\rfloor \quad \text{... (2)}
\end{aligned}$$</p>
<p>Adding equation (1) and (2)</p>
<p>$$\begin{aligned}
& 2 I=4 \int_\limits0^\pi\left(\frac{\pi \sin y}{1+\cos ^2 y}\right) d y \\
& I=2 \pi \int_\limits0^\pi \frac{\sin y}{1+\cos ^2 y} d y \\
& =2 \pi \times \frac{\pi}{2} \\
& =\pi^2
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-5th-april-morning-shift
|
lv9s205z
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$\beta(\mathrm{m}, \mathrm{n})=\int_\limits0^1 x^{\mathrm{m}-1}(1-x)^{\mathrm{n}-1} \mathrm{~d} x, \mathrm{~m}, \mathrm{n}>0$$. If $$\int_\limits0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x=\mathrm{a} \times \beta(\mathrm{b}, \mathrm{c})$$, then $$100(\mathrm{a}+\mathrm{b}+\mathrm{c})$$ equals _________.</p>
|
[{"identifier": "A", "content": "2012"}, {"identifier": "B", "content": "1021"}, {"identifier": "C", "content": "1120"}, {"identifier": "D", "content": "2120"}]
|
["D"]
| null |
<p>First, let's rewrite the given integral using the given form of the Beta function. The given integral is:</p>
<p>
<p>$$\int_\limits0^1\left(1-x^{10}\right)^{20} \mathrm{~d} x$$</p>
</p>
<p>To use the Beta function, let us make a substitution. Let $ x^{10} = t $. Then, $ dx = \frac{1}{10}t^{-\frac{9}{10}} dt $ or $ dx = \frac{1}{10} t^{-\frac{9}{10}} dt $. The limits of integration change as follows: when $ x = 0 $, $ t = 0 $, and when $ x = 1 $, $ t = 1 $.</p>
<p>Substituting these into the integral, we have:</p>
<p>
<p>$$\int_\limits0^1 (1 - t)^{20} \cdot \frac{1}{10} t^{-\frac{9}{10}} dt$$</p>
</p>
<p>which simplifies to:</p>
<p>
<p>$$\frac{1}{10} \int_\limits0^1 (1 - t)^{20} t^{-\frac{9}{10}} dt$$</p>
</p>
<p>We recognize this integral as a Beta function $ \beta(m, n) $ where $ m = 1 - \frac{9}{10} = \frac{1}{10} $ and $ n = 20 + 1 = 21 $.</p>
<p>Therefore, we can write this as:</p>
<p>
<p>$$\frac{1}{10} \beta \left( \frac{1}{10}, 21 \right)$$</p>
</p>
<p>Comparing this to $ a \times \beta(b, c) $, we have $ a = \frac{1}{10} $, $ b = \frac{1}{10} $, and $ c = 21 $.</p>
<p>Now we calculate $ 100(a + b + c) $:</p>
<p>
<p>$$100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{10} + \frac{1}{10} + 21 \right) = 100 \left( \frac{1}{5} + 21 \right) = 100 \left( \frac{1}{5} + \frac{105}{5} \right) = 100 \left( \frac{106}{5} \right) = 100 \times 21.2 = 2120$$</p>
</p>
<p>So, the answer is Option D, 2120.</p>
|
mcq
|
jee-main-2024-online-5th-april-evening-shift
|
lv9s20mx
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>If $$f(t)=\int_\limits0^\pi \frac{2 x \mathrm{~d} x}{1-\cos ^2 \mathrm{t} \sin ^2 x}, 0<\mathrm{t}<\pi$$, then the value of $$\int_\limits0^{\frac{\pi}{2}} \frac{\pi^2 \mathrm{dt}}{f(\mathrm{t})}$$ equals __________.</p>
|
[]
| null |
1
|
<p>$$\begin{aligned}
& f(t)=\int_\limits0^\pi \frac{2 x d x}{1-\cos ^2 t \sin ^2 x} \\
& x \rightarrow \pi-x
\end{aligned}$$</p>
<p>$$\begin{aligned}
& f(t)=\int_\limits0^\pi \frac{2(\pi-x) d x}{1-\cos ^2 t \sin ^2 x}=2 \pi \int_\limits0^\pi \frac{d x}{1-\cos ^2 t \sin ^2 x}-f(t) \\
& \Rightarrow f(t)=\pi \int_\limits0^\pi \frac{d x}{1-\cos ^2 t \sin ^2 x} \\
& =2 \pi \int_\limits0^{\frac{\pi}{2}} \frac{d x}{1-\cos ^2 t \sin ^2 x} \\
& f(t)=2 \pi \int_\limits0^{\frac{\pi}{2}} \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t \tan ^2 x} \\
& I_1=\int \frac{\sec ^2 x d x}{\sec ^2 x-\cos ^2 t \tan ^2 x} \\
& \text { Put } \cos t \tan x=\lambda \Rightarrow \cos t \sec ^2 x d x=d \lambda \\
& I_1=\int \frac{d \lambda}{\cos t \cdot\left(1+\lambda^2 \sec ^2 t-\lambda^2\right)}=\int \frac{d \lambda}{\cos t\left(1+\lambda^2 \tan ^2 t\right)} \\
\end{aligned}$$</p>
<p>$$\begin{aligned}
=\frac{1}{\cos t \cdot \tan ^2 t} \cdot \int \frac{d \lambda}{\lambda^2+\cos ^2 t}= & \frac{1}{\cos t \tan ^2 t} \\
& \times \frac{1}{\cos t} \tan ^{-1}(\lambda \tan t)
\end{aligned}$$</p>
<p>$$\begin{aligned}
& =\frac{1}{\sin t} \tan ^{-1}(\sin t \tan x) \\
\Rightarrow & \left.f(t)=\frac{2 \pi}{\sin t} \tan ^{-1}(\sin t \tan x)\right]_0^{\frac{\pi}{2}}=\frac{\pi^2}{\sin t} \\
\Rightarrow & \int_\limits0^{\frac{\pi}{2}} \frac{\pi^2 d t}{f(t)}=\int_\limits0^{\frac{\pi}{2}} \sin t d t=1
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-5th-april-evening-shift
|
lvb294zl
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$[t]$$ denote the largest integer less than or equal to $$t$$. If $$\int_\limits0^3\left(\left[x^2\right]+\left[\frac{x^2}{2}\right]\right) \mathrm{d} x=\mathrm{a}+\mathrm{b} \sqrt{2}-\sqrt{3}-\sqrt{5}+\mathrm{c} \sqrt{6}-\sqrt{7}$$, where $$\mathrm{a}, \mathrm{b}, \mathrm{c} \in \mathbf{Z}$$, then $$\mathrm{a}+\mathrm{b}+\mathrm{c}$$ is equal to __________.</p>
|
[]
| null |
23
|
<p>$$\int_\limits0^3\left(\left[x^2\right]+\left[\frac{x^2}{2}\right]\right) d x \quad=\int_\limits0^1 0 d x+\int_1^{\sqrt{2}} 1 d x+\int_\limits{\sqrt{2}}^{\sqrt{3}} 3 d x+$$</p>
<p>$$
\begin{aligned}
& \int_\limits{\sqrt{3}}^2 4 d x+\int_\limits2^{\sqrt{5}} 6 d x+\int_\limits{\sqrt{5}}^{\sqrt{6}} 7 d x+\int_\limits{\sqrt{6}}^{\sqrt{7}} 9 d x+\int_\limits{\sqrt{7}}^{\sqrt{8}} 10 d x+\int_\limits{\sqrt{8}}^3 12 d x \\
& =31-6 \sqrt{2}-\sqrt{3}-\sqrt{5}-\sqrt{7}-2 \sqrt{6} \\
& \Rightarrow a=31, b=-6, c=-2 \\
& \Rightarrow a+b+c=23
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-6th-april-evening-shift
|
lvc57ayt
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>$$\int_\limits0^{\pi / 4} \frac{\cos ^2 x \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x \text { is equal to }$$</p>
|
[{"identifier": "A", "content": "1/9"}, {"identifier": "B", "content": "1/6"}, {"identifier": "C", "content": "1/3"}, {"identifier": "D", "content": "1/12"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& \int_\limits0^{\pi / 4} \frac{\cos ^2 x \cdot \sin ^2 x}{\left(\cos ^3 x+\sin ^3 x\right)^2} d x \\
& =\int_\limits0^{\pi / 4} \frac{\tan ^2 x \cdot \sec ^2 x}{\left(1+\tan ^3 x\right)^2} d x
\end{aligned}$$</p>
<p>Let $$\tan x=t$$</p>
<p>$$\int_\limits0^1 \frac{t^2 d t}{\left(1+t^3\right)^2}$$</p>
<p>Let $$1+t^3=\mathrm{z}$$</p>
<p>$$\begin{gathered}
3 t^2 d t=d z \\
\frac{1}{3} \int_\limits1^2 \frac{d z}{z^2}=\left.\frac{1}{3}\left(-\frac{1}{z}\right)\right|_1 ^2 \\
=-\frac{1}{3}\left(\frac{1}{2}-1\right)=\frac{1}{6}\end{gathered}$$</p>
|
mcq
|
jee-main-2024-online-6th-april-morning-shift
|
lvc58e3u
|
maths
|
definite-integration
|
properties-of-definite-integration
|
<p>Let $$r_k=\frac{\int_0^1\left(1-x^7\right)^k d x}{\int_0^1\left(1-x^7\right)^{k+1} d x}, k \in \mathbb{N}$$. Then the value of $$\sum_\limits{k=1}^{10} \frac{1}{7\left(r_k-1\right)}$$ is equal to _________.</p>
|
[]
| null |
65
|
<p>$$r_k=\frac{I_a}{I_b} \text {, where } I_a=\int_0^1\left(1-x^7\right)^k d x$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lwd3wwxo/4611eb8f-9da2-472d-abed-83945d5d0d49/093f3fc0-15a2-11ef-9ea4-3bb319c2f90f/file-1lwd3wwxp.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lwd3wwxo/4611eb8f-9da2-472d-abed-83945d5d0d49/093f3fc0-15a2-11ef-9ea4-3bb319c2f90f/file-1lwd3wwxp.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 Morning Shift Mathematics - Definite Integration Question 1 English Explanation"></p>
<p>$$\begin{aligned}
& \left.=\left(1-x^7\right)^{k+1} \cdot x\right]_0^1-\int_0^1(k+1)\left(1-x^7\right)^k\left(-7 x^6\right) \cdot x d x \\
& =-7(k+1) \int_0^1\left(1-x^7\right)^k\left(-1+1-x^7\right) \\
& I_b=-7(k+1)\left[-l_a+l_b\right] \\
& \Rightarrow r_k=\frac{l_a}{I_b}=\frac{7 k+8}{7 k+7}=1+\frac{1}{7(k+1)} \\
& \frac{1}{7\left(r_k-1\right)}=(k+1) \\
& \Rightarrow \sum_{r=-1}^{10} \frac{1}{7\left(r_K-1\right)}=\sum_{r=1}^{10}(k+1)=\frac{11.12}{2}-1=65
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-6th-april-morning-shift
|
fRLdYJcHGpcbyscR
|
maths
|
differential-equations
|
formation-of-differential-equations
|
The differential equation for the family of circle $${x^2} + {y^2} - 2ay = 0,$$ where a is an arbitrary constant is :
|
[{"identifier": "A", "content": "$$\\left( {{x^2} + {y^2}} \\right)y' = 2xy$$ "}, {"identifier": "B", "content": "$$2\\left( {{x^2} + {y^2}} \\right)y' = xy$$"}, {"identifier": "C", "content": "$$\\left( {{x^2} - {y^2}} \\right)y' =2 xy$$"}, {"identifier": "D", "content": "$$2\\left( {{x^2} - {y^2}} \\right)y' = xy$$"}]
|
["C"]
| null |
$${x^2} + {y^2} - 2ay = 0\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$$
<br><br>Differentiate,
<br><br>$$2x + 2y{{dy} \over {dx}} - 2a{{dy} \over {dx}} = 0$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow a = {{x + yy'} \over {y'}}$$
<br><br>Put in $$(1),$$ $${x^2} + {y^2} - 2\left( {{{x + yy'} \over {y'}}} \right) = y = 0$$
<br><br>$$ \Rightarrow \left( {{x^2} + {y^2}} \right)y' - 2xy - 2{y^2}y' = 0$$
<br><br>$$ \Rightarrow \left( {{x^2} - {y^2}} \right)y' = 2xy$$
|
mcq
|
aieee-2004
|
Po1lw9JbsWomUyZs
|
maths
|
differential-equations
|
formation-of-differential-equations
|
The differential equation of all circles passing through the origin and having their centres on the $$x$$-axis is :
|
[{"identifier": "A", "content": "$${y^2} = {x^2} + 2xy{{dy} \\over {dx}}$$ "}, {"identifier": "B", "content": "$${y^2} = {x^2} - 2xy{{dy} \\over {dx}}$$ "}, {"identifier": "C", "content": "$${x^2} = {y^2} + xy{{dy} \\over {dx}}$$"}, {"identifier": "D", "content": "$${x^2} = {y^2} + 3xy{{dy} \\over {dx}}$$"}]
|
["A"]
| null |
General equation of circles passing through origin
<br><br>and having their centres on the $$x$$-axis is
<br><br>$${x^2} + {y^2} + 2gx = 0\,\,\,\,....\left( i \right)$$
<br><br>On differentiating $$w.r.t.x,$$ we get
<br><br>$$2x + 2y.{{dy} \over {dx}} + 2g = 0$$
<br><br>$$ \Rightarrow g = - \left( {x + y{{dy} \over {dx}}} \right)$$
<br><br>$$\therefore$$ equation $$(i)$$ be
<br><br>$${x^2} + {y^2} + 2\left\{ { - \left( {x + y{{dy} \over {dx}}} \right)} \right\}x = 0$$
<br><br>$$ \Rightarrow {x^2} + {y^2} - 2{x^2} - 2x{{dy} \over {dx}}.y = 0$$
<br><br>$$ \Rightarrow {y^2} = {x^2} + 2xy{{dy} \over {dx}}$$
|
mcq
|
aieee-2007
|
YmDJDelJwPq7xlxD
|
maths
|
differential-equations
|
formation-of-differential-equations
|
The differential equation which represents the family of curves $$y = {c_1}{e^{{c_2}x}},$$ where $${c_1}$$ , and $${c_2}$$ are arbitrary constants, is
|
[{"identifier": "A", "content": "$$y'' = y'y$$ "}, {"identifier": "B", "content": "$$yy'' = y'$$ "}, {"identifier": "C", "content": "$$yy'' = {\\left( {y'} \\right)^2}$$ "}, {"identifier": "D", "content": "$$y' = {y^2}$$ "}]
|
["C"]
| null |
We have $$y = {c_1}{e^{{c_2}x}}$$
<br><br>$$ \Rightarrow y' = {c_1}{c_2}{e^{{c_2}x}} = {c_2}y$$
<br><br>$$ \Rightarrow {{y'} \over y} = {c_2}$$
<br><br>$$ \Rightarrow {{y''y\left( {y'} \right){}^2} \over {{y^2}}} = 0$$
<br><br>$$ \Rightarrow y''y = {\left( {y'} \right)^2}$$
|
mcq
|
aieee-2009
|
9MRXujl4STa4xPl5c4rEH
|
maths
|
differential-equations
|
formation-of-differential-equations
|
The differential equation representing the family of ellipse having foci eith on the x-axis or on the $$y$$-axis, center at the origin and passing through the point (0, 3) is :
|
[{"identifier": "A", "content": "xy y'' + x (y')<sup>2</sup> $$-$$ y y' = 0"}, {"identifier": "B", "content": "x + y y'' = 0"}, {"identifier": "C", "content": "xy y'+ y<sup>2</sup> $$-$$ 9 = 0"}, {"identifier": "D", "content": "xy y' $$-$$ y<sup>2</sup> + 9 = 0"}]
|
["D"]
| null |
Equation of ellipse,
<br><br>$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$
<br><br>As ellipse passes through (0, 3)
<br><br>$$\therefore\,\,\,$$ $${{{0^2}} \over {{a^2}}} + {{{3^2}} \over {{b^2}}} = 1$$
<br><br>$$ \Rightarrow $$ b<sup>2</sup> = 9
<br><br>$$\therefore\,\,\,$$ Equation of ellipse becomes,
<br><br>$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9} = 1$$
<br><br>Differentiating w.r.t x, we get,
<br><br>$${{2x} \over {a{}^2}}$$ + $${{2y} \over 9}$$ . $${{dy} \over {dx}} = 0$$
<br><br>$$ \Rightarrow $$ $${x \over {{a^2}}}$$ = $$-$$ $${y \over 9}.{{dy} \over {da}}$$
<br><br>$$ \Rightarrow $$ $${x \over {{a^2}}} = - {y \over 9}.y'......$$ (1)
<br><br>We got earlier,
<br><br>$${{{x^2}} \over {{a^2}}} + {{{y^2}} \over 9}$$ = 1
<br><br>$$ \Rightarrow $$ $${x \over {{a^2}}}.x + {{{y^2}} \over 9} = 1$$
<br><br>putting value of equation (1) here,
<br><br>$$ {-{y\,y'} \over 9}.x + {{{y^2}} \over 9} = 1$$
<br><br>$$ \Rightarrow $$ $$-$$ xyy' + y<sup>2</sup> = 9
<br><br>$$ \Rightarrow $$ xyy' $$-$$ y<sup>2</sup> + 9 = 0
|
mcq
|
jee-main-2018-online-16th-april-morning-slot
|
Wv8Zsmh00NPdsxTb9E7k9k2k5hk6uoo
|
maths
|
differential-equations
|
formation-of-differential-equations
|
The differential equation of the family of
curves, x<sup>2</sup> = 4b(y + b), b $$ \in $$ R, is :
|
[{"identifier": "A", "content": "x(y')<sup>2</sup> = x \u2013 2yy'"}, {"identifier": "B", "content": "x(y')<sup>2</sup> = 2yy' \u2013 x"}, {"identifier": "C", "content": "xy\" = y'"}, {"identifier": "D", "content": "x(y')<sup>2</sup> = x + 2yy'"}]
|
["D"]
| null |
x<sup>2</sup> = 4b(y + b)
<br><br>$$ \Rightarrow $$ 2x = 4by'
<br><br>$$ \Rightarrow $$ b = $${x \over {2y'}}$$
<br><br>$$ \therefore $$ differential equation is
<br><br>x<sup>2</sup> = 4.y.$${x \over {2y'}}$$ + 4$${\left( {{x \over {2y'}}} \right)^2}$$
<br><br>$$ \Rightarrow $$ x<sup>2</sup> = $${{2xy} \over {y'}}$$ + $${{{x^2}} \over {{{\left( {y'} \right)}^2}}}$$
<br><br>$$ \Rightarrow $$ x = $${{2y} \over {y'}}$$ + $${x \over {{{\left( {y'} \right)}^2}}}$$
<br><br>$$ \Rightarrow $$ x(y')<sup>2</sup> = x + 2yy'
|
mcq
|
jee-main-2020-online-8th-january-evening-slot
|
FT0YVlJ8yJJqRUDqM9jgy2xukg0c7il3
|
maths
|
differential-equations
|
formation-of-differential-equations
|
If $$y = \left( {{2 \over \pi }x - 1} \right) cosec\,x$$ is the solution of the
differential equation, <br/><br>$${{dy} \over {dx}} + p\left( x \right)y = {2 \over \pi } cosec\,x$$,<br/><br/>
$$0 < x < {\pi \over 2}$$, then the function p(x) is equal to :</br>
|
[{"identifier": "A", "content": "cot x"}, {"identifier": "B", "content": "sec x"}, {"identifier": "C", "content": "tan x"}, {"identifier": "D", "content": "cosec x"}]
|
["A"]
| null |
$$y = \left( {{2 \over \pi }x - 1} \right) cosec\,x$$
<br>Differentiate w.r.t x
<br><br>$${{dy} \over {dx}} = {2 \over \pi }$$cosec x - $$\left( {{2 \over \pi }x - 1} \right)$$cosec x.cot x
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}}$$ + $$\left( {{2 \over \pi }x - 1} \right)$$cosec x.cot x = $${2 \over \pi }$$cosec x
<br><br>$$ \Rightarrow $$ $${{dy} \over {dx}}$$ + ycot x = $${2 \over \pi }$$cosec x
<br><br>Compare this differential equation with given differential equation, we get
<br><br>p(x) = cot x
|
mcq
|
jee-main-2020-online-6th-september-evening-slot
|
Fh1ueAAgvs7uHquWTM1klrk5zd2
|
maths
|
differential-equations
|
formation-of-differential-equations
|
If a curve y = f(x) passes through the point (1, 2) and satisfies $$x {{dy} \over {dx}} + y = b{x^4}$$, then for what value of b, $$\int\limits_1^2 {f(x)dx = {{62} \over 5}} $$?
|
[{"identifier": "A", "content": "$${{31} \\over 5}$$"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "$${{62} \\over 5}$$"}]
|
["B"]
| null |
$${{dy} \over {dx}} + {y \over x} = b{x^3}$$, $$I.F. = {e^{\int {{{dx} \over x}} }} = x$$<br><br>$$ \therefore $$ $$yx = \int {b{x^4}dx} = {{b{x^5}} \over 5} + C$$<br><br>Passes through (1, 2), we get<br><br>$$2 = {b \over 5} + C$$ ........ (i)<br><br>Also, $$\int\limits_1^2 {\left( {{{b{x^4}} \over 5} + {c \over x}} \right)dx = {{62} \over 5}} $$<br><br>$$ \Rightarrow {b \over {25}} \times 32 + C\ln 2 - {b \over {25}} = {{62} \over 5} \Rightarrow C = 0$$ & $$b = 10$$
|
mcq
|
jee-main-2021-online-24th-february-evening-slot
|
9Fv9AhxXoba7dLVrr21kmlhwgmb
|
maths
|
differential-equations
|
formation-of-differential-equations
|
The differential equation satisfied by the system of parabolas <br/><br/>y<sup>2</sup> = 4a(x + a) is :
|
[{"identifier": "A", "content": "$$y{\\left( {{{dy} \\over {dx}}} \\right)^2} - 2x\\left( {{{dy} \\over {dx}}} \\right) - y = 0$$"}, {"identifier": "B", "content": "$$y{\\left( {{{dy} \\over {dx}}} \\right)^2} - 2x\\left( {{{dy} \\over {dx}}} \\right) + y = 0$$"}, {"identifier": "C", "content": "$$y{\\left( {{{dy} \\over {dx}}} \\right)^2} + 2x\\left( {{{dy} \\over {dx}}} \\right) - y = 0$$"}, {"identifier": "D", "content": "$$y\\left( {{{dy} \\over {dx}}} \\right) + 2x\\left( {{{dy} \\over {dx}}} \\right) - y = 0$$"}]
|
["C"]
| null |
$${y^2} = 4ax + 4{a^2}$$<br><br>differentiate with respect to x<br><br>$$ \Rightarrow 2y{{dy} \over {dx}} = 4a$$<br><br>$$ \Rightarrow a = \left( {{y \over 2}{{dy} \over {dx}}} \right)$$<br><br>So, required differential equation is <br><br>$${y^2} = \left( {4 \times {y \over 2}{{dy} \over {dx}}} \right)x + 4{\left( {{y \over 2}{{dy} \over {dx}}} \right)^2}$$<br><br>$$ \Rightarrow {y^2}{\left( {{{dy} \over {dx}}} \right)^2} + 2xy\left( {{{dy} \over {dx}}} \right) - {y^2} = 0$$<br><br>$$ \Rightarrow y{\left( {{{dy} \over {dx}}} \right)^2} + 2x\left( {{{dy} \over {dx}}} \right) - y = 0$$
|
mcq
|
jee-main-2021-online-18th-march-morning-shift
|
1krzrdrnv
|
maths
|
differential-equations
|
formation-of-differential-equations
|
Let a curve y = f(x) pass through the point (2, (log<sub>e</sub>2)<sup>2</sup>) and have slope $${{2y} \over {x{{\log }_e}x}}$$ for all positive real value of x. Then the value of f(e) is equal to ______________.
|
[]
| null |
1
|
$$y' = {{2y} \over {x\ln x}}$$<br><br>$$ \Rightarrow {{dy} \over y} = {{2dx} \over {x\ln x}}$$<br><br>$$ \Rightarrow \ln |y| = 2\ln |\ln x| + C$$<br><br>put x = 2, y = (ln2)<sup>2</sup><br><br>$$\Rightarrow$$ c = 0<br><br>$$\Rightarrow$$ y = (lnx)<sup>2</sup><br><br>$$\Rightarrow$$ f(e) = 1
|
integer
|
jee-main-2021-online-25th-july-evening-shift
|
1ktepvi10
|
maths
|
differential-equations
|
formation-of-differential-equations
|
If $${y^{1/4}} + {y^{ - 1/4}} = 2x$$, and <br/><br/>$$({x^2} - 1){{{d^2}y} \over {d{x^2}}} + \alpha x{{dy} \over {dx}} + \beta y = 0$$, then | $$\alpha$$ $$-$$ $$\beta$$ | is equal to __________.
|
[]
| null |
17
|
$${y^{{1 \over 4}}} + {1 \over {{y^{{1 \over 4}}}}} = 2x$$<br><br>$$ \Rightarrow {\left( {{y^{{1 \over 4}}}} \right)^2} - 2x{y^{\left( {{1 \over 4}} \right)}} + 1 = 0$$<br><br>$$ \Rightarrow {y^{{1 \over 4}}} = x + \sqrt {{x^2} - 1} $$ or $$x - \sqrt {{x^2} - 1} $$<br><br>So, $${1 \over 4}{1 \over {{y^{{3 \over 4}}}}}{{dy} \over {dx}} = 1 + {x \over {\sqrt {{x^2} - 1} }}$$<br><br>$$ \Rightarrow {1 \over 4}{1 \over {{y^{{3 \over 4}}}}}{{dy} \over {dx}} = {{{y^{{1 \over 4}}}} \over {\sqrt {{x^2} - 1} }}$$<br><br>$$ \Rightarrow {{dy} \over {dx}} = {{4y} \over {\sqrt {{x^2} - 1} }}$$ .... (1)<br><br>Hence, $${{{d^2}y} \over {d{x^2}}} = 4{{\left( {\sqrt {{x^2} - 1} } \right)y' - {{yx} \over {\sqrt {{x^2} - 1} }}} \over {{x^2} - 1}}$$<br><br>$$ \Rightarrow ({x^2} - 1)y'' = 4{{({x^2} - 1)y' - xy} \over {\sqrt {{x^2} - 1} }}$$<br><br>$$ \Rightarrow ({x^2} - 1)y'' = 4\left( {\sqrt {{x^2} - 1} y' - {{xy} \over {\sqrt {{x^2} - 1} }}} \right)$$<br><br>$$ \Rightarrow ({x^2} - 1)y'' = 4\left( {4y - {{xy'} \over 4}} \right)$$ (from I)<br><br>$$ \Rightarrow ({x^2} - 1)y'' + xy' - 16y = 0$$<br><br>So, | $$\alpha$$ $$-$$ $$\beta$$ | = 17
|
integer
|
jee-main-2021-online-27th-august-morning-shift
|
1ktfzg9uq
|
maths
|
differential-equations
|
formation-of-differential-equations
|
A differential equation representing the family of parabolas with axis parallel to y-axis and whose length of latus rectum is the distance of the point (2, $$-$$3) from the line 3x + 4y = 5, is given by :
|
[{"identifier": "A", "content": "$$10{{{d^2}y} \\over {d{x^2}}} = 11$$"}, {"identifier": "B", "content": "$$11{{{d^2}x} \\over {d{y^2}}} = 10$$"}, {"identifier": "C", "content": "$$10{{{d^2}x} \\over {d{y^2}}} = 11$$"}, {"identifier": "D", "content": "$$11{{{d^2}y} \\over {d{x^2}}} = 10$$"}]
|
["D"]
| null |
Length of latus rectum<br><br>$$= {{|3(2) + 4( - 3) - 5|} \over 5} = {{11} \over 5}$$<br><br>$${(x - h)^2} = {{11} \over 5}(y - k)$$<br><br>differentiate w.r.t. 'x' :-<br><br>$$2(x - h) = {{11} \over 5}{{dy} \over {dx}}$$<br><br>again differentiate<br><br>$$2 = {{11} \over 5}{{{d^2}y} \over {d{x^2}}}$$<br><br>$${{11{d^2}y} \over {d{x^2}}} = 10$$
|
mcq
|
jee-main-2021-online-27th-august-evening-shift
|
1l57o2hul
|
maths
|
differential-equations
|
formation-of-differential-equations
|
<p>Let $${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}$$, where a, b, c are constants, represent a circle passing through the point (2, 5). Then the shortest distance of the point (11, 6) from this circle is :</p>
|
[{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "8"}, {"identifier": "C", "content": "7"}, {"identifier": "D", "content": "5"}]
|
["B"]
| null |
<p>$${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}$$</p>
<p>$$ = bx\,dy + cy\,dy + a\,dy = ax\,dx - by\,dx + a\,dx$$</p>
<p>$$ = cy\,dy + a\,dy - ax\,dx - a\,dx + b(x\,dy + y\,dx) = 0$$</p>
<p>$$ = c\int {y\,dy + a\int {x\,dx - a\int {dx + b\int {d(xy) = 0} } } } $$</p>
<p>$$ = {{c{y^2}} \over 2} + ay - {{a{x^2}} \over 2} - ax + bxy = k$$</p>
<p>$$ = a{x^2} - c{y^2} + 2ax - 2ay - 2bxy = k$$</p>
<p>Above equation is circle</p>
<p>$$\Rightarrow$$ a = $$-$$ c and b = 0</p>
<p>$$a{x^2} + a{y^2} + 2ax - 2ay = k$$</p>
<p>$$ \Rightarrow {x^2} + {y^2} + 2x - 2y = \lambda \,\,\,\,\,\,\,\left[ {\lambda = {k \over a}} \right]$$</p>
<p>Passes through (2, 5)</p>
<p>$$4 + 25 + 4 - 10 = \lambda \Rightarrow \lambda = 23$$</p>
<p>Circle $$ \equiv {x^2} + {y^2} + 2x - 2y - 23 = 0$$</p>
<p>Centre ($$-$$1, 1) $$r = \sqrt {{{( - 1)}^2} + {1^2} + 23} = 5$$</p>
<p>Shortest distance of $$(11,6) = \sqrt {{{12}^2} + {5^2}} - 5$$</p>
<p>$$ = 13 - 5$$</p>
<p>$$ = 8$$</p>
|
mcq
|
jee-main-2022-online-27th-june-morning-shift
|
1l5vzx4yi
|
maths
|
differential-equations
|
formation-of-differential-equations
|
<p>Let $${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}},\,a,b,c \in R$$, represents a circle with center ($$\alpha$$, $$\beta$$). Then, $$\alpha$$ + 2$$\beta$$ is equal to :</p>
|
[{"identifier": "A", "content": "$$-$$1"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}]
|
["C"]
| null |
<p>Given,</p>
<p>$${{dy} \over {dx}} = {{ax - by + a} \over {bx + cy + a}}$$</p>
<p>$$ \Rightarrow bxdy + cydy + ady = axdx - bydx + adx$$</p>
<p>$$ \Rightarrow bxdy + bydx + (cy + a)dy = (ax + a)dx$$</p>
<p>$$ \Rightarrow b(xdy + ydx) + (cy + a)dy = (ax + a)dx$$</p>
<p>$$ \Rightarrow bd(xy) + (cy + a)dy = (ax + a)dx$$</p>
<p>Integrating both sides, we get</p>
<p>$$ \Rightarrow b\int {d(xy) + \int {(cy + a)dy = \int {(ax + a)dx} } } $$</p>
<p>$$ \Rightarrow b\,.\,xy + c\,.\,{{{y^2}} \over 2} + ay = {{a{x^2}} \over 2} + ax + k$$</p>
<p>$$ \Rightarrow {{a{x^2}} \over 2} - {{c{y^2}} \over 2} - bxy + ax - ay + k = 0$$</p>
<p>For equation of circle,</p>
<p>Coefficient of x<sup>2</sup> = Coefficient of y<sup>2</sup></p>
<p>$$\therefore$$ $${a \over 2} = - {c \over 2}$$</p>
<p>$$ \Rightarrow a = - c$$</p>
<p>And coefficient of $$xy = 0$$</p>
<p>$$\therefore$$ $$ - b = 0$$</p>
<p>$$ \Rightarrow b = 0$$</p>
<p>$$\therefore$$ Circle equation becomes,</p>
<p>$${{a{x^2}} \over 2} + {{a{y^2}} \over 2} + ax - ay + k = 0$$</p>
<p>$$ \Rightarrow {x^2} + {y^2} + 2x - 2y + {{2k} \over a} = 0$$</p>
<p>$$\therefore$$ Center $$ = ( - g, - f) = ( - 1,1) = (\alpha ,\beta )$$</p>
<p>$$\therefore$$ $$\alpha = - 1$$ and $$\beta = 1$$</p>
<p>$$\therefore$$ $$\alpha + 2\beta = - 1 + 2 \times (1) = 1$$</p>
|
mcq
|
jee-main-2022-online-30th-june-morning-shift
|
1l6f1ihu7
|
maths
|
differential-equations
|
formation-of-differential-equations
|
<p>Let a smooth curve $$y=f(x)$$ be such that the slope of the tangent at any point $$(x, y)$$ on it is directly proportional to $$\left(\frac{-y}{x}\right)$$. If the curve passes through the points $$(1,2)$$ and $$(8,1)$$, then $$\left|y\left(\frac{1}{8}\right)\right|$$ is equal to</p>
|
[{"identifier": "A", "content": "$$2 \\log _{e} 2$$"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "$$4 \\log _{e} 2$$"}]
|
["B"]
| null |
<p>$${{dy} \over {dx}} \propto {{ - y} \over x}$$</p>
<p>$${{dy} \over {dx}} = {{ - ky} \over x} \Rightarrow \int {{{dy} \over y} = - K\int {{{dx} \over x}} } $$</p>
<p>$$\ln |y| = - K\ln |x| + C$$</p>
<p>If the above equation satisfy (1, 2) and (8, 1)</p>
<p>$$\ln 2 = - K \times 0 + C \Rightarrow C = \ln 2$$</p>
<p>$$\ln 1 = - K\ln 8 + \ln 2 \Rightarrow K = {1 \over 3}$$</p>
<p>So, at $$x = {1 \over 8}$$</p>
<p>$$\ln |y| = - {1 \over 3}\ln \left( {{1 \over 8}} \right) + \ln 2 = 2\ln 2$$</p>
<p>$$|y| = 4$$</p>
|
mcq
|
jee-main-2022-online-25th-july-evening-shift
|
1l6gjjaoy
|
maths
|
differential-equations
|
formation-of-differential-equations
|
<p>Let a curve $$y=y(x)$$ pass through the point $$(3,3)$$ and the area of the region under this curve, above the $$x$$-axis and between the abscissae 3 and $$x(>3)$$ be $$\left(\frac{y}{x}\right)^{3}$$. If this curve also passes through the point $$(\alpha, 6 \sqrt{10})$$ in the first quadrant, then $$\alpha$$ is equal to ___________.</p>
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[]
| null |
6
|
<p>$$\int\limits_3^x {f(x)dx = {{\left( {{{f(x)} \over x}} \right)}^3}} $$</p>
<p>$${x^3}\,.\,\int\limits_3^x {f(x)dx = {f^3}(x)} $$</p>
<p>Differentiate w.r.t. x</p>
<p>$${x^3}f(x) + 3{x^2}\,.\,{{{f^3}(x)} \over {{x^3}}} = 3{f^2}(x)f'(x)$$</p>
<p>$$ \Rightarrow 3{y^2}{{dy} \over {dx}} = {x^3}y + {{3{y^3}} \over x}$$</p>
<p>$$3xy{{dy} \over {dx}} = {x^4} + 3{y^2}$$</p>
<p>Let $${y^2} = t$$</p>
<p>$${3 \over 2}{{dt} \over {dx}} = {x^3} + {{3t} \over x}$$</p>
<p>$${{dt} \over {dx}} - {{2t} \over x} = {{2{x^3}} \over 3}$$</p>
<p>$$I.F. = \,.\,{e^{\int { - {2 \over x}dx} }} = {1 \over {{x^2}}}$$</p>
<p>Solution of differential equation</p>
<p>$$t\,.\,{1 \over {{x^2}}} = \int {{2 \over 3}x\,dx} $$</p>
<p>$${{{y^2}} \over {{x^2}}} = {{{x^2}} \over 3} + C$$</p>
<p>$${y^2} = {{{x^4}} \over 3} + C{x^2}$$</p>
<p>Curve passes through $$(3,3) \Rightarrow C = - 2$$</p>
<p>$${y^2} = {{{x^4}} \over 3} - 2{x^2}$$</p>
<p>Which passes through $$\left( {\alpha ,6\sqrt {10} } \right)$$</p>
<p>$${{{\alpha ^4} - 6{\alpha ^2}} \over 3} = 360$$</p>
<p>$${\alpha ^4} - 6{\alpha ^2} - 1080 = 0$$</p>
<p>$$\alpha = 6$$</p>
|
integer
|
jee-main-2022-online-26th-july-morning-shift
|
1l6nn544i
|
maths
|
differential-equations
|
formation-of-differential-equations
|
<p>The differential equation of the family of circles passing through the points $$(0,2)$$ and $$(0,-2)$$ is :</p>
|
[{"identifier": "A", "content": "$$2 x y \\frac{d y}{d x}+\\left(x^{2}-y^{2}+4\\right)=0$$"}, {"identifier": "B", "content": "$$2 x y \\frac{d y}{d x}+\\left(x^{2}+y^{2}-4\\right)=0$$"}, {"identifier": "C", "content": "$$2 x y \\frac{d y}{d x}+\\left(y^{2}-x^{2}+4\\right)=0$$"}, {"identifier": "D", "content": "$$2 x y \\frac{d y}{d x}-\\left(x^{2}-y^{2}+4\\right)=0$$"}]
|
["A"]
| null |
<p>Family of circles passing through the points (0, 2) and (0, $$-$$2)</p>
<p>$${x^2} + (y - 2)(y + 2) + \lambda x = 0,\,\lambda \in R$$</p>
<p>$${x^2} + {y^2} + \lambda x - 4 = 0$$ ...... (1)</p>
<p>Differentiate w.r.t x</p>
<p>$$2x + 2y{{dy} \over {dx}} + \lambda = 0$$ ....... (2)</p>
<p>Using (1) and (2), eliminate $$\lambda$$</p>
<p>$${x^2} + {y^2} - \left( {2x + 2y{{dy} \over {dx}}} \right)x - 4 = 0$$</p>
<p>$$2xy{{dy} \over {dx}} + {x^2} - {y^2} + 4 = 0$$</p>
|
mcq
|
jee-main-2022-online-28th-july-evening-shift
|
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