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stringclasses 32
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stringclasses 178
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stringlengths 26
9.64k
| options
stringlengths 2
1.63k
| correct_option
stringclasses 5
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|---|---|---|---|---|---|---|---|---|---|---|
1l6rdzvet
|
maths
|
limits-continuity-and-differentiability
|
continuity
|
<p>$$
\text { Let the function } f(x)=\left\{\begin{array}{cl}
\frac{\log _{e}(1+5 x)-\log _{e}(1+\alpha x)}{x} & ;\text { if } x \neq 0 \\
10 & ; \text { if } x=0
\end{array} \text { be continuous at } x=0 .\right.
$$</p>
<p>Then $$\alpha$$ is equal to</p>
|
[{"identifier": "A", "content": "10"}, {"identifier": "B", "content": "$$-$$10"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "$$-$$5"}]
|
["D"]
| null |
<p>$$f(x)$$ is continuous at $$x = 0$$</p>
<p>$$\therefore$$ $$f(0) = \mathop {\lim }\limits_{x \to 0} f(x)$$</p>
<p>$$ \Rightarrow 10 = \mathop {\lim }\limits_{x \to 0} {{{{\log }_e}(1 + 5x) - {{\log }_e}(1 + \alpha x)} \over x}$$</p>
<p>$$ = \mathop {\lim }\limits_{x \to 0} {{\log (1 + 5x)} \over {5x}} \times 5 - {{{{\log }_e}(1 + \alpha x)} \over {\alpha x}} \times \alpha $$</p>
<p>$$ = 1 \times 5 - \alpha $$</p>
<p>$$ \Rightarrow \alpha = 5 - 10 = - 5$$</p>
|
mcq
|
jee-main-2022-online-29th-july-evening-shift
|
1ldu4jonv
|
maths
|
limits-continuity-and-differentiability
|
continuity
|
<p>If the function $$f(x) = \left\{ {\matrix{
{(1 + |\cos x|)^{\lambda \over {|\cos x|}}} & , & {0 < x < {\pi \over 2}} \cr
\mu & , & {x = {\pi \over 2}} \cr
e^{{{\cot 6x} \over {{}\cot 4x}}} & , & {{\pi \over 2} < x < \pi } \cr
} } \right.$$</p>
<p>is continuous at $$x = {\pi \over 2}$$, then $$9\lambda + 6{\log _e}\mu + {\mu ^6} - {e^{6\lambda }}$$ is equal to</p>
|
[{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "2e$$^4$$ + 8"}]
|
["B"]
| null |
$$
\mathop {\lim }\limits_{x \rightarrow \frac{\pi^{-}}{2}}(1+|\cos x|)^ \frac{\lambda}{|\cos x|}=e^\lambda
$$
<br/><br/>
And,
<br/><br/>$$
\begin{aligned}
&\mathop {\lim }\limits_{x \rightarrow \frac{\pi^{+}}{2}} e^{\frac{\cot 6 x}{\cot 4 x}}\\\\ &= \mathop {\lim }\limits_{x \to {{{\pi ^ + }} \over 2}} {e^{{{\sin 4x.\cos 6x} \over {\sin 6x.\cos 4x}}}} \\\\
& =e^{\frac{2}{3}}
\end{aligned}
$$
<br/><br/>
Also, $$
f(\pi / 2)=\mu
$$
<br/><br/>For continuous function,
<br/><br/>$ \mathrm{e}^{2 / 3}=\mathrm{e}^\lambda=\mu$
<br/><br/>$$
\lambda=\frac{2}{3}, \mu=\mathrm{e}^{2 / 3}
$$
<br/><br/>
$$ \therefore $$ $9 \lambda+6 \ln \mu+\mu^{6}-e^{6 \lambda}$
<br/><br/>
$=6+4+e^{4}-e^{4}=10$
|
mcq
|
jee-main-2023-online-25th-january-evening-shift
|
1lgrgpa4y
|
maths
|
limits-continuity-and-differentiability
|
continuity
|
<p>Let $$[x]$$ be the greatest integer $$\leq x$$. Then the number of points in the interval $$(-2,1)$$, where the function $$f(x)=|[x]|+\sqrt{x-[x]}$$ is discontinuous, is ___________.</p>
|
[]
| null |
2
|
The function $f(x) = |[x]| + \sqrt{x-[x]}$ is composed of two parts : the greatest integer function $[x]$, and the fractional part function $x-[x]$.
<br/><br/>1. The greatest integer function $[x]$ is discontinuous at every integer, since it jumps from one integer to the next without taking any values in between. The absolute value does not affect where this function is continuous or discontinuous. So within the interval $(-2,1)$, $[x]$ is discontinuous at $-2, -1, 0$ and $1$. However, because the interval is open $(-2,1)$, the endpoints $-2$ and $1$ are not included.
<br/><br/>2. The function $\sqrt{x-[x]}$ is the square root of the fractional part of $x$. The fractional part function $x-[x]$ is continuous everywhere, as it always takes a value between $0$ and $1$ (inclusive of $0$, exclusive of $1$). However, the square root function $\sqrt{x}$ is only defined for $x \geq 0$, and so $\sqrt{x-[x]}$ is discontinuous wherever $x-[x] < 0$. This happens exactly at the points where $x$ is a negative integer, as the fractional part of a negative integer is $1$ (considering that the "fractional part" is defined as the part of the number to the right of the decimal point, which for negative numbers works a bit differently). Within the interval $(-2,1)$, this is the case for $-2$ and $-1$. However, since the interval is open $(-2,1)$, the endpoint $-2$ is not included.
<br/><br/>Now, let's analyze the discontinuities within the given interval $(-2,1)$:
<br/><br/>At $x=-1$: $f(-1^{+})=1+0=1$ and $f(-1^{-})=2+1=3$. Since these two values are different, $f(x)$ is discontinuous at $x=-1$.
<br/><br/>At $x=0$: $f(0^{+})=0+0=0$ and $f(0^{-})=1+1=2$. Again, these two values are different, so $f(x)$ is discontinuous at $x=0$.
<br/><br/>At $x=1$: $f(1^{+})=1+0=1$ and $f(1^{-})=0+1=1$. These two values are the same, so $f(x)$ is continuous at $x=1$. However, this point is not within the open interval $(-2,1)$.
<br/><br/>So within the interval $(-2,1)$, the function $f(x) = |[x]| + \sqrt{x-[x]}$ is discontinuous at the points $-1$ and $0$ (2 points in total).
|
integer
|
jee-main-2023-online-12th-april-morning-shift
|
1lguuf2hr
|
maths
|
limits-continuity-and-differentiability
|
continuity
|
<p>Let $$f(x)=\left[x^{2}-x\right]+|-x+[x]|$$, where $$x \in \mathbb{R}$$ and $$[t]$$ denotes the greatest integer less than or equal to $$t$$. Then, $$f$$ is :</p>
|
[{"identifier": "A", "content": "continuous at $$x=0$$, but not continuous at $$x=1$$"}, {"identifier": "B", "content": "continuous at $$x=0$$ and $$x=1$$"}, {"identifier": "C", "content": "continuous at $$x=1$$, but not continuous at $$x=0$$"}, {"identifier": "D", "content": "not continuous at $$x=0$$ and $$x=1$$"}]
|
["C"]
| null |
We have,
<br/><br/>$$\begin{aligned}
f(x) & =\left[x^2-x\right]+|-x+[x]| \\\\
& =[x(x-1)]+\{x\}
\end{aligned}
$$
<br/><br/>$$
f(x)=\left\{\begin{array}{ccc}
x+1 & ; & -0.5 < x < 0 \\
0 & ; & x=0 \\
-1+x & ; & 0 < x <1 \\
0 & ; & x=1 \\
x-1 & ; & 1 < x < 1.5
\end{array}\right.
$$
<br/><br/>At $x=0$,
<br/><br/>$$
\begin{array}{r}
\text { LHL }=\lim\limits_{x \rightarrow 0^{-}} f(x)=1 \\\\
\text { RHL } \lim\limits_{x \rightarrow 0^{+}} f(x)=-1 \\\\
f(0)=0
\end{array}
$$
<br/><br/>$\therefore f(x)$ is not continuous at $x=0$
<br/><br/>At $x=1$
<br/><br/>$$
\begin{aligned}
\mathrm{LHL} & =\lim _{x \rightarrow 1^{-}} f(x)=-1+1=0 \\\\
\mathrm{RHL} & =\lim _{x \rightarrow 1^{+}} f(x)=1-1=0 \\\\
f(1) & =0
\end{aligned}
$$
<br/><br/>$\therefore f(x)$ is continuous at $x=1$
<br/><br/>Hence, $f(x)$ is continuous at $x=1$, but not continuous at $x=0$.
|
mcq
|
jee-main-2023-online-11th-april-morning-shift
|
lsbkoryy
|
maths
|
limits-continuity-and-differentiability
|
continuity
|
Consider the function.
<br/><br/>$$
f(x)=\left\{\begin{array}{cc}
\frac{\mathrm{a}\left(7 x-12-x^2\right)}{\mathrm{b}\left|x^2-7 x+12\right|} & , x<3 \\\\
2^{\frac{\sin (x-3)}{x-[x]}} & , x>3 \\\\
\mathrm{~b} & , x=3,
\end{array}\right.
$$
<br/><br/>where $[x]$ denotes the greatest integer less than or equal to $x$. If $\mathrm{S}$ denotes the set of all ordered pairs (a, b) such that $f(x)$ is continuous at $x=3$, then the number of elements in $\mathrm{S}$ is :
|
[{"identifier": "A", "content": "Infinitely many"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}]
|
["D"]
| null |
<p>$$f\left(3^{-}\right)=\frac{a}{b} \frac{\left(7 x-12-x^2\right)}{\left|x^2-7 x+12\right|} \quad$$ (for $$f(x)$$ to be cont.)</p>
<p>$$\Rightarrow \mathrm{f}\left(3^{-}\right)=\frac{-\mathrm{a}}{\mathrm{b}} \frac{(\mathrm{x}-3)(\mathrm{x}-4)}{(\mathrm{x}-3)(\mathrm{x}-4)} ; \mathrm{x}<3 \Rightarrow \frac{-\mathrm{a}}{\mathrm{b}}$$</p>
<p>Hence $$f\left(3^{-}\right)=\frac{-a}{b}$$</p>
<p>Then $$f\left(3^{+}\right)=2^{\lim _\limits{x \rightarrow 3^{+}}\left(\frac{\sin (x-3)}{x-3}\right)}=2$$ and $$f(3)=b$$.</p>
<p>Hence $$\mathrm{f}(3)=\mathrm{f}\left(3^{+}\right)=\mathrm{f}\left(3^{-}\right)$$</p>
<p>$$\begin{aligned}
& \Rightarrow \mathrm{b}=2=-\frac{\mathrm{a}}{\mathrm{b}} \\
& \mathrm{b}=2, \mathrm{a}=-4
\end{aligned}$$</p>
<p>Hence only 1 ordered pair $$(-4,2)$$.</p>
|
mcq
|
jee-main-2024-online-27th-january-morning-shift
|
jaoe38c1lse5crz0
|
maths
|
limits-continuity-and-differentiability
|
continuity
|
<p>Let $$g(x)$$ be a linear function and $$f(x)=\left\{\begin{array}{cl}g(x) & , x \leq 0 \\ \left(\frac{1+x}{2+x}\right)^{\frac{1}{x}} & , x>0\end{array}\right.$$, is continuous at $$x=0$$. If $$f^{\prime}(1)=f(-1)$$, then the value $$g(3)$$ is</p>
|
[{"identifier": "A", "content": "$$\\log _e\\left(\\frac{4}{9}\\right)-1$$\n"}, {"identifier": "B", "content": "$$\\frac{1}{3} \\log _e\\left(\\frac{4}{9 e^{1 / 3}}\\right)$$\n"}, {"identifier": "C", "content": "$$\\log _e\\left(\\frac{4}{9 e^{1 / 3}}\\right)$$\n"}, {"identifier": "D", "content": "$$\\frac{1}{3} \\log _e\\left(\\frac{4}{9}\\right)+1$$"}]
|
["C"]
| null |
<p>Let $$g(x)=a x+b$$</p>
<p>Now function $$\mathrm{f}(\mathrm{x})$$ in continuous at $$\mathrm{x}=0$$</p>
<p>$$\begin{aligned}
& \therefore \lim _\limits{\mathrm{x} \rightarrow 0^{+}} \mathrm{f}(\mathrm{x})=\mathrm{f}(0) \\
& \Rightarrow \lim _\limits{\mathrm{x} \rightarrow 0}\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}}=\mathrm{b} \\
& \Rightarrow 0=\mathrm{b} \\
& \therefore \mathrm{g}(\mathrm{x})=\mathrm{ax}
\end{aligned}$$</p>
<p>Now, for $$\mathrm{x}>0$$</p>
<p>$$\begin{aligned}
& \mathrm{f}^{\prime}(\mathrm{x})=\frac{1}{\mathrm{x}} \cdot\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}-1} \cdot \frac{1}{(2+\mathrm{x})^2} \\
& +\left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right)^{\frac{1}{\mathrm{x}}} \cdot \ln \left(\frac{1+\mathrm{x}}{2+\mathrm{x}}\right) \cdot\left(-\frac{1}{\mathrm{x}^2}\right) \\
& \therefore \mathrm{f}^{\prime}(1)=\frac{1}{9}-\frac{2}{3} \cdot \ln \left(\frac{2}{3}\right) \\
& \text { And } \mathrm{f}(-1)=\mathrm{g}(-1)=-\mathrm{a} \\
& \therefore \mathrm{a}=\frac{2}{3} \ln \left(\frac{2}{3}\right)-\frac{1}{9} \\
& \therefore \mathrm{g}(3)=2 \ln \left(\frac{2}{3}\right)-\frac{1}{3} \\
& =\ln \left(\frac{4}{9 \cdot \mathrm{e}^{1 / 3}}\right)
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-31st-january-morning-shift
|
luy9clcn
|
maths
|
limits-continuity-and-differentiability
|
continuity
|
<p>Let $$f:(0, \pi) \rightarrow \mathbf{R}$$ be a function given by $$f(x)=\left\{\begin{array}{cc}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}, & 0< x<\frac{\pi}{2} \\ \mathrm{a}-8, & x=\frac{\pi}{2} \\ (1+\mid \cot x)^{\frac{\mathrm{b}}{\mathrm{a}}|\tan x|}, & \frac{\pi}{2} < x < \pi\end{array}\right.$$</p>
<p>where $$\mathrm{a}, \mathrm{b} \in \mathbf{Z}$$. If $$f$$ is continuous at $$x=\frac{\pi}{2}$$, then $$\mathrm{a}^2+\mathrm{b}^2$$ is equal to _________.</p>
|
[]
| null |
81
|
<p>$$\lim _\limits{x \rightarrow \frac{\pi^{-}}{2}} f(x)=f\left(\frac{\pi}{2}\right)=\lim _\limits{x \rightarrow \frac{\pi^{+}}{2}} f(x) \text { for continuity at } x=\frac{\pi}{2}$$</p>
<p>$$\begin{aligned}
& \Rightarrow \quad \lim _{x \rightarrow \frac{\pi^{-}}{2}}\left(\frac{8}{7}\right)^{\left(\frac{\tan 8 x}{\tan 7 x}\right)} \quad \text { Let } x=\frac{\pi}{2}-h \\
& \Rightarrow \quad \lim _{h \rightarrow 0}\left(\frac{8}{7}\right)^{\frac{\tan (4 \pi-8 h)}{\tan \left(3 \pi+\frac{\pi}{2}-7 h\right)}}=\lim _{h \rightarrow 0}\left(\frac{8}{7}\right)^{\frac{\tan (-8 h)}{\cot (7 h)}}=\left(\frac{8}{7}\right)^0=1 \\
& \Rightarrow \quad a-8=1 \Rightarrow a=9
\end{aligned}$$</p>
<p>$$\lim _\limits{x \rightarrow \frac{\pi^{+}}{2}}(1+|\cot x|)^{\frac{b}{a}|\tan x|}, x=\frac{\pi}{2}+h$$</p>
<p>$$\lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h}=\lim _\limits{h \rightarrow 0}(1-\tan h)^{-\frac{b}{9} \cot h}$$</p>
<p>$$\begin{aligned}
& =\lim _\limits{h \rightarrow 0}(1-\tan h)^{\left(\frac{-1}{\tan h}\right) \cdot(-\tan h) \cdot\left(\frac{-b}{9} \cot h\right)} \\
& =e^{\frac{b}{9}}=1 \quad \Rightarrow b=0 \\
& \Rightarrow a^2+b^2=81+0=81
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-9th-april-morning-shift
|
lv0vxdd3
|
maths
|
limits-continuity-and-differentiability
|
continuity
|
<p>Let $$f: \mathbf{R} \rightarrow \mathbf{R}$$ be a function given by</p>
<p>$$f(x)= \begin{cases}\frac{1-\cos 2 x}{x^2}, & x < 0 \\ \alpha, & x=0, \\ \frac{\beta \sqrt{1-\cos x}}{x}, & x>0\end{cases}$$</p>
<p>where $$\alpha, \beta \in \mathbf{R}$$. If $$f$$ is continuous at $$x=0$$, then $$\alpha^2+\beta^2$$ is equal to :</p>
|
[{"identifier": "A", "content": "48"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "12"}]
|
["D"]
| null |
<p>$$f(x)=\left\{\begin{array}{cl}
\frac{1-\cos 2 x}{x^2}, & x< \\
\alpha, & x=0 \\
\frac{\beta \sqrt{1-\cos x}}{x}, & x>0
\end{array}\right.$$</p>
<p>$$f(x)$$ is continuous at $$x=0$$</p>
<p>$$\Rightarrow f(0)=\lim _\limits{x \rightarrow 0^{-}} f(x)=\lim _\limits{x \rightarrow 0^{+}} f(x)$$</p>
<p>$$\begin{aligned}
&\begin{aligned}
& \lim _{x \rightarrow 0^{-}} f(x)=\alpha \\
& \lim _{x \rightarrow 0^{-}}\left(\frac{1-\cos 2 x}{x^2}\right)=\alpha \\
& \Rightarrow \lim _{x \rightarrow 0^{-}} \frac{2 \sin ^2 h}{x^2}=\alpha \\
& \Rightarrow \lim _{h \rightarrow 0} \frac{2 \sin ^2}{h^2} h=\alpha \\
& \Rightarrow \alpha=2 \\
&
\end{aligned}\\
&\begin{aligned}
& \text { Also, } \lim _{x \rightarrow 0^{+}} f(x)=f(0) \\
& \Rightarrow \quad \lim _{x \rightarrow 0^{+}} \frac{\beta \sqrt{1-\cos x}}{x}=2
\end{aligned}
\end{aligned}$$</p>
<p>$$\Rightarrow \lim _\limits{h \rightarrow 0} \frac{\beta \sqrt{\frac{1-\cos h}{h^2}} h^2}{h}=2$$</p>
<p>$$\begin{aligned}
& \Rightarrow \quad \frac{\beta}{\sqrt{2}}=2 \\
& \Rightarrow \quad \beta=2 \sqrt{2} \\
& \Rightarrow \quad \alpha^2+\beta^2=4+8 \\
& \quad=12
\end{aligned}$$</p>
<p>$$\therefore$$ Option (1) is correct</p>
|
mcq
|
jee-main-2024-online-4th-april-morning-shift
|
lv2eqvf0
|
maths
|
limits-continuity-and-differentiability
|
continuity
|
<p>If the function</p>
<p>$$f(x)= \begin{cases}\frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}}, & x \neq 0 \\ a \log _e 2 \log _e 3 & , x=0\end{cases}$$</p>
<p>is continuous at $$x=0$$, then the value of $$a^2$$ is equal to</p>
|
[{"identifier": "A", "content": "968"}, {"identifier": "B", "content": "1250"}, {"identifier": "C", "content": "1152"}, {"identifier": "D", "content": "746"}]
|
["C"]
| null |
<p>$$f(x) = \left\{ \matrix{
{{{{72}^x} - {9^x} - {8^x} + 1} \over {\sqrt 2 - \sqrt {1 + \cos x} }},\,x \ne 0 \hfill \cr
a{\log _e}2{\log _e}3\,\,\,\,\,\,,\,\,x = 0 \hfill \cr} \right.$$</p>
<p>$$\because f(x)$$ is continuous at $$x=0$$</p>
<p>$$\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 0} \frac{72^x-9^x-8^x+1}{\sqrt{2}-\sqrt{1+\cos x}} \\
& \lim _{x \rightarrow 0} \frac{\left(9^x-1\right)\left(8^x-1\right)(\sqrt{2}+\sqrt{1+\cos x})}{\frac{(1-\cos x)}{x^2} \times x^2} \\
& =(\ln 9 \cdot \ln 8)(2 \sqrt{2}) \times 2 \\
& =4 \sqrt{2} \times 2 \times 3 \ln 2 \cdot \ln 3 \\
& 24 \sqrt{2} \cdot \ln 2 \cdot \ln 3 \\
& \Rightarrow \quad a=24 \sqrt{2} \\
& \quad a^2=1152
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-4th-april-evening-shift
|
lv3ve5z1
|
maths
|
limits-continuity-and-differentiability
|
continuity
|
<p>For $$\mathrm{a}, \mathrm{b}>0$$, let $$f(x)= \begin{cases}\frac{\tan ((\mathrm{a}+1) x)+\mathrm{b} \tan x}{x}, & x< 0 \\ 3, & x=0 \\ \frac{\sqrt{\mathrm{a} x+\mathrm{b}^2 x^2}-\sqrt{\mathrm{a} x}}{\mathrm{~b} \sqrt{\mathrm{a}} x \sqrt{x}}, & x> 0\end{cases}$$
be a continuous function at $$x=0$$. Then $$\frac{\mathrm{b}}{\mathrm{a}}$$ is equal to :</p>
|
[{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "6"}]
|
["D"]
| null |
<p>$$f(x)=\left\{\begin{array}{cc}
\frac{\tan ((a+1) x)+b \tan x}{x}, & x<0 \\
3 & x=0 \\
\frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}, & x>0
\end{array}\right.$$</p>
<p>$$f(x)$$ is continuous at $$x=0$$</p>
<p>$$\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x) \\
& \lim _{x \rightarrow 0^{-}} f(x)=3 \\
& \Rightarrow \lim _{x \rightarrow 0^{-}} \frac{\tan ((a+1) x)+b+a x}{x}=3 \\
& \Rightarrow a+1+b=3 \\
& \Rightarrow a+b=2 \quad \text{..... (1)}
\end{aligned}$$</p>
<p>also, $$\lim _\limits{x \rightarrow 0^{+}} \frac{\sqrt{a x+b^2 x^2}-\sqrt{a x}}{b \sqrt{a} x \sqrt{x}}=3$$</p>
<p>$$\begin{aligned}
& =\lim _{x \rightarrow 0^{+}} \frac{\sqrt{a h+b^2 h^2}-\sqrt{a h}}{b \sqrt{a} \times h \sqrt{h}}=3 \\
& =\lim _{h \rightarrow 0} \frac{\sqrt{a+b^2 h^2}-\sqrt{a}}{b \sqrt{a} h} \times \frac{\sqrt{a+b^2 h}+\sqrt{a}}{\sqrt{a+b^2 h}+\sqrt{a}}=3 \\
& =\lim _{h \rightarrow 0} \frac{a+b^2 h-a}{b \sqrt{a} h\left(\sqrt{a+b^2 h}+\sqrt{a}\right)}=3 \\
& \Rightarrow \frac{b^2}{b \sqrt{a}(2 \sqrt{a})}=3 \\
& \Rightarrow \frac{b}{2 a}=3 \\
& \Rightarrow \frac{b}{a}=6 \quad \text{..... (2)}
\end{aligned}$$</p>
|
mcq
|
jee-main-2024-online-8th-april-evening-shift
|
lv7v3k5y
|
maths
|
limits-continuity-and-differentiability
|
continuity
|
<p>If the function $$f(x)=\frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3}, x \in \mathbf{R}$$, is continuous at $$x=0$$, then $$f(0)$$ is equal to :</p>
|
[{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "$$-$$2"}, {"identifier": "C", "content": "$$-$$4"}, {"identifier": "D", "content": "2"}]
|
["C"]
| null |
<p>$$\begin{aligned}
& \lim _\limits{x \rightarrow 0} f(x)=f(0) \quad \text { (continuous at } x=0) \\
& \lim _{x \rightarrow 0} \frac{\sin 3 x+\alpha \sin x-\beta \cos 3 x}{x^3}
\end{aligned}$$</p>
<p>For limit to exist $$\beta=0$$</p>
<p>$$\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 0} \frac{\sin 3 x+\alpha \sin x}{x^3} \\
& \Rightarrow \lim _{x \rightarrow 0} \frac{(3+\alpha) \sin x-4 \sin ^3 x}{x^3}
\end{aligned}$$</p>
<p>For limit to exist $$\alpha+3=0 \Rightarrow \alpha=-3$$</p>
<p>$$\Rightarrow \lim _\limits{x \rightarrow 0} \frac{-4 \sin ^3 x}{x^3}=-4=f(0)$$</p>
|
mcq
|
jee-main-2024-online-5th-april-morning-shift
|
lv9s207h
|
maths
|
limits-continuity-and-differentiability
|
continuity
|
<p>Let ,$$f:[-1,2] \rightarrow \mathbf{R}$$ be given by $$f(x)=2 x^2+x+\left[x^2\right]-[x]$$, where $$[t]$$ denotes the greatest integer less than or equal to $$t$$. The number of points, where $$f$$ is not continuous, is :</p>
|
[{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "4"}, {"identifier": "D", "content": "3"}]
|
["C"]
| null |
<p>$$\begin{aligned}
& f(x)=2 x^2+x+\left[x^2\right]-[x]=2 x^2+\left[x^2\right]+\{x\} \\
& f(-1)=2+1+0=3 \\
& f\left(-1^{+}\right)=2+0+0=2 \\
& f\left(0^{-}\right)=0+1=1 \\
& f\left(0^{+}\right)=0+0+0=0 \\
& f\left(1^{+}\right)=2+1+0=3
\end{aligned}$$</p>
<p>$$\begin{aligned}
& f\left(1^{-}\right)=2+0+1=3 \\
& f\left(2^{-}\right)=8+3+1=12 \\
& f\left(2^{+}\right)=8+4+0=12
\end{aligned}$$</p>
<p>$$\therefore$$ discontinuous at $$x=0, \sqrt{2}, \sqrt{3},-1$$</p>
|
mcq
|
jee-main-2024-online-5th-april-evening-shift
|
lvb294ya
|
maths
|
limits-continuity-and-differentiability
|
continuity
|
<p>Let $$[t]$$ denote the greatest integer less than or equal to $$t$$. Let $$f:[0, \infty) \rightarrow \mathbf{R}$$ be a function defined by $$f(x)=\left[\frac{x}{2}+3\right]-[\sqrt{x}]$$. Let $$\mathrm{S}$$ be the set of all points in the interval $$[0,8]$$ at which $$f$$ is not continuous. Then $$\sum_\limits{\text {aes }} a$$ is equal to __________.</p>
|
[]
| null |
17
|
<p>$$\begin{aligned}
f(x) & =\left[\frac{x}{3}+3\right]-[\sqrt{x}] \\
& =\left[\frac{x}{2}\right]-[\sqrt{x}]+3
\end{aligned}$$</p>
<p>Critical points where $$f(x)$$ might change behaviours when $$\frac{x}{2} \in$$ integer and $$\sqrt{x} \in$$ integer</p>
<p>$$\Rightarrow$$ Critical points,</p>
<p>$$\begin{aligned}
& f(0)=3 \\
& f\left(0^{+}\right)=3 \\
& f\left(1^{-}\right)=3 \\
& f\left(1^{+}\right)=2 \\
& f\left(2^{-}\right)=2 \\
& f\left(2^{+}\right)=3 \\
& f\left(3^{+}\right)=f\left(3^{-}\right)=3=f\left(4^{+}\right)=f\left(4^{-}\right)=f\left(5^{+}\right)=f\left(5^{-}\right) \\
& f\left(6^{-}\right)=3 \\
& f\left(6^{+}\right)=4 \\
& f\left(7^{-}\right)=f\left(7^{+}\right)=4 \\
& f\left(8^{-}\right)=4 \\
& f(8)=5
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow f(x) \text { is not continuous at } \\
& x=1,2,6,8 \\
& \Rightarrow \sum_{a \in s} a=1+2+6+8 \\
& \quad=17
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-6th-april-evening-shift
|
bX8g2hnm95fhLFjI
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
f(x) and g(x) are two differentiable functions on [0, 2] such that
<br/><br/>f''(x) - g''(x) = 0, f'(1) = 2, g'(1) = 4, f(2) = 3, g(2) = 9
<br/><br/>then f(x) - g(x) at x = $${3 \over 2}$$ is
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "10"}, {"identifier": "D", "content": "-5"}]
|
["D"]
| null |
<p>To find the value of $$f(x) - g(x)$$ at $$x = \frac{3}{2}$$, we need to use the given conditions and properties of differentiable functions.</p>
<p>First, we are told that:</p>
<p>$$f''(x) - g''(x) = 0$$</p>
<p>This implies that:</p>
<p>$$f''(x) = g''(x)$$</p>
<p>Since the second derivatives of both functions are equal, their difference, $$f'(x) - g'(x)$$, must be a linear function. Let’s denote it as:</p>
<p>$$f'(x) - g'(x) = k$$</p>
<p>We'll find the constant $$k$$ using the initial conditions of the derivatives:</p>
<p>$$f'(1) = 2$$</p>
<p>$$g'(1) = 4$$</p>
<p>Thus,</p>
<p>$$f'(1) - g'(1) = 2 - 4 = -2$$</p>
<p>Therefore,</p>
<p>$$f'(x) - g'(x) = -2$$</p>
<p>Integrating the above result, we get:</p>
<p>$$f(x) - g(x) = -2x + C$$</p>
<p>To determine the constant $$C$$, we use the values of the functions at $$x = 2$$:</p>
<p>$$f(2) = 3$$</p>
<p>$$g(2) = 9$$</p>
<p>Thus,</p>
<p>$$f(2) - g(2) = 3 - 9 = -6$$</p>
<p>Therefore,</p>
<p>$$-2 \cdot 2 + C = -6$$</p>
<p>$$-4 + C = -6$$</p>
<p>$$C = -2$$</p>
<p>So the expression for $$f(x) - g(x)$$ is:</p>
<p>$$f(x) - g(x) = -2x - 2$$</p>
<p>We need to find $$f\left( \frac{3}{2} \right) - g\left( \frac{3}{2} \right)$$:</p>
<p>$$f\left( \frac{3}{2} \right) - g\left( \frac{3}{2} \right) = -2 \cdot \frac{3}{2} - 2$$</p>
<p>$$= -3 - 2$$</p>
<p>$$= -5$$</p>
<p>Thus, the value of $$f(x) - g(x)$$ at $$x = \frac{3}{2}$$ is -5. </p>
|
mcq
|
aieee-2002
|
3Ip6iYJnQvKQ2klv
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
If f(x + y) = f(x).f(y) $$\forall $$ x, y and f(5) = 2, f'(0) = 3, then
<br/>f'(5) is
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "6"}, {"identifier": "D", "content": "2"}]
|
["C"]
| null |
$$f\left( {x + y} \right) = f\left( x \right) \times f\left( y \right)$$
<br><br>Differeniate with respect to $$x,$$ treating $$y$$ as constant
<br><br>$$f'\left( {x + y} \right) = f'\left( x \right)f\left( y \right)$$
<br><br>Putting $$x=0$$ and $$y=x$$, we get
<br><br>$$f'\left( x \right) = f'\left( 0 \right)f\left( x \right);$$
<br><br>$$ \Rightarrow f'\left( 5 \right) = 3f\left( 5 \right) = 3 \times 2 = 6.$$
|
mcq
|
aieee-2002
|
7X0bs7xPIHP0Q5wn
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let $$f(a) = g(a) = k$$ and their n<sup>th</sup> derivatives
<br/>$${f^n}(a)$$, $${g^n}(a)$$ exist and are not equal for some n. Further if
<br/><br/>$$\mathop {\lim }\limits_{x \to a} {{f(a)g(x) - f(a) - g(a)f(x) + f(a)} \over {g(x) - f(x)}} = 4$$
<br/><br/>then the value of k is
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}]
|
["B"]
| null |
$$\mathop {\lim }\limits_{x \to a} {{f\left( a \right)g'\left( x \right) - g\left( a \right)f'\left( x \right)} \over {g'\left( x \right) - f'\left( x \right)}}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,$$ (By $$L'$$ Hospital rule)
<br><br>$$\mathop {\lim }\limits_{x \to a} {{k\,\,g'\left( x \right) - k\,\,f'\left( x \right)} \over {g'\left( x \right) - f'\left( x \right)}} = 4$$
<br><br>$$\therefore$$ $$k=4.$$
|
mcq
|
aieee-2003
|
PmgTOqs8pMxk3diI
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
If $$f(x) = \left\{ {\matrix{
{x{e^{ - \left( {{1 \over {\left| x \right|}} + {1 \over x}} \right)}}} & {,x \ne 0} \cr
0 & {,x = 0} \cr
} } \right.$$
<br/><br/>then $$f(x)$$ is
|
[{"identifier": "A", "content": "discontinuous everywhere"}, {"identifier": "B", "content": "continuous as well as differentiable for all x"}, {"identifier": "C", "content": "continuous for all x but not differentiable at x = 0"}, {"identifier": "D", "content": "neither differentiable nor continuous at x = 0"}]
|
["C"]
| null |
$$f\left( 0 \right) = 0;\,\,f\left( x \right) = x{e^{ - \left( {{1 \over {\left| x \right|}} + {1 \over x}} \right)}}$$
<br><br>$$R.H.L.\,\,$$ $$\,\,\,\mathop {\lim }\limits_{h \to 0} \left( {0 + h} \right){e^{ - 2/h}}$$
<br><br>$$ = \,\mathop {\lim }\limits_{h \to 0} {h \over {{e^{2/h}}}} = 0$$
<br><br>$$L.H.L.$$ $$\,\,\,\mathop {\lim }\limits_{h \to 0} \left( {0 - h} \right){e^{ - \left( {{1 \over h} - {1 \over h}} \right)}} = 0$$
<br><br>therefore, $$f(x)$$ is continuous,
<br><br>$$R.H.D=$$ $$\,\,\,\mathop {\lim }\limits_{h \to 0} {{\left( {0 + h} \right){e^{ - \left( {{1 \over h} + {1 \over h}} \right)}} - 0} \over h} = 0$$
<br><br>$$L.H.D.$$ $$\,\,\, = \mathop {\lim }\limits_{h \to 0} {{\left( {0 - h} \right){e^{ - \left( {{1 \over h} - {1 \over h}} \right)}} - 0} \over { - h}} = 1$$
<br><br>therefore, $$L.H.D. \ne R.H.D.$$
<br><br>$$f(x)$$ is not differentiable at $$x=0.$$
|
mcq
|
aieee-2003
|
U9v6NjxKosuCkmhk
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Suppose $$f(x)$$ is differentiable at x = 1 and
<br/><br/>$$\mathop {\lim }\limits_{h \to 0} {1 \over h}f\left( {1 + h} \right) = 5$$, then $$f'\left( 1 \right)$$ equals
|
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "5"}, {"identifier": "D", "content": "6"}]
|
["C"]
| null |
$$f'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right) - f\left( 1 \right)} \over h};$$
<br><br>As function is differentiable so it is continuous as it
<br><br>is given that $$\mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right)} \over h} = 5$$ and hence $$f(1)=0$$
<br><br>Hence $$f'(1)$$ $$ = \mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right)} \over h} = 5$$
|
mcq
|
aieee-2005
|
err9ct1D8q5A5Tge
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
If $$f$$ is a real valued differentiable function satisfying
<br/><br/>$$\left| {f\left( x \right) - f\left( y \right)} \right|$$ $$ \le {\left( {x - y} \right)^2}$$, $$x, y$$ $$ \in R$$
<br/>and $$f(0)$$ = 0, then $$f(1)$$ equals
|
[{"identifier": "A", "content": "-1"}, {"identifier": "B", "content": "0"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}]
|
["B"]
| null |
$$f'\left( x \right) = \mathop {\lim }\limits_{h \to 0} {{f\left( {x + h} \right) - f\left( x \right)} \over h}$$
<br><br>$$\,\,\,\,\,\,\,\,\,\left| {f'\left( x \right)} \right| = \mathop {\lim }\limits_{h \to 0} \left| {{{f\left( {x + h} \right) - f\left( x \right)} \over h}} \right| \le \mathop {\lim }\limits_{h \to 0} \left| {{{{{\left( h \right)}^2}} \over h}} \right|$$
<br><br>$$ \Rightarrow \left| {f'\left( x \right)} \right| \le 0 \Rightarrow f'\left( x \right) = 0$$
<br><br>$$ \Rightarrow f\left( x \right) = $$ constant
<br><br>As $$f\left( 0 \right) = 0 \Rightarrow f\left( 1 \right) = 0$$
|
mcq
|
aieee-2005
|
20IRL1dIRMRqkslc
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
The set of points where $$f\left( x \right) = {x \over {1 + \left| x \right|}}$$ is differentiable is
|
[{"identifier": "A", "content": "$$\\left( { - \\infty ,0} \\right) \\cup \\left( {0,\\infty } \\right)$$ "}, {"identifier": "B", "content": "$$\\left( { - \\infty ,1} \\right) \\cup \\left( { - 1,\\infty } \\right)$$ "}, {"identifier": "C", "content": "$$\\left( { - \\infty ,\\infty } \\right)$$ "}, {"identifier": "D", "content": "$$\\left( {0,\\infty } \\right)$$ "}]
|
["C"]
| null |
$$f\left( x \right) = \left\{ {\matrix{
{{x \over {1 - x}},} & {x < 0} \cr
{{x \over {1 + x}},} & {x \ge 0} \cr
} } \right.$$
<br><br>$$ \Rightarrow f'\left( x \right) = \left\{ {\matrix{
{{x \over {{{\left( {1 - x} \right)}^2}}},} & {x < 0} \cr
{{x \over {{{\left( {1 + x} \right)}^2}}}} & {x \ge 0} \cr
} } \right.$$
<br><br>$$\therefore$$ $$f'\left( x \right)$$ exist at everywhere.
|
mcq
|
aieee-2006
|
O9Pfkq9XuIbELScm
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let $$f:R \to R$$ be a function defined by
<br/><br/>$$f(x) = \min \left\{ {x + 1,\left| x \right| + 1} \right\}$$, then which of the following is true?
|
[{"identifier": "A", "content": "$$f(x)$$ is differentiale everywhere"}, {"identifier": "B", "content": "$$f(x)$$ is not differentiable at x = 0"}, {"identifier": "C", "content": "$$f(x) > 1$$ for all $$x \\in R$$"}, {"identifier": "D", "content": "$$f(x)$$ is not differentiable at x = 1"}]
|
["A"]
| null |
$$f\left( x \right) = \min \left\{ {x + 1,\left| x \right| + 1} \right\}$$
<br><br>$$ \Rightarrow f\left( x \right) = x + 1\,\forall \,x \in R$$
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266784/exam_images/jxar1vgdorciobrxdspe.webp" loading="lazy" alt="AIEEE 2007 Mathematics - Limits, Continuity and Differentiability Question 212 English Explanation">
<br><br>Hence, $$f(x)$$ is differentiable everywhere for all $$x \in R.$$
|
mcq
|
aieee-2007
|
01SDTRQZ8xp9Z4WJ
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let $$f\left( x \right) = \left\{ {\matrix{
{\left( {x - 1} \right)\sin {1 \over {x - 1}}} & {if\,x \ne 1} \cr
0 & {if\,x = 1} \cr
} } \right.$$
<br/><br/>Then which one of the following is true?
|
[{"identifier": "A", "content": "$$f$$ is neither differentiable at x = 0 nor at x = 1"}, {"identifier": "B", "content": "$$f$$ is differentiable at x = 0 and at x = 1"}, {"identifier": "C", "content": "$$f$$ is differentiable at x = 0 but not at x = 1"}, {"identifier": "D", "content": "$$f$$ is differentiable at x = 1 but not at x = 0"}]
|
["C"]
| null |
We have $$f\left( x \right) = \left\{ {\matrix{
{\left( {x - 1} \right)\sin \left( {{1 \over {x - 1}}} \right),} & {if\,\,x \ne 1} \cr
{0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} & {if\,\,x = 1} \cr
} } \right.$$
<br><br>$$Rf'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0} {{f\left( {1 + h} \right) - f\left( 1 \right)} \over h}$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} {{h\,\sin {1 \over h} - 0} \over h} = \mathop {\lim }\limits_{h \to 0} \,\,\sin {1 \over h} = a$$ finite number
<br><br>Let this finite number be $$l$$
<br><br>$$L$$ $$f'(1)$$ $$ = \mathop {\lim }\limits_{h \to 0} {{f\left( {1 - h} \right) - f\left( 1 \right)} \over { - h}}$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} {{ - h\,\sin \left( {{1 \over { - h}}} \right)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} \,\,\sin \left( {{1 \over { - h}}} \right)$$
<br><br>$$ = - \mathop {\lim }\limits_{h \to 0} \sin \left( {{1 \over h}} \right) = - (\,$$ a finite number $$\,)$$ $$=-l$$
<br><br>Thus $$Rf'\left( 1 \right) \ne Lf'\left( 1 \right)$$
<br><br>$$\therefore$$ $$f$$ is not differentiable at $$x=1$$
<br><br>Also, $$f'\left( 0 \right) = \sin {1 \over {\left( {x - 1} \right)}} - {{x - 1} \over {{{\left( {x - 1} \right)}^2}}}\cos {\left. {\left( {{1 \over {x - 1}}} \right)} \right]_{x = 0}}$$
<br><br>$$ = - \sin 1 + \cos \,1$$
<br><br>$$\therefore$$ $$f$$ is differentiable at $$x=0$$
|
mcq
|
aieee-2008
|
0QeoBy8todI83IoW
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let $$f\left( x \right) = x\left| x \right|$$ and $$g\left( x \right) = \sin x.$$
<br/><b>Statement-1:</b> gof is differentiable at $$x=0$$ and its derivative is continuous at that point.
<br/><b>Statement-2:</b> gof is twice differentiable at $$x=0$$.
|
[{"identifier": "A", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1."}, {"identifier": "B", "content": "Statement-1 is true, Statement-2 is false "}, {"identifier": "C", "content": "Statement-1 is false, Statement-2 is true "}, {"identifier": "D", "content": "Statement-1 is true, Statement-2 is true Statement-2 is a correct explanation for Statement-1"}]
|
["B"]
| null |
Given that $$f\left( x \right) = x\left| x \right|\,\,$$ and $$\,\,g\left( x \right) = \sin x$$
<br><br>So that go
<br><br>$$f\left( x \right) = g\left( {f\left( x \right)} \right)$$
<br><br>$$ = g\left( {x\left| x \right|} \right) = \sin x\left| x \right|$$
<br><br>$$ = \left\{ {\matrix{
{\sin \left( { - {x^2}} \right),} & {if\,\,\,x < 0} \cr
{\sin \left( {{x^2}} \right),} & {if\,\,\,x \ge 0} \cr
} } \right.$$
<br><br>$$ = \left\{ {\matrix{
{ - \sin \,{x^2},} & {if\,\,\,x < 0} \cr
{\sin \,\,{x^2},} & {if\,\,\,x \ge 0} \cr
} } \right.$$
<br><br>$$\therefore$$ $$\left( {go\,f} \right)'\,\,\left( x \right) = \left\{ {\matrix{
{ - 2x\,\,\cos \,{x^2},\,\,\,\,if\,\,\,\,x < 0} \cr
{2x\,\cos \,{x^2},\,\,\,if\,\,\,\,x \ge 0} \cr
} } \right.$$
<br><br>Here we observe
<br><br>$$L\left( {gof} \right)'\left( 0 \right) = 0 = R\left( {gof} \right)'\left( 0 \right)$$
<br><br>$$ \Rightarrow $$ go $$f$$ is differentiable at $$x=0$$
<br><br>and $$\left( {go\,f} \right)'$$ is continuous at $$x=0$$
<br><br>Now $$\left( {go\,f} \right)''\left( x \right) = \left\{ {\matrix{
{ - 2\cos {x^2} + 4{x^2}\sin {x^2},x < 0} \cr
{2\cos {x^2} - 4{x^2}\sin {x^2},x \ge 0} \cr
} } \right.$$
<br><br>Here $$L\left( {gof} \right)''\left( 0 \right) = - 2$$ and $$R\left( {go\,f} \right)''\left( 0 \right) = 2$$
<br><br>As $$L{\left( {go\,f} \right)^{''}}\left( 0 \right) \ne R\left( {go\,f} \right)''\,\,\left( 0 \right)$$
<br><br>$$ \Rightarrow go\,f\left( x \right)$$ is not twice differentiable at $$x=0.$$
<br><br>$$\therefore$$ Statement - $$1$$ is true but statement $$-2$$ is false.
|
mcq
|
aieee-2009
|
vgUhS2v8cEKHz8Ex
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Consider the function, $$f\left( x \right) = \left| {x - 2} \right| + \left| {x - 5} \right|,x \in R$$
<br/><br/><b>Statement - 1 :</b> $$f'\left( 4 \right) = 0$$
<br/><br/><b>Statement - 2 :</b> $$f$$ is continuous in [2, 5], differentiable in (2, 5) and $$f$$(2) = $$f$$(5)
|
[{"identifier": "A", "content": "Statement - 1 is false, statement - 2 is true "}, {"identifier": "B", "content": "Statement - 1 is true, statement - 2 is true; statement - 2 is a correct explanation for statement - 1"}, {"identifier": "C", "content": "Statement - 1 is true, statement - 2 is true; statement - 2 is not a correct explanation for statement - 1"}, {"identifier": "D", "content": "Statement - 1 is true, statement - 2 is false"}]
|
["C"]
| null |
$$f\left( x \right) = \left| {x - 2} \right| = \left\{ {\matrix{
{x - 2\,\,\,,} & {x - 2 \ge 0} \cr
{2 - x\,\,\,,} & {x - 2 \le 0} \cr
} } \right.$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left\{ {\matrix{
{x - 2\,\,\,,} & {x \ge 2} \cr
{2 - x\,\,\,,} & {x \le 2} \cr
} } \right.$$
<br><br>Similarly, $$f\left( x \right) = \left| {x - 5} \right| = \left\{ {\matrix{
{x - 5\,\,\,,} & {x \ge 5} \cr
{5 - x\,\,\,,} & {x \le 5} \cr
} } \right.$$
<br><br>$$\therefore$$ $$f\left( x \right) = \left| {x - 2} \right| + \left| {x - 5} \right|$$
<br><br>$$ = \left\{ {x - 2 + 5 - x = 3,2 \le x \le 5} \right\}$$
<br><br>Thus $$\,\,\,\,\,\,\,\,\,f\left( x \right) = 3,2 \le x \le 5$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( x \right) = 0,2 < x < 5$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,f'\left( 4 \right) = 0$$
<br><br>$$\therefore$$ Statement $$1$$ is true.
<br><br><img class="question-image" src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266217/exam_images/fghim8fqhfaqmfos46vi.webp" loading="lazy" alt="AIEEE 2012 Mathematics - Limits, Continuity and Differentiability Question 205 English Explanation">
<br><br>As $$f\left( 2 \right) = 0 + \left| {2 - 5} \right| = 3$$
<br><br>and $$f\left( 5 \right) = \left| {5 - 2} \right| + 0 = 3$$
<br><br>$$\therefore$$ Statement - $$2$$ is also true but not a correct explanation for statement $$1.$$
|
mcq
|
aieee-2012
|
RtZ9MK0tEqbtwJlp
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
If the function.
<br/><br/>$$g\left( x \right) = \left\{ {\matrix{
{k\sqrt {x + 1} ,} & {0 \le x \le 3} \cr
{m\,x + 2,} & {3 < x \le 5} \cr
} } \right.$$
<br/><br/>is differentiable, then the value of $$k+m$$ is :
|
[{"identifier": "A", "content": "$${{10} \\over 3}$$ "}, {"identifier": "B", "content": "$$4$$"}, {"identifier": "C", "content": "$$2$$ "}, {"identifier": "D", "content": "$${{16} \\over 5}$$"}]
|
["C"]
| null |
Since $$g(x)$$ is differentiable, -
<br><br>it will be continuous at $$x=3$$
<br><br>$$\therefore$$ $$\mathop {\lim }\limits_{x \to {3^ - }} g\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} g\left( x \right)$$
<br><br>$$2k = 3m + 2\,\,\,\,\,...\left( 1 \right)$$
<br><br>Also $$g(x)$$ is differentiable at $$x=0$$
<br><br>$$\therefore$$ $$\mathop {\lim }\limits_{x \to {3^ - }} g'\left( x \right) = \mathop {\lim }\limits_{x \to {3^ + }} g'\left( x \right)$$
<br><br>$${K \over {2\sqrt {3 + 1} }} = m$$
<br><br>$$k=4m$$ $$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$$
<br><br>Solving $$(1)$$ and $$(2)$$, we get
<br><br>$$m = {2 \over 5},\,\,k = {8 \over 5}$$
<br><br>$$\therefore$$ $$k+m=2$$
|
mcq
|
jee-main-2015-offline
|
dTkaSbOpd0NkpeiE
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
For $$x \in \,R,\,\,f\left( x \right) = \left| {\log 2 - \sin x} \right|\,\,$$
<br/><br/>and $$\,\,g\left( x \right) = f\left( {f\left( x \right)} \right),\,\,$$ then :
|
[{"identifier": "A", "content": " $$g$$ is not differentiable at $$x=0$$"}, {"identifier": "B", "content": "$$g'\\left( 0 \\right) = \\cos \\left( {\\log 2} \\right)$$ "}, {"identifier": "C", "content": "$$g'\\left( 0 \\right) = - \\cos \\left( {\\log 2} \\right)$$"}, {"identifier": "D", "content": "$$g$$ is differentiable at $$x=0$$ and $$g'\\left( 0 \\right) = - \\sin \\left( {\\log 2} \\right)$$"}]
|
["B"]
| null |
$$g\left( x \right) = f\left( {f\left( x \right)} \right)$$
<br><br>In the neighbourhood of $$x=0,$$
<br><br>$$f\left( x \right) = \left| {\log 2 - \sin \,x} \right| = \left( {\log 2 - \sin x} \right)$$
<br><br>$$\therefore$$ $$g\left( x \right) = \left| {\log 2 - \sin \left. {\left| {\log 2 - \sin x} \right|} \right|} \right.$$
<br><br>$$ = \left( {\log 2 - \sin \left( {\log 2 - \sin x} \right)} \right)$$
<br><br>$$\therefore$$ $$g(x)$$ is differentiable
<br><br>and $$g'\left( x \right) = - \cos \left( {\log 2 - \sin x} \right)\left( { - \cos x} \right)$$
<br><br>$$ \Rightarrow g'\left( 0 \right) = \cos \left( {\log 2} \right)$$
|
mcq
|
jee-main-2016-offline
|
hF8RmsEat5uIVIgrYz5HO
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
If the function
<br/><br/>f(x) = $$\left\{ {\matrix{
{ - x} & {x < 1} \cr
{a + {{\cos }^{ - 1}}\left( {x + b} \right),} & {1 \le x \le 2} \cr
} } \right.$$
<br/><br/>is differentiable at x = 1, then $${a \over b}$$ is equal to :
|
[{"identifier": "A", "content": "$${{\\pi - 2} \\over 2}$$"}, {"identifier": "B", "content": "$${{ - \\pi - 2} \\over 2}$$"}, {"identifier": "C", "content": "$${{\\pi + 2} \\over 2}$$"}, {"identifier": "D", "content": "$$ - 1 - {\\cos ^{ - 1}}\\left( 2 \\right)$$"}]
|
["C"]
| null |
As f(x) is differentiable at x = 1
<br><br>$$ \therefore $$ $$\mathop {\lim }\limits_{x \to {1^ - }} \left( { - x} \right) = \mathop {\lim }\limits_{x \to {1^ + }} \left( {a + {{\cos }^{ - 1}}\left( {x + b} \right)} \right) = f(1)$$
<br><br>$$ \Rightarrow $$ $$-$$1 $$=$$ a + cos<sup>$$-$$1</sup>(1 + b)
<br><br>$$ \Rightarrow $$ cos<sup>$$-$$1</sup> (1 + b) $$=$$ $$-$$1 $$-$$ a . . . . . .(1)
<br><br>As f(x) is differentiable, so,
<br><br>2 . H . D $$=$$ R . H . D
<br><br>Here, L . H . D $$=$$ $$\mathop {\lim }\limits_{h \to 0} $$ $${{f\left( {1 - h} \right) - f\left( 1 \right)} \over { - h}}$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} {{ - \left( {1 - h} \right) - \left( { - 1} \right)} \over { - h}}$$
<br><br>$$ = \mathop {\lim }\limits_{x \to 0} {{ - 1 + h + 1} \over { - h}}$$
<br><br>$$=$$ $$ = \mathop {\lim }\limits_{x \to 0} {h \over { - h}}$$
<br><br>$$=$$ $$-$$ 1
<br><br>R. H. D $$ = \mathop {\lim }\limits_{x \to 0} {{f\left( {1 + h} \right) - f\left( 1 \right)} \over h}$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} {{a + {{\cos }^{ - 1}}\left( {1 + h + b} \right) - \left[ {a + {{\cos }^{ - 1}}\left( {1 + b} \right)} \right]} \over h}$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} {{{{\cos }^{ - 1}}\left( {1 + h + b} \right) - {{\cos }^{ - 1}}\left( {1 + b} \right)} \over h}\left[ {{0 \over 0}\,\,form\left. \, \right]} \right.$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} {{ - 1} \over {\sqrt {1 - {{\left( {1 + h + b} \right)}^2}} }}$$ [ Using L' Hospital Rule]
<br><br>$$ = {{ - 1} \over {\sqrt {1 - {{\left( {1 + b} \right)}^2}} }}$$
<br><br>$$ \therefore $$ $$ - 1 = {{ - 1} \over {\sqrt {1 - {{\left( {1 + b} \right)}^2}} }}$$
<br><br>$$ \Rightarrow 1 - {\left( {1 + b} \right)^2} = 1$$
<br><br>$$ \Rightarrow $$ $${\left( {1 + b} \right)^2} = 0$$
<br><br>$$ \Rightarrow $$ $$b = - 1$$
<br><br>putting value of b in equation (1), we get,
<br><br>$${\cos ^{ - 1}}\left( {1 - 1} \right) = - 1 - a$$
<br><br>$$ \Rightarrow $$ $${\pi \over 2} = - 1 - a$$
<br><br>$$ \Rightarrow $$ $$a = - 1 - {\pi \over 2}$$
<br><br>$$ \therefore $$ $${a \over b} = {{ - 1 - {\pi \over 2}} \over { - 1}} = 1 + {\pi \over 2} = {{\pi + 2} \over 2}$$
|
mcq
|
jee-main-2016-online-9th-april-morning-slot
|
ppH8dXaEhhGCho4t
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let S = { t $$ \in R:f(x) = \left| {x - \pi } \right|.\left( {{e^{\left| x \right|}} - 1} \right)$$$$\sin \left| x \right|$$ is not differentiable at t}, then the set S is equal to
|
[{"identifier": "A", "content": "{0, $$\\pi $$}"}, {"identifier": "B", "content": "$$\\phi $$ (an empty set)"}, {"identifier": "C", "content": "{0}"}, {"identifier": "D", "content": "{$$\\pi $$}"}]
|
["B"]
| null |
Check differtiability at x = $$\pi $$ and x = 0
<br><br><b>at x = 0 : </b>
<br><br>We have L. H. D = $$\mathop {\lim }\limits_{h \to 0} {{f\left( {0 - h} \right) - f\left( 0 \right)} \over { - h}}$$
<br><br>= $$\mathop {\lim }\limits_{h \to 0} {{\left| { - h - \pi } \right|\left( {{e^{\left| { - h} \right|}} - 1} \right)\sin \left| h \right| - 0} \over { - h}} = 0$$
<br><br>R. H. D = $$\mathop {\lim }\limits_{h \to 0} {{f\left( {0 + h} \right) - f\left( 0 \right)} \over h}$$
<br><br>$$ = \mathop {\lim }\limits_{h \to 0} {{\left| {h - \pi } \right|\left( {{e^{\left| h \right|}} - 1} \right)\sin \left| h \right| - o} \over h}$$
<br><br>= 0
<br><br>$$\therefore,\,\,$$ LHD = RHD
<br><br>Therefore, function is differentiable at x = $$\pi $$.
<br><br><b>at x = $$\pi $$ : </b>
<br><br>L. H. D = $$\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi - h} \right) - f\left( \pi \right)} \over { - h}}$$
<br><br>= $$\mathop {\lim }\limits_{h \to 0} {{\left| {\pi - h - \pi } \right|\left( {{e^{\left| {\pi - h} \right|}} - 1} \right)\sin \left| {\pi - h} \right| - 0.} \over { - h}}$$
<br><br>= 0
<br><br>RH. D = $$\mathop {\lim }\limits_{h \to 0} {{f\left( {\pi + h} \right) - f\left( \pi \right)} \over h}$$
<br><br>= $$\mathop {\lim }\limits_{h \to 0} {{\left| {\pi + h - \pi } \right|\left( {{e^{\left| {\pi + h} \right|}} - 1} \right)\sin \left| {\pi + h} \right| - 0} \over h}$$
<br><br>= 0
<br><br>$$\therefore\,\,\,$$ L. H. D = R H D
<br><br>Therefore, function is differentiable at x = $$\pi $$,
<br><br>Since, the function f(x) is differentiable at all the points including 0 and $$\pi $$.
<br><br>So, f(x) is differentiable everywhere.
<br><br>Therefore, set S is empty set.
<br><br>S = $$\phi $$
|
mcq
|
jee-main-2018-offline
|
LyYVn5Xz5QYZhjEVYw7qG
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let S = {($$\lambda $$, $$\mu $$) $$ \in $$ <b>R</b> $$ \times $$ <b>R</b> : f(t) = (|$$\lambda $$| e<sup>|t|</sup> $$-$$ $$\mu $$). sin (2|t|), <b>t</b> $$ \in $$ <b>R</b>, is a differentiable function}. Then S is a subset of :
|
[{"identifier": "A", "content": "<b>R</b> $$ \\times $$ [0, $$\\infty $$)"}, {"identifier": "B", "content": "[0, $$\\infty $$) $$ \\times $$ <b>R</b>"}, {"identifier": "C", "content": "<b>R</b> $$ \\times $$ ($$-$$ $$\\infty $$, 0)"}, {"identifier": "D", "content": "($$-$$ $$\\infty $$, 0) $$ \\times $$ <b>R</b>"}]
|
["A"]
| null |
S = {($$\lambda $$, $$\mu $$) $$ \in $$ <b>R</b> $$ \times $$ <b>R</b> : f(t) = $$\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)$$ sin $$\left( {2\left| t \right|} \right),$$ t $$ \in $$ <b>R</b>
<br><br>f(t) = $$\left( {\left| \lambda \right|{e^{\left| t \right|}} - \mu } \right)\sin \left( {2\left| t \right|} \right)$$
<br><br>= $$\left\{ {\matrix{
{\left( {\left| \lambda \right|{e^t} - \mu } \right)\sin 2t,} & {t > 0} \cr
{\left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( { - \sin 2t} \right),} & {t < 0} \cr
} } \right.$$
<br><br>f'(t) = $$\left\{ {\matrix{
{\left( {\left| \lambda \right|{e^t}} \right)\sin 2t + \left( {\left| \lambda \right|{e^t} - \mu } \right)\left( {2\cos 2t} \right),\,\,t > 0} \cr
{\left| \lambda \right|{e^{ - t}}\sin 2t + \left( {\left| \lambda \right|{e^{ - t}} - \mu } \right)\left( {-2\cos 2t} \right),t < 0} \cr
} } \right.$$
<br><br>As, f(t) is differentiable
<br><br>$$ \therefore $$ <b>LHD</b> = <b>RHD</b> at t = 0
<br><br>$$ \Rightarrow $$ $$\left| \lambda \right|$$ . sin2(0) + $$\left( {\left| \lambda \right|{e^0} - \mu } \right)$$2cos(0)
<br><br> = $$\left| \lambda \right|{e^{ - 0}}\,$$ . sin 2(0) $$-$$ 2 cos (0) $$\left( {\left| \lambda \right|{e^{ - 0}} - \mu } \right)$$
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 0 + $$\left( {\left| \lambda \right| - \mu } \right)$$2 = 0 $$-$$ 2$$\left( {\left| \lambda \right| - \mu } \right)$$
<br><br>$$ \Rightarrow $$ 4$$\left( {\left| \lambda \right| - \mu } \right)$$ = 0
<br><br>$$ \Rightarrow $$ $$\left| \mu \right|$$ = $$\mu $$
<br><br>So, S $$ \equiv $$ ($$\lambda $$, $$\mu $$) = {$$\lambda $$ $$ \in $$ <b>R</b> & $$\mu $$ $$ \in $$ [0, $$\infty $$)}
<br><br>Therefore set S is subset of R $$ \times $$ [0, $$\infty $$)
|
mcq
|
jee-main-2018-online-15th-april-morning-slot
|
E1UPnoGMUo46JcFfe7wpS
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let ƒ(x) = 15 – |x – 10|; x $$ \in $$ R. Then the set
of all values of x, at which the function,
g(x) = ƒ(ƒ(x)) is not differentiable, is :
|
[{"identifier": "A", "content": "{10,15}"}, {"identifier": "B", "content": "{5,10,15,20}"}, {"identifier": "C", "content": "{10}"}, {"identifier": "D", "content": "{5,10,15}"}]
|
["D"]
| null |
ƒ(x) = 15 – |x – 10|
<br><br>g(x) = ƒ(ƒ(x)) = 15 – |ƒ(x) – 10|
<br><br>= 15 – |15 – |x – 10| – 10|
<br><br>= 15 – |5 – |x – 10||
<br><br>As this is a linear expression so it is non differentiable when value inside the modulus is zero.
<br><br>So non differentiable when
<br><br>x – 10 = 0 $$ \Rightarrow $$ x = 10
<br><br>and 5 – |x – 10| = 0
<br><br>$$ \Rightarrow $$ |x – 10| = 5
<br><br>$$ \Rightarrow $$ x - 10 = $$ \pm $$ 5
<br><br>$$ \Rightarrow $$ x = 5, 15
<br><br>$$ \therefore $$ g(x) is not differentiable at x = 5, 10, 15.
|
mcq
|
jee-main-2019-online-9th-april-morning-slot
|
OjgfQrxVeltiW0ZoZE3rsa0w2w9jwxut8on
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let f : R $$ \to $$ R be differentiable at c $$ \in $$ R and f(c) = 0. If g(x) = |f(x)| , then at x = c, g is :
|
[{"identifier": "A", "content": "differentiable if f '(c) = 0"}, {"identifier": "B", "content": "differentiable if f '(c) $$ \\ne $$ 0"}, {"identifier": "C", "content": "not differentiable"}, {"identifier": "D", "content": "not differentiable if f '(c) = 0"}]
|
["A"]
| null |
$$g'(c) = \mathop {\lim }\limits_{x \to c} {{g(x) - g(c)} \over {x - c}}$$<br><br>
$$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \right| - \left| {f(c)} \right|} \over {x - c}}$$<br><br>
$$ \therefore $$ f(c) = 0<br><br>
$$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \right|} \over {x - c}}$$<br><br>
$$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{f(x)} \over {x - c}}$$ if f(x) > 0<br><br>
and $$g'(c) = \mathop {\lim }\limits_{x \to c} {{ - f(x)} \over {x - c}}$$ if f(x) < 0<br><br>
$$ \Rightarrow g'(c) = f'(c) = - f'(c)$$<br><br>
$$ \Rightarrow $$ 2f'(c) = 0<br><br>
$$ \Rightarrow $$ f'(c) = 0
|
mcq
|
jee-main-2019-online-10th-april-morning-slot
|
5lu85xMvzqpVj7b8OfEQ8
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let ƒ : R $$ \to $$ R be a differentiable function
satisfying ƒ'(3) + ƒ'(2) = 0.<br/>
Then $$\mathop {\lim }\limits_{x \to 0} {\left( {{{1 + f(3 + x) - f(3)} \over {1 + f(2 - x) - f(2)}}} \right)^{{1 \over x}}}$$ is equal to
|
[{"identifier": "A", "content": "e"}, {"identifier": "B", "content": "e<sup>2</sup>"}, {"identifier": "C", "content": "e<sup>\u20131</sup>"}, {"identifier": "D", "content": "1"}]
|
["D"]
| null |
The general formula for indeterminate form 1<sup>$$\infty $$</sup> is
<br><br>$$\mathop {\lim }\limits_{x \to a} {\left( {f\left( x \right)} \right)^{g\left( x \right)}} = {e^{\mathop {\lim }\limits_{x \to a} g\left( x \right)\left( {f\left( x \right) - 1} \right)}}$$
<br><br>I = $$\mathop {\lim }\limits_{x \to 0} {\left( {{{1 + f(3 + x) - f(3)} \over {1 + f(2 - x) - f(2)}}} \right)^{{1 \over x}}}$$
<br><br>Here I is in 1<sup>$$\infty $$</sup> form.
<br><br>$$ \therefore $$ I = $${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 + f\left( {3 + x} \right) - f\left( 3 \right)} \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}} - 1} \right){1 \over x}}}$$
<br><br>= $${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 + f\left( {3 + x} \right) - f\left( 3 \right) - 1 - f\left( {2 - x} \right) + f\left( 2 \right)} \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right){1 \over x}}}$$
<br><br>= $${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{f\left( {3 + x} \right) - f\left( 3 \right) - f\left( {2 - x} \right) + f\left( 2 \right)} \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right){1 \over x}}}$$
<br><br>Here $${{{f\left( {3 + x} \right) - f\left( 3 \right) - f\left( {2 - x} \right) + f\left( 2 \right)} \over x}}$$ is in $${0 \over 0}$$ form.
<br><br>So using L'Hopital rule we get
<br><br>= $${e^{\mathop {\lim }\limits_{x \to 0} \left( {{{f'\left( {3 + x} \right) + f'\left( {2 - x} \right)} \over 1}} \right).\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right)}}$$
<br><br>= $${e^{\left( {f'\left( 3 \right) + f'\left( 2 \right)} \right).1}}$$
<br><br>= e<sup>0</sup> [ as given ƒ'(3) + ƒ'(2) = 0 ]
<br><br>= 1
|
mcq
|
jee-main-2019-online-8th-april-evening-slot
|
fG7JGryJ49wVQOWIzE1lB
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let f be a differentiable function such that f(1) = 2 and f '(x) = f(x) for all x $$ \in $$ R R. If h(x) = f(f(x)), then h'(1) is equal to :
|
[{"identifier": "A", "content": "4e"}, {"identifier": "B", "content": "2e<sup>2</sup>"}, {"identifier": "C", "content": "4e<sup>2</sup>"}, {"identifier": "D", "content": "2e"}]
|
["A"]
| null |
$${{f'(x)} \over {f(x)}} = 1\forall x \in R$$
<br><br>Intergrate & use f(1) = 2
<br><br>f(x) = 2e<sup>x-1</sup> $$ \Rightarrow $$ f '(x) = 2e<sup>x$$-$$1</sup>
<br><br>h(x) = f(f(x)) $$ \Rightarrow $$ h'(x) = f '(f(x)) f'(x)
<br><br>h'(1) = f '(f(1)) f'(1)
<br><br>= f '(2) f '(1)
<br><br>= 2e . 2 = 4e
|
mcq
|
jee-main-2019-online-12th-january-evening-slot
|
00UBMw9xgFLk3qTNH0S2d
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let S be the set of all points in (–$$\pi $$, $$\pi $$) at which the function, f(x) = min{sin x, cos x} is not differentiable. Then S is a subset of which of the following ?
|
[{"identifier": "A", "content": "$$\\left\\{ { - {\\pi \\over 2}, - {\\pi \\over 4},{\\pi \\over 4},{\\pi \\over 2}} \\right\\}$$"}, {"identifier": "B", "content": "$$\\left\\{ { - {{3\\pi } \\over 4}, - {\\pi \\over 2},{\\pi \\over 2},{{3\\pi } \\over 4}} \\right\\}$$"}, {"identifier": "C", "content": "$$\\left\\{ { - {\\pi \\over 4},0,{\\pi \\over 4}} \\right\\}$$"}, {"identifier": "D", "content": "$$\\left\\{ { - {{3\\pi } \\over 4}, - {\\pi \\over 4},{{3\\pi } \\over 4},{\\pi \\over 4}} \\right\\}$$"}]
|
["D"]
| null |
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264785/exam_images/rhqvelsmxghhl3cztv4v.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Mathematics - Limits, Continuity and Differentiability Question 172 English Explanation">
|
mcq
|
jee-main-2019-online-12th-january-morning-slot
|
ei04ZJBaNytj0HKOk0XL4
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let K be the set of all real values of x where the function f(x) = sin |x| – |x| + 2(x – $$\pi $$) cos |x| is not differentiable. Then the set K is equal to :
|
[{"identifier": "A", "content": "{0, $$\\pi $$}"}, {"identifier": "B", "content": "$$\\phi $$ (an empty set)"}, {"identifier": "C", "content": "{ r }"}, {"identifier": "D", "content": "{0}"}]
|
["B"]
| null |
f(x) = sin$$\left| x \right| - \left| x \right|$$ + 2(x $$-$$ $$\pi $$) cosx
<br><br>$$ \because $$ sin$$\left| x \right|$$ $$-$$ $$\left| x \right|$$ is differentiable function at c = 0
<br><br>$$ \therefore $$ k = $$\phi $$
|
mcq
|
jee-main-2019-online-11th-january-evening-slot
|
iZbnjQ958X1XbdDcn9Kdv
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let $$f\left( x \right) = \left\{ {\matrix{
{ - 1} & { - 2 \le x < 0} \cr
{{x^2} - 1,} & {0 \le x \le 2} \cr
} } \right.$$ and
<br/><br/>$$g(x) = \left| {f\left( x \right)} \right| + f\left( {\left| x \right|} \right).$$
<br/><br/>Then, in the interval (–2, 2), g is :
|
[{"identifier": "A", "content": "non continuous"}, {"identifier": "B", "content": "differentiable at all points"}, {"identifier": "C", "content": "not differentiable at two points"}, {"identifier": "D", "content": "not differentiable at one point"}]
|
["D"]
| null |
$$\left| {f\left( x \right)} \right| = \left\{ {\matrix{
1 & , & { - 2 \le x < 0} \cr
{1 - {x^2}} & , & {0 \le x < 1} \cr
{{x^2} - 1} & , & {1 \le x \le 2} \cr
} } \right.$$
<br><br>and $$f\left( {\left| x \right|} \right) = {x^2} - 1,x \in \left[ { - 2,2} \right]$$
<br><br>Hence $$g(x) = \left\{ {\matrix{
{{x^2}} & , & {x \in \left[ { - 2,0} \right]} \cr
0 & , & {x \in \left[ {0,1} \right)} \cr
{2\left( {{x^2} - 1} \right)} & , & {x \in \left[ {1,2} \right]} \cr
} } \right.$$
<br><br>It is not differentiable at x = 1
|
mcq
|
jee-main-2019-online-11th-january-morning-slot
|
UWIY66U2e4Fst3ZMSEOx2
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let f : ($$-$$1, 1) $$ \to $$ R be a function defined by f(x) = max $$\left\{ { - \left| x \right|, - \sqrt {1 - {x^2}} } \right\}.$$ If K be the set of all points at which f is not differentiable, then K has exactly -
|
[{"identifier": "A", "content": "one element"}, {"identifier": "B", "content": "three elements"}, {"identifier": "C", "content": "five elements"}, {"identifier": "D", "content": "two elements"}]
|
["B"]
| null |
f : ($$-$$ 1, 1) $$ \to $$ R
<br><br>f(x) = max {$$-$$ $$\left| x \right|, - \sqrt {1 - {x^2}} $$}
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264960/exam_images/j9hfstro1mtiidz6xe3y.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Evening Slot Mathematics - Limits, Continuity and Differentiability Question 178 English Explanation">
<br>Non-derivable at 3 points in ($$-$$1, 1)
|
mcq
|
jee-main-2019-online-10th-january-evening-slot
|
txkv94d31U1ZqKVvm4wJ2
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let $$f\left( x \right) = \left\{ {\matrix{
{\max \left\{ {\left| x \right|,{x^2}} \right\}} & {\left| x \right| \le 2} \cr
{8 - 2\left| x \right|} & {2 < \left| x \right| \le 4} \cr
} } \right.$$
<br/><br/>Let S be the set of points in the interval (– 4, 4) at which f is not differentiable. Then S
|
[{"identifier": "A", "content": "equals $$\\left\\{ { - 2, - 1,1,2} \\right\\}$$"}, {"identifier": "B", "content": "equals $$\\left\\{ { - 2, - 1,0,1,2} \\right\\}$$"}, {"identifier": "C", "content": "equals $$\\left\\{ { - 2,2} \\right\\}$$"}, {"identifier": "D", "content": "is an empty set"}]
|
["B"]
| null |
$$f\left( x \right) = \left\{ {\matrix{
{8 + 2x,} & { - 4 \le x \le - 2} \cr
{{x^2},} & { - 2 \le x \le - 1} \cr
{\left| x \right|,} & { - 1 < x < 1} \cr
{{x^2},} & {1 \le x \le 2} \cr
{8 - 2x,} & {2 < x \le 4} \cr
} } \right.$$
<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266646/exam_images/sowb6jz8ffftgdwa0zm0.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 10th January Morning Slot Mathematics - Limits, Continuity and Differentiability Question 179 English Explanation">
<br>f(x) is not differentiable at
<br><br>x = $$\left\{ { - 2, - 1,0,1,2} \right\}$$
<br><br>$$ \Rightarrow $$ S = {$$-$$2, $$-$$ 1, 0, 1, 2}
<br><br>
|
mcq
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jee-main-2019-online-10th-january-morning-slot
|
Q0KAqKgD9KG5tr0MYv7k9k2k5e4n2c9
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let S be the set of points where the function, ƒ(x) = |2-|x-3||, x $$ \in $$ R is not differentiable. Then $$\sum\limits_{x \in S} {f(f(x))} $$ is equal to_____.
|
[]
| null |
3
|
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267767/exam_images/g4hsbmpaxkynaavxvqy0.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 7th January Morning Slot Mathematics - Limits, Continuity and Differentiability Question 155 English Explanation">
<br><br>f(x) is non-differentiable at x = 1, 3, 5
<br><br>$$ \therefore $$ S is {1, 3, 5}
<br><br>$$\sum\limits_{x \in S} {f(f(x))} $$
<br><br>= f(f(1)) + f(f(3)) + f(f (5))
<br><br>= f(0) + f(2) + f(0)
<br><br> = 1 + 1 + 1 = 3
|
integer
|
jee-main-2020-online-7th-january-morning-slot
|
hdTiOWFqbqx1gGwexE7k9k2k5hiwqze
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let S be the set of all functions ƒ : [0,1] $$ \to $$ R,
which are continuous on [0,1] and differentiable
on (0,1). Then for every ƒ in S, there exists a
c $$ \in $$ (0,1), depending on ƒ, such that
|
[{"identifier": "A", "content": "$$\\left| {f(c) - f(1)} \\right| < \\left| {f'(c)} \\right|$$"}, {"identifier": "B", "content": "$$\\left| {f(c) + f(1)} \\right| < \\left( {1 + c} \\right)\\left| {f'(c)} \\right|$$"}, {"identifier": "C", "content": "$$\\left| {f(c) - f(1)} \\right| < \\left( {1 - c} \\right)\\left| {f'(c)} \\right|$$"}, {"identifier": "D", "content": "None"}]
|
["D"]
| null |
<p>If we consider the case where f(x) is a constant function, then its derivative f'(x) is equal to 0 for all x in the interval (0,1). </p>
<p>Therefore, if we substitute this into the expressions provided in Options A, B and C, we would have :</p>
<p>Option A : |f(c) - f(1)| < |f'(c)| would become |constant - constant| < |0|, which is 0 < 0. This is not true.</p>
<p>Option B : |f(c) + f(1)| < (1 + c)|f'(c)| would become |constant + constant| < (1 + c)$$ \times $$0, which is a positive number < 0. This is not true.</p>
<p>Option C : |f(c) - f(1)| < (1 - c)|f'(c)| would become |constant - constant| < (1 - c)$$ \times $$0, which is 0 < 0. This is not true.</p>
<p>Hence, for the case where f(x) is a constant function, none of the options A, B and C are correct.</p>
<p>So, the correct answer would be Option D : None.</p>
|
mcq
|
jee-main-2020-online-8th-january-evening-slot
|
QQ2Lw8UucYICcFIsyS7k9k2k5ita4xf
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let ƒ be any function continuous on [a, b] and
twice differentiable on (a, b). If for all x $$ \in $$ (a, b),
ƒ'(x) > 0 and ƒ''(x) < 0, then for any c $$ \in $$ (a, b),
$${{f(c) - f(a)} \over {f(b) - f(c)}}$$ is greater than :
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$${{b - c} \\over {c - a}}$$"}, {"identifier": "C", "content": "$${{b + a} \\over {b - a}}$$"}, {"identifier": "D", "content": "$${{c - a} \\over {b - c}}$$"}]
|
["D"]
| null |
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264640/exam_images/thggkxyle8cgcwavqqvd.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 9th January Morning Slot Mathematics - Limits, Continuity and Differentiability Question 150 English Explanation">
<br><br>It is clear from graph that, slope of AC $$>$$ slope of CB
<br><br>$$ \Rightarrow $$ $${{f\left( c \right) - f\left( a \right)} \over {c - a}}$$ $$>$$ $${{f\left( b \right) - f\left( c \right)} \over {b - c}}$$
<br><br>$$ \Rightarrow $$ $${{f\left( c \right) - f\left( a \right)} \over {f\left( b \right) - f\left( c \right)}}$$ $$ > {{c - a} \over {b - c}}$$
|
mcq
|
jee-main-2020-online-9th-january-morning-slot
|
hojNfgyrlIYlVlbyLXjgy2xukf8zuozg
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Suppose a differentiable function f(x) satisfies the identity <br/>f(x+y) = f(x) + f(y) + xy<sup>2</sup> + x<sup>2</sup>y, for all real x and y.<br/>
$$\mathop {\lim }\limits_{x \to 0} {{f\left( x \right)} \over x} = 1$$, then f'(3) is equal to ______.
|
[]
| null |
10
|
Given, f(x + y) = f(x) + f(y) + xy<sup>2</sup> + x<sup>2</sup>y ...(1)<br><br>differentiating partially with respect to x,<br><br>f'(x+y) = f'(x) + 0 + y<sup>2</sup> + y(2x) [y = constant]<br><br>Put x = 0 and y = x<br><br>$$ \therefore $$ f'(x) = f'(0) + x<sup>2</sup> ....(2)<br><br>putting x = y = 0 at equation (1),<br><br>f(0) = 2f(0)<br><br>$$ \Rightarrow $$ f(0) = 0<br><br>Given, $$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over x} = 1$$<br><br>This is in $$ \frac{0}{0} $$ form, so we can apply L' hospital rule.<br><br>$$\mathop {\lim }\limits_{x \to 0} {{f'(x)} \over 1} = 1$$<br><br>$$ \Rightarrow f'(0) = 1$$<br><br>Putting value of f'(0) at equation (2), we get<br><br>f'(x) = 1 + x<sup>2</sup><br><br>$$ \therefore $$ f'(3) = 1 + 3<sup>2</sup> = 10
|
integer
|
jee-main-2020-online-4th-september-morning-slot
|
NQ1pUMPuqzxQ3JslnJjgy2xukfah3nal
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let $$f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)$$ be a differentiable function such that f(1) = e and <br/>$$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$$. If f(x) = 1, then x is equal to :
|
[{"identifier": "A", "content": "$${1 \\over e}$$"}, {"identifier": "B", "content": "e"}, {"identifier": "C", "content": "$${1 \\over 2e}$$"}, {"identifier": "D", "content": "2e"}]
|
["A"]
| null |
$$\mathop {\lim }\limits_{t \to x} {{{t^2}{f^2}(x) - {x^2}{f^2}(t)} \over {t - x}} = 0$$
<br><br>(Using L'Hospital's Rule)<br><br>$$ \Rightarrow \mathop {\lim }\limits_{t \to x} {{2t{f^2}(x) - 2{x^2}f(t).f'(t)} \over 1} = 0$$
<br><br>$$ \Rightarrow $$ 2xf<sup>2</sup>(x) - 2x<sup>2</sup>.f(x).f'(x) = 0
<br><br>$$ \Rightarrow $$ 2xf(x){f(x) - xf'(x)} = 0
<br><br>[x $$ \ne $$ 0, f(x) $$ \ne $$ 0 as given <br>function $$f:\left( {0,\infty } \right) \to \left( {0,\infty } \right)$$ only takes positive value as input and output]
<br><br>$$ \Rightarrow f(x) = xf'(x) $$
<br><br>$$\Rightarrow {{f'(x)} \over {f(x)}} = {1 \over x}$$<br><br>Integrating w.r.t x, we get<br><br>$$ \Rightarrow ln\,f(x) = ln\,x + ln\,C$$<br><br>$$ \Rightarrow f(x) = Cx$$<br><br>$$ \because $$ f(1) = e<br><br>$$ \Rightarrow C = e;\,so\,f(x) = ex$$<br><br>When f(x) = 1 = ex<br><br>$$ \Rightarrow x = {1 \over e}$$
|
mcq
|
jee-main-2020-online-4th-september-evening-slot
|
fskTSdIfHZmkcWjazCjgy2xukfak0lbu
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
The function <br/>$$f(x) = \left\{ {\matrix{
{{\pi \over 4} + {{\tan }^{ - 1}}x,} & {\left| x \right| \le 1} \cr
{{1 \over 2}\left( {\left| x \right| - 1} \right),} & {\left| x \right| > 1} \cr
} } \right.$$ is :
|
[{"identifier": "A", "content": "continuous on R\u2013{\u20131} and differentiable on R\u2013{\u20131, 1}"}, {"identifier": "B", "content": "both continuous and differentiable on R\u2013{1}\n"}, {"identifier": "C", "content": "both continuous and differentiable on R\u2013{\u20131}"}, {"identifier": "D", "content": "continuous on R\u2013{1} and differentiable on R\u2013{\u20131, 1}\n"}]
|
["D"]
| null |
$$f\left( x \right) = \left\{ {\matrix{
{{\pi \over 4} + {{\tan }^{ - 1}}x,} & {x \in \left[ { - 1,1} \right]} \cr
{{1 \over 2}\left( {x - 1} \right),} & {x > 1} \cr
{{1 \over 2}\left( { - x - 1} \right),} & {x < - 1} \cr
} } \right.$$
<br><br>At x = 1
<br><br>L.H.L = $$\mathop {\lim }\limits_{x \to {1^ - }} \left( {{\pi \over 4} + {{\tan }^{ - 1}}x} \right)$$ = $${{\pi \over 4} + {\pi \over 4}}$$ = $${{\pi \over 2}}$$
<br><br>f(1) = $${{\pi \over 4} + {{\tan }^{ - 1}}x}$$ = $${{\pi \over 4} + {\pi \over 4}}$$ = $${{\pi \over 2}}$$
<br><br>R.H.L = $$\mathop {\lim }\limits_{x \to {1^ + }} \left( {{1 \over 2}\left( {x - 1} \right)} \right)$$ = 0
<br><br>As L.H.L $$ \ne $$ R.H.L so function is discontinuous $$ \Rightarrow $$ non differentiable.
<br><br>At x = -1
<br><br>L.H.L = $$\mathop {\lim }\limits_{x \to - {1^ - }} \left( {{1 \over 2}\left( { - x - 1} \right)} \right)$$ = $${{1 \over 2}\left( { - \left( { - 1} \right) - 1} \right)}$$ = 0
<br><br>f(-1) = $${\pi \over 4} + {\tan ^{ - 1}}\left( { - 1} \right)$$ = $${\pi \over 4} - {\pi \over 4}$$ = 0
<br><br>R.H.L = $$\mathop {\lim }\limits_{x \to - {1^ + }} \left( {{\pi \over 4} + {{\tan }^{ - 1}}x} \right)$$ <br><br>= $${\pi \over 4} + {\tan ^{ - 1}}\left( { - 1} \right)$$ = $${\pi \over 4} - {\pi \over 4}$$ = 0
<br><br>As L.H.L = f(-1) = R.H.L so function is continuous.
<br><br>$$f'\left( x \right) = \left\{ {\matrix{
{{1 \over {1 + {x^2}}},} & {x \in \left[ { - 1,1} \right]} \cr
{{1 \over 2},} & {x > 1} \cr
{ - {1 \over 2},} & {x < - 1} \cr
} } \right.$$
<br><br>For differentiability at x = –1
<br><br>L.H.D = $${ - {1 \over 2}}$$
<br><br>R.H.D. = $${{1 \over 2}}$$
<br><br>So, non differentiable at x = –1
|
mcq
|
jee-main-2020-online-4th-september-evening-slot
|
Qnulz8PdgvmSS14jDPjgy2xukfg792co
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
If the function <br/>$$f\left( x \right) = \left\{ {\matrix{
{{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \cr
{{k_2}\cos x,} & {x > \pi } \cr
} } \right.$$ is<br/> twice differentiable, then the ordered pair (k<sub>1</sub>, k<sub>2</sub>) is equal to :
|
[{"identifier": "A", "content": "$$\\left( {{1 \\over 2},-1} \\right)$$"}, {"identifier": "B", "content": "(1, 1)"}, {"identifier": "C", "content": "(1, 0)"}, {"identifier": "D", "content": "$$\\left( {{1 \\over 2},1} \\right)$$"}]
|
["D"]
| null |
Given, $$f\left( x \right) = \left\{ {\matrix{
{{k_1}{{\left( {x - \pi } \right)}^2} - 1,} & {x \le \pi } \cr
{{k_2}\cos x,} & {x > \pi } \cr
} } \right.$$
<br><br>Differentiating one time,
<br><br>$$f'\left( x \right) = \left\{ {\matrix{
{2{k_1}\left( {x - \pi } \right),} & {x \le \pi } \cr
{ - {k_2}\sin x,} & {x > \pi } \cr
} } \right.$$
<br><br>Differentiating one more time,
<br><br>$$f''\left( x \right) = \left\{ {\matrix{
{2{k_1},} & {x \le \pi } \cr
{ - {k_2}\cos x,} & {x > \pi } \cr
} } \right.$$
<br><br>As f''(x) is differentiable so
<br><br>f''($$\pi $$<sup>+</sup>) = f''($$\pi $$<sup>-</sup>)
<br><br>$$ \Rightarrow $$ -k<sub>2</sub>(-1) = 2k<sub>1</sub>
<br><br>$$ \Rightarrow $$ 2k<sub>1</sub> = k<sub>2</sub>
<br><br>$$ \therefore $$ (k<sub>1</sub>, k<sub>2</sub>) = $$\left( {{1 \over 2},1} \right)$$
|
mcq
|
jee-main-2020-online-5th-september-morning-slot
|
k2z9XqI4UQVKoU9SK7jgy2xukfxgleez
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let f : R $$ \to $$ R be defined as
<br/>$$f\left( x \right) = \left\{ {\matrix{
{{x^5}\sin \left( {{1 \over x}} \right) + 5{x^2},} & {x < 0} \cr
{0,} & {x = 0} \cr
{{x^5}\cos \left( {{1 \over x}} \right) + \lambda {x^2},} & {x > 0} \cr
} } \right.$$
<br/><br/>The value of $$\lambda $$ for which f ''(0) exists, is _______.
|
[]
| null |
5
|
If g(x) = x<sup>5</sup>sin$$\left( {{1 \over x}} \right)$$
<br><br>and h(x) = x<sup>5</sup>cos$$\left( {{1 \over x}} \right)$$
<br><br>then g''(0) = 0 and h''(0) = 0
<br><br>So, f''(0<sup>+</sup>
) = g''(0<sup>+</sup>
) + 10 = 10
<br><br>and f''(0<sup>–</sup>) = h''(0<sup>–</sup>) + 2$$\lambda $$ = f''(0<sup>+</sup>)
<br><br>$$ \Rightarrow $$ 2$$\lambda $$ = 10
<br><br>$$ \Rightarrow $$ $$\lambda $$ = 5
|
integer
|
jee-main-2020-online-6th-september-morning-slot
|
hI9jZChLQ9ZIKlSkxWjgy2xukg38e1e8
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let f : R $$ \to $$ R be a function defined by<br/> f(x) = max {x, x<sup>2</sup>}. Let S denote the set of all points in R, where f is not differentiable.
Then :
|
[{"identifier": "A", "content": "{0, 1}"}, {"identifier": "B", "content": "{0}"}, {"identifier": "C", "content": "$$\\phi $$(an empty set)"}, {"identifier": "D", "content": "{1}"}]
|
["A"]
| null |
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266844/exam_images/nsxa64efsvtbdkfgqlui.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Evening Slot Mathematics - Limits, Continuity and Differentiability Question 133 English Explanation">
<br>From graph you can see,
<br>(1) when x < 0 then y = x<sup>2</sup> is greater than y = x. That is why for f(x) that curved part is chosen.
<br>(2) when 0 $$ \le $$ x < 1 then y = x is greater than y = x<sup>2</sup>. That is why for f(x) part of that straight line is chosen.
<br>(3) when x $$ \ge $$ 1 then y = x<sup>2</sup> is greater than y = x. That is why for f(x) that curved part is chosen.
<br><br>Here on the graph of f(x) there is two sharp corner at x = 0 and x = 1. As we know no function is differentiable at the sharp corner. So f(x) is not differentiable at those two sharp corner.
|
mcq
|
jee-main-2020-online-6th-september-evening-slot
|
tmzgtioYKrhGyobz5kjgy2xukg38nmd4
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
For all twice differentiable functions f : R $$ \to $$ R,<br/>
with f(0) = f(1) = f'(0) = 0
|
[{"identifier": "A", "content": "f''(x) $$ \\ne $$ 0, at every point x $$ \\in $$ (0, 1)\n"}, {"identifier": "B", "content": "f''(x) = 0, for some x $$ \\in $$ (0, 1)"}, {"identifier": "C", "content": "f''(0) = 0\n"}, {"identifier": "D", "content": "f''(x) = 0, at every point x $$ \\in $$ (0, 1)"}]
|
["B"]
| null |
f : R $$ \to $$ R, with f(0) = f(1) = 0
<br>and f'(0) = 0
<br>$$ \because $$ f(x) is differentiable and continuous
<br>and f(0) = f(1) = 0
<br><br>Applying Rolle’s theorem in [0, 1] for function f(x)
<br>f'(c) = 0, c $$ \in $$ (0, 1)
<br><br>Now again
<br>$$ \because $$ f'(c) = 0, f'(0) = 0
<br>again applying Rolles theorem in [0, c] for function f'(x)
<br>f''(c<sub>1</sub>) = 0 for some c<sub>1</sub> $$ \in $$ (0, c) $$ \in $$ (0, 1)
|
mcq
|
jee-main-2020-online-6th-september-evening-slot
|
7CStJ13dzNzxBm9hXE1kls5qoi3
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
The number of points, at which the function <br/>f(x) = | 2x + 1 | $$-$$ 3| x + 2 | + | x<sup>2</sup> + x $$-$$ 2 |, x$$\in$$R is not differentiable, is __________.
|
[]
| null |
2
|
$$f(x) = |2x + 1| - 3|x + 2| + |{x^2} + x - 2|$$<br><br>$$f(x) = \left\{ {\matrix{
{{x^2} - 7;} & {x > 1} \cr
{ - {x^2} - 2x - 3;} & { - {1 \over 2} < x < 1} \cr
{ - {x^2} - 6x - 5;} & { - 2 < x < {{ - 1} \over 2}} \cr
{{x^2} + 2x + 3;} & {x < - 2} \cr
} } \right.$$<br><br> $$ \therefore $$ $$f'(x) = \left\{ {\matrix{
{2x;} & {x > 1} \cr
{2x - 3;} & { - {1 \over 2} < x < 1} \cr
{ - 2x - 6;} & { - 2 < x < {{ - 1} \over 2}} \cr
{2x + 2;} & {x < - 2} \cr
} } \right.$$<br><br>Check at 1, $$-$$2 and $${{ - 1} \over 2}$$<br><br>Non. differentiable at x = 1 and $${{ - 1} \over 2}$$
|
integer
|
jee-main-2021-online-25th-february-morning-slot
|
G51JVSYGYUdeZKBExi1klt9xprc
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
A function f is defined on [$$-$$3, 3] as<br/><br/>$$f(x) = \left\{ {\matrix{
{\min \{ |x|,2 - {x^2}\} ,} & { - 2 \le x \le 2} \cr
{[|x|],} & {2 < |x| \le 3} \cr
} } \right.$$ where [x] denotes the greatest integer $$ \le $$ x. The number of points, where f is not differentiable in ($$-$$3, 3) is ___________.
|
[]
| null |
5
|
<img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263292/exam_images/hlto1ikhz1kx3bmvaaoo.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 25th February Evening Shift Mathematics - Limits, Continuity and Differentiability Question 127 English Explanation">
<br>Points of non-differentiability in ($$-$$3, 3) are at x = $$-$$2, $$-$$1, 0, 1, 2.<br><br>i.e. 5 points.
|
integer
|
jee-main-2021-online-25th-february-evening-slot
|
ZnqJ1UOMHCkoMDoY6E1kluvnyb9
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let f(x) be a differentiable function at x = a with f'(a) = 2 and f(a) = 4. <br/><br/>Then $$\mathop {\lim }\limits_{x \to a} {{xf(a) - af(x)} \over {x - a}}$$ equals :
|
[{"identifier": "A", "content": "4 $$-$$ 2a"}, {"identifier": "B", "content": "2a + 4"}, {"identifier": "C", "content": "a + 4"}, {"identifier": "D", "content": "2a $$-$$ 4"}]
|
["A"]
| null |
$$L = \mathop {\lim }\limits_{x \to a} {{xf(a) - af(x)} \over {x - a}}$$ [$${0 \over 0}$$ form] <br><br>Using L' Hospital rule we get<br><br>$$L = \mathop {\lim }\limits_{x \to a} {{f(a) - af'(x)} \over 1}$$<br><br>$$f(a) - af'(a) = 4 - 2a$$
|
mcq
|
jee-main-2021-online-26th-february-evening-slot
|
VY62ViQI22I2zXz9wn1kmhxcjph
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let the functions f : R $$ \to $$ R and g : R $$ \to $$ R be defined as :<br/><br/>$$f(x) = \left\{ {\matrix{
{x + 2,} & {x < 0} \cr
{{x^2},} & {x \ge 0} \cr
} } \right.$$ and <br/><br/>$$g(x) = \left\{ {\matrix{
{{x^3},} & {x < 1} \cr
{3x - 2,} & {x \ge 1} \cr
} } \right.$$<br/><br/>Then, the number of points in R where (fog) (x) is NOT differentiable is equal to :
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}]
|
["C"]
| null |
$$fog(x) = \left\{ {\matrix{
{{x^3} + 2,} & {x \le 0} \cr
{{x^6},} & {0 \le x \le 1} \cr
{{{(3x - 2)}^2},} & {x \ge 1} \cr
} } \right.$$<br><br>$$ \because $$ fog(x) is discontinuous at x = 0 then non-differentiable at x = 0<br><br>Now, <br><br>at x = 1<br><br>$$RHD = \mathop {\lim }\limits_{h \to 0} {{f(1 + h) - f(1)} \over h} = \mathop {\lim }\limits_{h \to 0} {{{{(3(1 + h) - 2)}^2} - 1} \over h} = 6$$<br><br>$$LHD = \mathop {\lim }\limits_{h \to 0} {{f(1 - h) - f(1)} \over { - h}} = \mathop {\lim }\limits_{h \to 0} {{{{(1 - h)}^6} - 1} \over { - h}} = 6$$<br><br>Number of points of non-differentiability = 1
|
mcq
|
jee-main-2021-online-16th-march-morning-shift
|
5A659Mvl7seLH4Q37Z1kmiwihi3
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let f : S $$ \to $$ S where S = (0, $$\infty $$) be a twice differentiable function such that f(x + 1) = xf(x). If g : S $$ \to $$ R be defined as g(x) = log<sub>e</sub> f(x), then the value of |g''(5) $$-$$ g''(1)| is equal to :
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "$${{187} \\over {144}}$$"}, {"identifier": "C", "content": "$${{197} \\over {144}}$$"}, {"identifier": "D", "content": "$${{205} \\over {144}}$$"}]
|
["D"]
| null |
$$f(x + 1) = xf(x)$$<br><br>$$\ln (f(x + 1)) = \ln x + \ln f(x)$$<br><br>$$g(x + 1) = \ln x + g(x)$$<br><br>$$g(x + 1) - g(x) = \ln x$$ ..... (i)<br><br>$$g'(x + 1) - g'(x) = {1 \over x}$$<br><br>$$g''(x + 1) - g''(x) = {{ - 1} \over {{x^2}}}$$<br><br>$$g''(2) - g'(1) = {{ - 1} \over 1}$$ .... (ii)<br><br>$$g''(3) - g''(2) = {{ - 1} \over 4}$$ .... (iii)<br><br>$$g''(4) - g''(3) = {{ - 1} \over 9}$$ ..... (iv)<br><br>$$g''(5) - g''(4) = {{ - 1} \over {16}}$$ ....(v)<br><br>Adding (ii), (iii), (iv) & (v)<br><br>$$g''(5) - g''(1) = - \left( {{1 \over 1} + {1 \over 4} + {1 \over 9} + {1 \over {16}}} \right) = {{ - 205} \over {144}}$$<br><br>$$|g''(5) - g''(1)|\, = {{205} \over {144}}$$
|
mcq
|
jee-main-2021-online-16th-march-evening-shift
|
nOdMVjSHQppilEI1PC1kmlj3ysh
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
If $$f(x) = \left\{ {\matrix{
{{1 \over {|x|}}} & {;\,|x|\, \ge 1} \cr
{a{x^2} + b} & {;\,|x|\, < 1} \cr
} } \right.$$ is differentiable at every point of the domain, then the values of a and b are respectively :
|
[{"identifier": "A", "content": "$${1 \\over 2},{1 \\over 2}$$"}, {"identifier": "B", "content": "$${1 \\over 2}, - {3 \\over 2}$$"}, {"identifier": "C", "content": "$${5 \\over 2}, - {3 \\over 2}$$"}, {"identifier": "D", "content": "$$ - {1 \\over 2},{3 \\over 2}$$"}]
|
["D"]
| null |
$$f(x) = \left\{ {\matrix{
{{1 \over {|x|}},} & {|x| \ge 1} \cr
{a{x^2} + b,} & {|x| < 1} \cr
} } \right.$$<br><br>$$ = \left\{ {\matrix{
{ - {1 \over x};} & {x \le - 1} \cr
{a{x^2} + b;} & { - 1 < x < 1} \cr
{{1 \over x};} & {x \ge 1} \cr
} } \right.$$<br><br>As f(x) is differentiable so it is also continuous,<br><br> at x = 1,<br><br>$$\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} f(x)$$<br><br>$$ \Rightarrow a + b = {1 \over 1}$$<br><br>$$ \Rightarrow a + b = 1$$ ...... (1)<br><br>As f(x) is differentiable, so at x = 1<br><br>L.H.D. = R.H.D.<br><br>$$ \Rightarrow 2ax = - {1 \over {{x^2}}}$$<br><br>$$ \Rightarrow 2a = - 1$$<br><br>$$ \Rightarrow a = - {1 \over 2}$$<br><br>From (1), $$b = {3 \over 2}$$
|
mcq
|
jee-main-2021-online-18th-march-morning-shift
|
Xf9fFqIdYJZD5on2nB1kmm49510
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let f : R $$ \to $$ R satisfy the equation f(x + y) = f(x) . f(y) for all x, y $$\in$$R and f(x) $$\ne$$ 0 for any x$$\in$$R. If the function f is differentiable at x = 0 and f'(0) = 3, then <br/><br/>$$\mathop {\lim }\limits_{h \to 0} {1 \over h}(f(h) - 1)$$ is equal to ____________.
|
[]
| null |
3
|
<p>Given, $$f(x + y) = f(x)\,.\,f(y)\,\forall x,y \in R$$</p>
<p>$$\therefore$$ $$f(x) = {a^x} \Rightarrow f'(x) = {a^x}\,.\,\log (a)$$</p>
<p>Now, $$f'(0) = \log (a) \Rightarrow 3 = \log (a) \Rightarrow a = {e^3}$$</p>
<p>$$\therefore$$ $$f(x) = {({e^3})^x} = {e^{3x}}$$</p>
<p>$$\therefore$$ $$f(h) = {e^{3h}}$$</p>
<p>Now, $$\mathop {\lim }\limits_{h \to 0} \left( {{{f(h) - 1} \over h}} \right) = \mathop {\lim }\limits_{h \to 0} \left( {{{{e^{3h}} - 1} \over h}} \right)$$</p>
<p>$$ = \mathop {\lim }\limits_{h \to 0} \left( {{{{e^{3h}} - 1} \over {3h}} \times 3} \right) = 3 \times 1 = 3$$</p>
|
integer
|
jee-main-2021-online-18th-march-evening-shift
|
1krrwrbii
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let a function g : [ 0, 4 ] $$\to$$ R be defined as <br/><br/>$$g(x) = \left\{ {\matrix{
{\mathop {\max }\limits_{0 \le t \le x} \{ {t^3} - 6{t^2} + 9t - 3),} & {0 \le x \le 3} \cr
{4 - x,} & {3 < x \le 4} \cr
} } \right.$$, then the number of points in the interval (0, 4) where g(x) is NOT differentiable, is ____________.
|
[]
| null |
1
|
$$f(x) = {x^3} - 6{x^2} + 9x - 3$$<br><br>$$f(x) = 3{x^2} - 12x + 9 = 3(x - 1)(x - 3)$$<br><br>$$f(1) = 1$$, $$f(3) = 3$$<br><br>$$g(x) = \left[ {\matrix{
{f(9x)} & {0 \le x \le 1} \cr
0 & {1 \le x \le 3} \cr
{ - 1} & {3 < x \le 4} \cr
} } \right.$$<br><br>g(x) is continuous<br><br>$$g'(x) = \left[ {\matrix{
{3(x - 1)(x - 3)} & {0 \le x \le 1} \cr
0 & {1 \le x \le 3} \cr
{ - 1} & {3 < x \le 4} \cr
} } \right.$$<br><br>g(x) is non-differentiable at x = 3
|
integer
|
jee-main-2021-online-20th-july-evening-shift
|
1krubclxp
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let f : R $$\to$$ R be a function defined as $$f(x) = \left\{ {\matrix{
{3\left( {1 - {{|x|} \over 2}} \right)} & {if} & {|x|\, \le 2} \cr
0 & {if} & {|x|\, > 2} \cr
} } \right.$$<br/><br/>Let g : R $$\to$$ R be given by $$g(x) = f(x + 2) - f(x - 2)$$. If n and m denote the number of points in R where g is not continuous and not differentiable, respectively, then n + m is equal to ______________.
|
[]
| null |
4
|
<p>$$f(x) = \left\{ {\matrix{
{3\left( {{{1 - \left| x \right|} \over 2}} \right)} & {if\,\left| x \right| \le 2} \cr
0 & {if\,\left| x \right| > 2} \cr
} } \right.$$</p>
<p>$$g(x) = f(x + 2) - f(x - 2)$$</p>
<p>$$f(x) = \left\{ {\matrix{
{0,} & {x < - 2} \cr
{{3 \over 2}(1 + x),} & { - 2 \le x < 0} \cr
{{3 \over 2}(1 - x),} & {0 \le x < 2} \cr
{0,} & {x > 2} \cr
} } \right.$$</p>
<p>$$f(x + 2) = \left\{ {\matrix{
{0,} & {x < - 4} \cr
{{3 \over 2}( 3 + x),} & { - 4 \le x < - 2} \cr
{{3 \over 2}( - 1 - x),} & { - 2 \le x < 0} \cr
{0,} & {x > 4} \cr
} } \right.$$</p>
<p>$$f(x - 2) = \left\{ {\matrix{
{0,} & {x < 0} \cr
{{3 \over 2}(x - 1),} & {0 \le x < 2} \cr
{{3 \over 2}( - 1 - x),} & {2 \le x < 4} \cr
{0,} & {x > 4} \cr
} } \right.$$</p>
<p>$$g(x) = f(x + 2) + f(x - 2)$$</p>
<p>$$ = \left\{ {\matrix{
{{{3x} \over 2} + 6,} & { - 4 \le x \le 2} \cr
{ - {{3x} \over 2},} & { - 2 < x < 2} \cr
{{{3x} \over 2} - 6,} & {2 \le x \le 4} \cr
{0,} & {\left| x \right| > 4} \cr
} } \right.$$</p>
<p>So, n = 0 and m = 4</p>
<p>$$\therefore$$ m + n = 4</p>
|
integer
|
jee-main-2021-online-22th-july-evening-shift
|
1krxlb5jm
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let $$f:[0,\infty ) \to [0,3]$$ be a function defined by <br/><br/>$$f(x) = \left\{ {\matrix{
{\max \{ \sin t:0 \le t \le x\} ,} & {0 \le x \le \pi } \cr
{2 + \cos x,} & {x > \pi } \cr
} } \right.$$<br/><br/>Then which of the following is true?
|
[{"identifier": "A", "content": "f is continuous everywhere but not differentiable exactly at one point in (0, $$\\infty$$)"}, {"identifier": "B", "content": "f is differentiable everywhere in (0, $$\\infty$$)"}, {"identifier": "C", "content": "f is not continuous exactly at two points in (0, $$\\infty$$)"}, {"identifier": "D", "content": "f is continuous everywhere but not differentiable exactly at two points in (0, $$\\infty$$)"}]
|
["B"]
| null |
Graph of $$\max \{ \sin t:0 \le t \le x\} $$ in $$x \in [0,\pi ]$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267003/exam_images/njxdcddt27d4labpnhcm.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Evening Shift Mathematics - Limits, Continuity and Differentiability Question 99 English Explanation 1"><br><br>& graph of cos x for $$x \in [\pi ,\infty )$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266177/exam_images/mlkrr07qnsgtcedtgrri.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Evening Shift Mathematics - Limits, Continuity and Differentiability Question 99 English Explanation 2"><br><br>So graph of <br><br>$$f(x) = \left\{ {\matrix{
{\max \{ \sin t:0 \le t \le x\} ,} & {0 \le x \le \pi } \cr
{2 + \cos x,} & {x > \pi} \cr
} } \right.$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267341/exam_images/xx4dogq2jwbvxilj1i8d.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Evening Shift Mathematics - Limits, Continuity and Differentiability Question 99 English Explanation 3"><br><br>f(x) is differentiable everywhere in (0, $$\infty$$)
|
mcq
|
jee-main-2021-online-27th-july-evening-shift
|
1ks0dcadh
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let $$f:[0,3] \to R$$ be defined by $$f(x) = \min \{ x - [x],1 + [x] - x\} $$ where [x] is the greatest integer less than or equal to x. Let P denote the set containing all x $$\in$$ [0, 3] where f i discontinuous, and Q denote the set containing all x $$\in$$ (0, 3) where f is not differentiable. Then the sum of number of elements in P and Q is equal to ______________.
|
[]
| null |
5
|
<picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264944/exam_images/avmgbcbdz7wsmx5hhnpg.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264951/exam_images/ot72zwnlvx1tkyc9kohr.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267638/exam_images/uj4pdcgnkf2sk9ywcaok.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Morning Shift Mathematics - Limits, Continuity and Differentiability Question 95 English Explanation 1"></picture> <br><br>1 $$-$$ {x} = 1 $$-$$ x; 0 $$\le$$ x < 1<br><br> <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734265349/exam_images/oa2ttguxwfbldlp3zpq0.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264167/exam_images/riljabrvycs6ypoits0f.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267197/exam_images/qgjuuqtt5dvgetesk2np.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266706/exam_images/ojrovdzbjtospfgtjkfj.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Morning Shift Mathematics - Limits, Continuity and Differentiability Question 95 English Explanation 2"></picture> <br><br>Non differentiable at<br><br>$$x = {1 \over 2},1,{3 \over 2},2,{5 \over 2}$$
|
integer
|
jee-main-2021-online-27th-july-morning-shift
|
1ktcxxctq
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let [t] denote the greatest integer less than or equal to t. Let <br/>f(x) = x $$-$$ [x], g(x) = 1 $$-$$ x + [x], and h(x) = min{f(x), g(x)}, x $$\in$$ [$$-$$2, 2]. Then h is :
|
[{"identifier": "A", "content": "continuous in [$$-$$2, 2] but not differentiable at more than <br>four points in ($$-$$2, 2)"}, {"identifier": "B", "content": "not continuous at exactly three points in [$$-$$2, 2]"}, {"identifier": "C", "content": "continuous in [$$-$$2, 2] but not differentiable at exactly <br>three points in ($$-$$2, 2)"}, {"identifier": "D", "content": "not continuous at exactly four points in [$$-$$2, 2]"}]
|
["A"]
| null |
min{x $$-$$ [x], 1 $$-$$ x + [x]}<br><br>h(x) = min{x $$-$$ [x], 1 $$-$$ [x $$-$$ [x])}<br><br> <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266549/exam_images/ownvykvbglm0alabpgjc.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264579/exam_images/tump60qgktboye2c0orm.webp"><source media="(max-width: 680px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267659/exam_images/xysrrqavu5scrrxfshfc.webp"><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734263692/exam_images/yav6eilmb28mctcjaeyf.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 26th August Evening Shift Mathematics - Limits, Continuity and Differentiability Question 91 English Explanation"></picture> <br><br>$$\Rightarrow$$ always continuous in [$$-$$2, 2] but not differentiable at 7 points.
|
mcq
|
jee-main-2021-online-26th-august-evening-shift
|
1ktipaq58
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
The function <br/><br/>$$f(x) = \left| {{x^2} - 2x - 3} \right|\,.\,{e^{\left| {9{x^2} - 12x + 4} \right|}}$$ is not differentiable at exactly :
|
[{"identifier": "A", "content": "four points"}, {"identifier": "B", "content": "three points"}, {"identifier": "C", "content": "two points "}, {"identifier": "D", "content": "one point"}]
|
["C"]
| null |
$$f(x) = \left| {(x - 3)(x + 1)} \right|\,.\,{e^{{{(3x - 2)}^2}}}$$<br><br>$$f(x) = \left\{ {\matrix{
{(x - 3)(x + 1).\,{e^{{{(3x - 2)}^2}}}} & ; & {x \in (3,\infty )} \cr
{ - (x - 3)(x + 1).\,{e^{{{(3x - 2)}^2}}}} & ; & {x \in [ - 1,3]} \cr
{(x - 3)\,.\,(x + 1).\,{e^{{{(3x - 2)}^2}}}} & ; & {x \in ( - \infty , - 1)} \cr
} } \right.$$<br><br>Clearly, non-differentiable at x = $$-$$1 & x = 3.
|
mcq
|
jee-main-2021-online-31st-august-morning-shift
|
1ktk9w00p
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let f be any continuous function on [0, 2] and twice differentiable on (0, 2). If f(0) = 0, f(1) = 1 and f(2) = 2, then
|
[{"identifier": "A", "content": "f''(x) = 0 for all x $$\\in$$ (0, 2)"}, {"identifier": "B", "content": "f''(x) = 0 for some x $$\\in$$ (0, 2)"}, {"identifier": "C", "content": "f'(x) = 0 for some x $$\\in$$ [0, 2]"}, {"identifier": "D", "content": "f''(x) > 0 for all x $$\\in$$ (0, 2)"}]
|
["B"]
| null |
<p>f(0) = 0, f(1) = 1 and f(2) = 2</p>
<p>Let h(x) = f(x) $$-$$ x</p>
<p>Clearly h(x) is continuous and twice differentiable on (0, 2)</p>
<p>Also, h(0) = h(1) = h(2) = 0</p>
<p>$$\therefore$$ h(x) satisfies all the condition of Rolle's theorem.</p>
<p>$$\therefore$$ there exist C<sub>1</sub> $$\in$$(0, 1) such that h'(c<sub>1</sub>) = 0</p>
<p>$$\Rightarrow$$ f'(<sub>1</sub>) $$-$$ 1 = 0 $$\Rightarrow$$ f'(c<sub>1</sub>) = 1</p>
<p>also there exist c<sub>2</sub> $$\in$$(1, 2) such that h'(c<sub>2</sub>) = 0</p>
<p>$$\Rightarrow$$ f'(c<sub>2</sub>) = 1</p>
<p>Now, using Rolle's theorem on [c<sub>1</sub>, c<sub>2</sub>] for f'(x)</p>
<p>We have f''(c) = 0, c$$\in$$(c<sub>1</sub>, c<sub>2</sub>)</p>
<p>Hence, f''(x) = 0 for some x$$\in$$(0, 2).</p>
|
mcq
|
jee-main-2021-online-31st-august-evening-shift
|
1l589h9sz
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>Let f, g : R $$\to$$ R be two real valued functions defined as $$f(x) = \left\{ {\matrix{
{ - |x + 3|} & , & {x < 0} \cr
{{e^x}} & , & {x \ge 0} \cr
} } \right.$$ and $$g(x) = \left\{ {\matrix{
{{x^2} + {k_1}x} & , & {x < 0} \cr
{4x + {k_2}} & , & {x \ge 0} \cr
} } \right.$$, where k<sub>1</sub> and k<sub>2</sub> are real constants. If (gof) is differentiable at x = 0, then (gof) ($$-$$ 4) + (gof) (4) is equal to :</p>
|
[{"identifier": "A", "content": "$$4({e^4} + 1)$$"}, {"identifier": "B", "content": "$$2(2{e^4} + 1)$$"}, {"identifier": "C", "content": "$$4{e^4}$$"}, {"identifier": "D", "content": "$$2(2{e^4} - 1)$$"}]
|
["D"]
| null |
<p>$$\because$$ gof is differentiable at x = 0</p>
<p>So R.H.D = L.H.D</p>
<p>$${d \over {dx}}(4{e^x} + {k_2}) = {d \over {dx}}\left( {{{( - |x + 3|)}^2} - {k_1}|x + 3|} \right)$$</p>
<p>$$ \Rightarrow 4 = 6 - {k_1} \Rightarrow {k_1} = 2$$</p>
<p>Also $$f(f({0^ + })) = g(f({0^ - }))$$</p>
<p>$$ \Rightarrow 4 + {k_2} = 9 - 3{k_1} \Rightarrow {k_2} = - 1$$</p>
<p>Now $$g(f( - 4)) + g(f(4))$$</p>
<p>$$ = g( - 1) + g({e^4}) = (1 - {k_1}) + (4{e^4} + {k_2})$$</p>
<p>$$ = 4{e^4} - 2$$</p>
<p>$$ = 2(2{e^4} - 1)$$</p>
|
mcq
|
jee-main-2022-online-26th-june-morning-shift
|
1l58f5paa
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>Let f(x) = min {1, 1 + x sin x}, 0 $$\le$$ x $$\le$$ 2$$\pi $$. If m is the number of points, where f is not differentiable and n is the number of points, where f is not continuous, then the ordered pair (m, n) is equal to</p>
|
[{"identifier": "A", "content": "(2, 0)"}, {"identifier": "B", "content": "(1, 0)"}, {"identifier": "C", "content": "(1, 1)"}, {"identifier": "D", "content": "(2, 1)"}]
|
["B"]
| null |
<p>$$f(x) = \min \{ 1,\,1 + x\sin x\} $$, $$0 \le x \le x$$</p>
<p>$$f(x) = \left\{ {\matrix{
{1,} & {0 \le x < \pi } \cr
{1 + x\sin x,} & {\pi \le x \le 2\pi } \cr
} } \right.$$</p>
<p>Now at $$x = \pi ,\,\,\mathop {\lim }\limits_{x \to {\pi ^ - }} f(x) = 1 = \mathop {\lim }\limits_{x \to {\pi ^ + }} f(x)$$</p>
<p>$$\therefore$$ f(x) is continuous in [0, 2$$\pi$$]</p>
<p>Now, at x = $$\pi$$ $$L.H.D = \mathop {\lim }\limits_{h \to 0} {{f(\pi - h) - f(\pi )} \over { - h}} = 0$$</p>
<p>$$R.H.D = \mathop {\lim }\limits_{h \to 0} {{f(\pi + h) - f(\pi )} \over h} = 1 - {{(\pi + h)\sin \,h - 1} \over h} = - \pi $$</p>
<p>$$\therefore$$ f(x) is not differentiable at x = $$\pi$$</p>
<p>$$\therefore$$ (m, n) = (1, 0)</p>
|
mcq
|
jee-main-2022-online-26th-june-evening-shift
|
1l5b7z89d
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>Let $$f(x) = \left\{ {\matrix{
{{{\sin (x - [x])} \over {x - [x]}}} & {,\,x \in ( - 2, - 1)} \cr
{\max \{ 2x,3[|x|]\} } & {,\,|x| < 1} \cr
1 & {,\,otherwise} \cr
} } \right.$$</p>
<p>where [t] denotes greatest integer $$\le$$ t. If m is the number of points where $$f$$ is not continuous and n is the number of points where $$f$$ is not differentiable, then the ordered pair (m, n) is :</p>
|
[{"identifier": "A", "content": "(3, 3)"}, {"identifier": "B", "content": "(2, 4)"}, {"identifier": "C", "content": "(2, 3)"}, {"identifier": "D", "content": "(3, 4)"}]
|
["C"]
| null |
<p>$$f(x) = \left\{ {\matrix{
{{{\sin (x - [x])} \over {x[x]}}} & , & {x \in ( - 2, - 1)} \cr
{\max \{ 2x,3[|x|]\} } & , & {|x| < 1} \cr
1 & , & {otherwise} \cr
} } \right.$$</p>
<p>$$f(x) = \left\{ {\matrix{
{{{\sin (x + 2)} \over {x + 2}}} & , & {x \in ( - 2, - 1)} \cr
0 & , & {x \in ( - 1,0]} \cr
{2x} & , & {x \in (0,1)} \cr
1 & , & {otherwise} \cr
} } \right.$$</p>
<p>It clearly shows that f(x) is discontinuous</p>
<p>At x = $$-$$1, 1 also non differentiable</p>
<p>and at $$x = 0$$, $$L.H.D = \mathop {\lim }\limits_{h \to 0} {{f(0 - h) - f(0)} \over { - h}} = 0$$</p>
<p>$$R.H.D = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h} = 2$$</p>
<p>$$\therefore$$ f(x) is not differentiable at x = 0</p>
<p>$$\therefore$$ m = 2, n = 3</p>
|
mcq
|
jee-main-2022-online-24th-june-evening-shift
|
1l6dxbjjs
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>Let $$f(x)=\left\{\begin{array}{l}\left|4 x^{2}-8 x+5\right|, \text { if } 8 x^{2}-6 x+1 \geqslant 0 \\ {\left[4 x^{2}-8 x+5\right], \text { if } 8 x^{2}-6 x+1<0,}\end{array}\right.$$ where $$[\alpha]$$ denotes the greatest integer less than or equal to $$\alpha$$. Then the number of points in $$\mathbf{R}$$ where $$f$$ is not differentiable is ___________. </p>
|
[]
| null |
3
|
$f(x)= \begin{cases}\left|4 x^{2}-8 x+5\right|, & \text { if } 8 x^{2}-6 x+1 \geq 0 \\ {\left[4 x^{2}-8 x+5\right],} & \text { if } 8 x^{2}-6 x+1<0\end{cases}$
<br><br>
$$
= \begin{cases}4 x^{2}-8 x+5, & \text { if } x \in\left[-\infty, \frac{1}{4}\right] \cup\left[\frac{1}{2}, \infty\right) \\ {\left[4 x^{2}-8 x+5\right]} & \text { if } x \in\left(\frac{1}{4}, \frac{1}{2}\right)\end{cases}
$$
<br><br>
$$f(x)=\left\{\begin{array}{cc}
4 x^2-8 x+5 & \text { if } x \in\left(-\infty, \frac{1}{4}\right] \cup\left[\frac{1}{2}, \infty\right) \\
3 & x \in\left(\frac{1}{4}, \frac{2-\sqrt{2}}{2}\right) \\
2 & x \in\left[\frac{2-\sqrt{2}}{2}, \frac{1}{2}\right)
\end{array}\right.$$<br><br>
<img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l97st8z9/f16176cf-a04a-4df8-9ae6-30f66a17f773/b9ee7e50-4b5d-11ed-bfde-e1cb3fafe700/file-1l97st8za.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l97st8z9/f16176cf-a04a-4df8-9ae6-30f66a17f773/b9ee7e50-4b5d-11ed-bfde-e1cb3fafe700/file-1l97st8za.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 - Limits, Continuity and Differentiability Question 63 English Explanation"><br>
$\therefore \quad$ Non-diff at $x=\frac{1}{4}, \frac{2-\sqrt{2}}{2}, \frac{1}{2}$
|
integer
|
jee-main-2022-online-25th-july-morning-shift
|
1l6m6jj6x
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>Let $$f:[0,1] \rightarrow \mathbf{R}$$ be a twice differentiable function in $$(0,1)$$ such that $$f(0)=3$$ and $$f(1)=5$$. If the line $$y=2 x+3$$ intersects the graph of $$f$$ at only two distinct points in $$(0,1)$$, then the least number of points $$x \in(0,1)$$, at which $$f^{\prime \prime}(x)=0$$, is ____________.</p>
|
[]
| null |
2
|
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7rb5odf/8441c3d2-e56e-4faa-a526-4c7b6253ae3c/ed702f30-2e7f-11ed-8702-156c00ced081/file-1l7rb5odg.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7rb5odf/8441c3d2-e56e-4faa-a526-4c7b6253ae3c/ed702f30-2e7f-11ed-8702-156c00ced081/file-1l7rb5odg.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2022 (Online) 28th July Morning Shift Mathematics - Limits, Continuity and Differentiability Question 54 English Explanation"></p>
<p>If a graph cuts $$y = 2x + 5$$ in (0, 1) twice then its concavity changes twice.</p>
<p>$$\therefore$$ $$f'(x) = 0$$ at at least two points.</p>
|
integer
|
jee-main-2022-online-28th-july-morning-shift
|
1l6p2shrs
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>The number of points, where the function $$f: \mathbf{R} \rightarrow \mathbf{R}$$,</p>
<p>$$f(x)=|x-1| \cos |x-2| \sin |x-1|+(x-3)\left|x^{2}-5 x+4\right|$$, is NOT differentiable, is :</p>
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "4"}]
|
["B"]
| null |
<p>$$f:R \to R$$.</p>
<p>$$f(x) = |x - 1|\cos |x - 2|\sin |x - 1| + (x - 3)|{x^2} - 5x + 4|$$</p>
<p>$$ = |x - 1|\cos |x - 2|\sin |x - 1| + (x - 3)|x - 1||x - 4|$$</p>
<p>$$ = |x - 1|[\cos |x - 2|\sin |x - 1| + (x - 3)|x - 4|]$$</p>
<p>Sharp edges at $$x = 1$$ and $$x = 4$$</p>
<p>$$\therefore$$ Non-differentiable at $$x = 1$$ and $$x = 4$$</p>
|
mcq
|
jee-main-2022-online-29th-july-morning-shift
|
1l6rfufia
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>If $$[t]$$ denotes the greatest integer $$\leq t$$, then the number of points, at which the function $$f(x)=4|2 x+3|+9\left[x+\frac{1}{2}\right]-12[x+20]$$ is not differentiable in the open interval $$(-20,20)$$, is __________.</p>
|
[]
| null |
79
|
$f(x)=4|2 x+3|+9\left[x+\frac{1}{2}\right]-12[x+20]$
<br/><br/>$$
=4|2 x+3|+9\left[x+\frac{1}{2}\right]-12[x]-240
$$
<br/><br/>$f(x)$ is non differentiable at $x=-\frac{3}{2}$
<br/><br/>and $f(x)$ is discontinuous at $\{-19,-18, \ldots ., 18,19\}$ <br/><br/>as well as $\left\{-\frac{39}{2},-\frac{37}{2}, \ldots,-\frac{3}{2},-\frac{1}{2}, \frac{1}{2}, \ldots, \frac{39}{2}\right\}$, <br/><br/>at same point they are also non differentiable
<br/><br/>$$
\begin{aligned}
\therefore & \text { Total number of points of non differentiability } \\
&=39+40 \\
&=79
\end{aligned}
$$
|
integer
|
jee-main-2022-online-29th-july-evening-shift
|
1ldr770hb
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>Suppose $$f: \mathbb{R} \rightarrow(0, \infty)$$ be a differentiable function such that $$5 f(x+y)=f(x) \cdot f(y), \forall x, y \in \mathbb{R}$$. If $$f(3)=320$$, then $$\sum_\limits{n=0}^{5} f(n)$$ is equal to :</p>
|
[{"identifier": "A", "content": "6875"}, {"identifier": "B", "content": "6525"}, {"identifier": "C", "content": "6575"}, {"identifier": "D", "content": "6825"}]
|
["D"]
| null |
<p>$$5f(x + y) = f(x).f(y)$$</p>
<p>$$5f(3) = f(1).f(2)$$</p>
<p>$$5f(2) = {(f(1))^2}$$</p>
<p>$$f(10) = 5$$</p>
<p>$$f(1) = 20$$</p>
<p>$$ \Rightarrow f(1).{{{{(f(1))}^2}} \over 5} = 1600$$</p>
<p>$$\sum\limits_{n = 0}^5 {f(n) = f(0) + 20 + 80 + 320 + 1280 + 5120} $$</p>
<p>$$ = 1750 + 5120 = 6825$$</p>
|
mcq
|
jee-main-2023-online-30th-january-morning-shift
|
1ldybn6xt
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>Let $$f(x) = \left\{ {\matrix{
{{x^2}\sin \left( {{1 \over x}} \right)} & {,\,x \ne 0} \cr
0 & {,\,x = 0} \cr
} } \right.$$</p>
<p>Then at $$x=0$$</p>
|
[{"identifier": "A", "content": "$$f$$ is continuous but $$f'$$ is not continuous"}, {"identifier": "B", "content": "$$f$$ and $$f'$$ both are continuous"}, {"identifier": "C", "content": "$$f$$ is continuous but not differentiable"}, {"identifier": "D", "content": "$$f'$$ is continuous but not differentiable"}]
|
["A"]
| null |
<p>Given,</p>
<p>$$f(x) = \left\{ {\matrix{
{{x^2}\sin \left( {{1 \over x}} \right),} & {x \ne 0} \cr
{0,} & {x = 0} \cr
} } \right.$$</p>
<p>$$\therefore$$ $$f'(x) = 2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)$$</p>
<p>Now,</p>
<p>$$\mathop {\lim }\limits_{x \to 0} f'(x) = \mathop {\lim }\limits_{x \to 0} \left[ {2x\sin \left( {{1 \over x}} \right) - \cos \left( {{1 \over x}} \right)} \right]$$</p>
<p>$$ = 0 - \mathop {\lim }\limits_{x \to 0} \cos \left( {{1 \over x}} \right)$$</p>
<p>= Does not exist</p>
<p>$$\therefore$$ $$f'(x)$$ is discontinuous function at $$x = 0$$.</p>
<p>L.H.D. $$ = \mathop {\lim }\limits_{h \to {0^ + }} {{f(0 - h) - f(0)} \over { - h}}$$</p>
<p>$$ = \mathop {\lim }\limits_{h \to {0^ + }} {{f( - h)} \over { - h}}$$</p>
<p>$$ = \mathop {\lim }\limits_{h \to {0^ + }} {{ - {h^2}\sin \left( {{1 \over h}} \right)} \over { - h}} = 0$$</p>
<p>R.H.D. $$ = \mathop {\lim }\limits_{h \to {0^ + }} {{f(0 + h) - f(0)} \over h}$$</p>
<p>$$ = \mathop {\lim }\limits_{h \to {0^ + }} {{f(h)} \over h}$$</p>
<p>$$ = \mathop {\lim }\limits_{h \to {0^ + }} {{{h^2}\sin \left( {{1 \over h}} \right)} \over h} = 0$$</p>
<p>$$\therefore$$ L.H.D. = R.H.D.</p>
<p>$$ \Rightarrow f(x)$$ is differentiable at $$x = 0$$. So, f(x) is continuous.</p>
|
mcq
|
jee-main-2023-online-24th-january-morning-shift
|
lgnx6vt2
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let $[x]$ denote the greatest integer function and <br/><br/>$f(x)=\max \{1+x+[x], 2+x, x+2[x]\}, 0 \leq x \leq 2$. Let $m$ be the number of <br/><br/>points in $[0,2]$, where $f$ is not continuous and $n$ be the number of points in <br/><br/>$(0,2)$, where $f$ is not differentiable. Then $(m+n)^{2}+2$ is equal to :
|
[{"identifier": "A", "content": "3"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "11"}]
|
["A"]
| null |
$$
\begin{aligned}
& \text { Let } g(x)=1+x+[x]=\left\{\begin{array}{cc}
1+x ; & x \in[0,1) \\\
2+x ; & x \in[1,2) \\
5 ; & x=2
\end{array}\right. \\\\
& \lambda(x)=x+2[x]=\left\{\begin{array}{cc}
x ; & x \in[0,1) \\
x+2 ; & x \in[1,2) \\
6 ; & x=2
\end{array}\right. \\\\
& r(x)=2+x \\\\
& f(x)=\left\{\begin{array}{cc}
2+x ; & x \in[0,2) \\
6 ; & x=2
\end{array}\right.
\end{aligned}
$$
<br/><br/>$\mathrm{f}(\mathrm{x})$ is discontinuous only at $x=2 \Rightarrow \mathrm{m}=1$ <br/><br/>$\mathrm{f}(\mathrm{x})$ is differentiable in $(0,2) \Rightarrow \mathrm{n}=0$
<br/><br/>$$
(m+n)^2+2=3
$$
|
mcq
|
jee-main-2023-online-15th-april-morning-shift
|
1lgsvmum4
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>Let $$f$$ and $$g$$ be two functions defined by</p>
<p>$$f(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\ |x-1|, & x \geq 0\end{array}\right.$$ and $$\mathrm{g}(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\ 1, & x \geq 0\end{array}\right.$$</p>
<p>Then $$(g \circ f)(x)$$ is :</p>
|
[{"identifier": "A", "content": "continuous everywhere but not differentiable at $$x=1$$"}, {"identifier": "B", "content": "differentiable everywhere"}, {"identifier": "C", "content": "not continuous at $$x=-1$$"}, {"identifier": "D", "content": "continuous everywhere but not differentiable exactly at one point"}]
|
["D"]
| null |
$$
\begin{aligned}
& \text { Sol. } f(x)=\left\{\begin{array}{c}
x+1, x<0 \\\
1-x, 0 \leq x<1 \\
x-1,1 \leq x
\end{array}\right. \\\\
& g(x)=\left\{\begin{array}{c}
x+1, x<0 \\
1, x \geq 0
\end{array}\right. \\\\
& g(f(x))=\left\{\begin{array}{c}
x+2, x<-1 \\
1, x \geq-1
\end{array}\right.
\end{aligned}
$$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lifi9pn2/0d3d2e66-f8f7-4969-8143-a77c378e96f7/b6eee2e0-01c8-11ee-91de-e7d2360ea0c8/file-1lifi9pn3.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lifi9pn2/0d3d2e66-f8f7-4969-8143-a77c378e96f7/b6eee2e0-01c8-11ee-91de-e7d2360ea0c8/file-1lifi9pn3.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh;" alt="JEE Main 2023 (Online) 11th April Evening Shift Mathematics - Limits, Continuity and Differentiability Question 36 English Explanation">
<br>$\therefore \mathrm{g}(\mathrm{f}(\mathrm{x}))$ is continuous everywhere
<br><br>$\mathrm{g}(\mathrm{f}(\mathrm{x}))$ is not differentiable at $\mathrm{x}=-1$
Differentiable everywhere else.
|
mcq
|
jee-main-2023-online-11th-april-evening-shift
|
1lgxw8rjx
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>Let $$f:( - 2,2) \to R$$ be defined by $$f(x) = \left\{ {\matrix{
{x[x],} & { - 2 < x < 0} \cr
{(x - 1)[x],} & {0 \le x \le 2} \cr
} } \right.$$ where $$[x]$$ denotes the greatest integer function. If m and n respectively are the number of points in $$( - 2,2)$$ at which $$y = |f(x)|$$ is not continuous and not differentiable, then $$m + n$$ is equal to ____________.</p>
|
[]
| null |
4
|
Given function is $f(x)=\left\{\begin{array}{cc}x[x], & -2 < x < 0 \\ (x-1)[x], & 0 \leq x<2\end{array}\right.$
<br><br>When $[x]$ is denotes greatest integer function
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lnd9enbh/f19f578f-55e0-486b-a43d-c6ee1891f168/919e63d0-6389-11ee-b38e-7becbda30131/file-6y3zli1lnd9enbi.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lnd9enbh/f19f578f-55e0-486b-a43d-c6ee1891f168/919e63d0-6389-11ee-b38e-7becbda30131/file-6y3zli1lnd9enbi.png" loading="lazy" style="max-width: 100%; height: auto; display: block; margin: 0px auto; max-height: 40vh; vertical-align: baseline;" alt="JEE Main 2023 (Online) 10th April Morning Shift Mathematics - Limits, Continuity and Differentiability Question 34 English Explanation">
<br><br>Clearly, $|f(x)|$ remains same.
<br><br>Given that, $m$ and $n$ respectively are the number points in $(-2,2)$ at which $y=|f(x)|$ is not continuous and not differentiable
<br><br>So, $m=1$ where $y=|f(x)|$ not continuous
<br><br>and $n=3$ where $|f(x)|$ is not differentiable.
<br><br>Thus, $m+n=4$
|
integer
|
jee-main-2023-online-10th-april-morning-shift
|
1lgyorpkh
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>Let $$\mathrm{k}$$ and $$\mathrm{m}$$ be positive real numbers such that the function $$f(x)=\left\{\begin{array}{cc}3 x^{2}+k \sqrt{x+1}, & 0 < x < 1 \\ m x^{2}+k^{2}, & x \geq 1\end{array}\right.$$ is differentiable for all $$x > 0$$. Then $$\frac{8 f^{\prime}(8)}{f^{\prime}\left(\frac{1}{8}\right)}$$ is equal to ____________.</p>
|
[]
| null |
309
|
Here, $$f(x)=\left\{\begin{array}{cc}3 x^{2}+k \sqrt{x+1}, & 0 < x < 1 \\ m x^{2}+k^{2}, & x \geq 1\end{array}\right.$$
<br/><br/>$\because f(x)$ is differentiable at $x>0$
<br/><br/>So, $f(x)$ is differentiable at $x=1$
<br/><br/>$$
\begin{gathered}
f\left(1^{-}\right)=f(1)=f\left(1^{+}\right) \\\\
3+k \sqrt{2}=m+k^2 ......(i)
\end{gathered}
$$
<br/><br/>$$
\begin{aligned}
& f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right) \\\\
& 6(1)+\frac{k}{2 \sqrt{1+1}}=2 m(1) \\\\
& \Rightarrow 6+\frac{k}{2 \sqrt{2}}=2 m ......(ii)
\end{aligned}
$$
<br/><br/>Using (i) and (ii),
<br/><br/>$$
\begin{aligned}
& 3+k \sqrt{2}=3+\frac{k}{4 \sqrt{2}}+k^2 \\\\
& \Rightarrow k^2+k\left[\frac{1}{4 \sqrt{2}}-\sqrt{2}\right]=0
\end{aligned}
$$
<br/><br/>$$
\Rightarrow k\left[k+\frac{1-8}{4 \sqrt{2}}\right]=0 \Rightarrow k=0, \frac{7}{4 \sqrt{2}}
$$
<br/><br/>As the problem specifies k to be a positive real number, we can rule out k = 0. Hence, k = $\frac{7}{4 \sqrt{2}}$
<br/><br/>$$
\begin{aligned}
& \text { for } k=\frac{7}{4 \sqrt{2}}, m=3+\frac{\frac{7}{4 \sqrt{2}}}{4 \sqrt{2}} \\\\
& =3+\frac{7}{32}=\frac{96+7}{32}=\frac{103}{32}
\end{aligned}
$$
<br/><br/>$$
\text { So, } \frac{8 f^{\prime}(8)}{f^{\prime}\left(\frac{1}{8}\right)}=\frac{8 \times\left[2 \times \frac{103}{32} \times 8\right]}{6 \times \frac{1}{8}+\frac{7}{4 \sqrt{2}} \times 2 \sqrt{918}}
$$
<br/><br/>$$
=\frac{412}{\frac{3}{4}+\frac{7}{12}}=\frac{412}{\frac{9+7}{12}}=\frac{412 \times 12}{16}=309
$$
<br/><br/><b>Concept :</b>
<br/><br/>$f(x)$ is differentiable at $x=a$, if $f^{\prime}\left(a^{-}\right)=f'\left(a^{+}\right)$
|
integer
|
jee-main-2023-online-8th-april-evening-shift
|
1lh23sso3
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>Let $$a \in \mathbb{Z}$$ and $$[\mathrm{t}]$$ be the greatest integer $$\leq \mathrm{t}$$. Then the number of points, where the function $$f(x)=[a+13 \sin x], x \in(0, \pi)$$ is not differentiable, is __________.</p>
|
[]
| null |
25
|
<p>Given that $ f(x) = [a + 13\sin(x)] $, where $[t]$ is the greatest integer function and $ x \in (0, \pi) $.</p>
<p>The function $[t]$ is not differentiable wherever $ t $ is an integer because at these points, the function has a jump discontinuity.</p>
<p>For $ f(x) $ to have a point of non-differentiability, the value inside the greatest integer function, i.e., $ a + 13\sin(x) $, should be an integer. </p>
<p>So, we need to find the values of $ x $ in the interval $ (0, \pi) $ for which $ a + 13\sin(x) $ is an integer.</p>
<p>Now, $ \sin(x) $ varies from 0 to 1 in the interval $ (0, \pi) $, so the maximum value of $ 13\sin(x) $ in this interval is 13.</p>
<p>For each integer value of $ 13\sin(x) $ between 0 and 13, we'll have a corresponding value of $ x $. There will be two such values of $ x $ for each value of $ 13\sin(x) $ (because of the periodic and symmetric nature of sine function over the interval $ (0, \pi) $), except for the maximum value, 13, which will have only one corresponding value of $ x $ (namely $ x = \frac{\pi}{2} $).</p>
<p>Thus, the total number of integer values of $ 13\sin(x) $ between 0 and 13 is 13. Excluding the maximum value, we have 12 integer values, each giving rise to two values of $ x $. Including the maximum value, which gives one value of $ x $, we have :</p>
<p>$ 12 \times 2 + 1 = 25 $</p>
<p>Thus, $ f(x) $ is not differentiable at 25 points in the interval $ (0, \pi) $.</p>
|
integer
|
jee-main-2023-online-6th-april-morning-shift
|
lsamhqd8
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let $f(x)=\left|2 x^2+5\right| x|-3|, x \in \mathbf{R}$. If $\mathrm{m}$ and $\mathrm{n}$ denote the number of points where $f$ is not continuous and not differentiable respectively, then $\mathrm{m}+\mathrm{n}$ is equal to :
|
[{"identifier": "A", "content": "5"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "0"}]
|
["B"]
| null |
$\begin{aligned} & f(x)=\left|2 x^2+5\right| x|-3| \\\\ & \text { Graph of } y=\left|2 x^2+5 x-3\right|\end{aligned}$
<br><br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsq9xozp/ba71121b-98fa-43fb-8b6d-a25522270b09/e89ef840-cdae-11ee-8e42-2f10126c4c2c/file-6y3zli1lsq9xozq.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsq9xozp/ba71121b-98fa-43fb-8b6d-a25522270b09/e89ef840-cdae-11ee-8e42-2f10126c4c2c/file-6y3zli1lsq9xozq.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) 1st February Evening Shift Mathematics - Limits, Continuity and Differentiability Question 28 English Explanation 1">
<br><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsqa8zwn/f12f5f83-13c0-4aa6-8f79-7a22074474aa/22f45480-cdb0-11ee-8e42-2f10126c4c2c/file-6y3zli1lsqa8zwo.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsqa8zwn/f12f5f83-13c0-4aa6-8f79-7a22074474aa/22f45480-cdb0-11ee-8e42-2f10126c4c2c/file-6y3zli1lsqa8zwo.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) 1st February Evening Shift Mathematics - Limits, Continuity and Differentiability Question 28 English Explanation 2">
<br><br>Now $f(x)$ is continuous $\forall x \in R$
<br><br>but non- differentiable at $x=\frac{-1}{2}, \frac{1}{2}, 0$
<br><br>$$
\begin{aligned}
\therefore m & =0 \\\\
n & =3 \\\\
m+n & =3
\end{aligned}
$$
|
mcq
|
jee-main-2024-online-1st-february-evening-shift
|
lsaowab9
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
Let $f: \mathbf{R} \rightarrow \mathbf{R}$ be defined as :
<br/><br/>$$
f(x)= \begin{cases}\frac{a-b \cos 2 x}{x^2} ; & x<0 \\\\ x^2+c x+2 ; & 0 \leq x \leq 1 \\\\ 2 x+1 ; & x>1\end{cases}
$$
<br/><br/>If $f$ is continuous everywhere in $\mathbf{R}$ and $m$ is the number of points where $f$ is NOT differential then $\mathrm{m}+\mathrm{a}+\mathrm{b}+\mathrm{c}$ equals :
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "2"}]
|
["D"]
| null |
At $\mathrm{x}=1, \mathrm{f}(\mathrm{x})$ is continuous therefore, <br/><br/>$\mathrm{f}\left(1^{-}\right)=\mathrm{f}(1)=\mathrm{f}\left(1^{+}\right)$
<br/><br/>$$
f(1)=3+c
$$ .........(1)
<br/><br/>$$
\begin{aligned}
& \mathrm{f}\left(1^{+}\right)=\lim _{h \rightarrow 0} 2(1+\mathrm{h})+1 \\\\
& \mathrm{f}\left(1^{+}\right)=\lim _{h \rightarrow 0} 3+2 \mathrm{~h}=3 .........(2)
\end{aligned}
$$
<br/><br/>from (1) and (2)
<br/><br/>$$
\mathrm{c}=0
$$
<br/><br/>at $\mathrm{x}=0, \mathrm{f}(\mathrm{x})$ is continuous therefore,
<br/><br/>$$
\begin{aligned}
& \mathrm{f}\left(0^{-}\right)=\mathrm{f}(0)=\mathrm{f}\left(0^{+}\right) ........(3) \\\\
& \mathrm{f}(0)=\mathrm{f}\left(0^{+}\right)=2 ...........(4)
\end{aligned}
$$
<br/><br/>$\mathrm{f}\left(0^{-}\right)$has to be equal to 2
<br/><br/>$\lim\limits_{h \rightarrow 0} \frac{a-b \cos (2 h)}{h^2}$
<br/><br/>= $\lim\limits_{h \rightarrow 0} \frac{a-b\left\{1-\frac{4 h^2}{2 !}+\frac{16 h^4}{4 !}+\ldots\right\}}{h^2}$
<br/><br/>= $\lim\limits_{h \rightarrow 0} \frac{a-b+b\left\{2 h^2-\frac{2}{3} h^4 \ldots\right\}}{h^2}$
<br/><br/>for limit to exist $a-b=0$ and limit is $2 b$ from (3), (4) and (5)
<br/><br/>$$
\mathrm{a}=\mathrm{b}=1
$$
<br/><br/>checking differentiability at $\mathrm{x}=0$
<br/><br/>$$
\begin{aligned}
& \text { LHD : } \lim _{h \rightarrow 0} \frac{\frac{1-\cos 2 h}{h^2}-2}{-h} \\\\
& \lim _{h \rightarrow 0} \frac{1-\left(1-\frac{4 h^2}{2 !}+\frac{16 h^4}{4 !} \ldots\right)-2 h^2}{-h^3}=0 \\\\
& \text { RHD : } \lim _{h \rightarrow 0} \frac{(0+h)^2+2-2}{h}=0
\end{aligned}
$$
<br/><br/>Function is differentiable at every point in its domain
<br/><br/>$$
\therefore \mathrm{m}=0
$$
<br/><br/>m + a + b + c = 0 + 1 + 1 + 0 = 2
|
mcq
|
jee-main-2024-online-1st-february-morning-shift
|
jaoe38c1lscn03is
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>Consider the function $$f:(0,2) \rightarrow \mathbf{R}$$ defined by $$f(x)=\frac{x}{2}+\frac{2}{x}$$ and the function $$g(x)$$ defined by</p>
<p>$$g(x)=\left\{\begin{array}{ll}
\min \lfloor f(t)\}, & 0<\mathrm{t} \leq x \text { and } 0 < x \leq 1 \\
\frac{3}{2}+x, & 1 < x < 2
\end{array} .\right. \text { Then, }$$</p>
|
[{"identifier": "A", "content": "$$g$$ is continuous but not differentiable at $$x=1$$\n"}, {"identifier": "B", "content": "$$g$$ is continuous and differentiable for all $$x \\in(0,2)$$\n"}, {"identifier": "C", "content": "$$g$$ is not continuous for all $$x \\in(0,2)$$\n"}, {"identifier": "D", "content": "$$g$$ is neither continuous nor differentiable at $$x=1$$"}]
|
["A"]
| null |
<p>$$\begin{aligned}
& f:(0,2) \rightarrow R ; f(x)=\frac{x}{2}+\frac{2}{x} \\
& f^{\prime}(x)=\frac{1}{2}-\frac{2}{x^2}
\end{aligned}$$</p>
<p>$$\therefore \mathrm{f}(\mathrm{x})$$ is decreasing in domain.</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1qvsrr/82daf6e5-d235-46d9-8017-461f6f164b34/ce430770-d3fd-11ee-acd6-03dea0d2c848/file-1lt1qvsrs.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1qvsrr/82daf6e5-d235-46d9-8017-461f6f164b34/ce430770-d3fd-11ee-acd6-03dea0d2c848/file-1lt1qvsrs.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 Evening Shift Mathematics - Limits, Continuity and Differentiability Question 22 English Explanation 1"></p>
<p>$$g(x)= \begin{cases}\frac{x}{2}+\frac{2}{x} & 0 < x \leq 1 \\ \frac{3}{2}+x & 1 < x < 2\end{cases}$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1lt1qwxt1/0909b125-954a-4e40-9ed4-1383ae43484b/edf5f550-d3fd-11ee-acd6-03dea0d2c848/file-1lt1qwxt2.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1lt1qwxt1/0909b125-954a-4e40-9ed4-1383ae43484b/edf5f550-d3fd-11ee-acd6-03dea0d2c848/file-1lt1qwxt2.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 Evening Shift Mathematics - Limits, Continuity and Differentiability Question 22 English Explanation 2"></p>
|
mcq
|
jee-main-2024-online-27th-january-evening-shift
|
jaoe38c1lsd4ro0j
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>Consider the function $$f:(0, \infty) \rightarrow \mathbb{R}$$ defined by $$f(x)=e^{-\left|\log _e x\right|}$$. If $$m$$ and $$n$$ be respectively the number of points at which $$f$$ is not continuous and $$f$$ is not differentiable, then $$m+n$$ is</p>
|
[{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "1"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "3"}]
|
["B"]
| null |
<p>$$\begin{aligned}
& f:(0, \infty) \rightarrow R \\
& f(x)=e^{-\left|\log _e x\right|}
\end{aligned}$$</p>
<p>$$\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{e}^{|\ln \mathrm{x}|}}=\left\{\begin{array}{l}
\frac{1}{\mathrm{e}^{-\ln \mathrm{x}}} ; 0<\mathrm{x}<1 \\
\frac{1}{\mathrm{e}^{\ln \mathrm{x}}} ; \mathrm{x} \geq 1
\end{array}\right.$$</p>
<p>$$\left\{\begin{array}{l}
\frac{1}{\frac{1}{x}}=x ; 0< x<1 \\
\frac{1}{x}, x \geq 1
\end{array}\right.$$</p>
<p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/6y3zli1lsjwnla3/94c1149f-a7b9-41cc-b180-4bf0cc496ba7/451674b0-ca2e-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwnla4.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/6y3zli1lsjwnla3/94c1149f-a7b9-41cc-b180-4bf0cc496ba7/451674b0-ca2e-11ee-8854-3b5a6c9e9092/file-6y3zli1lsjwnla4.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) 31st January Evening Shift Mathematics - Limits, Continuity and Differentiability Question 20 English Explanation"></p>
<p>$$\mathrm{m}=0$$ (No point at which function is not continuous)</p>
<p>$$\mathrm{n}=1$$ (Not differentiable)</p>
<p>$$\therefore \mathrm{m}+\mathrm{n}=1$$</p>
|
mcq
|
jee-main-2024-online-31st-january-evening-shift
|
jaoe38c1lsflbgha
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>Let $$f(x)=\sqrt{\lim _\limits{r \rightarrow x}\left\{\frac{2 r^2\left[(f(r))^2-f(x) f(r)\right]}{r^2-x^2}-r^3 e^{\frac{f(r)}{r}}\right\}}$$ be differentiable in $$(-\infty, 0) \cup(0, \infty)$$ and $$f(1)=1$$. Then the value of ea, such that $$f(a)=0$$, is equal to _________.</p>
|
[]
| null |
2
|
<p>$$\begin{aligned}
& f(1)=1, f(a)=0 \\
& {f^2}(x) = \mathop {\lim }\limits_{r \to x} \left( {{{2{r^2}({f^2}(r) - f(x)f(r))} \over {{r^2} - {x^2}}} - {r^3}{e^{{{f(r)} \over r}}}} \right) \\
& = \mathop {\lim }\limits_{r \to x} \left( {{{2{r^2}f(r)} \over {r + x}}{{(f(r) - f(x))} \over {r - x}} - {r^3}{e^{{{f(r)} \over r}}}} \right) \\
& f^2(x)=\frac{2 x^2 f(x)}{2 x} f^{\prime}(x)-x^3 e^{\frac{f(x)}{x}} \\
& y^2=x y \frac{d y}{d x}-x^3 e^{\frac{y}{x}} \\
& \frac{y}{x}=\frac{d y}{d x}-\frac{x^2}{y} e^{\frac{y}{x}}
\end{aligned}$$</p>
<p>Put $$y=v x \Rightarrow \frac{d y}{d x}=v+x \frac{d v}{d x}$$</p>
<p>$$\begin{aligned}
& v=v+x \frac{d v}{d x}-\frac{x}{v} e^v \\
& \frac{d v}{d x}=\frac{e^v}{v} \Rightarrow e^{-v} v d v=d x
\end{aligned}$$</p>
<p>Integrating both side</p>
<p>$$\begin{aligned}
& \mathrm{e}^{\mathrm{v}}(\mathrm{x}+\mathrm{c})+1+\mathrm{v}=0 \\
& \mathrm{f}(1)=1 \Rightarrow \mathrm{x}=1, \mathrm{y}=1
\end{aligned}$$</p>
<p>$$\begin{aligned}
& \Rightarrow c=-1-\frac{2}{e} \\
& e^v\left(-1-\frac{2}{e}+x\right)+1+v=0 \\
& e^{\frac{y}{x}}\left(-1-\frac{2}{e}+x\right)+1+\frac{y}{x}=0 \\
& x=a, y=0 \Rightarrow a=\frac{2}{e} \\
& a e=2
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-29th-january-evening-shift
|
1lsgcplix
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>If the function</p>
<p>$$f(x)= \begin{cases}\frac{1}{|x|}, & |x| \geqslant 2 \\ \mathrm{a} x^2+2 \mathrm{~b}, & |x|<2\end{cases}$$</p>
<p>is differentiable on $$\mathbf{R}$$, then $$48(a+b)$$ is equal to __________.</p>
|
[]
| null |
15
|
<p>$$\mathrm{f}(\mathrm{x})\left\{\begin{array}{c}
\frac{1}{\mathrm{x}} ; \mathrm{x} \geq 2 \\
\mathrm{ax}^2+2 \mathrm{~b} ;-2<\mathrm{x}<2 \\
-\frac{1}{\mathrm{x}} ; \mathrm{x} \leq-2
\end{array}\right.$$</p>
<p>Continuous at $$\mathrm{x}=2 \quad \Rightarrow \frac{1}{2}=\frac{\mathrm{a}}{4}+2 \mathrm{~b}$$</p>
<p>Continuous at $$\mathrm{x}=-2 \quad \Rightarrow \frac{1}{2}=\frac{\mathrm{a}}{4}+2 \mathrm{~b}$$</p>
<p>Since, it is differentiable at $$\mathrm{x}=2$$</p>
<p>$$-\frac{1}{x^2}=2 \mathrm{ax}$$</p>
<p>Differentiable at $$x=2 \quad \Rightarrow \frac{-1}{4}=4 a \Rightarrow a=\frac{-1}{16}, b =\frac{3}{8}$$</p>
|
integer
|
jee-main-2024-online-30th-january-morning-shift
|
lv7v3v83
|
maths
|
limits-continuity-and-differentiability
|
differentiability
|
<p>Let $$f$$ be a differentiable function in the interval $$(0, \infty)$$ such that $$f(1)=1$$ and $$\lim _\limits{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{t-x}=1$$ for each $$x>0$$. Then $$2 f(2)+3 f(3)$$ is equal to _________.</p>
|
[]
| null |
24
|
<p>$$\lim _\limits{t \rightarrow x} \frac{t^2 f(x)-x^2 f(t)}{(t-x)}=1 \quad\left(\frac{0}{0} \text { form }\right)$$</p>
<p>$$\begin{aligned}
& \lim _{t \rightarrow x} \frac{2 \operatorname{tf}(x)-x^2 f^{\prime}(t)}{1}=1 \\
& \Rightarrow 2 x f(x)-x^2 f'(x)=1 \\
& \frac{d y}{d x}-\frac{2 x y}{x^2}=\frac{-1}{x^2} \\
& \Rightarrow \frac{d y}{d x}-\left(\frac{2}{x}\right) y=\frac{-1}{x^2} \\
& \Rightarrow \text { I.F. }=e^{\int \frac{-2}{x} d x}=e^{-2 \ln x}=x^{-2}=\frac{1}{x^2}
\end{aligned}$$</p>
<p>$$\begin{aligned}
\Rightarrow & y\left(\frac{1}{x^2}\right)=\int\left(\frac{-1}{x^2}\right)\left(\frac{1}{x^2}\right) d x+C \\
& \frac{y}{x^2}=\frac{1}{3 x^3}+C \text { at } x=1, y=1 \\
\Rightarrow & 1=\frac{1}{3}+C \Rightarrow C=\frac{2}{3} \\
\Rightarrow & y=\frac{1}{3 x}+\frac{2}{3} x^2=f(x) \\
\Rightarrow & 2 f(2)+3 f(3)=24
\end{aligned}$$</p>
|
integer
|
jee-main-2024-online-5th-april-morning-shift
|
wcEtMRsaibecOeOmm3jgy2xukf0x5uc1
|
maths
|
limits-continuity-and-differentiability
|
existance-of-limits
|
Let [t] denote the greatest integer
$$ \le $$ t. If for some
<br/>$$\lambda $$ $$ \in $$ R - {1, 0}, $$\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + \left[ x \right]}}} \right|$$ = L, then L is
equal to :
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "2"}, {"identifier": "C", "content": "0"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}]
|
["B"]
| null |
Here $$\mathop {\lim }\limits_{x \to 0} \left| {{{1 - x + \left| x \right|} \over {\lambda - x + [x]}}} \right| = L$$<br><br>Here L.H.L. $$\mathop {\lim }\limits_{h \to 0^-} \left| {{{1 + h + h} \over {\lambda + h - 1}}} \right| = \left| {{1 \over {\lambda - 1}}} \right|$$<br><br>R.H.L. = $$\mathop {\lim }\limits_{h \to 0^+} \left| {{{1 - h + h} \over {\lambda + h + 0}}} \right| = \left| {{1 \over \lambda }} \right|$$<br><br>$$ \because $$ Limit exists. Hence L.H.L. = R.H.L.<br><br>$$ \Rightarrow $$ $$\left| {\lambda - 1} \right| = \left| \lambda \right|$$<br><br>$$ \Rightarrow $$ $$\lambda = {1 \over 2}$$
<br><br>$$ \therefore $$ L = $${1 \over {\left| \lambda \right|}}$$ = 2
|
mcq
|
jee-main-2020-online-3rd-september-morning-slot
|
1krrxiqlj
|
maths
|
limits-continuity-and-differentiability
|
existance-of-limits
|
If $$\mathop {\lim }\limits_{x \to 0} {{\alpha x{e^x} - \beta {{\log }_e}(1 + x) + \gamma {x^2}{e^{ - x}}} \over {x{{\sin }^2}x}} = 10,\alpha ,\beta ,\gamma \in R$$, then the value of $$\alpha$$ + $$\beta$$ + $$\gamma$$ is _____________.
|
[]
| null |
3
|
$$\mathop {\lim }\limits_{x \to 0} {{\alpha x\left( {1 + x + {{{x^2}} \over x}} \right) - \beta \left( {x - {{{x^2}} \over 2} + {{{x^3}} \over 3}} \right) + \gamma {x^2}(1 - x)} \over {{x^3}}}$$<br><br>$$\mathop {\lim }\limits_{x \to 0} {{x(\alpha - \beta ) + {x^2}\left( {\alpha + {\beta \over 2} + \gamma } \right) + {x^3}\left( {{\alpha \over 2} - {\beta \over 3} - \gamma } \right)} \over {{x^3}}} = 10$$<br><br>For limit to exist<br><br>$$\alpha - \beta = 0,\alpha + {\beta \over 2} + \gamma = 0$$$${\alpha \over 2} - {\beta \over 3} - \gamma = 10$$ ..... (i)<br><br>$$\beta = \alpha ,\gamma = - 3{\alpha \over 2}$$<br><br>Put in (i)<br><br>$${\alpha \over 2} - {\alpha \over 3} + {{3\alpha } \over 2} = 10$$<br><br>$${\alpha \over 6} + {{3\alpha } \over 2} = 10 \Rightarrow {{\alpha + 9\alpha } \over 6} = 10$$<br><br>$$ \Rightarrow \alpha = 6$$<br><br>$$\alpha$$ = 6, $$\beta$$ = 6, $$\gamma$$ = $$-$$9<br><br>$$\alpha$$ + $$\beta$$ + $$\gamma$$ = 3
|
integer
|
jee-main-2021-online-20th-july-evening-shift
|
6EivwZ6DrQWjLqAR
|
maths
|
limits-continuity-and-differentiability
|
limits-of-algebric-function
|
If $$f\left( 1 \right) = 1,{f'}\left( 1 \right) = 2,$$ then
<br/>$$\mathop {\lim }\limits_{x \to 1} {{\sqrt {f\left( x \right)} - 1} \over {\sqrt x - 1}}$$ is
|
[{"identifier": "A", "content": "$$2$$"}, {"identifier": "B", "content": "$$4$$"}, {"identifier": "C", "content": "$$1$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}]
|
["A"]
| null |
$$\mathop {\lim }\limits_{x \to 1} {{\sqrt {f\left( x \right)} - 1 } \over {\sqrt x - 1}}\,\,\left( {{0 \over 0}} \right)$$ form using $$L'$$ Hospital's rule
<br><br>$$ = \mathop {\lim }\limits_{x \to 1} {{{1 \over {2\sqrt {f\left( x \right)} }}f'\left( x \right)} \over {1/2\sqrt x }}$$
<br><br>$$ = {{f'\left( 1 \right)} \over {\sqrt {f\left( 1 \right)} }} = {2 \over 1} = 2.$$
|
mcq
|
aieee-2002
|
j2lxIWGAe1k1pqXc
|
maths
|
limits-continuity-and-differentiability
|
limits-of-algebric-function
|
Let $$f(2) = 4$$ and $$f'(x) = 4.$$
<br/><br/>Then $$\mathop {\lim }\limits_{x \to 2} {{xf\left( 2 \right) - 2f\left( x \right)} \over {x - 2}}$$ is given by
|
[{"identifier": "A", "content": "$$2$$"}, {"identifier": "B", "content": "$$- 2$$"}, {"identifier": "C", "content": "$$- 4$$"}, {"identifier": "D", "content": "$$3$$"}]
|
["C"]
| null |
Given,
<br><br>$$\mathop {\lim }\limits_{x \to 2} {{xf\left( 2 \right) - 2f\left( x \right)} \over {x - 2}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 2} {{xf\left( 2 \right) - 2f\left( 2 \right) + 2f\left( 2 \right) - 2f\left( x \right)} \over {x - 2}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 2} {{f\left( 2 \right)\left( {x - 2} \right) - 2\left\{ {f\left( x \right) - f\left( 2 \right)} \right\}} \over {x - 2}}$$
<br><br>= $$f\left( 2 \right) - 2\mathop {\lim }\limits_{x \to 2} {{f\left( x \right) - f\left( 2 \right)} \over {x - 2}}$$
<br><br> $$\left[ {As\,f'\left( x \right) = \mathop {\lim }\limits_{x \to 2} {{f\left( x \right) - f\left( 2 \right)} \over {x - 2}}} \right]$$
<br><br>= $$f\left( 2 \right) - 2f'\left( x \right)$$
<br><br>= 4 - 2 $$ \times $$ 4
<br><br>= - 4
|
mcq
|
aieee-2002
|
lGj7y8njpmls6ye2
|
maths
|
limits-continuity-and-differentiability
|
limits-of-algebric-function
|
Let $$f:R \to R$$ be a positive increasing function with
<br/><br/>$$\mathop {\lim }\limits_{x \to \infty } {{f(3x)} \over {f(x)}} = 1$$. Then $$\mathop {\lim }\limits_{x \to \infty } {{f(2x)} \over {f(x)}} = $$
|
[{"identifier": "A", "content": "$${2 \\over 3}$$"}, {"identifier": "B", "content": "$${3 \\over 2}$$"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "1"}]
|
["D"]
| null |
$$f(x)$$ is a positive increasing function.
<br><br>$$\therefore$$ $$0 < f\left( x \right) < f\left( {2x} \right) < f\left( {3x} \right)$$
<br><br>$$ \Rightarrow 0 < 1 < {{f\left( {2x} \right)} \over {f\left( x \right)}} < {{f\left( {3x} \right)} \over {f\left( x \right)}}$$
<br><br>$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } 1 \le \mathop {\lim }\limits_{x \to \infty } {{f\left( {2x} \right)} \over {f\left( x \right)}} \le \mathop {\lim }\limits_{x \to \infty } {{f\left( {3x} \right)} \over {f\left( x \right)}}$$
<br><br>By Sandwich Theorem.
<br><br>$$ \Rightarrow \mathop {\lim }\limits_{x \to \infty } {{f\left( {2x} \right)} \over {f\left( x \right)}} = 1$$
|
mcq
|
aieee-2010
|
wLgGeNX26Y861NCnho4cC
|
maths
|
limits-continuity-and-differentiability
|
limits-of-algebric-function
|
$$\mathop {\lim }\limits_{x \to 3} $$ $${{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$$ is equal to :
|
[{"identifier": "A", "content": "$$\\sqrt 3 $$ "}, {"identifier": "B", "content": "$${1 \\over {\\sqrt 2 }}$$ "}, {"identifier": "C", "content": "$${{\\sqrt 3 } \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over {2\\sqrt 2 }}$$"}]
|
["B"]
| null |
Given,
<br><br> $$\mathop {\lim }\limits_{x \to 3} $$ $${{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$$
<br><br>Here if you put x = 3 in $${{\sqrt {3x} - 3} \over {\sqrt {2x - 4 - \sqrt 2 } }}$$
<br><br>you will get $${0 \over 0}$$ form.
<br><br>So, we can apply L' Hospital rule
<br><br>$$\therefore\,\,\,$$ $$\mathop {\lim }\limits_{x \to 3} {{\sqrt {3x} - 3} \over {\sqrt {2x - 4} - \sqrt 2 }}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 3} $$ $${{\sqrt 3 .{1 \over {2\sqrt x }}} \over {{2 \over {2\sqrt {2x - 4} }}}}$$ (applying L' Hospital rule)
<br><br>= $${{\sqrt 3 .{1 \over {2\sqrt 3 }}} \over {{1 \over {\sqrt 6 - 4}}}}$$
<br><br>= $${1 \over 2}$$ $$ \times $$ $$\sqrt 2 $$
<br><br>= $${1 \over {\sqrt 2 }}$$
|
mcq
|
jee-main-2017-online-8th-april-morning-slot
|
WtrrL787R3wCQZqP
|
maths
|
limits-continuity-and-differentiability
|
limits-of-algebric-function
|
For each t $$ \in R$$, let [t] be the greatest integer less than or equal to t.
<br/><br/>Then $$\mathop {\lim }\limits_{x \to {0^ + }} x\left( {\left[ {{1 \over x}} \right] + \left[ {{2 \over x}} \right] + ..... + \left[ {{{15} \over x}} \right]} \right)$$
|
[{"identifier": "A", "content": "does not exist in R"}, {"identifier": "B", "content": "is equal to 0"}, {"identifier": "C", "content": "is equal to 15"}, {"identifier": "D", "content": "is equal to 120"}]
|
["D"]
| null |
Given,
<br><br>$$\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left( {\left[ {{1 \over x}} \right] + \left[ {{2 \over x}} \right]} \right. + $$ $$\left. {\,.\,.\,.\,.\,.\, + \left[ {{{15} \over x}} \right]} \right)$$
<br><br>as we know that
<br><br>$${1 \over x} = \left[ {{1 \over x}} \right] + \left\{ {{1 \over x}} \right\}$$
<br><br>$$ \Rightarrow \,\,\,\,\left[ {{1 \over x}} \right] = {1 \over x} - \left\{ {{1 \over x}} \right\}$$
<br><br>$$ = \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\,x\left[ {{1 \over x} - \left\{ {{1 \over x}} \right\} + {2 \over 2} - \left\{ {{2 \over x}} \right\} + ........{{15} \over x} - \left\{ {{{15} \over x}} \right\}} \right]$$
<br><br>$$ = \,\,\,\,\mathop {\lim }\limits_{x \to {0^ + }} \,\left[ {x.{1 \over x} + x.{2 \over x} + .....x.{{15} \over x}} \right]$$
<br><br>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, - \,\left[ {x.\left\{ {{1 \over 2}} \right\} + .. + x.\left\{ {{{15} \over x}} \right\}} \right]$$
<br><br>We know $$\left\{ {{1 \over x}} \right\}$$ is fractional part of $${1 \over x}.$$
<br><br>So, the range of $$\,\left\{ {{1 \over x}} \right\}$$ is $$0 \le \left\{ {{1 \over x}} \right\} < 1$$
<br><br>So, $$\mathop {lim}\limits_{x \to {0^ + }} \,\,x\,\left\{ {{1 \over x}} \right\} = 0.$$ (finite no) $$=0$$
<br><br>Similarly $$\mathop {\lim }\limits_{x \to {0^ + }} x.\left\{ {{2 \over x}} \right\} = 0$$
<br><br>$$ = \,\,\,\,\left( {1 + 2 + ... + 15} \right) - \left( {0 + 0...} \right)$$
<br><br>$$ = \,\,\,\,{{15 \times 16} \over 2}$$
<br><br>$$ = \,\,\,\,120$$
|
mcq
|
jee-main-2018-offline
|
L9hsA4UrY7uaEooFI7Xug
|
maths
|
limits-continuity-and-differentiability
|
limits-of-algebric-function
|
Let f(x) be a polynomial of degree $$4$$ having extreme values at $$x = 1$$ and $$x = 2.$$
<br/><br/>If $$\mathop {lim}\limits_{x \to 0} \left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right) = 3$$ then f($$-$$1) is equal to :
|
[{"identifier": "A", "content": "$${9 \\over 2}$$"}, {"identifier": "B", "content": "$${5 \\over 2}$$"}, {"identifier": "C", "content": "$${3 \\over 2}$$"}, {"identifier": "D", "content": "$${1 \\over 2}$$"}]
|
["A"]
| null |
$$ \because $$ f(x) has extremum values at x = 1, and x = 2
<br><br>$$ \because $$ f'(1) = 0 and f'(2) = 0
<br><br>As, f(x) is a polynomial of degree 4.
<br><br>Suppose f(x) = Ax<sup>4</sup> + Bx<sup>3</sup> + cx<sup>2</sup> + Dx + E
<br><br>$$ \because $$ $$\mathop {\lim }\limits_{x \to 0} $$ $$\left( {{{f\left( x \right)} \over {{x^2}}} + 1} \right)$$ = 3
<br><br>$$ \Rightarrow $$$$\,\,\,$$$$\mathop {\lim }\limits_{x \to 0} \left( {{{A{x^4} + B{x^3} + C{x^2} + Dx + E} \over {{x^2}}} + 1} \right)$$ = 3
<br><br>$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 0} $$ $$\left( {A{x^2} + Bx + C + {D \over x} + {E \over {{x^2}}} + 1} \right)$$ = 3
<br><br>As limit has finite value, so D = 0 and E = 0
<br><br>Now A(0)<sup>2</sup> + B(0) + C + 0 + 0 + 1 = 3
<br><br>$$ \Rightarrow $$ c + 1 = 3 $$ \Rightarrow $$ c = 2
<br><br>f'(x) = 4Ax<sup>3</sup> + 3Bx<sup>2</sup> + 2Cx + D
<br><br>f'(1) = 0 $$ \Rightarrow $$ 4A(1) + 3B(1) + 2C(1) + D = 0
<br><br>$$ \Rightarrow $$$$\,\,\,$$ 4A + 3B = $$-$$ 4 . . . .(1)
<br><br>f'(2) = 0 $$ \Rightarrow $$ 4A(8) + 3B(4) + 2C(2) + D = 0
<br><br>$$ \Rightarrow $$ 8A + 3B = $$-$$ 2 . . . . .(2)
<br><br>From equations (1) and (2), we get
<br><br>A = $${1 \over 2}$$ and B = $$-$$ 2
<br><br>So, f(x) = $${{{x_4}} \over 2} - 2{x^3} + 2x{}^2$$
<br><br>Therefore, f($$-$$ 1) = $${{{{( - 1)}^4}} \over 2} - 2{( - 1)^3} + 2{( - 1)^2}$$
<br><br>= $${1 \over 2} + 2 + 2 = {9 \over 2}$$
<br><br>Hence f($$-$$1) = $${9 \over 2}$$
|
mcq
|
jee-main-2018-online-15th-april-evening-slot
|
wztfeePIP7DfsJ4eiSWl3
|
maths
|
limits-continuity-and-differentiability
|
limits-of-algebric-function
|
$$\mathop {\lim }\limits_{x \to 0} \,\,{{{{\left( {27 + x} \right)}^{{1 \over 3}}} - 3} \over {9 - {{\left( {27 + x} \right)}^{{2 \over 3}}}}}$$ equals.
|
[{"identifier": "A", "content": "$${1 \\over 3}$$ "}, {"identifier": "B", "content": "$$-$$ $${1 \\over 3}$$"}, {"identifier": "C", "content": "$$-$$ $${1 \\over 6}$$"}, {"identifier": "D", "content": "$${1 \\over 6}$$"}]
|
["C"]
| null |
Given,
<br><br>$$\mathop {lim}\limits_{x \to 0} \,{{{{\left( {27 + x} \right)}^{{1 \over 3}}} - 3} \over {9 - {{\left( {27 + x} \right)}^{{2 \over 3}}}}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \,{{3\left[ {{{\left( {1 + {x \over {27}}} \right)}^{{1 \over 3}}} - 1} \right]} \over {9\left[ {1 - {{\left( {1 + {x \over {27}}} \right)}^{{2 \over 3}}}} \right]}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \,\,{{\left( {1 + {x \over {3 \times 27}}} \right) - 1} \over {3\left[ {1 - \left( {1 + {{2x} \over {3 \times 27}}} \right)} \right]}}$$
<br><br>= $$\mathop {\lim }\limits_{x \to 0} \,\,{{{x \over {81}}} \over {3\left( { - {{2x} \over {81}}} \right)}} = - {1 \over 6}$$
|
mcq
|
jee-main-2018-online-16th-april-morning-slot
|
sX1MfZeJPUdVnpCXpDYX5
|
maths
|
limits-continuity-and-differentiability
|
limits-of-algebric-function
|
$$\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$$
|
[{"identifier": "A", "content": "exists and equals $${1 \\over {2\\sqrt 2 }}$$"}, {"identifier": "B", "content": "exists and equals $${1 \\over {4\\sqrt 2 }}$$"}, {"identifier": "C", "content": "exists and equals $${1 \\over {2\\sqrt 2 (1 + \\sqrt {2)} }}$$"}, {"identifier": "D", "content": "does not exists"}]
|
["B"]
| null |
$$\mathop {\lim }\limits_{y \to 0} {{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$$
<br><br>If you put y = 0 at $${{\sqrt {1 + \sqrt {1 + {y^4}} } - \sqrt 2 } \over {{y^4}}}$$ it is in $${0 \over 0}$$ form. So we can use L' Hospital's Rule.
<br><br>= $$\mathop {\lim }\limits_{y \to 0} {{{1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times \left( {{1 \over {2\sqrt {1 + {y^4}} }}} \right) \times 4{y^3}} \over {4{y^3}}}$$
<br><br>= $$\mathop {\lim }\limits_{y \to 0} {1 \over {2\sqrt {1 + \sqrt {1 + {y^4}} } }} \times {1 \over {2\sqrt {1 + {y^4}} }}$$
<br><br>= $${1 \over {4\sqrt 2 }}$$
|
mcq
|
jee-main-2019-online-9th-january-morning-slot
|
3hyv1yKOaRDf7sWOv23rsa0w2w9jwxrvaa6
|
maths
|
limits-continuity-and-differentiability
|
limits-of-algebric-function
|
If $$\mathop {\lim }\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \mathop {\lim }\limits_{x \to k} {{{x^3} - {k^3}} \over {{x^2} - {k^2}}}$$, then k is :
|
[{"identifier": "A", "content": "$${3 \\over 2}$$"}, {"identifier": "B", "content": "$${8 \\over 3}$$"}, {"identifier": "C", "content": "$${4 \\over 3}$$"}, {"identifier": "D", "content": "$${3 \\over 8}$$"}]
|
["B"]
| null |
If $$\mathop {\lim }\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \mathop {\lim }\limits_{x \to K} \left( {{{{x^3} - {k^3}} \over {{x^2} - {k^2}}}} \right)$$<br><br>
<b>L·H·S·</b><br><br>
$$\mathop {Lt}\limits_{x \to 1} {{{x^4} - 1} \over {x - 1}} = \left( {{0 \over 0}form} \right)$$<br><br>
$$ \Rightarrow \mathop {Lt}\limits_{x \to 1} {{4{x^3}} \over 1} = 4$$<br><br>
Now, $$\mathop {\lim }\limits_{x \to K} \left( {{{{x^3} - {k^3}} \over {{x^2} - {k^2}}}} \right)$$ = 4<br><br>
$$ \Rightarrow \mathop {\lim }\limits_{x \to K} {{3{x^2}} \over {2x}} = 4$$<br><br>
$$ \Rightarrow {3 \over 2}k = 4 \Rightarrow k = {8 \over 3}$$
|
mcq
|
jee-main-2019-online-10th-april-morning-slot
|
BRO7yDGmrfgKvmK5Aa3rsa0w2w9jx1yg7ul
|
maths
|
limits-continuity-and-differentiability
|
limits-of-algebric-function
|
If $$\mathop {\lim }\limits_{x \to 1} {{{x^2} - ax + b} \over {x - 1}} = 5$$, then a + b is equal to :
|
[{"identifier": "A", "content": "1"}, {"identifier": "B", "content": "- 4"}, {"identifier": "C", "content": "- 7"}, {"identifier": "D", "content": "5"}]
|
["C"]
| null |
$$\mathop {\lim }\limits_{x \to 1} {{{x^2} - ax + b} \over {x - 1}} = 5$$<br><br>
$$ \Rightarrow $$ $${(1)^2} - a(1) + b = 0$$<br><br>
$$ \Rightarrow $$$$1 - a + b = 0$$<br><br>
$$ \Rightarrow $$$$a - b = 1\,\,......(1)$$<br><br>
Now 'L' hospital rule<br><br>
2x - a = 5<br><br>
$$ \Rightarrow $$2 - a = 5 ($$ \because $$ x = 1)<br><br>
$$ \Rightarrow $$ a = - 3<br><br>
By putting a = -3 in (1)<br><br>
$$ \Rightarrow $$ b = -4<br><br>
$$ \therefore $$ a + b = -7
|
mcq
|
jee-main-2019-online-10th-april-evening-slot
|
fyMDR3N15sswpQy8CW3rsa0w2w9jx5crbn3
|
maths
|
limits-continuity-and-differentiability
|
limits-of-algebric-function
|
If $$\alpha $$ and $$\beta $$ are the roots of the equation 375x<sup>2</sup>
– 25x – 2 = 0, then $$\mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{\alpha ^r}} + \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n {{\beta ^r}} $$ is equal to :
|
[{"identifier": "A", "content": "$${7 \\over {116}}$$"}, {"identifier": "B", "content": "$${{29} \\over {348}}$$"}, {"identifier": "C", "content": "$${1 \\over {12}}$$"}, {"identifier": "D", "content": "$${{21} \\over {346}}$$"}]
|
["B"]
| null |
$$\alpha $$ and $$\beta $$ are the two root of 375x<sup>2</sup> - 25x - 2 = 0<br><br>
Both of the roots are lie in (-1, 1) hence sum of given series is finite<br><br>
$$\mathop {\lim }\limits_{n \to \infty } \left( {{\alpha \over {1 - \alpha }} + {\beta \over {1 - \beta }}} \right) = {{\alpha (1 - \beta ) + \beta (1 - \alpha )} \over {(1 - \alpha )(1 - \beta )}}$$<br><br>
$$ \Rightarrow {{\left( {\alpha + \beta } \right) - 2\alpha \beta } \over {1 - (\alpha + \beta ) + \alpha \beta }} = {{25 - 2( - 2)} \over {375 - 25 - 2}} = {{29} \over {348}}$$
|
mcq
|
jee-main-2019-online-12th-april-morning-slot
|
RJPBYHj7M347weF0yn3rsa0w2w9jxb3yvjl
|
maths
|
limits-continuity-and-differentiability
|
limits-of-algebric-function
|
Let f(x) = 5 – |x – 2| and g(x) = |x + 1|, x $$ \in $$ R. If f(x) attains maximum value at $$\alpha $$ and g(x) attains
minimum value at $$\beta $$, then
$$\mathop {\lim }\limits_{x \to -\alpha \beta } {{\left( {x - 1} \right)\left( {{x^2} - 5x + 6} \right)} \over {{x^2} - 6x + 8}}$$ is equal to :
|
[{"identifier": "A", "content": "$${1 \\over 2}$$"}, {"identifier": "B", "content": "$$-{1 \\over 2}$$"}, {"identifier": "C", "content": "$${3 \\over 2}$$"}, {"identifier": "D", "content": "$$-{3 \\over 2}$$"}]
|
["A"]
| null |
From f(x) = 5 - | x - 2 |<br>
maximum value of f(x) is at x = 2<br><br>
From g(x) = | x + 1 |<br>
minimum value of g(x) is at x = -1<br><br>
$$ \therefore $$ $$\alpha \beta $$ = - 2<br><br>
$$ \Rightarrow $$ $$\mathop {\lim }\limits_{x \to 2} {{(x - 1)(x - 2)(x - 3)} \over {(x - 2)(x - 4)}}$$<br><br>
$$ \Rightarrow $$ $${{(2 - 1)(2 - 3)} \over {(2 - 4)}} = {1 \over 2}$$
|
mcq
|
jee-main-2019-online-12th-april-evening-slot
|
N2VlnkwaNhHCXerce6jgy2xukewticxv
|
maths
|
limits-continuity-and-differentiability
|
limits-of-algebric-function
|
If $$\mathop {\lim }\limits_{x \to 1} {{x + {x^2} + {x^3} + ... + {x^n} - n} \over {x - 1}}$$ = 820,
<br/>(n $$ \in $$ N) then
the value of n is equal to _______.
|
[]
| null |
40
|
$$\mathop {\lim }\limits_{x \to 1} {{x + {x^2} + {x^3} + ... + {x^n} - n} \over {x - 1}}$$ = 820
<br><br>As it is $$\left( {{0 \over 0}} \right)$$ form, Apply L'Hospital's Rule.
<br><br>$$\mathop {\lim }\limits_{x \to 1} \left( {{{1 + 2x + 3{x^2} + ... + n{x^{n - 1}}} \over 1}} \right)$$ = 820
<br><br>$$ \Rightarrow $$ 1 + 2 + 3 + .....+ n = 820
<br><br>$$ \Rightarrow $$ $${{n\left( {n + 1} \right)} \over 2}$$ = 820
<br><br>$$ \Rightarrow $$ n<sup>2</sup> + n – 1640 = 0
<br><br>$$ \Rightarrow $$ (n – 40)(n + 41) = 0
<br><br>Since n $$ \in $$ N, so n = 40.
|
integer
|
jee-main-2020-online-2nd-september-morning-slot
|
IJisEY7lKJW1ewvjbNjgy2xukf4999tx
|
maths
|
limits-continuity-and-differentiability
|
limits-of-algebric-function
|
$$\mathop {\lim }\limits_{x \to a} {{{{\left( {a + 2x} \right)}^{{1 \over 3}}} - {{\left( {3x} \right)}^{{1 \over 3}}}} \over {{{\left( {3a + x} \right)}^{{1 \over 3}}} - {{\left( {4x} \right)}^{{1 \over 3}}}}}$$ ($$a$$ $$ \ne $$ 0) is equal to :
|
[{"identifier": "A", "content": "$$\\left( {{2 \\over 9}} \\right){\\left( {{2 \\over 3}} \\right)^{{1 \\over 3}}}$$"}, {"identifier": "B", "content": "$$\\left( {{2 \\over 3}} \\right){\\left( {{2 \\over 9}} \\right)^{{1 \\over 3}}}$$"}, {"identifier": "C", "content": "$${\\left( {{2 \\over 3}} \\right)^{{4 \\over 3}}}$$"}, {"identifier": "D", "content": "$${\\left( {{2 \\over 9}} \\right)^{{4 \\over 3}}}$$"}]
|
["B"]
| null |
L = $$\mathop {\lim }\limits_{x \to a} {{{{\left( {a + 2x} \right)}^{{1 \over 3}}} - {{\left( {3x} \right)}^{{1 \over 3}}}} \over {{{\left( {3a + x} \right)}^{{1 \over 3}}} - {{\left( {4x} \right)}^{{1 \over 3}}}}}$$
<br><br>$$= \mathop {\lim }\limits_{h \to 0} {{{{(a + 2(a + h))}^{1/3}} - {{(3(a + h))}^{1/3}}} \over {{{(3a + a + h)}^{1/3}} - {{(4(a + h))}^{1/3}}}}$$<br><br>= $$\mathop {\lim }\limits_{h \to 0} {{{{(3a)}^{1/3}}{{\left( {1 + {{2h} \over {3a}}} \right)}^{1/3}} - {{(3a)}^{1/3}}{{\left( {1 + {h \over a}} \right)}^{1/3}}} \over {{{(4a)}^{1/3}}{{\left( {1 + {h \over {4a}}} \right)}^{1/3}} - {{(4a)}^{1/3}}{{\left( {1 + {h \over a}} \right)}^{1/3}}}}$$<br><br>= $$\mathop {\lim }\limits_{h \to 0} \left( {{{{3^{1/3}}} \over {{4^{1/3}}}}} \right)\left[ {{{\left( {1 + {{2h} \over {9a}}} \right) - \left( {1 + {h \over {3a}}} \right)} \over {\left( {1 + {h \over {12a}}} \right) - \left( {1 + {h \over {3a}}} \right)}}} \right]$$<br><br>$$ = {\left( {{3 \over 4}} \right)^{1/3}}{{\left( {{2 \over 9} - {1 \over 3}} \right)} \over {\left( {{1 \over {12}} - {1 \over 3}} \right)}} = {\left( {{3 \over 4}} \right)^{1/3}}\left( {{{8 - 12} \over {3 - 12}}} \right)$$<br><br>$$ = {\left( {{3 \over 4}} \right)^{1/3}}\left( {{{ - 4} \over { - 9}}} \right) = {{{4^{1 - {1 \over 3}}}} \over {{3^{2 - {1 \over 3}}}}} = {{{4^{2/3}}} \over {{3^{5/3}}}}$$<br><br>$$ = {{{{(8 \times 2)}^{1/3}}} \over {{{(27 \times 9)}^{1/3}}}} = {2 \over 3}{\left( {{2 \over 9}} \right)^{1/3}}$$
|
mcq
|
jee-main-2020-online-3rd-september-evening-slot
|
zig2fGHakn5vCGZ9cI1kluxa7u1
|
maths
|
limits-continuity-and-differentiability
|
limits-of-algebric-function
|
Let $$f(x) = {\sin ^{ - 1}}x$$ and $$g(x) = {{{x^2} - x - 2} \over {2{x^2} - x - 6}}$$. If $$g(2) = \mathop {\lim }\limits_{x \to 2} g(x)$$, then the domain of the function fog is :
|
[{"identifier": "A", "content": "$$( - \\infty , - 2] \\cup \\left[ { - {4 \\over 3},\\infty } \\right)$$"}, {"identifier": "B", "content": "$$( - \\infty , - 2] \\cup [ - 1,\\infty )$$"}, {"identifier": "C", "content": "$$( - \\infty , - 2] \\cup \\left[ { - {3 \\over 2},\\infty } \\right)$$"}, {"identifier": "D", "content": "$$( - \\infty , - 1] \\cup [2,\\infty )$$"}]
|
["A"]
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$$g(2) = \mathop {\lim }\limits_{x \to 2} {{(x - 2)(x + 1)} \over {(2x + 3)(x - 2)}} = {3 \over 7}$$<br><br>Domain of $$fog(x) = {\sin ^{ - 1}}(g(x))$$<br><br>$$ \Rightarrow |g(x)|\, \le 1$$<br><br>$$\left| {{{{x^2} - x - 2} \over {2{x^2} - x - 6}}} \right| \le 1$$<br><br>$$\left| {{{(x + 1)(x - 2)} \over {(2x + 3)(x - 2)}}} \right| \le 1$$<br><br>$${{x + 1} \over {2x + 3}} \le 1$$ and $${{x + 1} \over {2x + 3}} \ge - 1$$<br><br>$${{x + 1 - 2x - 3} \over {2x + 3}} \le 0$$ and $${{x + 1 + 2x + 3} \over {2x + 3}} \ge 0$$<br><br>$${{x + 2} \over {2x + 3}} \ge 0$$ and $${{3x + 4} \over {2x + 3}} \ge 0$$<br><br>$$x \in ( - \infty , - 2] \cup \left[ { - {4 \over 3},\infty } \right)$$
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mcq
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jee-main-2021-online-26th-february-evening-slot
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