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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>$$ = \mathop {\lim }\limits_{x \to 0} {{\log (1 + 5x)} \over {5x}} \times 5 - {{{... | mcq | jee-main-2022-online-29th-july-evening-shift |
1ldu4jonv | maths | limits-continuity-and-differentiability | continuity | <p>If the function $$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.$$</p>
<p>is continuou... | [{"identifier": "A", "content": "11"}, {"identifier": "B", "content": "10"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "2e$$^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.... | 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... | 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/>$$
\begin{... | mcq | jee-main-2023-online-11th-april-morning-shift |
lsbkoryy | maths | limits-continuity-and-differentiability | continuity | Consider the function.
<br/><br/>$$
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.
$$
<br/><br/>where $[x]$ denotes the greatest integer less than o... | [{"identifier": "A", "content": "Infinitely many"}, {"identifier": "B", "content": "4"}, {"identifier": "C", "content": "2"}, {"identifier": "D", "content": "1"}] | ["D"] | null | <p>$$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}}... | 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": "$$\\frac{1}{3} \\log ... | ["C"] | null | <p>Let $$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}}\ri... | 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 $$f(x)=\left\{\begin{array}{cc}\left(\frac{8}{7}\right)^{\frac{\tan 8 x}{\tan 7 x}}, & 0< x<\frac{\pi}{2} \\ \mathrm{a}-8, & x=\frac{\pi}{2} \\ (1+\mid \cot x)^{\frac{\mathrm{b}}{\mathrm{a}}|\tan x|}, & \frac{\pi}{2} < x < \pi\e... | [] | null | 81 | <p>$$\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 ... | 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 $$\alpha^2+... | [{"identifier": "A", "content": "48"}, {"identifier": "B", "content": "6"}, {"identifier": "C", "content": "3"}, {"identifier": "D", "content": "12"}] | ["D"] | null | <p>$$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}
&\... | 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>$$\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 0} \frac{72^x-9^... | mcq | jee-main-2024-online-4th-april-evening-shift |
lv3ve5z1 | maths | limits-continuity-and-differentiability | continuity | <p>For $$\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$$... | [{"identifier": "A", "content": "4"}, {"identifier": "B", "content": "5"}, {"identifier": "C", "content": "8"}, {"identifier": "D", "content": "6"}] | ["D"] | null | <p>$$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^{... | 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>$$\begin{aligned}
& \Rightarrow \lim _{x \rightarrow 0} \frac{\sin 3 x+\alpha \s... | mcq | jee-main-2024-online-5th-april-morning-shift |
lv9s207h | maths | limits-continuity-and-differentiability | continuity | <p>Let ,$$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... | 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 $$\sum_\limits{\text ... | [] | null | 17 | <p>$$\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(... | 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 a... | 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>$$f'\left( x \right) = f'\left( 0 \right)f\left(... | mcq | aieee-2002 |
7X0bs7xPIHP0Q5wn | maths | limits-continuity-and-differentiability | differentiability | Let $$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'\... | 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>$$L.H.L.$$ $$\,... | mcq | aieee-2003 |
U9v6NjxKosuCkmhk | maths | limits-continuity-and-differentiability | differentiability | Suppose $$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)... | 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>$$\,\,\,\,\,\,\,\,\,\left| {f'\left( x \right)} \right| = \mathop {\lim }\limits_{h \to 0} \left| {{{f\left( {x + h} \right) - f\left( x \right)} \over h}} \right| \le \mathop {\lim }\limits_{h \to ... | mcq | aieee-2005 |
20IRL1dIRMRqkslc | maths | limits-continuity-and-differentiability | differentiability | The set of points where $$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)$$ "}, {"identifier": "D"... | ["C"] | null | $$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)... | 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 Ma... | 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... | ["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>$$Rf'\left( 1 \right) = \mathop {\lim }\limits_{h \to 0... | mcq | aieee-2008 |
0QeoBy8todI83IoW | maths | limits-continuity-and-differentiability | differentiability | Let $$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$$. | [{"identifier": "A", "content": "Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1."}, {"identifier": "B", "content": "Statement-1 is true, Statement-2 is false "}, {"identifier": "C", "content": "Statement-1 is false, Statement-2 is true "}, {"identifier": "D", "content... | ["B"] | null | Given that $$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}} \r... | 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 ... | ["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>$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left\{ {\matrix{
{x - 2\,\,\,,} & {x \ge 2} \cr
... | mcq | aieee-2012 |
RtZ9MK0tEqbtwJlp | maths | limits-continuity-and-differentiability | differentiability | If the function.
<br/><br/>$$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... | 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 different... | ["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>$$\therefore$$ $$g\left( x \right) = \left| {\log 2 - \sin \left. {\left| {\log 2 - \sin x} \right|} \right|... | mcq | jee-main-2016-offline |
hF8RmsEat5uIVIgrYz5HO | maths | limits-continuity-and-differentiability | differentiability | If the function
<br/><br/>f(x) = $$\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 $$ $$-$$1 $$=$$ a ... | mcq | jee-main-2016-online-9th-april-morning-slot |
ppH8dXaEhhGCho4t | maths | limits-continuity-and-differentiability | differentiability | Let S = { t $$ \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>= $$\mathop {\lim }\limits_{h \to 0} {{\left| { - h - \pi } \right|\left( {{e^{\left| { - h} \right|}} - 1} \righ... | mcq | jee-main-2018-offline |
LyYVn5Xz5QYZhjEVYw7qG | maths | limits-continuity-and-differentiability | differentiability | Let S = {($$\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... | 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><b... | 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>
$$ \Rightarrow g'(c) = \mathop {\lim }\limits_{x \to c} {{\left| {f(x)} \... | mcq | jee-main-2019-online-10th-april-morning-slot |
5lu85xMvzqpVj7b8OfEQ8 | maths | limits-continuity-and-differentiability | differentiability | Let ƒ : R $$ \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} {\le... | 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": "$$\\left\... | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264785/exam_images/rhqvelsmxghhl3cztv4v.webp" style="max-width: 100%; height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2019 (Online) 12th January Morning Slot Mathematics - Limits, Continuity and Differentiability Question 172 En... | mcq | jee-main-2019-online-12th-january-morning-slot |
ei04ZJBaNytj0HKOk0XL4 | maths | limits-continuity-and-differentiability | differentiability | Let K be the set of all real values of x where the function f(x) = sin |x| – |x| + 2(x – $$\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} \righ... | 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 201... | 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. The... | [{"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.clo... | mcq | 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 Engli... | 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": "$$\\left| {f(c) - f(1)} \\right| < \\left( {1 - c} \\right)\... | ["D"] | null | <p>If we consider the case where f(x) is a constant function, then its derivative f'(x) is equal to 0 for all x in the interval (0,1). </p>
<p>Therefore, if we substitute this into the expressions provided in Options A, B and C, we would have :</p>
<p>Option A : |f(c) - f(1)| < |f'(c)| would become |constant... | mcq | jee-main-2020-online-8th-january-evening-slot |
QQ2Lw8UucYICcFIsyS7k9k2k5ita4xf | maths | limits-continuity-and-differentiability | differentiability | Let ƒ be any function continuous on [a, b] and
twice differentiable on (a, b). If for all x $$ \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}}$$"}] | ["D"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264640/exam_images/thggkxyle8cgcwavqqvd.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 9th January Morning Slot Mathematics - Limits, Continuity and Differentiability Question 150 Engli... | mcq | jee-main-2020-online-9th-january-morning-slot |
hojNfgyrlIYlVlbyLXjgy2xukf8zuozg | maths | limits-continuity-and-differentiability | differentiability | Suppose a differentiable function f(x) satisfies the identity <br/>f(x+y) = f(x) + f(y) + xy<sup>2</sup> + x<sup>2</sup>y, for all real x and y.<br/>
$$\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 $$ f'(x) = f'(0) + x<sup>2</sup> ....(2)<br><br>putting x = y = 0 at equation (1),<... | integer | jee-main-2020-online-4th-september-morning-slot |
NQ1pUMPuqzxQ3JslnJjgy2xukfah3nal | maths | limits-continuity-and-differentiability | differentiability | Let $$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 $$ 2xf<sup>2</sup>(x) - 2x<sup>2</sup>.f(x).f'(x) = 0
<br><br>$$ \Right... | 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": "con... | ["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 = $$\mathop {\li... | mcq | jee-main-2020-online-4th-september-evening-slot |
Qnulz8PdgvmSS14jDPjgy2xukfg792co | maths | limits-continuity-and-differentiability | differentiability | If the function <br/>$$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 } ... | 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 $$\lambda $... | [] | null | 5 | If g(x) = x<sup>5</sup>sin$$\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 $$ = f''(0<sup>+</sup>)
<... | integer | jee-main-2020-online-6th-september-morning-slot |
hI9jZChLQ9ZIKlSkxWjgy2xukg38e1e8 | maths | limits-continuity-and-differentiability | differentiability | Let f : R $$ \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 $$(an empty set)"}, {"identifier": "D", "content": "{1}"}] | ["A"] | null | <img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266844/exam_images/nsxa64efsvtbdkfgqlui.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2020 (Online) 6th September Evening Slot Mathematics - Limits, Continuity and Differentiability Question 133 Eng... | mcq | jee-main-2020-online-6th-september-evening-slot |
tmzgtioYKrhGyobz5kjgy2xukg38nmd4 | maths | limits-continuity-and-differentiability | differentiability | For all twice differentiable functions f : R $$ \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 $$ f'(c) = 0, f'(0) = 0
<br>again applying Rolles th... | mcq | jee-main-2020-online-6th-september-evening-slot |
7CStJ13dzNzxBm9hXE1kls5qoi3 | maths | limits-continuity-and-differentiability | differentiability | The number of points, at which the function <br/>f(x) = | 2x + 1 | $$-$$ 3| x + 2 | + | x<sup>2</sup> + x $$-$$ 2 |, x$$\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... | 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 En... | 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/>$$g(x) = \left\{ {\matrix{
{{x^3},} & {x < 1} \cr
{3x - 2,} & {x \ge 1} \cr
} } \right.$... | [{"identifier": "A", "content": "0"}, {"identifier": "B", "content": "3"}, {"identifier": "C", "content": "1"}, {"identifier": "D", "content": "2"}] | ["C"] | null | $$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... | 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>$$g''(3) - g'... | mcq | jee-main-2021-online-16th-march-evening-shift |
nOdMVjSHQppilEI1PC1kmlj3ysh | maths | limits-continuity-and-differentiability | differentiability | If $$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.$$<br><br>... | mcq | jee-main-2021-online-18th-march-morning-shift |
Xf9fFqIdYJZD5on2nB1kmm49510 | maths | limits-continuity-and-differentiability | differentiability | Let f : R $$ \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}}$$</p>
<p... | integer | jee-main-2021-online-18th-march-evening-shift |
1krrwrbii | maths | limits-continuity-and-differentiability | differentiability | Let a function g : [ 0, 4 ] $$\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.$$, then the number of points in the interval (0, 4) where g(x) is NOT diffe... | [] | null | 1 | $$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'(... | 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 poin... | [] | 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>$$f(x) = \left\{ {\matrix{
{0,} & {x < - 2} \cr
{{3 \over 2}(1 + x),} & { - 2 \le x < 0... | integer | jee-main-2021-online-22th-july-evening-shift |
1krxlb5jm | maths | limits-continuity-and-differentiability | differentiability | Let $$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$$)"}, {"identifi... | ["B"] | null | Graph of $$\max \{ \sin t:0 \le t \le x\} $$ in $$x \in [0,\pi ]$$<br><br><img src="https://res.cloudinary.com/dckxllbjy/image/upload/v1734267003/exam_images/njxdcddt27d4labpnhcm.webp" style="max-width: 100%;height: auto;display: block;margin: 0 auto;" loading="lazy" alt="JEE Main 2021 (Online) 27th July Evening Shift ... | mcq | jee-main-2021-online-27th-july-evening-shift |
1ks0dcadh | maths | limits-continuity-and-differentiability | differentiability | Let $$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$$ (0, 3) where f is not differentiable. Then the sum of numbe... | [] | null | 5 | <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264944/exam_images/avmgbcbdz7wsmx5hhnpg.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734264951/exam_images/ot72zwnlvx1tkyc9kohr.webp"><img src="https://res.c... | integer | jee-main-2021-online-27th-july-morning-shift |
1ktcxxctq | maths | limits-continuity-and-differentiability | differentiability | Let [t] denote the greatest integer less than or equal to t. Let <br/>f(x) = x $$-$$ [x], g(x) = 1 $$-$$ x + [x], and h(x) = min{f(x), g(x)}, x $$\in$$ [$$-$$2, 2]. Then h is : | [{"identifier": "A", "content": "continuous in [$$-$$2, 2] but not differentiable at more than <br>four points in ($$-$$2, 2)"}, {"identifier": "B", "content": "not continuous at exactly three points in [$$-$$2, 2]"}, {"identifier": "C", "content": "continuous in [$$-$$2, 2] but not differentiable at exactly <br>three ... | ["A"] | null | min{x $$-$$ [x], 1 $$-$$ x + [x]}<br><br>h(x) = min{x $$-$$ [x], 1 $$-$$ [x $$-$$ [x])}<br><br> <picture><source media="(max-width: 320px)" srcset="https://res.cloudinary.com/dckxllbjy/image/upload/v1734266549/exam_images/ownvykvbglm0alabpgjc.webp"><source media="(max-width: 500px)" srcset="https://res.cloudinary.com/d... | mcq | jee-main-2021-online-26th-august-evening-shift |
1ktipaq58 | maths | limits-continuity-and-differentiability | differentiability | The function <br/><br/>$$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}}}} &... | 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'... | 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 $$g(x) = \left\{ {\matrix{
{{x^2} + {k_1}x} & , & {x < 0} \cr
{4x + {k_2}} & , & {x \ge 0} ... | [{"identifier": "A", "content": "$$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} = ... | 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 $$x = \pi ,\,\,\mathop {\lim }\limits_{x \to {\pi ^ - }} f(x) = 1 = \mathop {\lim }\limits_{x \to {\pi ^ + }} f(x)... | mcq | jee-main-2022-online-26th-june-evening-shift |
1l5b7z89d | maths | limits-continuity-and-differentiability | differentiability | <p>Let $$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$$ is not continuo... | [{"identifier": "A", "content": "(3, 3)"}, {"identifier": "B", "content": "(2, 4)"}, {"identifier": "C", "content": "(2, 3)"}, {"identifier": "D", "content": "(3, 4)"}] | ["C"] | null | <p>$$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]... | 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 $$\mathbf{R}$$ wher... | [] | 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>
$$
= \begin{cases}4 x^{2}-8 x+5, & \text { if } x \in\left[-\infty, \frac{1}{4}\right] \cup\left[\frac{1}{2}, \infty\right) \\ {\left[4 ... | integer | jee-main-2022-online-25th-july-morning-shift |
1l6m6jj6x | maths | limits-continuity-and-differentiability | differentiability | <p>Let $$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$$, is _________... | [] | null | 2 | <p><img src="https://app-content.cdn.examgoal.net/fly/@width/image/1l7rb5odf/8441c3d2-e56e-4faa-a526-4c7b6253ae3c/ed702f30-2e7f-11ed-8702-156c00ced081/file-1l7rb5odg.png?format=png" data-orsrc="https://app-content.cdn.examgoal.net/image/1l7rb5odf/8441c3d2-e56e-4faa-a526-4c7b6253ae3c/ed702f30-2e7f-11ed-8702-156c00ced081... | integer | jee-main-2022-online-28th-july-morning-shift |
1l6p2shrs | maths | limits-continuity-and-differentiability | differentiability | <p>The number of points, where the function $$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$$ Non-differentiable at $... | mcq | jee-main-2022-online-29th-july-morning-shift |
1l6rfufia | maths | limits-continuity-and-differentiability | differentiability | <p>If $$[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
$$
<br/><br/>$f(x)$ is non differentiable at $x=-\frac{3}{2}$
<br/><br/>and $f(x)$ is discontinuous at $\{-19,-18, \ldots ., 18,19\}$ <br/><br/>as well as $\left\{-\frac{39}{2},-\frac{37}{2}, \ldots,-\fr... | integer | jee-main-2022-online-29th-july-evening-shift |
1ldr770hb | maths | limits-continuity-and-differentiability | differentiability | <p>Suppose $$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) = \math... | 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 | $$
\begin{aligned}
& \text { Let } g(x)=1+x+[x]=\left\{\begin{array}{cc}
1+x ; & x \in[0,1) \\\
2+x ; & x \in[1,2) \\
5 ; & x=2
\end{array}\right. \\\\
& \lambda(x)=x+2[x]=\left\{\begin{array}{cc}
x ; & x \in[0,1) \\
x+2 ; & x \in[1,2) \\
6 ; & x=2
\end{array}\right. \\\\
& r(x)=2+x \\\\
& f(x)=\left\{\begin{array}{cc}... | mcq | jee-main-2023-online-15th-april-morning-shift |
1lgsvmum4 | maths | limits-continuity-and-differentiability | differentiability | <p>Let $$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{... | 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 n... | [] | 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-... | 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 $$\frac{8 f^{\prime}(8)}{f^{\prime}\left(\frac{1}{8}\right... | [] | null | 309 | Here, $$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 ... | 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 va... | 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=... | 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 di... | [{"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/>$$
\begin{aligned}
& \mathrm{f}\left(1^{+}\right)=\lim _{h \rightarrow 0} 2(1+\mathrm{h})+1 \\\\
& \mathrm{f}\left(1^{+}\... | mcq | jee-main-2024-online-1st-february-morning-shift |
jaoe38c1lscn03is | maths | limits-continuity-and-differentiability | differentiability | <p>Consider the function $$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.... | [{"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 ne... | ["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-... | 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>$$\mathrm{f}(\mathrm{x})=\frac{1}{\mathrm{e}^{|\ln \mathrm{x}|}}=\left\{\begin{array}{l}
\frac{1}{\mathrm{e}^{-\ln \mathrm{x}}} ; 0<\mathrm{x}<1 \\
\frac{1}{\mathrm{e}^{\ln \mathrm{x}}} ; \mathr... | mcq | jee-main-2024-online-31st-january-evening-shift |
jaoe38c1lsflbgha | maths | limits-continuity-and-differentiability | differentiability | <p>Let $$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... | 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>Cont... | 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>$$\begin{aligned}
& \lim _{t \rightarrow x} \frac{2 \operatorname{tf}(x)-x^2 f^{\prime}(t)}{1}=1 \\
& \Rightarrow 2 x f(x)-x^2 f'(x)=1 \\
& \frac{d y}{d x}-\frac{2 x y}{x^2}=\frac{-1}{x^2} \\
& \R... | integer | jee-main-2024-online-5th-april-morning-shift |
wcEtMRsaibecOeOmm3jgy2xukf0x5uc1 | maths | limits-continuity-and-differentiability | existance-of-limits | Let [t] denote the greatest integer
$$ \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... | 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>$$\mathop {\lim }\limits_{x \to 0} {{x(\alpha - \beta ) + {x^2}\left( {\alpha + {\beta \over 2} + \gamma } \right) ... | integer | jee-main-2021-online-20th-july-evening-shift |
6EivwZ6DrQWjLqAR | maths | limits-continuity-and-differentiability | limits-of-algebric-function | If $$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 \righ... | 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>= $$\mathop {\lim }\limits_{x \to 2} {{f\left( 2 \right... | mcq | aieee-2002 |
lGj7y8njpmls6ye2 | maths | limits-continuity-and-differentiability | limits-of-algebric-function | Let $$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 \m... | 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' Hospita... | 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>$${1 \over x} = \left[ {{1 \over x}} \right] + \left\{ {{1 \over x}} \r... | mcq | jee-main-2018-offline |
L9hsA4UrY7uaEooFI7Xug | maths | limits-continuity-and-differentiability | limits-of-algebric-function | Let f(x) be a polynomial of degree $$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... | 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>= $$\mathop {\lim }\limits_{x \to 0} \,{{3\left[ {{{\left( {1 + {x \over {27}}} \right)}^{{1 \over 3}}} - 1} \right]} \over {9\left[ {1 - {{\le... | mcq | jee-main-2018-online-16th-april-morning-slot |
sX1MfZeJPUdVnpCXpDYX5 | maths | limits-continuity-and-differentiability | limits-of-algebric-function | $$\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\sqr... | 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>
$$ \Rightarrow \mathop {L... | mcq | jee-main-2019-online-10th-april-morning-slot |
BRO7yDGmrfgKvmK5Aa3rsa0w2w9jx1yg7ul | maths | limits-continuity-and-differentiability | limits-of-algebric-function | If $$\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 ($$ \because $... | mcq | jee-main-2019-online-10th-april-evening-slot |
fyMDR3N15sswpQy8CW3rsa0w2w9jx5crbn3 | maths | limits-continuity-and-differentiability | limits-of-algebric-function | If $$\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 ... | 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>
$$ \Rightarrow $$... | 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>$$ \Rightarrow $$... | 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": "$${\\left( {{2 \\over 3}} \\right)^{{4 \\over 3}}... | ["B"] | null | L = $$\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 {... | 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", "co... | ["A"] | null | $$g(2) = \mathop {\lim }\limits_{x \to 2} {{(x - 2)(x + 1)} \over {(2x + 3)(x - 2)}} = {3 \over 7}$$<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 -... | mcq | jee-main-2021-online-26th-february-evening-slot |
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