Question 511 Mark
If $y=\log \left(\cos e^x\right)$, then find $\frac{d y}{d x}$.
Answer(d): Given, $y=\log \left(\cos e^x\right)$
On differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=\frac{1}{\cos e^x}\left(-\sin e^x \cdot e^x\right)=-e^x \tan e^x
$
View full question & answer→Question 521 Mark
The derivative of $e^{x^3}$ with respect to $\log x$ is
Answer$(c) :$ Let $y=e^{x^3}, z=\log x$
Differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=e^{x^3}\left(3 x^2\right)=3 x^2 e^{x^3} \text { and } \frac{d z}{d x}=\frac{1}{x}$
$\therefore \frac{d y}{d z}=\frac{\frac{d y}{d x}}{\frac{d z}{d x}}=\frac{3 x^2 e^{x^3}}{\left(\frac{1}{x}\right)}=3 x^3 e^{x^3}$
View full question & answer→Question 531 Mark
Determine the value of $k$ for which the function $f(x)$ is continuous at $x=4$.
$f(x)=\left\{\begin{array}{ll} \frac{x^2-16}{x-4}, & x \neq 4 \\ k, & x=4 \end{array}\right. $
Answer$(d):$ Since $f(x)$ is continuous at $x=4$. Therefore,
$\lim _{x \rightarrow 4} f(x)=f(4)$
$\Rightarrow \lim _{x \rightarrow 4} f(x)=k \quad[\because f(4)=k]$
$\Rightarrow \lim _{x \rightarrow 4} \frac{x^2-16}{x-4}=k \Rightarrow \lim _{x \rightarrow 4} \frac{(x-4)(x+4)}{x-4}=k$
$\Rightarrow \lim _{x \rightarrow 4}(x+4)=k \Rightarrow k=8$
View full question & answer→Question 541 Mark
If the function $f(x)=\left\{\begin{array}{l}\frac{\left(e^{k x}-1\right) \tan k x}{4 x^2}, x \neq 0 \\ 16, x=0\end{array}\right.$ is continuous at $x=0$, then $k=$
AnswerSince $,f(x)$ is continuous at $x=0$.
$\therefore \lim _{x \rightarrow 0} f(x)=f(0) $
$\Rightarrow \lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right) \tan k x}{4 x^2}=16$
$\Rightarrow \frac{1}{4} \lim _{x \rightarrow 0} \frac{\left(e^{k x}-1\right)}{k x} \times \frac{\tan k x}{k x} \times k^2=16$
$ \Rightarrow \frac{k^2}{4} \times 1 \times 1=16$
$\Rightarrow k^2=64 $
$\Rightarrow k= \pm 8$
View full question & answer→Question 551 Mark
If $x^y=e^{x-y}$, then $\frac{d y}{d x}$ is
Answer(c) : We have, $x^y=e^{x-y}$
Taking $\log$ on both sides, we get
$
y \log x=(x-y) \log e \Rightarrow y=\frac{x}{1+\log x}
$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=\frac{(1+\log x)-x \cdot \frac{1}{x}}{(1+\log x)^2}=\frac{\log x}{(1+\log x)^2}
$
View full question & answer→Question 561 Mark
If $y=\cos ^2\left(\frac{3 x}{2}\right)-\sin ^2\left(\frac{3 x}{2}\right)$, then $\frac{d^2 y}{d x^2}$ is equal to
Answer$\text { (c) : Given, } y=\cos ^2\left(\frac{3 x}{2}\right)-\sin ^2\left(\frac{3 x}{2}\right)$
$\Rightarrow y=\cos 3 x \Rightarrow \frac{d y}{d x}=-3 \sin 3 x$
$\Rightarrow \frac{d^2 y}{d x^2}=-3 \times 3 \cos 3 x=-9 \cos 3 x=-9 y$
View full question & answer→Question 571 Mark
If $f(x)=x^2 g(x)$ and $g(x)$ is twice differentiable, then $f^{\prime \prime}(x)$ is equal to
Answer$(c) : f^{\prime}(x)=\frac{d}{d x}\left(x^2 g(x)\right)=x^2 g^{\prime}(x)+2 x g(x)$
Now, $f^{\prime \prime}(x)=\frac{d}{d x}\left(x^2 g^{\prime}(x)+2 x g(x)\right)$
$=x^2 g^{\prime \prime}(x)+g^{\prime}(x) 2 x+2\left\{x g^{\prime}(x)+g(x) \cdot 1\right\}$
$=x^2 g^{\prime \prime}(x)+4 x g^{\prime}(x)+2 g(x)$
View full question & answer→Question 581 Mark
If $y=5 \cos x-3 \sin x$, then $\frac{d^2 y}{d x^2}$ is equal to
Answer$\text { (a) : We have, } y=5 \cos x-3 \sin x$
$\Rightarrow \frac{d y}{d x}=-5 \sin x-3 \cos x$
$\Rightarrow \frac{d^2 y}{d x^2}=-5 \cos x+3 \sin x=-y$
View full question & answer→Question 591 Mark
If $y=\frac{\ln x}{x}$, then the value of $y^{\prime \prime}(e)$ is
Answer$(d) :$ Given, $y=\frac{\ln x}{x}$
Differentiating w.r.t. $x$, we get
$y^{\prime}=\frac{(1-\ln x)}{x^2} \Rightarrow y^{\prime \prime}=\frac{x^2\left(-\frac{1}{x}\right)-(1-\ln x) 2 x}{x^4}$
$\therefore y^{\prime \prime}(e)=\frac{-e-0}{e^4}=-\frac{1}{e^3}$
View full question & answer→Question 601 Mark
If $f(x)=-\sqrt{25-x^2}$, then $\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}$ is equal to
Answer(d) : $\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}=f^{\prime}(1)$
Now, $f^{\prime}(x)=-\frac{1}{2} \cdot \frac{-2 x}{\sqrt{25-x^2}}=\frac{x}{\sqrt{25-x^2}}$
$
\therefore \quad f^{\prime}(1)=\frac{1}{\sqrt{25-1^2}}=\frac{1}{\sqrt{24}}
$
View full question & answer→Question 611 Mark
The function $f(x)=\left\{\begin{array}{cc}|x-3|, & x \geq 1 \\ \frac{x^2}{4}-\frac{3 x}{2}+\frac{13}{4}, & x<1\end{array}\right.$ is
Answer$(d) : |x-3|$ is continuous at $x=3,$ but not differentiable.
$f\left(1^{-}\right)=f\left(1^{+}\right)=f(1)=2$
$f^{\prime}\left(1^{-}\right)=f^{\prime}\left(1^{+}\right)=-1=f^{\prime}(1)$
$\therefore f(x)$ is continuous and differentiable at $x=1$.
View full question & answer→Question 621 Mark
Let $f$ be defined on $[-5,5]$ as $f(x)=\left\{\begin{array}{l} x \text { if } x \text { is rational } \\ -x \text { if } x \text { is irrational }\end{array}\right.$ Then $f(x)$ is
AnswerAs $x \rightarrow 0$ both $x$ and $-x$ tend to zero, $f(0)=0$
$\therefore f(x)$ is continuous at $x=0$.
For $x \neq 0, x \neq-x, f(x)$ is discontinuous.
View full question & answer→Question 631 Mark
If $f(x) f(y)=f(x+y)$ for all $x, y$; suppose $f(5)=2$ and $f^{\prime}(0)=3$, then $f^{\prime}(5)$ is equal to
Answer(b) : $f(x+y)=f(x) f(y) \Rightarrow f(0+5)=f(0) f(5)$
Also $f^{\prime}(5)=\lim _{h \rightarrow 0} \frac{f(5+h)-f(5)}{h}$
$=\lim _{h \rightarrow 0} \frac{f(5) f(h)-f(5+0)}{h}=\lim _{h \rightarrow 0} \frac{f(5) f(h)-f(5) f(0)}{h}$
$=f(5) \lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=f(5) f^{\prime}(0)=2 \times 3=6$
View full question & answer→Question 641 Mark
If $y=\log _7(\log x)$, then find $\frac{d y}{d x}$.
Answer(a) : $y=\log _7(\log x)=\frac{\log (\log x)}{\log 7}$
$
\therefore \frac{d y}{d x}=\frac{1}{\log 7} \cdot \frac{1}{\log x} \cdot \frac{1}{x} \Rightarrow \frac{d y}{d x}=\frac{1}{x \log 7 \log x}
$
View full question & answer→Question 651 Mark
If $y=e^{\frac{1}{2} \log \left(1+\tan ^2 x\right)},$ then $\frac{d y}{d x}$ is equal to
Answer$\text { (c) : } y=e^{\frac{1}{2} \log \left(1+\tan ^2 x\right)}$
$=\left(\sec ^2 x\right)^{1 / 2}$
$=\sec x$
$\therefore \frac{d y}{d x}=\sec x \tan x$
View full question & answer→Question 661 Mark
The number of discontinuous functions $y(x)$ on $[-2,2]$ satisfying $x^2+y^2=4$ is
Answer(a) : Functions which satisfy the relation $x^2+y^2=4$ are $y(x)=\sqrt{4-x^2}$ and $y(x)=-\sqrt{4-x^2}$. And both functions are continuous in $[-2,2]$.
View full question & answer→Question 671 Mark
If $y=a x^2+b$, then $\frac{d y}{d x}$ at $x=2$ is equal to
Answer$(a) :$ We have, $y=a x^2+b$
$\Rightarrow \frac{d y}{d x}=2 a x$
$\left.\frac{d y}{d x}\right|_{x=2}=2 a \times 2=4 a$
View full question & answer→Question 681 Mark
For what value of $k$, the function $f(x)=\left\{\begin{aligned} k x^2 & \text { if } x \leq 2 \\ 3 & \text { if } x>2\end{aligned}\right.$ is continuous at $x=2$ ?
Answer(c) : $\because f(x)$ is continuous at $x=2$,
$
\therefore \lim _{x \rightarrow 2^{-}} f(x)=\lim _{x \rightarrow 2^{+}} f(x) . \quad \therefore \quad k(2)^2=3 \Rightarrow k=\frac{3}{4}
$
View full question & answer→Question 691 Mark
The point $(s),$ at which the function $f$ given by $f(x)=\left\{\begin{array}{l}\frac{x}{|x|}, x<0 \\ -1, x \geq 0\end{array}\right.$ is continuous, is$/$are
AnswerWe have $,f(x)=\left\{\begin{array}{ll}\frac{x}{|x|}, & x<0 \\ -1, & x \geq 0\end{array}\right.$
$ \Rightarrow f(x)=\left\{\begin{array}{cc} \frac{x}{-x}=-1, & x<0 \\ -1, & x \geq 0 \end{array} \right.$
$\Rightarrow f(x)=-1 \forall x \in R$
$\Rightarrow f(x)$ is continuous $\forall x \in R$ as it is a constant function.
View full question & answer→Question 701 Mark
If $y=\tan ^{-1}\left[\frac{\sin x+\cos x}{\cos x-\sin x}\right]$, then $\frac{d y}{d x}$ is equal to
Answer(d): We have, $y=\tan ^{-1}\left[\frac{\sin x+\cos x}{\cos x-\sin x}\right]$
$\Rightarrow y=\tan ^{-1}\left[\frac{1+\tan x}{1-\tan x}\right]=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}+x\right)\right]$
$\Rightarrow \quad y=\frac{\pi}{4}+x$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=1
$
View full question & answer→Question 711 Mark
The function $f(x)=\cot x$ is discontinuous on the set
Answer(a) : $f(x)=\cot x$ is discontinuous if $\cot x \rightarrow \infty$
$\Rightarrow \cot x=\cot 0 \Rightarrow x=n \pi \forall n \in Z$.
View full question & answer→Question 721 Mark
If $u=x^2+y^2$ and $x=s+3 t, y=2 s-t,$ then $\frac{d^2 u}{d s^2}$ is equal to
Answer$(d)$ : Given $,u=x^2+y^2, x$
$=s+3 t, y=2 s-t$
$\Rightarrow \frac{d x}{d s}=1, \frac{d y}{d s}=2$
Now $, u=x^2+y^2 $
$\Rightarrow \frac{d u}{d s}=2 x \frac{d x}{d s}+2 y \frac{d y}{d s}$
$=2 x+4 y$
$\Rightarrow \frac{d^2 u}{d s^2}$
$=2\left(\frac{d x}{d s}\right)+4\left(\frac{d y}{d s}\right) $
$\Rightarrow \frac{d^2 u}{d s^2}=2(1)+4(2)=10$
View full question & answer→Question 731 Mark
If $y=\log _a x+\log _x a+\log _x x+\log _a a$, then $\frac{d y}{d x}$ is equal to
Answer$\text { (d): } \because y=\log _a x+\frac{\log a}{\log x}+1+1 \quad\left\{\because \log _x x=1\right\}$
$\Rightarrow \frac{d y}{d x}=\frac{1}{x} \log _a e-\log a\left(\frac{1}{\log x}\right)^2 \frac{1}{x}=\frac{1}{x \log a}-\frac{\log a}{x(\log x)^2}$
View full question & answer→Question 741 Mark
If $y=a e^x+b e^{-x}+c$, where $a, b, c$ are parameters, then $y^{\prime}$ is equal to
Answer(a) : $y=a e^x+b e^{-x}+c$
Differentiating w.r.t. $x$, we get
$
y^{\prime}=a e^x-b e^{-x}
$
View full question & answer→Question 751 Mark
If $y=e^{3 x+7}$, then the value of $\left[\frac{d y}{d x}\right]_{x=0}$ is equal to
Answer(d) : $\because y=e^{3 x+7} \Rightarrow \frac{d y}{d x}=3 e^{3 x+7}$
$\therefore\left[\frac{d y}{d x}\right]_{x=0}=3 e^{3 \times 0+7}=3 e^7$
View full question & answer→Question 761 Mark
If $f$ is a real-valued differentiable function satisfying $|f(x)-f(y)| \leq(x-y)^2, x, y \in R$ and $f(0)=0$, then $f(1)$ equals
Answer(c) : Since, $\lim _{x \rightarrow y} \frac{|f(x)-f(y)|}{|x-y|} \leq \lim _{x \rightarrow y}|x-y|$
$\Rightarrow\left|f^{\prime}(y)\right| \leq 0 \Rightarrow f^{\prime}(y)=0 \Rightarrow f(y)=$ constant
As $f(0)=0 \Rightarrow f(y)=0 \Rightarrow f(1)=0$
View full question & answer→Question 771 Mark
If $f(x)=\sqrt{1+\cos ^2\left(x^2\right)}$, then the value of $f^{\prime}\left(\frac{\sqrt{\pi}}{2}\right)$ is
Answer$(b) :$ We have, $f(x)=\sqrt{1+\cos ^2\left(x^2\right)}$
$\Rightarrow f^{\prime}(x)=\frac{1}{2} \times \frac{-2 \sin x^2 \cos x^2}{\sqrt{1+\cos ^2 x^2}}(2 x)=\frac{-\sin 2 x^2}{\sqrt{1+\cos ^2 x^2}}(x)$
$\therefore f^{\prime}\left(\frac{\sqrt{\pi}}{2}\right)=-\frac{\sqrt{\pi}}{2} \cdot \frac{\sin 2\left(\frac{\pi}{4}\right)}{\sqrt{1+\cos ^2\left(\frac{\pi}{4}\right)}}=-\sqrt{\frac{\pi}{6}}$
View full question & answer→MCQ 781 Mark
The derirative of $\sin ^2 x$ with respect to $x$ is-
- A
$\cos ^2 x$
- B
$-\cos ^2 x$
- ✓
$\sin 2 x$
- D
$\cos 2 x$
AnswerCorrect option: C. $\sin 2 x$
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