Question 11 Mark
The value of $k(k<0)$ for which the function $f$ defined as $f(x)=\left\{\begin{array}{cc}\frac{1-\cos k x}{x \sin x} & , x \neq 0 \\ \frac{1}{2} & , x=0\end{array}\right.$ is continuous at $x=0$ is
AnswerWe have, $f(x)=\left\{\begin{array}{cc}\frac{1-\cos k x}{x \sin x}, & x \neq 0 \\ \frac{1}{2}, & x=0\end{array}\right.$
$\because f(x)$ is continuous at $x=0$.
$\therefore \lim _{x \rightarrow 0} \frac{1-\cos k x}{x \sin x}=\frac{1}{2} $
$\Rightarrow \lim _{x \rightarrow 0} \frac{2 \sin ^2 \frac{k x}{2}}{x^2 \frac{\sin x}{x}}=\frac{1}{2}$
$\Rightarrow \lim _{x \rightarrow 0} 2 \cdot \frac{k^2}{4}\left\{\frac{\sin \left(\frac{k x}{2}\right)}{\frac{k x}{2}}\right\}^2 \frac{1}{\frac{(\sin x)}{x}}=\frac{1}{2}$
$\Rightarrow \frac{k^2}{2}=\frac{1}{2} $
$\Rightarrow k= \pm 1$
But $k < 0 $
$\therefore k=-1$
View full question & answer→Question 21 Mark
The derivative of $\sin ^{-1}\left(2 x \sqrt{1-x^2}\right) w.r.t. \sin ^{-1} x \frac{1}{\sqrt{2}} < x <1,$ is
AnswerLet $u=\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)$
and $v=\sin ^{-1} x, \frac{1}{\sqrt{2}}$
$\Rightarrow \sin v=x$
From $(i)$ and $(ii),$ we get
$\Rightarrow u=\sin ^{-1}(2 \sin v \cos v)$
$=\sin ^{-1}(\sin 2 v)$
$ u=2 v$
Differentiating with respect to $v$ both sides, we get $\frac{d u}{d v}=2$
View full question & answer→Question 31 Mark
If $y=\tan ^{-1}\left(e^{2 x}\right)$, then $\frac{d y}{d x}$ is equal to
AnswerGiven, $y=\tan ^{-1}\left(e^{2 x}\right)$
$\therefore \frac{d y}{d x}=\frac{1}{1+e^{4 x}} \times 2 e^{2 x}=\frac{2 e^{2 x}}{1+e^{4 x}}$
View full question & answer→Question 41 Mark
If $y=5 \cos x-3 \sin x$, then $\frac{d^2 y}{d x^2}$ is equal to
Answer$\text {We have, } y=5 \cos x-3 \sin x$
$\Rightarrow \frac{d y}{d x}=-5 \sin x-3 \cos x$
$\Rightarrow \frac{d^2 y}{d x^2}=-5 \cos x+3 \sin x=-y$
View full question & answer→Question 51 Mark
If $y=\log \left(\cos e^x\right)$, then $\frac{d y}{d x}$ is
AnswerWe have, $y=\log \left(\cos e^x\right)$
Differentiating both sides w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{1}{\cos e^x} \cdot\left(-\sin e^x\right) \cdot e^x$
$\Rightarrow \frac{d y}{d x}=-e^x \tan e^x$
View full question & answer→Question 61 Mark
If $e^x+e^y=e^{x+y}$, then $\frac{d y}{d x}$ is
AnswerWe have, $e^x+e^y=e^{x+y}$
$
\Rightarrow e^{-y}+e^{-x}=1
$
Differentiating w.r.t. $x$, we get
$
-e^{-y} \frac{d y}{d x}-e^{-x}=0 \Rightarrow \frac{d y}{d x}=-e^{y-x}
$
View full question & answer→Question 71 Mark
If $x=a \sec \theta, y=b \tan \theta$, then $\frac{d^2 y}{d x^2}$ at $\theta=\frac{\pi}{6}$ is
AnswerWe have, $x=a \sec \theta$
$\Rightarrow \frac{d x}{d \theta}=a \tan \theta \sec \theta \text { and } y=b \tan \theta$
$\Rightarrow \frac{d y}{d \theta}=b \sec ^2 \theta$
$\therefore \frac{d y}{d x}=\frac{b \sec ^2 \theta}{a \tan \theta \sec \theta}=\frac{b}{a} \operatorname{cosec} \theta$
$\Rightarrow \frac{d^2 y}{d x^2}=\frac{-b}{a} \operatorname{cosec} \theta \cot \theta \frac{d \theta}{d x}$
$=\frac{-b}{a} \operatorname{cosec} \theta \cot \theta \frac{1}{a \tan \theta \sec \theta}=\frac{-b}{a^2} \cot ^3 \theta$
$\therefore\left[\frac{d^2 y}{d x^2}\right]_{\theta=\frac{\pi}{6}}=\frac{-3 \sqrt{3} b}{a^2}$
View full question & answer→Question 81 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 $,f(x)=\left\{\begin{array}{ll}\frac{x}{|x|}, & x<0 \\ -1, & x \geq 0\end{array}\right.$
$\begin{array}{l} \Rightarrow f(x)=\left\{\begin{array}{cc} \frac{x}{-x}=-1, & x<0 \\ -1, & x \geq 0 \end{array}\right. \\\end{array}$
$\Rightarrow f(x)=-1 \forall x \in R$
$\Rightarrow f(x)$ is continuous $\forall x \in R$ as it is a constant function.
View full question & answer→Question 91 Mark
If $y=e^{-x}$, then $\frac{d^2 y}{d x^2}$ is equal to
AnswerGiven, $y=e^{-x}$
$\Rightarrow \quad \frac{d y}{d x}=-e^{-x} \Rightarrow \frac{d^2 y}{d x^2}=e^{-x}=y$
View full question & answer→Question 101 Mark
If $y^2(2-x)=x^3$, then $\left(\frac{d y}{d x}\right)_{(1,1)}$ is equal to
Answer$\text {Given, } y^2(2-x)=x^3$
$\Rightarrow y^2=\frac{x^3}{2-x}$
$\Rightarrow 2 y \cdot \frac{d y}{d x}=\frac{(2-x) \times 3 x^2-x^3(-1)}{(2-x)^2}$
$\Rightarrow \frac{d y}{d x}=\frac{6 x^2-2 x^3}{2 y(2-x)^2}$
$\Rightarrow\left(\frac{d y}{d x}\right)_{(1,1)}=\frac{6-2}{2 \times 1}=2$
View full question & answer→Question 111 Mark
If $y=\sin \left(2 \sin ^{-1} x\right)$, then $\left(1-x^2\right) y_2$ is equal to
Answer$\text { We have, } y=\sin \left(2 \sin ^{-1} x\right)$
$\Rightarrow y=\sin \left[\sin ^{-1}\left(2 x \sqrt{1-x^2}\right)\right]$
$\Rightarrow y=2 x \sqrt{1-x^2}.........(i)$
$\Rightarrow y_1=2 x \times \frac{-2 x}{2 \sqrt{1-x^2}}+2 \sqrt{1-x^2}=\frac{-4 x^2+2}{\sqrt{1-x^2}}.......(ii)$
$\therefore y_2=\frac{\sqrt{1-x^2}(-8 x)-\left(-4 x^2+2\right) \times \frac{-2 x}{2 \sqrt{1-x^2}}}{1-x^2}$
$\quad=\frac{4 x^3-6 x}{\left(1-x^2\right) \sqrt{1-x^2}} \Rightarrow\left(1-x^2\right) y_2=\frac{4 x^3-6 x}{\sqrt{1-x^2}}$
Now, consider $x y_1-4 y$
$=\frac{-4 x^3+2 x}{\sqrt{1-x^2}}-8 x \sqrt{1-x^2}$
$=\frac{4 x^3-6 x}{\sqrt{1-x^2}}
$[Using $(i)$ and $(ii)]$
Thus, $\left(1-x^2\right) y_2=x y_1-4 y$
View full question & answer→Question 121 Mark
If the function $f(x)=\left\{\begin{array}{cc}3 x-8, & \text { if } x \leq 5 \\ 2 k, & \text { if } x>5\end{array}\right.$ is continuous, then the value of $k$ is
Answer Since $f(x) $ is continuous at $ x=5 \text {, }$
$\Rightarrow \lim _{x \rightarrow 5} f(x)=\lim _{x \rightarrow 5^{+}} f(x)=f(5)$
$\Rightarrow 3(5)-8=2 k $
$\Rightarrow 7=2 k $
$\Rightarrow k=\frac{7}{2}$
View full question & answer→Question 131 Mark
The function $f(x)=\left\{\begin{array}{cc}x^2 & \text { for } x<1 \\ 2-x & \text { for } x \geq 1\end{array}\right.$ is
AnswerAt $x=1 \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} x^2=1$
And $\lim _{x \rightarrow 1^{+}} f(x)=\lim _{x \rightarrow 1} 2-x=1$
Also, $f(1)=2-1=1 \because \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)=f(x)$
$\therefore f(x)$ is continuous at $x=1$
Now, L.H.D. $=\lim _{x \rightarrow 1^{-}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1} \frac{x^2-1}{x-1}=\lim _{x \rightarrow 1}(x+1)=2$
R.H.D. $=\lim _{x \rightarrow 1^{+}} \frac{f(x)-f(1)}{x-1}=\lim _{x \rightarrow 1} \frac{(2-x)-1}{x-1}=-1$
$\because \quad$ L.H.D. $\neq$ R.H.D. $\therefore f(x)$ is not differentiable at $x=1$.
View full question & answer→Question 141 Mark
If $x=t^2+1, y=2 a t$, then $\frac{d^2 y}{d x^2}$ at $t=a$ is
AnswerGiven, $x=t^2+1$ and $y=2 a t$
$\Rightarrow \frac{d x}{d t}=2 t$
$\Rightarrow \frac{d y}{d t}=2 a$
$\therefore \frac{d y}{d x}=\frac{a}{t}$
$\Rightarrow \frac{d^2 y}{d x^2}=\frac{-a}{t^2} \cdot \frac{d t}{d x}=\frac{-a}{2 t^3}$
$\therefore\left(\frac{d^2 y}{d x^2}\right)_{a t t=a}=\frac{-a}{2 a^3}=\frac{-1}{2 a^2}$
View full question & answer→Question 151 Mark
Derivative of $e^{2 x}$ with respect to $e^x$, is
AnswerLet $u=e^{2 x}$ and $v=e^x$
$
\Rightarrow \quad \frac{d u}{d x}=2 e^{2 x}, \frac{d v}{d x}=e^x \therefore \quad \frac{d u}{d v}=\frac{d u / d x}{d v / d x}=\frac{2 e^{2 x}}{e^x}=2 e^x
$
View full question & answer→Question 161 Mark
If $y=\cos ^{-1}\left(e^x\right)$, then $\frac{d y}{d x}$ is :
AnswerWe have, $y=\cos ^{-1}\left(e^x\right)$
$\Rightarrow \frac{d y}{d x}=\frac{-1}{\sqrt{1-\left(e^x\right)^2}} e^x=\frac{-e^x}{\sqrt{1-e^{2 x}}}=\frac{-e^x}{e^x \sqrt{e^{-2 x}-1}}=\frac{-1}{\sqrt{e^{-2 x}-1}}$
View full question & answer→Question 171 Mark
The number of points of discontinuity of
$f(x)=\left\{\begin{array}{ll}|x|+3, & \text { if } x \leq-3 \\ -2 x, & \text { if }-3<x<3 is\\ 6 x+2, & \text { if } x \geq 3\end{array}\right.$
Answerwe have
$f(x)=\left\{\begin{array}{ll}|x|+3, & \text { if } x \leq-3 \\ -2 x, & \text { if }-3<x<3 is\\ 6 x+2, & \text { if } x \geq 3\end{array}\right.$
Now, $\lim _{x \rightarrow-3^{-}} f(x)=-(-3)+3=6$ and $\lim _{x \rightarrow-3^{+}} f(x)=-2(-3)=6$
Also, $f(-3)=-(-3)+3=3+3=6$
As $\lim _{x \rightarrow-3^{-}} f(x)=\lim _{x \rightarrow-3^{+}} f(x)=f(-3)$
$\therefore f(x)$ is discontinuous at $x=-3$
Again $\lim _{x \rightarrow 3^{-}} f(x)=-2(3)=-6$ and $\lim _{x \rightarrow 3^{+}} f(x)=6(3)+2=20 \neq-6$
$\therefore f(x)$ is discontinuous at $x=3$.
So, only one point of discontinuity.
View full question & answer→Question 181 Mark
Derivative of $e^{\sin ^2 x}$ with respect to $\cos x$ is
AnswerLet $P=e^{\sin ^2 x}$ and $Q=\cos x$
Differentiating both sides w.r.t. $x$, we get $\frac{d P}{d x}=e^{\sin ^2 x} \cdot 2 \sin x \cdot \cos x$ and $\frac{d Q}{d x}=-\sin x$
Now, $\frac{d p}{d Q}=\frac{\frac{d P}{d x}}{\frac{d Q}{d x}}=\frac{2 e^{\sin ^2 x} \sin x \cos x}{-\sin x}=-2 e^{\sin ^2 x} \cos x$
View full question & answer→Question 191 Mark
The derivative of $\tan ^{-1}\left(x^2\right)$ w.r.t. $x$ is :
AnswerLet $y=\tan ^{-1}\left(x^2\right)$
$\Rightarrow \frac{d y}{d x}=\frac{d}{d x}\left(\tan ^{-1}\left(x^2\right)\right)=\frac{1}{1+\left(x^2\right)^2} \times 2 x=\frac{2 x}{1+x^4}$
View full question & answer→Question 201 Mark
For what value of $k,$ the function given below is continuous at $x=0$ ? $f(x)=\left\{\begin{array}{cc}\frac{\sqrt{4+x}-2}{x} & , x \neq 0 \\ k & , x=0\end{array}\right.$
AnswerAs, $f(x)=\left\{\begin{array}{cc}\frac{\sqrt{4+x-2}}{x}, & x \neq 0 \\ k, & x=0\end{array}\right.$ is continuous at $x=0$
$ \Rightarrow \ce{LHL=RHL}=f(0) $ or $ \lim _{x \rightarrow 0} f(x)=f(0)$
$ \Rightarrow \lim _{x \rightarrow 0} \frac{\sqrt{4+x}-2}{x} \times \frac{\sqrt{4+x}+2}{\sqrt{4+x}+2}=k$
$ \Rightarrow \lim _{x \rightarrow 0} \frac{4+x-4}{x(\sqrt{4+x}+2)}=k $
$\Rightarrow k=\lim _{x \rightarrow 0} \frac{1}{(\sqrt{4+x}+2)}$
$ \therefore k=\frac{1}{4}$
View full question & answer→Question 211 Mark
The function $f(x)=[x]$, where $[x]$ denotes the greatest integer less than or equal to $x$, is continuous at
AnswerLet $x=1.5$
$\therefore$ L.H.L. $=\operatorname{Lt}_{x \rightarrow 1.5^{-}} f(x)=\underset{h \rightarrow 0}{\operatorname{Lt}}[1.5-h]=1$
and R.H.L. $=\operatorname{Lt}_{x \rightarrow 1.5^{+}} f(x)=\underset{h \rightarrow 0}{\operatorname{Lt}}[1.5+h]=1$
$\because \quad$ L.H.L. = R.H.L.
$\therefore f(x)$ is continuous at $x=1.5$
Also, greatest integer function is discontinuous at all integral values of $x$.
View full question & answer→Question 221 Mark
A function $f: R \rightarrow R$ is defined by:
$f(x)=\left\{\begin{array}{cc}e^{-2 x}, & x<\ln \frac{1}{2} \\ 4, & \ln \frac{1}{2} \leq x \leq 0 \\ e^{-2 x}, & x>0\end{array}\right.$
Which of the following statements is true about the function at the point $x=\ln \frac{1}{2}$ ?
Answer$f(x)$ is continuous but not differentiable.
View full question & answer→Question 231 Mark
If $f(x)=\left\{\begin{array}{l}\frac{k x}{|x|} \text {, if } x<0 \\ 3, \text { if } x \geq 0\end{array}\right.$ is continuous at $x=0$, then the value of $k$ is
Answer$ f(x)=\left\{\begin{array}{l}\frac{k x}{|x|}, \text { if } x<0 \\ 3, \text { if } x \geq 0\end{array}\right. $
Since $, f $ is continuous at $x=0,$
$\Rightarrow \text { L.H.L }=\text { R.H.L. }=f(0)$
$\Rightarrow \lim _{x \rightarrow 0^{-}} f(x)=\lim _{x \rightarrow 0^{+}} f(x)=f(0)$
$\Rightarrow \lim _{x \rightarrow 0^{-}} \frac{-k x}{x}=\lim _{x \rightarrow 0^{+}} 3=3 $
$\Rightarrow k=-3 .$
View full question & answer→Question 241 Mark
The set of all points where the function $f(x)=x+|x|$ is differentiable, is
View full question & answer→Question 251 Mark
If $f(x)=\cos ^{-1} \sqrt{x}$, 0 < x < 1, which of the following is aqual to $f^{\prime}(x) ?$
Answer$\frac{-1}{2 \sqrt{x(1-x)}}$
View full question & answer→Question 261 Mark
If $y=\log \left(\sin e^x\right)$, then $\frac{d y}{d x}$ is
Answer$y=\log \left(\sin e^x\right)$
Differentiating w.r.t. $x$, we get
$\begin{aligned}
\frac{d y}{d x}= & \frac{1}{\sin e^x} \cdot \frac{d}{d x}\left(\sin e^x\right)=\frac{1}{\sin e^x} \cos e^x \cdot \frac{d}{d x} e^x=\frac{1}{\sin e^x} \cos e^x \cdot e^x \\
& =e^x \cot e^x
\end{aligned}
$
View full question & answer→Question 271 Mark
If $(\cos x)^y=(\cos y)^x$, then $\frac{d y}{d x}$ is equal to:
AnswerGiven, $(\cos x)^y=(\cos y)^x$
Taking \log on both sides, we get
$\log \left[(\cos x)^y\right]=\log \left[(\cos y)^x\right] \Rightarrow y \log (\cos x)=x \log (\cos y)$
Differentiate w.r.t. $x$, we get
$\frac{d y}{d x} \log (\cos x)+\frac{y}{\cos x}(-\sin x)=\log (\cos y)+\frac{x}{\cos y}(-\sin y) \cdot \frac{d y}{d x}$
$\Rightarrow \frac{d y}{d x}(\log (\cos x)+x \tan y)=\log (\cos y)+y \tan x$
$\Rightarrow \frac{d y}{d x}=\frac{y \tan x+\log (\cos y)}{x \tan y+\log (\cos x)}$
View full question & answer→Question 281 Mark
The function $f(x)=|x|$ is
Answer$f(x)=|x|$ = $\left\{ x, x\} > 0 -x, x<0\right.$
The function $f(x)$ is continuous everywhere but not differentiable at $x=0$ as at $x=0$
$L f^{\prime}(0)=\lim _{x \rightarrow 0^{-}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{x \rightarrow 0} \frac{-x-0}{x}=-1$
$R f^{\prime}(0)=\lim _{x \rightarrow 0^{+}} \frac{f(x)-f(0)}{x-0}$
$=\lim _{x \rightarrow 0} \frac{x-0}{x}=1$
$\therefore L f^{\prime}(0) \neq R f^{\prime}(0)$, so $f(x)$ is not differentiable at $x=0$. View full question & answer→Question 291 Mark
The derivative of $x^{2 x}$ w.r.t. $x$ is
AnswerLet $y=x^{2 x}$
Taking log on both sides, we get
$
\log y=2 x \log x
$
Differentiating both sides w.r.t. $x$, we get
$
\begin{aligned}
& \frac{1}{y} \frac{d y}{d x}=2\left\{x \cdot \frac{1}{x}+\log x \cdot 1\right\} \\
\Rightarrow & \frac{d y}{d x}=2 y\{1+\log x\}=2 x^{2 x}(1+\log x)
\end{aligned}
$
View full question & answer→Question 301 Mark
If $x=a \cos \theta+b \sin \theta, y=a \sin \theta-b \cos \theta$, then which one of the following is true?
AnswerGiven, $x=a \cos \theta+b \sin \theta$ and $y=a \sin \theta-b \cos \theta$ Differentiate w.r.t $\theta$, we get
$
\frac{d x}{d \theta}=-a \sin \theta+b \cos \theta \text { and } \frac{d y}{d \theta}=a \cos \theta+b \sin \theta
$
Now, $\frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}=\frac{a \cos \theta+b \sin \theta}{-(a \sin \theta-b \cos \theta)}=\frac{x}{-y}$
$
\Rightarrow \frac{d y}{d x}=\frac{-x}{y}
$
Differentiate (i) w.r.t. $x$, we get
$
\begin{aligned}
& \frac{d^2 y}{d x^2}=\frac{y(-1)+x(d y / d x)}{y^2} \\
\Rightarrow & y^2 \frac{d^2 y}{d x^2}=-y+\frac{x d y}{d x} \Rightarrow y^2 \frac{d^2 y}{d x^2}-\frac{x d y}{d x}+y=0
\end{aligned}
$
View full question & answer→Question 311 Mark
If $x=A \cos 4 t+B \sin 4 t$, then $\frac{d^2 x}{d t^2}$ is equal to
AnswerWe have, $x=A \cos 4 t+B \sin 4 t$
Differentiating both sides $\text{w.r.t. t}$, we get
$\frac{d x}{d t}=A \cdot(-\sin 4 t) \cdot 4+B \cos 4 t \cdot 4$
Again differentiating both sides of $(i) \ \text{w.r.t. t}$ we get
$\frac{d^2 x}{d t^2}=-4 A(\cos 4 t) \cdot 4+4 B(-\sin 4 t) \cdot 4$
$=-16 A \cos 4 t-16 B \sin 4 t=-16(A \cos 4 t+B \sin 4 t)=-16 x$
View full question & answer→Question 321 Mark
The value of $k$ for which $f(x)=\left\{\begin{array}{cc}3 x+5, & x \geq 2 \\ k x^2, & x<2\end{array}\right.$ is a continuous functions, is:
AnswerGiven Function is
$f(x)=\left\{\begin{array}{cc}
3 x+5, & x \geq 2 \\
k x^2, & x<2
\end{array}\right.
$
For a function $f(x)$ to be continuous at $x=a$,
L.H.L of $f(x)$ at $a=$ R.H.L. of $f(x)$ at $a$
i.e., $\lim _{x \rightarrow a^{-}} f(x)=\lim _{x \rightarrow a^{+}} f(x)$
$\Rightarrow \quad \lim _{h \rightarrow 0} k(2-h)^2=\lim _{h \rightarrow 0} 3(2+h)+5 \Rightarrow 4 \times k=11 \Rightarrow k=\frac{11}{4}
$
View full question & answer→Question 331 Mark
The value of ' $k$ ' for which the function $f(x)=\left\{\begin{array}{cll}\frac{1-\cos 4 x}{8 x^2}, & \text { if } & x \neq 0 \\ k, & \text { if } & x=0\end{array}\right.$ is continuous at $x=0$ is
AnswerGiven, the function $f$ is continuous at $x=0$.
$\therefore \lim _{x \rightarrow 0} f(x)=f(0)$.
Now, $\lim _{x \rightarrow 0} f(x)$
$=\lim _{x \rightarrow 0} \frac{1-\cos 4 x}{8 x^2}$
$=\lim _{x \rightarrow 0} \frac{2 \sin ^2 2 x}{8 x^2}$
$=\lim _{x \rightarrow 0} \frac{\sin ^2 2 x}{4 x^2}$
$=\lim _{x \rightarrow 0}\left(\frac{\sin 2 x}{2 x}\right)^2=1$
Also, $f(0)=k$
Hence $k=1$
View full question & answer→Question 341 Mark
If $\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a$, then $\frac{d y}{d x}$ is equal to
AnswerGiven, $\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a \Rightarrow \sec a=\frac{1+x}{1-y}$
On differentiating, we get
$\frac{(1-y)+(1+x) \frac{d y}{d x}}{(1-y)^2}=0 \Rightarrow(1+x) \frac{d y}{d x}=y-1 \Rightarrow \frac{d y}{d x}=\frac{y-1}{1+x}
$
View full question & answer→Question 351 Mark
If $y=\log _e\left(\frac{x^2}{e^2}\right)$, then $\frac{d^2 y}{d x^2}$ equals
Answer$\text { 84. (d) : We have, } y=\log _e\left(\frac{x^2}{e^2}\right)$
$\therefore \frac{d y}{d x}=\frac{e^2}{x^2} \cdot \frac{1}{e^2} \cdot 2 x=\frac{2}{x}$
$\Rightarrow \frac{d^2 y}{d x^2}=-\frac{2}{x^2}$
View full question & answer→Question 361 Mark
If $f(x)=\left\{\begin{array}{l}x, \text { if } x \leq 1 \\ 7, \text { if } x>1\end{array}\right.,$ then
Answer$f(1)=1$
$\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} x=1, $
$\lim _{x \rightarrow 1^{+}} f(x)$
$=\lim _{x \rightarrow 1} 7=7$
Since, $f(1) \neq \lim _{x \rightarrow 1^{+}} f(x) $
$\therefore f$ is discontinuous at $x=1$
View full question & answer→Question 371 Mark
If $y=(\tan x)^{\sin x}$, then $\frac{d y}{d x}$ is equal to
Answer(d) : We have, $y=(\tan x)^{\sin x}$
Taking $\log$ on both sides, we get
$\log y=\sin x \log (\tan x)$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{1}{y} \frac{d y}{d x} & =\frac{\sin x}{\tan x} \cdot \sec ^2 x+\cos x \log (\tan x) \\
& =(\tan x)^{\sin x}[\sec x+\cos x(\log \tan x)]
\end{aligned}
$
View full question & answer→Question 381 Mark
If $f(x)=\frac{5 x}{(1-x)^{2 / 3}}+\cos ^2(2 x+1),$ then $f^{\prime}(0)=$
Answer$\text { (c) : } f(x)=5 x(1-x)^{-\frac{2}{3}}+\cos ^2(2 x+1)$
$ \Rightarrow f^{\prime}(x)=5\left\{x \times \frac{-2}{3}(1-x)^{-5 / 3}(-1)+(1-x)^{-2 / 3} \times 1\right\}$
$ +2 \cos \ (2 x+1)\{-\sin (2 x+1) \times 2\}$
$\Rightarrow f^{\prime}(x)=5(1-x)^{-\frac{2}{3}}+\frac{10 x}{3}(1-x)^{-\frac{5}{3}}-2 \sin (4 x+2)$
$\therefore f^{\prime}(0)=5-2 \sin 2$
View full question & answer→Question 391 Mark
Let $y=t^{10}+1$ and $x=t^8+1$, then $\frac{d^2 y}{d x^2}$ is equal to
Answer$(b) :$ We have, $y=t^{10}+1, x=t^8+1$
$\Rightarrow \frac{d y}{d t}=10 t^9, \frac{d x}{d t}=8 t^7$
$\therefore \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{10 t^9}{8 t^7}=\frac{5}{4} t^2$
$\Rightarrow \frac{d^2 y}{d x^2}=\frac{5}{4}(2 t) \frac{d t}{d x}=\frac{5}{4} \times 2 t \times \frac{1}{8 t^7}=\frac{5}{16 t^6}$
View full question & answer→Question 401 Mark
If $y=\tan ^{-1}(\sqrt{x})-\tan ^{-1}(x)$, then $y^{\prime}(1)$ is equal to
Answer(d) : $y=\tan ^{-1}(\sqrt{x})-\tan ^{-1}(x)$
Differentiating w.r.t. $x$, we get
$
y^{\prime}(x)=\frac{1}{1+x} \cdot \frac{1}{2 \sqrt{x}}-\frac{1}{1+x^2} \Rightarrow y^{\prime}(1)=\frac{1}{2} \cdot \frac{1}{2}-\frac{1}{2}=-\frac{1}{4}
$
View full question & answer→Question 411 Mark
$\frac{d}{d x}\left[\sin ^{-1} x-\sin ^{-1} \sqrt{x}\right]$ is equal to
Answer$\text { (c) : Let } y=\frac{d}{d x}\left[\sin ^{-1} x-\sin ^{-1} \sqrt{x}\right]$
$=\frac{1}{\sqrt{1-x^2}}-\frac{1}{2 \sqrt{x} \sqrt{1-x}}=\frac{1}{\sqrt{1-x^2}}-\frac{1}{2 \sqrt{x(1-x)}}$
View full question & answer→Question 421 Mark
If $a x^2+2 h x y+b y^2=1$, then $\frac{d y}{d x}$ equals
Answer(d) : Given, $a x^2+2 h x y+b y^2=1$
Differentiating w.r.t. $x$, we get
$
2 a x+2 h\left(x \frac{d y}{d x}+y\right)+2 b y \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\left(\frac{a x+h y}{h x+b y}\right)
$
View full question & answer→Question 431 Mark
Differential coefficient of $\sqrt{\sec \sqrt{x}}$ is
Answer$(b) :$ Let $y=\sqrt{\sec \sqrt{x}}$
Differentiating w.r.t. $x$, we get
$\frac{d y}{d x}=\frac{1}{2 \sqrt{\sec \sqrt{x}}} \cdot \sec \sqrt{x} \cdot \tan \sqrt{x} \cdot \frac{1}{2 \sqrt{x}}$
$=\frac{1}{4 \sqrt{x}}(\sec \sqrt{x})^{1 / 2} \frac{\sin \sqrt{x}}{\cos \sqrt{x}}=\frac{1}{4 \sqrt{x}}(\sec \sqrt{x})^{3 / 2} \cdot \sin \sqrt{x}$
View full question & answer→Question 441 Mark
If $y=\sqrt{\sin x+y}$, then $\frac{d y}{d x}$ is equal to
Answer(a) : $y=\sqrt{\sin x+y} \Rightarrow y^2=\sin x+y$
Differentiating w.r.t. $x$, we get
$
2 y \frac{d y}{d x}=\cos x+\frac{d y}{d x} \Rightarrow \frac{d y}{d x}=\frac{\cos x}{2 y-1}
$
View full question & answer→Question 451 Mark
If $y=\log \left(\frac{1-x^2}{1+x^2}\right)$, then $\frac{d y}{d x}$ is equal to
Answer(b) : $y=\log \left(\frac{1-x^2}{1+x^2}\right)$
$
\Rightarrow y=\log \left(1-x^2\right)-\log \left(1+x^2\right)
$
Differentiating w.r.t. $x$, we get
$
\frac{d y}{d x}=\frac{1}{1-x^2}(-2 x)-\frac{1}{1+x^2}(2 x)=\frac{-4 x}{1-x^4}
$
View full question & answer→Question 461 Mark
The function $f(x)=\frac{4-x^2}{4 x-x^3}$ is
Answer(c) : $f(x)=\frac{4-x^2}{4 x-x^3}=\frac{4-x^2}{x(2-x)(2+x)}$
So, $f(x)$ is discontinuous at $x=0,2,-2$.
View full question & answer→Question 471 Mark
The derivative of $\cos ^{-1}\left(2 x^2-1\right)$ w.r.t. $\cos ^{-1} x$ is
Answer(a): Let $y=\cos ^{-1}\left(2 x^2-1\right)=2 \cos ^{-1} x$
Differentiating w.r.t. $\cos ^{-1} x$, we get
$
\frac{d y}{d\left(\cos ^{-1} x\right)}=\frac{2 d\left(\cos ^{-1} x\right)}{d\left(\cos ^{-1} x\right)}=2
$
View full question & answer→Question 481 Mark
If $x=f(t)$ and $y=g(t)$ are differentiable functions of $t$, then $\frac{d^2 y}{d x^2}$ is
Answer(a) : We have, $x=f(t), y=g(t)$
$
\Rightarrow \quad \frac{d x}{d t}=f^{\prime}(t) \text { and } \frac{d y}{d t}=g^{\prime}(t)
$
Now, $\frac{d y}{d x}=\frac{\frac{d y}{d t}}{\frac{d x}{d t}}=\frac{g^{\prime}(t)}{f^{\prime}(t)}$
$
\begin{aligned}
\Rightarrow \quad \frac{d^2 y}{d x^2} & =\frac{f^{\prime}(t) \cdot g^{\prime \prime}(t)-g^{\prime}(t) \cdot f^{\prime \prime}(t)}{\left(f^{\prime}(t)\right)^2} \cdot \frac{d t}{d x} \\
& =\frac{f^{\prime}(t) \cdot g^{\prime \prime}(t)-g^{\prime}(t) \cdot f^{\prime \prime}(t)}{\left(f^{\prime}(t)\right)^2} \cdot \frac{1}{f^{\prime}(t)} \\
& =\frac{f^{\prime}(t) \cdot g^{\prime \prime}(t)-g^{\prime}(t) f^{\prime \prime}(t)}{\left(f^{\prime}(t)\right)^3}
\end{aligned}
$
View full question & answer→Question 491 Mark
If the function $f(x)=\left\{\begin{array}{cc}\frac{\sin x^2}{x} ; & x \neq 0 \\ 0 ; & x=0\end{array}\right.$, is differentiable at $x=0$, then right hand derivative of $f(x)$ at $x=0$ is
Answer(c) : At $x=0$, right hand derivative
$
\begin{aligned}
f^{\prime}(0) & =\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h} \\
& =\lim _{h \rightarrow 0} \frac{f(h)-f(0)}{h}=\lim _{h \rightarrow 0} \frac{\frac{\sin h^2}{h}-0}{h}=\lim _{h \rightarrow 0} \frac{\sin h^2}{h^2}=1
\end{aligned}
$
View full question & answer→Question 501 Mark
If $f(x)=x+1$, find $\frac{d}{d x}(f o f)(x)$.
Answer(b) : Given, $f(x)=x+1$
Now, $(f o f)(x)=f(f(x))=f(x+1)=(x+1)+1=x+2$
$\therefore \quad \frac{d}{d x}(f o f)(x)=\frac{d}{d x}(x+2)=1$
View full question & answer→