MCQ 511 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
- A
continuous at $x=1$
- B
differentiable at $x=1$
- C
continuous at $x=3$
- ✓
Answer(d): $|x-3|$ is continuous at $x=3$, but not differentiable.
$
\begin{array}{l}
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)
\end{array}
$
$\therefore \quad f(x)$ is continuous and differentiable at $x=1$.
View full question & answer→MCQ 521 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
AnswerCorrect option: B. discontinuous at every $x$ except $x=0$
(b) : As $x \rightarrow 0$ 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→MCQ 531 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→MCQ 541 Mark
If $y=\log _7(\log x)$, then find $\frac{d y}{d x}$.
AnswerCorrect option: A. $\frac{1}{x \log 7 \log x}$
(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→MCQ 551 Mark
If $y=e^{\frac{1}{2} \log \left(1+\tan ^2 x\right)}$, then $\frac{d y}{d x}$ is equal to
Answer$\begin{array}{l}\text { (c) : } y=e^{\frac{1}{2} \log \left(1+\tan ^2 x\right)}=\left(\sec ^2 x\right)^{1 / 2}=\sec x \\ \therefore \quad \frac{d y}{d x}=\sec x \tan x\end{array}$
View full question & answer→MCQ 561 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→MCQ 571 Mark
If $y=a x^2+b$, then $\frac{d y}{d x}$ at $x=2$ is equal to
Answer(a) : We have, $y=a x^2+b$
$
\begin{array}{l}
\Rightarrow \frac{d y}{d x}=2 a x \\
\left.\frac{d y}{d x}\right|_{x=2}=2 a \times 2=4 a
\end{array}
$
View full question & answer→MCQ 581 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 $x=2$ ?
- A
$\frac{1}{2}$
- B
$\frac{1}{4}$
- ✓
$\frac{3}{4}$
- D
$\frac{3}{2}$
AnswerCorrect option: C. $\frac{3}{4}$
(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→MCQ 591 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
- ✓
$x \in R$
- B
$x=0$
- C
$x \in R-\{0\}$
- D
$x=-1$ and 1
AnswerCorrect option: A. $x \in R$
(a) : We 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} \Rightarrow f(x)=-1 \forall x \in R\right.
$
$\Rightarrow f(x)$ is continuous $\forall x \in R$ as it is a constant function.
View full question & answer→MCQ 601 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
- A
$\frac{1}{2}$
- B
$\frac{\pi}{4}$
- C
- ✓
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→MCQ 611 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→MCQ 621 Mark
If $y=a e^x+b e^{-x}+c$, where $a, b, c$ are parameters, then $y^{\prime}$ is equal to
AnswerCorrect option: A. $a e^x-b e^{-x}$
(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→MCQ 631 Mark
If $y=e^{3 x+7}$, then the value of $\left[\frac{d y}{d x}\right]_{x=0}$ is equal to
AnswerCorrect option: D. $3 e^7$
(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→MCQ 641 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→MCQ 651 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 \quad \frac{d x}{d s}=1, \frac{d y}{d s}=2
$
$
\begin{array}{l}
\text { 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 \quad \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
\end{array}
$
View full question & answer→MCQ 661 Mark
If $f(x)=\left\{\begin{array}{l}x, \text { if } x \leq 1 \\ 7, \text { if } x>1\end{array}\right.$, then
AnswerCorrect option: D. $f$ is discontinuous at $x=1$
(d): $f(1)=1$
$
\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1} x=1, \quad \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→MCQ 671 Mark
If $y=(\tan x)^{\sin x}$, then $\frac{d y}{d x}$ is equal to
- A
$\sec x+\cos x$
- B
$\sec x+\log \tan x$
- C
$(\tan x)^{\sin x}$
- ✓
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→MCQ 681 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
- A
$\frac{\sqrt{\pi}}{6}$
- ✓
$-\sqrt{\frac{\pi}{6}}$
- C
$\frac{1}{\sqrt{6}}$
- D
$\frac{\pi}{\sqrt{6}}$
AnswerCorrect option: B. $-\sqrt{\frac{\pi}{6}}$
(b) : We have, $f(x)=\sqrt{1+\cos ^2\left(x^2\right)}$
$
\begin{array}{l}
\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}}
\end{array}
$
View full question & answer→MCQ 691 Mark
If $f(x)=\frac{5 x}{(1-x)^{2 / 3}}+\cos ^2(2 x+1)$, then $f^{\prime}(0)=$
- A
$5+2 \sin 2$
- B
$5+2 \cos 2$
- C
$5-2 \sin 2$
- D
$5-2 \cos 2$
Answer$\begin{array}{l}\text { (c) : } f(x)=5 x(1-x)^{-\frac{2}{3}}+\cos ^2(2 x+1) \\ \begin{aligned} \Rightarrow & f^{\prime}(x)=5\left\{x \times \frac{-2}{3}(1-x)^{-5 / 3}(-1)+(1-x)^{-2 / 3} \times 1\right\} \\ & +2 \cos (2 x+1)\{-\sin (2 x+1) \times 2\} \\ \Rightarrow & f^{\prime}(x)=5(1-x)^{-\frac{2}{3}}+\frac{10 x}{3}(1-x)^{-\frac{5}{3}}-2 \sin (4 x+2)\end{aligned} \\ \therefore \quad f^{\prime}(0)=5-2 \sin 2\end{array}$
View full question & answer→MCQ 701 Mark
$\frac{d}{d x}\left[\sin ^{-1} x-\sin ^{-1} \sqrt{x}\right]$ is equal to
- A
$\frac{1}{2 \sqrt{x(1-x)}}-\frac{1}{\sqrt{1-x^2}}$
- B
$\frac{1}{\sqrt{1-\left\{x \sqrt{1-x}-\sqrt{\left.x\left(1-x^2\right)\right\}^2}\right.}}$
- C
$\frac{1}{\sqrt{1-x^2}}-\frac{1}{2 \sqrt{x(1-x)}}$
- D
$\frac{1}{\sqrt{x(1-x)(1-x)^2}}$
Answer$\begin{array}{l}\text { (c) : Let } y=\frac{d}{d x}\left[\sin ^{-1} x-\sin ^{-1} \sqrt{x}\right] \\ =\frac{1}{\sqrt{1-x^2}}-\frac{1}{2 \sqrt{x} \sqrt{1-x}}=\frac{1}{\sqrt{1-x^2}}-\frac{1}{2 \sqrt{x(1-x)}}\end{array}$
View full question & answer→MCQ 711 Mark
If $a x^2+2 h x y+b y^2=1$, then $\frac{d y}{d x}$ equals
- A
$\frac{h x+b y}{a x+h y}$
- B
$\frac{a x+h y}{h x+b y}$
- C
$\frac{a x+h x}{h y+b y}$
- ✓
$\frac{-(a x+h y)}{h x+b y}$
AnswerCorrect option: D. $\frac{-(a x+h y)}{h x+b y}$
(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→MCQ 721 Mark
Differential coefficient of $\sqrt{\sec \sqrt{x}}$ is
- A
$\frac{1}{4 \sqrt{x}} \sec \sqrt{x} \sin \sqrt{x}$
- ✓
$\frac{1}{4 \sqrt{x}}(\sec \sqrt{x})^{3 / 2} \cdot \sin \sqrt{x}$
- C
$\frac{1}{2} \sqrt{x} \sec \sqrt{x} \sin \sqrt{x}$
- D
$\frac{1}{2} \sqrt{x}(\sec \sqrt{x})^{3 / 2} \cdot \sin \sqrt{x}$
AnswerCorrect option: B. $\frac{1}{4 \sqrt{x}}(\sec \sqrt{x})^{3 / 2} \cdot \sin \sqrt{x}$
(b) : Let $y=\sqrt{\sec \sqrt{x}}$
Differentiating w.r.t. $x$, we get
$
\begin{array}{l}
\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}
\end{array}
$
View full question & answer→MCQ 731 Mark
If $y=\sqrt{\sin x+y}$, then $\frac{d y}{d x}$ is equal to
- ✓
$\frac{\cos x}{2 y-1}$
- B
$\frac{\cos x}{1-2 y}$
- C
$\frac{\sin x}{1-2 y}$
- D
$\frac{\sin x}{2 y-1}$
AnswerCorrect option: A. $\frac{\cos x}{2 y-1}$
(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→MCQ 741 Mark
If $y=\log \left(\frac{1-x^2}{1+x^2}\right)$, then $\frac{d y}{d x}$ is equal to
- A
- ✓
$\frac{-4 x}{1-x^4}$
- C
- D
$\frac{-4 x^3}{1-x^4}$
AnswerCorrect option: B. $\frac{-4 x}{1-x^4}$
(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→MCQ 751 Mark
The function $f(x)=\frac{4-x^2}{4 x-x^3}$ is
- A
discontinuous at only one point
- B
discontinuous at exactly two points
- ✓
discontinuous at exactly three points
- D
AnswerCorrect option: C. discontinuous at exactly three points
(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→MCQ 761 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→MCQ 771 Mark
If $y=\log \left(\cos e^x\right)$, then find $\frac{d y}{d x}$.
- A
$-e^x \tan ^2 x$
- B
$e^x \tan x$
- C
$e^x \sec x$
- ✓
$-e^x \tan x$
AnswerCorrect option: D. $-e^x \tan x$
(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→