datavorous/entrance-exam-dataset huggingface dataset:
If $\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc}1+\sin ^{2} \mathrm{x} & \cos ^{2} \mathrm{x} & 4 \sin 2 \mathrm{x} \\ \sin ^{2} \mathrm{x} & 1+\cos ^{2} \mathrm{x} & 4 \sin 2 \mathrm{x} \\ \sin ^{2} \mathrm{x} & \cos ^{2} \mathrm{x} & 1+4 \sin 2 \mathrm{x}\end{array}\right|$
What is the maximum value of $\mathrm{f}(\mathrm{x})$?

Options:
A 2
B 4
C 6
D 8

The correct answer is:
6

Solution:
$\mathrm{f}(\mathrm{x})=\left|\begin{array}{ccc}1+\sin ^{2} \mathrm{x} & \cos ^{2} x & 4 \sin 2 \mathrm{x} \\ \sin ^{2} \mathrm{x} & 1+\cos ^{2} \mathrm{x} & 4 \sin 2 \mathrm{x} \\ \sin ^{2} \mathrm{x} & \cos ^{2} \mathrm{x} & 1+4 \sin 2 \mathrm{x}\end{array}\right|$
Applying $\mathrm{C}_{1} \rightarrow \mathrm{C}_{1}+\mathrm{C}_{2}$
$=\left|\begin{array}{ccc}2 & \cos ^{2} x & 4 \sin 2 x \\ 2 & 1+\cos ^{2} x & 4 \sin 2 x \\ 1 & \cos ^{2} x & 1+4 \sin 2 x\end{array}\right|$
(Applying $\mathrm{R}_{2} \rightarrow \mathrm{R}_{2}-\mathrm{R}_{1}$ and $\left.\mathrm{R}_{3} \rightarrow \mathrm{R}_{3}-\mathrm{R}_{1}\right)$
$=\left|\begin{array}{ccc}2 & \cos ^{2} x & 4 \sin 2 x \\ 0 & 1 & 0 \\ -1 & 0 & 1\end{array}\right|$
$\mathrm{f}(\mathrm{x})=2+4 \sin 2 \mathrm{x}$
$\therefore \quad-1 \leq \sin 2 x \leq 1$, maximum value of $\sin 2 x=1$
Thus, maximum value of $f(x)=2+4=6$

Topic: Mathematics, Determinants, NDA, NDA 2006 (Phase 1)
If $\mathrm{f}(\mathrm{x})=(1+\mathrm{x})^{5 / \mathrm{x}}$ is continuous at $\mathrm{x}=0$, then what is the value of $\mathrm{f}(0)$ ?

Options:
A 0
B 1
C $\infty$
D $\mathrm{e}^{5}$

The correct answer is:
$\mathrm{e}^{5}$

Solution:
Given that $\mathrm{f}(\mathrm{x})=(1+\mathrm{x})^{5 / \mathrm{x}}$ and $\mathrm{f}(\mathrm{x})$ is continuous at $\mathrm{x}=0$. Value of function at $\mathrm{x}=0$ is same as limit of the function at $\mathrm{x}=0$.
$\mathrm{f}(0)=\lim _{\mathrm{x} \rightarrow 0}(1+\mathrm{x})^{5 / \mathrm{x}}=\left\{\lim _{\mathrm{x} \rightarrow 0}(1+\mathrm{x})^{1 / \mathrm{x}}\right\}^{5}=\mathrm{e}^{5}$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2006 (Phase 2)
If $f(x)=2^{\sin x}$, then what is the derivative of $f(x)$?

Options:
A $2^{\sin x} \ln 2$
B $(\sin x) 2^{2 \sin x-1}$
C $(\cos x) 2^{2 \sin x-1}$
D None of the above

The correct answer is:
None of the above

Solution:
Let $f(x)=2^{\sin x}$.
$\mathrm{f}^{\prime}(\mathrm{x})=2^{\sin \mathrm{x}} . \ln 2 \cos \mathrm{x}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2012 (Phase 2)
If $f(\mathrm{x})=\frac{\mathrm{x}}{2}-1$, then on the interval $[0, \pi]$ which one of the following is correct?

Options:
A $\tan [f(\mathrm{x})]$, where $[\cdot]$ is the greatest integer function, and $\frac{1}{f(\mathrm{x})}$ are both continuous.
B $\tan [f(\mathrm{x})]$, where $[\cdot]$ is the greatest integer functin, and $f^{-1}(\mathrm{x})$ are both continuous.
C $\tan [f(\mathrm{x})]$, where $[\cdot]$ is the greatest integer function, and $\frac{1}{f(\mathrm{x})}$ are both discontinuous.
D tan $[f(\mathrm{x})]$, where $[\cdot]$ is the greatest integer function, is discontinuous but $\frac{1}{f(\mathrm{x})}$ is continuous.

The correct answer is:
$\tan [f(\mathrm{x})]$, where $[\cdot]$ is the greatest integer function, and $\frac{1}{f(\mathrm{x})}$ are both discontinuous.

Solution:
$\begin{aligned} & \mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{2}-1, \quad[0, \pi] \\ & \tan \cdot \mathrm{f}(\mathrm{x})=\tan \left(\frac{\mathrm{x}}{2}-1\right) \end{aligned}$
$\frac{1}{\mathrm{f}(\mathrm{x})}=\frac{1}{\frac{\mathrm{x}}{2}-1}$ is discontinuous at $\mathrm{x}=2$
tan. $\mathrm{f}(\mathrm{x})$ is discontinuous for $\mathrm{x}=2$ in $[0, \pi]$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2017 (Phase 2)
If $f(x)=2 x^{2}+3 x-5$, then what is $f^{\prime}(0)+3 f^{\prime}(-1)$ equal to?

Options:
A $-1$
B 0
C 1
D 2

The correct answer is:
0

Solution:
$\mathrm{F}(\mathrm{x})=2 \mathrm{x}^{2}+3 \mathrm{x}-5$
$\mathrm{~F}^{\prime}(\mathrm{x})=4 \mathrm{x}+3$
$\mathrm{~F}^{\prime}(0)+3 \mathrm{~F}^{\prime}(-1)=3+3(-4+3)=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2013 (Phase 2)
If $f(x)=2 x+7$ and $g(x)=x^{2}+7, x \in R$, then what are values of $x$ for which $\operatorname{fog}(x)=25 ?$

Options:
A $-1,1$
B $-2,2$
C $-\sqrt{2}, \sqrt{2}$
D None of these

The correct answer is:
$-\sqrt{2}, \sqrt{2}$

Solution:
Let, $\mathrm{f}(\mathrm{x})=2 \mathrm{x}+7$ and $\mathrm{g}(\mathrm{x})=\mathrm{x}^{2}+7, \mathrm{x} \in \mathrm{R}$
Now fog $(x)=f[g(x)]=f\left(x^{2}+7\right)$
$=2\left(\mathrm{x}^{2}+7\right)+7=2 \mathrm{x}^{2}+14+7$
But fog $(x)=25$ $\Rightarrow \quad 2 x^{2}+21=25$
$\Rightarrow \quad x^{2}=2$
$\Rightarrow \quad \mathrm{x}=\pm \sqrt{2}$

Topic: Mathematics, Functions, NDA, NDA 2010 (Phase 2)
If $f(x)=2^{x}$, then what is $f^{\prime \prime}(x)$ equal to?

Options:
A $2^{x}(\ln 2)^{2}$
B $x(x-1) 2^{x-2}$
C $2^{x+1}(\ln 2)$
D $2^{x}\left(\log _{10} 2\right)^{2}$

The correct answer is:
$2^{x}(\ln 2)^{2}$

Solution:
Let $f(x)=2^{x}$
On differentiating w.r.t. $x$, we get
$f^{\prime}(x)=2^{x}(\ln 2)$
On again differentiating w.r.t. $x$, we get
$f^{\prime \prime}(x)=2^{x}(\ln 2)^{2}$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2011 (Phase 1)
If $\mathrm{f}(\mathrm{x})=3^{1+\mathrm{x}}$, then $\mathrm{f}(\mathrm{x}) \mathrm{f}(\mathrm{y}) \mathrm{f}(\mathrm{z})$ is equal to

Options:
A $\mathrm{f}(\mathrm{x}+\mathrm{y}+\mathrm{z})$
B $\mathrm{f}(\mathrm{x}+\mathrm{y}+\mathrm{z}+1)$
C $f(x+y+z+2)$
D $f(x+y+z+3)$

The correct answer is:
$f(x+y+z+2)$

Solution:
$\mathrm{f}(\mathrm{x})=3^{1+\mathrm{x}}$
$\mathrm{f}(\mathrm{x}) \cdot \mathrm{f}(\mathrm{y}) \cdot \mathrm{f}(\mathrm{z})=3^{1+\mathrm{x}} \cdot 3^{1+\mathrm{y}} \cdot 3^{1+\mathrm{z}}$
$=3^{1+\mathrm{x}+1+\mathrm{y}+1+\mathrm{z}}=3^{3+\mathrm{x}+\mathrm{y}+\mathrm{z}}=3^{1+\mathrm{x}+\mathrm{y}+\mathrm{z}+2}$
$=\mathrm{f}(\mathrm{x}+\mathrm{y}+\mathrm{z}+2)$

Topic: Mathematics, Functions, NDA, NDA 2019 (Phase 1)
If $f(x)=\frac{2}{3} x+\frac{3}{2}, x \in R, \quad$ [2010-II]
then what is $f^{-1}(x)$ equal to?

Options:
A $\frac{3}{2} x+\frac{2}{3}$
B $\frac{3}{2} x-\frac{9}{4}$
C $\frac{2}{3} x-\frac{4}{9}$
D $\frac{2}{3} x-\frac{2}{3}$

The correct answer is:
$\frac{3}{2} x-\frac{9}{4}$

Solution:
Let $f(x)=\frac{2}{3} x+\frac{3}{2}=y($ say $)=\frac{4 x+9}{6}=y$
$\Rightarrow \quad 4 x+9=6 y$
$\Rightarrow \quad x=\frac{6 y-9}{4}$
$\quad x=f^{-1}(y)$
$\Rightarrow \quad f^{-1}(x)=\frac{6 x-9}{4}=\frac{3 x}{2}-\frac{9}{4}$

Topic: Mathematics, Functions, NDA, NDA 2010 (Phase 2)
If $f(x)=3 x^{2}+6 x-9$, then

Options:
A $f(x)$ is increasing in $(-1,3)$
B $f(x)$ is decreasing in $(3, \infty)$
C $f(x)$ is increasing in $(-\infty,-1)$
D $f(x)$ is decreasing in $(-\infty,-1)$

The correct answer is:
$f(x)$ is decreasing in $(-\infty,-1)$

Solution:
Given $f(x)=3 x^{2}+6 x-9$ On differentiating w.r.t. $x$, we get $f^{\prime}(x)=6 x+6$
$\begin{aligned} & f^{\prime}(x) < 0 \\ \Rightarrow & 6 x+6 < 0 \end{aligned}$
$\Rightarrow 6 \mathrm{x} < -6$
$\Rightarrow \mathrm{x} < -1$
Hence $f(x)$ is decreasing in $(-\infty,-1)$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 1)
If $f(x)=\left\{\begin{array}{ll}3 x-4,0 \leq x \leq 2 \\ 2 x+\lambda, 2 < x \leq 3\end{array} \right.$
is continuous at $x=2$, then what is the value of $\lambda$ ?

Options:
A 1
B -1
C 2
D -2

The correct answer is:
-2

Solution:
Given $f(x)=\left\{\begin{array}{l}3 x-4,0 \leq x \leq 2 \\ 2 x+\lambda, 2 < x \leq 3\end{array}\right.$
Also $f(x)$ is continuous at $x=2$
$\therefore \lim _{x \rightarrow 2} f(x)=f(2)$
Now, $f(2)=3 \times 2-4=6-4=2$
$\Rightarrow \lim _{x \rightarrow 2}(2 x+\lambda)=2$
$\Rightarrow 4+\lambda=2$
$\Rightarrow \quad \lambda=-2$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2009 (Phase 1)
If $\mathrm{F}(\mathrm{x})=\sqrt{9-\mathrm{x}^{2}}$, then what is $\lim _{\mathrm{x} \rightarrow 1} \frac{\mathrm{F}(\mathrm{x})-\mathrm{F}(1)}{\mathrm{x}-1}$ equal to?

Options:
A $-\frac{1}{4 \sqrt{2}}$
B $\frac{1}{8}$
C $-\frac{1}{2 \sqrt{2}}$
D $\frac{1}{2 \sqrt{2}}$

The correct answer is:
$-\frac{1}{2 \sqrt{2}}$

Solution:
$\lim _{x \rightarrow 1} \frac{F(x)-F(1)}{x-1}$
$\quad$ We know, $_{x \rightarrow a}{\lim } \frac{F(n)-F(a)}{x-a}=F^{\prime}(a)$
$\therefore F^{\prime}(x)=\frac{d}{d x}\left(\sqrt{9-x^{2}}\right)$
$=\frac{1(0-2 x)}{2 \sqrt{9-x^{2}}}=\frac{-x}{\sqrt{9-x^{2}}}$
$\quad F^{\prime}(1)=\frac{-1}{\sqrt{9-1}}=\frac{-1}{\sqrt{8}}=\frac{-1}{2 \sqrt{2}}$

Topic: Mathematics, Functions, NDA, NDA 2017 (Phase 1)
If $\mathrm{f}(\mathrm{x})=\sqrt{25-\mathrm{x}^{2}}$, then what is $\lim _{\mathrm{x} \rightarrow 1} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(1)}{\mathrm{x}-1}$ equal to

Options:
A $-\frac{1}{\sqrt{24}}$
B $\frac{1}{\sqrt{24}}$
C $-\frac{1}{4 \sqrt{3}}$
D $\frac{1}{4 \sqrt{3}}$

The correct answer is:
$-\frac{1}{\sqrt{24}}$

Solution:
$\lim _{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}=f^{\prime}(x)$
Now, $f^{\prime}(x)=\frac{-2 x}{2 \sqrt{25-x^{2}}}=\frac{-x}{\sqrt{25-x^{2}}}$
$f^{\prime}(1)=\frac{-1}{\sqrt{25-1}}=\frac{-1}{\sqrt{24}}$

Topic: Mathematics, Limits, NDA, NDA 2018 (Phase 2)
If $\mathrm{f}(\mathrm{x})=\sqrt{25-\mathrm{x}^{2}}$, then what is $\operatorname{Lim}_{\mathrm{x} \rightarrow 1} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(1)}{\mathrm{x}-1}$ equal to ?

Options:
A $\frac{1}{5}$
B $\frac{1}{24}$
C $\sqrt{24}$
D $-\frac{1}{\sqrt{24}}$

The correct answer is:
$-\frac{1}{\sqrt{24}}$

Solution:
$\mathrm{f}(\mathrm{x})=\sqrt{25-\mathrm{x}^{2}}$
$\mathrm{f}(1)=\sqrt{24}$
$\Rightarrow \lim _{\mathrm{x} \rightarrow 1} \frac{\mathrm{f}(\mathrm{x})-\mathrm{f}(1)}{\mathrm{x}-1}$
$\therefore$ It is $\frac{0}{0}($ undefined condition $)$ so using L'hospital's
rule
$\Rightarrow \lim _{x \rightarrow 1} \frac{\mathrm{f}^{\prime}(\mathrm{x})-0}{1}=\lim _{\mathrm{x} \rightarrow 1}\left(\sqrt{25-\mathrm{x}^{2}}\right)^{\prime}$
$\Rightarrow \lim _{x \rightarrow 1} \frac{1}{2} \times \frac{1}{\sqrt{25-x^{2}}}(-2 x)$
$=\frac{1}{2} \times \frac{1}{\sqrt{25-(1)^{2}}} \times(-2)$
$=-\frac{1}{\sqrt{24}}$

Topic: Mathematics, Limits, NDA, NDA 2015 (Phase 2)
If $f(x)=a+b x+c x^{2}$, then what is $\int_{0}^{1} f(x) d x$ equal to?

Options:
A $[f(0)+4 f(1 / 2)+f(1)] / 6$
B $[f(0)+4 f(1 / 2)+f(1)] / 3$
C $[f(0)+4 f(1 / 2)+f(1)]$
D $[f(0)+2 f(1 / 2)+f(1)] / 6$

The correct answer is:
$[f(0)+4 f(1 / 2)+f(1)] / 6$

Solution:
Given, $f(x)=a+b x+c x^{2}$
$\therefore \int_{0}^{1} f(x) d x=\int_{0}^{1}\left(a+b x+c x^{2}\right) d x$
$=\left[a x+\frac{b x^{2}}{2}+\frac{c x^{3}}{3}\right]_{0}^{1}$
$=a+\frac{b}{2}+\frac{c}{3}$
Here, $f(0)=a, f\left(\frac{1}{2}\right)=a+\frac{b}{2}+\frac{c}{4}$
and $f(1)=a+b+c$
Now, $\frac{f(0)+4 f\left(\frac{1}{2}\right)+f(1)}{6}$
$=\frac{a+4\left(a+\frac{b}{2}+\frac{c}{4}\right)+a+b+c}{6}$
$=\frac{a+4\left(\frac{4 a+2 b+c}{4}\right)+a+b+c}{6}$
$=\frac{a+4 a+2 b+c+a+b+c}{6}=\frac{6 a+3 b+2 c}{6}$
$=a+\frac{b}{2}+\frac{c}{3}$
$\therefore$ From Eqs. (i) and (ii), we get
$\int_{0}^{1} f(x) d x=\frac{f(0)+4 f\left(\frac{1}{2}\right)+f(1)}{6}$

Topic: Mathematics, Definite Integration, NDA, NDA 2009 (Phase 2)
$\quad$ If $f(x)=A \sin \left(\frac{\pi x}{2}\right)+B$ and $f^{\prime}\left(\frac{1}{2}\right)=\sqrt{2}$ and
$\int_{0}^{1} \mathrm{f}(\mathrm{x}) \mathrm{d} \mathrm{x}=\frac{2 \mathrm{~A}}{\pi}$, then what is the value of $\mathrm{B}$ ?

Options:
A $\frac{2}{\pi}$
B $\frac{4}{\pi}$
C 0
D 1

The correct answer is:
0

Solution:
Given function $\mathrm{f}(\mathrm{x})=\mathrm{A} \sin \left(\frac{\pi \mathrm{x}}{2}\right)+\mathrm{B}$
Differentiating w. r. t. x $f^{\prime}(x)=A \cos \left(\frac{\pi x}{2}\right) \cdot \frac{\pi}{2}$
$\mathrm{f}^{\prime}\left(\frac{1}{2}\right)=\sqrt{2}=\mathrm{A}\left(\cos \frac{\pi}{4}\right) \frac{\pi}{2}=\mathrm{A} \cdot \frac{1}{\sqrt{2}} \cdot \frac{\pi}{2}$
$\Rightarrow \quad A=\frac{(\sqrt{2} \times \sqrt{2}) \times 2}{\pi}=\frac{4}{\pi}$
Now, $\int_{0}^{1} \mathrm{f}(\mathrm{x}) \mathrm{dx}=\frac{2 \mathrm{~A}}{\pi}$
$\Rightarrow \quad \int_{0}^{1}\left\{\mathrm{~A} \sin \left(\frac{\pi \mathrm{x}}{2}\right)+\mathrm{B}\right\} \mathrm{d} \mathrm{x}=\frac{2 \times 4}{\pi^{2}}$
$\Rightarrow\left[-\mathrm{A} \cos \frac{\pi \mathrm{x}}{2} \cdot \frac{2}{\pi}+\mathrm{Bx}\right]_{0}^{1}=\frac{8}{\pi^{2}}$
$\Rightarrow \quad-\frac{4}{\pi} \cdot \frac{2}{\pi} \cos \frac{\pi}{2}+\mathrm{B}+\frac{4}{\pi} \cdot \frac{2}{\pi} \cos 0=\frac{8}{\pi^{2}}$
$\Rightarrow \quad B+\frac{8}{\pi^{2}}=\frac{8}{\pi^{2}} \Rightarrow B=0$

Topic: Mathematics, Definite Integration, NDA, NDA 2006 (Phase 1)
If $f(\mathrm{x})$ and $\mathrm{g}(\mathrm{x})$ are continuous functions satisfying $f(x)=f(a-x)$ and $g(x)+g(a-x)=2$, then what is
$\int_{0}^{a} f(x) g(x) d x$ equal to?

Options:
A $\int_{0}^{a} g(x) d x$
B $\int_{0}^{a} f(x) d x$
C $2 \int_{0}^{a} f(x) d x$
D 0

The correct answer is:
$\int_{0}^{a} f(x) d x$

Solution:
$\begin{aligned} I &=\int_{0}^{a} f(x) \cdot g(x) d x \\ & I=\int_{0}^{a} f(a-x) \cdot g(a-x) d x \\ & I=\int_{0}^{a} f(x) \cdot[2-g(x)] d x \end{aligned}$
$\therefore f(a-x)=f(x)$ and $g(a-x)=2 g(x)$
$\mathrm{I}-\int_{0}^{\mathrm{a}} 2 . \mathrm{f}(\mathrm{x}) \cdot \mathrm{d} \mathrm{x}-\int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \cdot \mathrm{g}(\mathrm{s}) \cdot \mathrm{d} \mathrm{x}$
$\mathrm{I}=2 \cdot \int_{0}^{\mathrm{a}} \mathrm{f}(\mathrm{x})-\mathrm{I}$
$\Rightarrow \not z^{\prime} I=\not^{\prime} \int_{0}^{a} f(x) d x$
$\therefore I=\int_{0}^{a} f(x) d x$

Topic: Mathematics, Definite Integration, NDA, NDA 2017 (Phase 1)
If $f(x)=\cos x, g(x)=\log x$ and $y=($ gof $)(x)$, then what is the value of $\frac{d y}{d x}$ at $x=0 ?$

Options:
A 0
B 1
C $-1$
D 2

The correct answer is:
0

Solution:
Given $f(x)=\cos x$ and $g(x)=\log x$
Consider $\mathrm{y}=\operatorname{gof}(x)$
$\quad=g(f(x)\}$
$\quad=\log (f(x))$
$\quad=\log (\cos x)$
$\therefore \frac{d y}{d x}=\frac{1}{\cos x}(-\sin x)=-\tan x$
$\Rightarrow\left(\frac{d y}{d x}\right)_{x=0}=-\tan 0=0$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 1)
If $\mathrm{f}\{\mathrm{x}\}=\left[\begin{array}{lcl}\cos \mathrm{x} & -\sin x & 0 \\ \sin \mathrm{x} & \cos \mathrm{x} & 0 \\ 0 & 0 & 1\end{array}\right]$, then which of the following are
correct ?
$1.$ $\quad \mathrm{f}(\theta) \times \mathrm{f}(\phi)=\mathrm{f}(\theta+\phi)$
$2.$ The value of the determinant of the matrix $\mathrm{f}(\theta) \times \mathrm{f}(\phi)$ is 1 .
$3.$ The determinant of $f(x)$ is an even function.
Select the correct answer using the code given below:

Options:
A 1 and 2 only
B 2 and 3 only
C 1 and 3 only
D 1,2 and 3

The correct answer is:
1,2 and 3

Solution:
$\quad f(\theta)=\left[\begin{array}{ccc}\cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{array}\right] \& f(\phi)\left[\begin{array}{ccc}\cos \phi & -\sin \phi & 0 \\ \sin \phi & \cos \phi & 0 \\ 0 & 0 & 1\end{array}\right]$
$\mathbf{f}(\theta) \times \mathrm{f}(\phi)=\left[\begin{array}{ccc}\cos \theta \cos \phi-\sin \theta \sin \phi & -\cos \theta \sin \phi-\sin \theta \cos \phi & 0 \\ \sin \theta \cos \phi+\cos \theta \sin \phi & -\sin \theta \sin \phi+\cos \theta \cos \phi & 0 \\ 0 & 0 & 1\end{array}\right]$
(using Trigonometric Identities)
$\Rightarrow \quad \mathrm{f}(\theta) \times \mathrm{f}(\phi)=\mathrm{f}(\theta+\phi)$
Also, det, $[\mathrm{f}(\theta) \times \mathrm{f}(\phi)]=1\left[\cos ^{2}(\theta+\phi)-\left(-\sin ^{2}(\theta+\phi)\right)\right]$
$=\cos ^{2}(\theta+\phi)+\sin ^{2}(\theta+\phi)=1 .$
$\& \operatorname{det}(\mathrm{f}(x))=\left(\cos ^{2} x-\left(-\sin ^{2} x\right)=\cos ^{2} x+\sin ^{2} x=1 .\right.$
For $x=-x$
det. $(\mathrm{f}(-x))=\cos ^{2}(-x)+\operatorname{Sin}^{2}(-x)=1$
Hence, det. $(\mathrm{f}(-x))=\operatorname{det}(\mathrm{f}(x))$
Hence, $\operatorname{det} .(\mathrm{f}(\mathrm{x}))$ is even function.

Topic: Mathematics, Matrices, NDA, NDA 2016 (Phase 2)
If $\int \frac{\mathrm{dx}}{\mathrm{f}(\mathrm{x})}=\log \{\mathrm{f}(\mathrm{x})\}^{2}+\mathrm{c}$, then what is $\mathrm{f}(\mathrm{x})$ equal to?

Options:
A $2 x+\alpha$
B $x+\alpha$
C $\frac{x}{2}+\alpha$
D $x^{2}+\alpha$

The correct answer is:
$\frac{x}{2}+\alpha$

Solution:
We check from the given options one by one. Options
(a) and (b) do not satisfy. We check option (c).
Let $\mathrm{f}(\mathrm{x})=\frac{\mathrm{x}}{2}+\alpha$
$\therefore \int \frac{\mathrm{dx}}{\frac{\mathrm{x}}{2}+\alpha}=\int \frac{2 \mathrm{dx}}{(\mathrm{x}+2 \alpha)}$
$=2 \log (x+2 \alpha)+c_{1}=\log (x+2 \alpha)^{2}+c_{1}$
$=\log \left(\frac{x}{2}+\alpha\right)^{2}+\log 2^{2}+c_{1}=\log \left(\frac{x}{2}+\alpha\right)^{2}+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2008 (Phase 1)
If $f(x)=e^{\sin (\log \cos x)}$ and $g(x)=\log \cos x$, then what is the derivative of $\mathrm{f}(\mathrm{x})$ with respect to $\mathrm{g}(\mathrm{x})$?

Options:
A $\mathrm{f}(\mathrm{x}) \cos [\mathrm{g}(\mathrm{x})]$
B $\mathrm{f}(\mathrm{x}) \sin [\mathrm{g}(\mathrm{x})]$
C $g(x) \cos [f(x)]$
D $g(x) \sin [\mathrm{f}(\mathrm{x})]$

The correct answer is:
$\mathrm{f}(\mathrm{x}) \cos [\mathrm{g}(\mathrm{x})]$

Solution:
Given funcion is: $\mathrm{f}(\mathrm{x})=\mathrm{e}^{\sin (\log \cos \mathrm{x})}$
Differentiating w.r.t. $\mathrm{x}$
$f^{\prime}(x)=e^{\sin (\log \cos x)} \cdot \cos (\log \cos x) \cdot \frac{1}{\cos x}(-\sin x)$
$=-\mathrm{e}^{\sin (\log \cos x)} \cdot \cos (\log \cos x) \cdot \tan x$
and $\mathrm{g}(\mathrm{x})=\log \cos \mathrm{x}$
$\therefore g^{\prime}(x)=\frac{1}{\cos x}(-\sin x)=-\tan x$
Hence,
$\frac{f^{\prime}(x)}{g^{\prime}(x)}=\frac{-e^{\sin (\log \cos x)} \cdot \cos (\log \cos x) \cdot \tan x}{-\tan x}$
$=\mathrm{e}^{\sin (\log \cos x)} \cdot \cos (\log \cos x)$
$=\mathrm{f}(\mathrm{x}) \cdot \cos [\mathrm{g}(\mathrm{x})]$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2008 (Phase 1)
If $f(x)=e^{x}$ and $g(x)=\log x$, then what is the value of (gof)' $(x)$?

Options:
A 0
B 1
C e
D None of these

The correct answer is:
1

Solution:
Let $f(x)=e^{x}, g(x)=\log x$
Consider $($ g of $)(x)=g[f(x)]$
$=\log f(x) \quad($ By defn of $g(x))$
$=\log \left(e^{x}\right) \quad\left(\because f(x)=e^{x}\right)$
$=x \quad(\because \log e=1)$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2009 (Phase 2)
If $f(x)$ is an even function, then what is $\int_{0}^{\pi} f(\cos x) d x$ equal

Options:
A 0
B $\int_{0}^{\frac{\pi}{2}} f(\cos x) d x$
C $2 \int_{0}^{\frac{\pi}{2}} f(\cos x) d x$
D 1

The correct answer is:
$2 \int_{0}^{\frac{\pi}{2}} f(\cos x) d x$

Solution:
Since $f(x)$ is an even function therefore
$$
\begin{array}{l}
\int_{0}^{\pi} f(x) d x=2 \int_{0}^{\pi / 2} f(x) d x \\
\text { Hence, } \int_{0}^{\pi} f(\cos x) d x=2 \int_{0}^{\pi / 2} f(\cos x) d x
\end{array}
$$

Topic: Mathematics, Definite Integration, NDA, NDA 2011 (Phase 1)
If $\mathrm{f}(\mathrm{x})$ is an even function, where $\mathrm{f}(\mathrm{x}) \neq 0$, then which one of the following is correct?

Options:
A $\mathrm{f}^{\prime}(\mathrm{x})$ is an even function
B $\mathrm{f}^{\prime}(\mathrm{x})$ is an odd function
C $\mathrm{f}^{\prime}(\mathrm{x})$ may be an even or odd function depending on the type of function
D $\mathrm{f}^{\prime}(\mathrm{x})$ is a constant function

The correct answer is:
$\mathrm{f}^{\prime}(\mathrm{x})$ is an odd function

Solution:
$\mathrm{f}(\mathrm{x})$ is an even function. Let's see some examples
(1) If $\mathrm{f}(\mathrm{x})=\cos \mathrm{x}$, which is even function, $\mathrm{f}^{\prime}(\mathrm{x})=-\sin \mathrm{x}$, which is odd function.
(2) If $\mathrm{f}(\mathrm{x})=\mathrm{x}^{2}$, which is even function, $\mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}$, which is odd function.

Topic: Mathematics, Functions, NDA, NDA 2018 (Phase 1)
If $\mathrm{f}(\mathrm{x})$ is differentiable everywhere,then which one of the following is correct?

Options:
A $|\mathrm{f}|$ is differentiable everywhere
B $|\mathrm{f}|^{2}$ is differentiable everywhere
C $\mathrm{f}|\mathrm{f}|$ is not differentiable at some points
D $\text { None of the above }$

The correct answer is:
$\mathrm{f}|\mathrm{f}|$ is not differentiable at some points

Solution:
If $\mathrm{f}(\mathrm{x})$ is differential everywhere then $\mid \mathrm{f}$ | is not differentiable at some point, so, $\mathrm{f}|\mathrm{f}|$ is not differentiable at some point. [Example: $\mathrm{f}(\mathrm{x})=\mathrm{x}$ is differentiable everywhere but $|\mathrm{f}(\mathrm{x})|=|\mathrm{x}|$ is not differentiable at $\mathrm{x}=0]$

Topic: Mathematics, Continuity and Differentiability, NDA, NDA 2007 (Phase 1)
If $f(x)=k x^{3}-9 x^{2}+9 x+3$ is monotonically increasing in every interval, then which one of the following is correct?

Options:
A $k < 3$
B $k \leq 3$
C $k>3$
D $k \geq 3$

The correct answer is:
$k>3$

Solution:
Given $f(x)=k x^{3}-9 x^{2}+9 x+3$ On differentiating w.r.t. $x$, we get $f^{\prime}(x)=3 k x^{2}-18 x+9$
For a function to be monotonically increasing. $b^{2}-4 a c < 0$
Here, $a=3 k, b=-18, c=9$
$\therefore^{2}-4 a c=(-18)^{2}-4(3 k)(9)=(-18)(-18)-(3 k) 18 \times 2$
$\Rightarrow 36-12 k < 0$
$\Rightarrow k>3$

Topic: Mathematics, Application of Derivatives, NDA, NDA 2010 (Phase 1)
If $f(x)=\frac{\sin \left(e^{x-2}-1\right)}{\ln (x-1)}$, then $\lim _{x \rightarrow 2} f(x)$ is equal to

Options:
A -2
B -1
C 0
D 1

The correct answer is:
1

Solution:
$\mathrm{f}(\mathrm{x})=\frac{\sin \left(\mathrm{e}^{\mathrm{x}-2}-1\right)}{\ln (\mathrm{x}-1)}$
$\lim _{\mathrm{x} \rightarrow 2} \frac{\sin \left(\mathrm{e}^{\mathrm{x}-2}-1\right)}{\ln (\mathrm{x}-1)}=\mathrm{L}$
It is $\frac{0}{0}$ (undefined) condition so using L'hospital's rule
$\Rightarrow \mathrm{L}=\lim _{\mathrm{x} \rightarrow 2}\left[\frac{\left\{\sin \left(\mathrm{e}^{\mathrm{x}-2}-1\right)\right\}^{-}}{\{\ln (\mathrm{x}-1)\}^{-}}\right]$
$\Rightarrow \mathrm{L}=\lim _{\mathrm{x} \rightarrow 2} \frac{\cos \left(\mathrm{e}^{\mathrm{x}-2}-1\right) \cdot \mathrm{e}^{(\mathrm{x}-2)}}{1 /(\mathrm{x}-1)}$
$\Rightarrow \mathrm{L}=\lim _{\mathrm{x} \rightarrow 2} \cos \left(\mathrm{e}^{2-2}-1\right) \mathrm{e}^{2-2} \cdot(2-1)$
$\Rightarrow \mathrm{L}=\cos (0) \mathrm{e}^{0} \cdot 1$
$\Rightarrow \mathrm{L}=1$

Topic: Mathematics, Limits, NDA, NDA 2015 (Phase 2)
If $f(x)=\ln \left(x-\sqrt{1+x^{2}}\right)$, then what is $\int f^{\prime \prime}(x) d x$ equal to?

Options:
A $\frac{1}{\left(x-\sqrt{1+x^{2}}\right)}+c$
B $-\frac{1}{\sqrt{1+x^{2}}}+c$
C $-\sqrt{1+x^{2}}+c$
D $\ln \left(x-\sqrt{1+x^{2}}\right)+c$

The correct answer is:
$-\frac{1}{\sqrt{1+x^{2}}}+c$

Solution:
Given that $f(x)=\ln \left(x-\sqrt{1+x^{2}}\right)$
$\int \mathrm{f}^{\prime \prime}(\mathrm{x}) \mathrm{dx}=\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{c}$ where $\mathrm{c}$ is a constant
$=\frac{1}{\left(x-\sqrt{1+x^{2}}\right)} \cdot\left(1-\frac{2 x}{2 \sqrt{1+x^{2}}}\right)+c$
$=\frac{-\left(x-\sqrt{1+x^{2}}\right)}{\left(\sqrt{1+x^{2}}\right)\left(x-\sqrt{1+x^{2}}\right)}+c=-\frac{1}{\sqrt{1+x^{2}}}+c$

Topic: Mathematics, Indefinite Integration, NDA, NDA 2006 (Phase 1)
If $\mathrm{f}(\mathrm{x})=\log \left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)$, then what is $\mathrm{f}\left(\frac{2 \mathrm{x}}{1-\mathrm{x}^{2}}\right)$ equal to ?

Options:
A $(f(x))^{2}$
B 1
C $2 \mathrm{f}(\mathrm{x})$
D $\mathrm{f}\left(\frac{1-\mathrm{x}}{1+\mathrm{x}}\right)$

The correct answer is:
$2 \mathrm{f}(\mathrm{x})$

Solution:
Given that $\mathrm{f}(\mathrm{x})=\log \left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right)$
So, $f\left(\frac{2 x}{1+x^{2}}\right)=\log \left(\frac{1+\frac{2 x}{1+x^{2}}}{1-\frac{2 x}{1+x^{2}}}\right)$
$=\log \left(\frac{1+x^{2}+2 x}{1+x^{2}-2 x}\right)=\log \left(\frac{(1+x)^{2}}{(1-x)^{2}}\right)$
$=\log \left(\frac{1+x}{1-x}\right)^{2}=2 \log \left(\frac{1+x}{1-x}\right)$
$=2 \mathrm{f}(\mathrm{x})$
$\left[\right.$ since $\left.f(x)=\log \left(\frac{1+x}{1-x}\right)\right]$

Topic: Mathematics, Functions, NDA, NDA 2006 (Phase 2)
If $f(x)=\log _{10}(1+x)$, then what is $4 \mathrm{f}(4)+5 f(1)-\log _{10} 2$ equal
to ?

Options:
A 0
B 1
C 2
D 4

The correct answer is:
4

Solution:
$\mathrm{f}(\mathrm{x})=\log _{10}(1+\mathrm{x})$
$4 \cdot \mathrm{f}(4)+5 \cdot \mathrm{f}(1)-\log _{10} 2$
$=4 \cdot \log _{10}(1+4)+5 \cdot \log _{10}(1+1)-\log _{10} 2$
$=4 \log _{10} 5+5 \log _{10} 2-\log _{10} 2$
$=4 \log _{10} 5+4 \log _{10} 2$
$=4\left(\log _{10} 5+\log _{10} 2\right)$
$=4\left(\log _{10} 10\right)=$

Topic: Mathematics, Sets and Relations, NDA, NDA 2019 (Phase 1)