datavorous/entrance-exam-dataset huggingface dataset:
Let $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ be three non-zero vectors which are pairwise non-collinear. If $\vec{\alpha}+3 \vec{\beta}$ is collinear with $\vec{\gamma}$ and $\vec{\beta}+2 \vec{\gamma}$ is collinear with $\vec{\alpha}$, then $\vec{\alpha}+3 \vec{\beta}+6 \vec{\gamma}$ is

Options:
A $\vec{\gamma}$
B $\overrightarrow{0}$
C $\vec{\alpha}+\vec{\gamma}$
D $\vec{\alpha}$

The correct answer is:
$\overrightarrow{0}$

Solution:
$\vec{\alpha}+3 \vec{\beta}=\mathrm{k}_{1} \vec{\gamma} \Rightarrow \vec{\beta}=\frac{\mathrm{k}_{1}}{3} \vec{\gamma}-\frac{\vec{\alpha}}{3}$
$\vec{\beta}+2 \vec{\gamma}=\mathrm{k}_{2} \vec{\alpha} \Rightarrow \vec{\beta}=\mathrm{k}_{2} \vec{\alpha}-2 \vec{\gamma}$
$\Rightarrow \frac{\mathrm{k}_{1} \vec{\gamma}}{3}-\frac{\vec{\alpha}}{3}=\mathrm{k}_{2} \vec{\alpha}-2 \vec{\gamma} \Rightarrow \vec{\alpha}\left(\mathrm{k}_{2}+\frac{1}{3}\right)=\vec{\gamma}\left(\frac{\mathrm{k}_{1}}{3}+2\right) \Rightarrow \mathrm{k}_{2}=-\frac{1}{3}$ and $\frac{\mathrm{k}_{1}}{3}=-2 \Rightarrow \mathrm{k}_{1}=-6$
$\Rightarrow \vec{\alpha}+3 \vec{\beta}+6 \vec{\gamma}=\overrightarrow{0}$

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2021
Let $\hat{\alpha}, \hat{\beta}, \hat{\gamma}$ be three unit vectors such that $\hat{\alpha} \times (\hat{\beta} \times \hat{\gamma})=\frac{1}{2}(\hat{\beta}+\hat{\gamma})$ where $\hat{\alpha} \times(\hat{\beta} \times \hat{\gamma})=(\hat{\alpha} \cdot \hat{\gamma}) \hat{\beta}-(\hat{\alpha} \cdot \hat{\beta}) \gamma \cdot$ If $\hat{\beta}$ is not parallel to $\hat{\gamma},$ then the angle between $\hat{\alpha}$ and $\hat{\beta}$ is

Options:
A $\frac{5 \pi}{6}$
B $\frac{\pi}{6}$
C $\frac{\pi}{3}$
D $\frac{2 \pi}{3}$

The correct answer is:
$\frac{2 \pi}{3}$

Solution:
Given,
$|\hat{\alpha}|=|\hat{\beta}|=|\hat{\gamma}|=1$
and $\quad \hat{\alpha} \times(\hat{\beta} \times \hat{\gamma})=\frac{1}{2}(\hat{\beta}+\hat{\gamma})$
$\left.\operatorname{Now}, \hat{\alpha} \times(\hat{\beta} \times \hat{\gamma})=\frac{1}{2} \hat{(\hat{\beta}}+\hat{\gamma}\right)$
$\Rightarrow \quad(\hat{\alpha} \cdot \hat{\gamma}) \hat{\beta}-(\hat{\alpha} \cdot \hat{\beta}) \hat{\gamma}=\frac{1}{2} \hat{\beta}+\frac{1}{2} \hat{\gamma}$
On comparing we get, $-(\hat{\alpha} \cdot \hat{\beta})=\frac{1}{2}$
$\Rightarrow|\hat{\alpha}||\hat{\beta}| \cos \theta=-\frac{1}{2}$
$\Rightarrow \quad \cos \theta=-\frac{1}{2} \quad[\because|\hat{\alpha}|=|\hat{\beta}|=1]$
$\Rightarrow \quad \theta=120^{\circ}$ or $\frac{2 \pi}{3}$

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2019
Let $C_{1}$ and $C_{2}$ denote the centres of the circles $x^{2}+y^{2}=4$ and $(x-2)^{2}+y^{2}=1$ respectively and let $P$ and $Q$ be their points of intersection.Then, the areas of $\Delta C_{1} P Q$ and $C_{2} P Q$ are in the ratio

Options:
A 3: 1
B 5: 1
C 7: 1
D 9: 1

The correct answer is:
7: 1

Solution:
$x^{2}+y^{2}=4$
$$
\begin{array}{l}
(x-2)^{2}+y^{2}=1 \\
C_{1}=(0,0), C_{2}=(2,0)
\end{array}
$$
On solving Eqs. (i) and (ii), we get

$$
\begin{aligned}
x=\frac{7}{4}, y=\pm \frac{\sqrt{15}}{4} \\
C_{1} N=\frac{7}{4}, P Q=\frac{2 \sqrt{15}}{4} &=\frac{\sqrt{15}}{2} \\
\frac{\text { Area }\left(\Delta C_{1} P Q\right)}{\text { Area }\left(\Delta C_{2} P Q\right)}=& \frac{\frac{1}{2} \cdot P Q \cdot C_{1} N}{\frac{1}{2} \cdot P Q \cdot C_{2} N} \\
&=\frac{C_{1} N}{C_{2} N}=\frac{\frac{7}{4}}{2-\frac{7}{4}}=\frac{\frac{7}{4}}{\frac{1}{4}}=7: 1
\end{aligned}
$$
Hence, the required option is (c).

Topic: Mathematics, Circle, WBJEE, WBJEE 2012
Let $\mathrm{C}$ denote the set of all complex numbers.
Define $A=\{(z, w) \mid z, w \in C$ and $|z|=|w|\}, B=\left\{(z, w) \mid z, w \in C\right.$ and $\left.z^{2}=w^{2}\right\}$. Then

Options:
A $A=B$
B $\mathrm{A} \subset \mathrm{B}$
C $\mathrm{B} \subset \mathrm{A}$
D $A \cap B=\varphi$

The correct answer is:
$\mathrm{B} \subset \mathrm{A}$

Solution:
$z^{2}=w^{2}$
$(z-w)(z+w)$
$z=w,-w$
$|z|=|w|$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2021
Let $\left(\mathrm{C}_{\mathrm{rms}}\right)_{\mathrm{H}_2}$ is the r.m.s speed of $\mathrm{H}_2$ at $150 \mathrm{~K}$. At what temperature, the most probable speed of helium $\left[\left(\mathrm{C}_{\mathrm{mp}}\right)_{\mathrm{He}}\right]$ will be half of $\left(\mathrm{C}_{\mathrm{rms}}\right)_{\mathrm{H}_2}$ ?

Options:
A $75 \mathrm{~K}$
B $112.5 \mathrm{~K}$
C $225 \mathrm{~K}$
D $900 \mathrm{~K}$

The correct answer is:
$112.5 \mathrm{~K}$

Solution:
$C_{\mathrm{rms}}=\sqrt{\frac{3 R T}{M}} \quad C_{m p}=\sqrt{\frac{2 R T}{M}}$
As per question $\frac{1}{2} \sqrt{\frac{3 \times R \times 150}{2}}=\sqrt{\frac{2 R T}{4}}$ $\mathrm{T}=112.5 \mathrm{~K}$

Topic: Chemistry, States of Matter, WBJEE, WBJEE 2022
Let $\cos ^{-1}\left(\frac{y}{b}\right)=\log \left(\frac{x}{n}\right)^{n}$. Then

Options:
A $x^{2} y_{2}+x y_{1}+n^{2} y=0$
B $x y_{2}-x y_{1}+2 n^{2} y=0$
C $x^{2} y_{2}+3 x y_{1}-n^{2} y=0$
D $x y_{2}+5 x y_{1}-3 y=0$

The correct answer is:
$x^{2} y_{2}+x y_{1}+n^{2} y=0$

Solution:
Hint:
$\cos ^{-1}\left(\frac{y}{b}\right)=\log \left(\frac{x}{n}\right)^{n}=n \times \log \left(\frac{x}{n}\right)$
$\Rightarrow-\frac{1}{\sqrt{b^{2}-y^{2}}} \cdot y_{1}=\frac{n}{x}$
$\Rightarrow x^{2} y_{1}^{2}=n^{2}\left(b^{2}-y^{2}\right) \Rightarrow x^{2} y_{2}+x y_{1}+n^{2} y=0$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2020
Let \(\cos ^{-1}\left(\frac{y}{b}\right)=\log _e\left(\frac{x}{n}\right)^n\), then \(A y_2+B y_1+C y=0\) is possible for

Options:
A \(\mathrm{A}=2, \mathrm{~B}=\mathrm{x}^2, \mathrm{C}=\mathrm{n}\)
B \(A=x^2, B=x, C=n^2\)
C \(A=x, B=2 x, C=3 n+1\)
D \(A=x^2, B=3 x, C=2 n\)

The correct answer is:
\(A=x^2, B=x, C=n^2\)

Solution:
\(\begin{aligned}
& \text { Hint: } \frac{-1}{\sqrt{1-\frac{y^2}{b^2}}} \times \frac{1}{b} y_1=\left(\frac{n}{n \times \frac{x}{n}}\right) \\
& \Rightarrow \frac{-1 y_1}{\sqrt{b^2-y^2}}=\frac{n}{x} \\
& \Rightarrow y_1 x+n \sqrt{b^2-y^2}=0 \\
& \Rightarrow y_1+x y_2+\frac{n}{2 \sqrt{b^2-y^2}} \cdot(-2 y) y_1=0
\end{aligned}\)
\(\begin{aligned} & \Rightarrow y_1+x y_2+\frac{n^2}{x} y=0 \\ & \Rightarrow y_1 x+x^2 y_2+n^2 y=0 \\ & \Rightarrow x^2 y_2+x y_1+n^2 y=0 \\ & A=x^2, B=x, C=n^2\end{aligned}\)

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2023
Let $d_{1}$ and $d_{2}$ be the lengths of the perpendiculars drawn from any point of the line $7 x-9 y+10=0$ upon the lines $3 x+4 y=5$ and $12 x+5 y=7,$ respectively. Then,

Options:
A $d_{1}>d_{2}$
B $d_{1}=d_{2}$
C $d_{1} < d_{2}$
D $d_{1}=2 d_{2}$

The correct answer is:
$d_{1}=d_{2}$

Solution:
Let $(h, k)$ be any point on the line $7 x-9 y+10=0,$ then $7 h-9 k+10=0$
$\Rightarrow \quad 7 h=9 k-10$
$\Rightarrow \quad h=\frac{9 k-10}{7}$
Now, perpendicular distance from point
$(h, k)$ to the line $3 x+4 y=5$ is $d_{1}$
$d_{1}=\frac{3 h+4 k-5}{\sqrt{3^{2}+4^{2}}}$
$\Rightarrow \quad d_{1}=\frac{3 h+4 k-5}{5}$
and perpendicular distance from $(h, k)$ to the line $12 x+5 y=7$ is $\bar{d}_{2}$
$\therefore \quad d_{2}=\frac{12 h+5 k-7}{\sqrt{12^{2}+5^{2}}}$
$\Rightarrow \quad d_{2}=\frac{12 h+5 k-7}{13}$
Now, $d_{1}-d_{2}=\frac{3 h+4 k-5}{5}-\frac{12 h+5 k-7}{13}$
$\Rightarrow d_{1}-d_{2}$
$=\frac{13(3 h+4 k-5)-5(12 h+5 k-7)}{65}$
$=\frac{39 h+52 k-65-60 h-25 k+35}{65}$
$=\frac{-21 h+27 k-30}{65}$
$=\frac{-21\left(\frac{9 k-10}{7}\right)+27 k-30}{65} \quad[$ from Eq.(i) $]$
$=\frac{-27 k+30+27 k-30}{65}=0$
$\Rightarrow d_{1}-d_{2}=0 \Rightarrow d_{1}=d_{2}$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2017
Let $d(n)$ denotes the number of divisors of $n$ including 1 and itself. Then, $d(225), d(1125)$ and $d(640)$ are

Options:
A in AP
B in HP
C in GP
D consecutive integers

The correct answer is:
in GP

Solution:
Here, $225=3^{2} \times 5^{2}$
$\Rightarrow d(225)=3 \times 3=9$
$1126=3^{2} \times 5^{3}$
$\Rightarrow \quad d(1125)=3 \times 4=12$
and $640=2^{7} \times 5$
$\Rightarrow \quad d(640)=8 \times 2=16$
Hence, 9,12 and 16 are in GP.

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2015
Let $\alpha, \beta$ denote the cube roots of unity other than 1 and $\alpha \neq \beta .$ Let $S=\sum_{n=0}^{\infty}(-1)^{n}\left(\frac{\alpha}{\beta}\right)^{n} .$ Then the value of $S$ is

Options:
A either $-2 \omega$ or $-2 \omega^{2}$
B either $-2 \omega$ or $2 \omega^{2}$
C either $2 \omega$ or $-2 \omega^{2}$
D either $2 \omega$ or $2 \omega^{2}$

The correct answer is:
either $-2 \omega$ or $-2 \omega^{2}$

Solution:
Case I Let $\alpha=\omega$ and $\beta=\omega^{2}$
$S=\sum_{n=0}^{302}(-1)^{n}\left(\frac{\omega}{\omega^{2}}\right)^{n}$
$=\sum_{n=0}^{302}(-1)^{n}\left(\omega^{2}\right)^{n}$
$=1-\omega^{2}+\omega^{4}-\omega^{6}+\omega^{8}-\omega^{10}+\omega^{12}$
$\quad+\ldots+\omega^{600}-\omega^{602}+\omega^{604}$
$=1-\omega^{2}+\omega-1+\omega^{2}-\omega+1+\ldots$
$=0+\ldots+1-\omega^{2}+\omega$
$=-\omega^{2}-\omega^{2}=-2 \omega^{2} \quad\left[\because 1+\omega+\omega^{2}=0\right]$
Case II Let $\alpha=\omega^{2}$ and $\beta=\omega$
$\therefore S=\sum_{n=0}^{\infty_{2}}(-1)^{n}\left(\frac{\omega^{2}}{\omega}\right)^{n}$
$=\sum_{n=0}^{302}(-1)^{n}\left(\frac{\omega^{4}}{\omega^{3}}\right)^{n}$
$=\sum_{n=0}^{302}(-1)^{n}(\omega)$
$=1-\omega+\omega^{2}-\omega^{3}+\omega^{4}-\omega^{5}+\omega^{6}-\ldots$
$+\omega^{300}-\omega^{301}+\omega^{302}$
$=1-\omega+\omega^{2}-1+\omega-\omega^{2}+1$
$=0+\ldots+1+\omega^{2}-\omega$
$=-\omega-\omega=-2 \omega$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2014
Let each of the equations $x^{2}+2 x y+a y^{2}=0 \& a x^{2}+2 x y+y^{2}=0$ represent two straight lines passing through the origin. If they have a common line, then the other two lines are given by

Options:
A $x-y=0, x-3 y=0$
B $x+3 y=0,3 x+y=0$
C $3 x+y=0,3 x-y=0$
D $(3 x-2 y)=0, x+y=0$

The correct answer is:
$x+3 y=0,3 x+y=0$

Solution:
Hint:
$\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{2}+2\left(\frac{\mathrm{x}}{\mathrm{y}}\right)+\mathrm{a}=0 \& \mathrm{a}\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{2}+2\left(\frac{\mathrm{x}}{\mathrm{y}}\right)+1=0$ have exactly one root in common (taking $\frac{\mathrm{x}}{\mathrm{y}}$ as a single
variable).
By, $\left(a_{1} b_{2}-a_{2} b_{1}\right)\left(b_{1} c_{2}-b_{2} c_{1}\right)=\left(a_{1} c_{2}-a_{2} c_{1}\right)^{2}$
We get $: \Rightarrow \mathrm{a}=1$ or $-3$
a cannot be 1
Taking $\mathrm{a}=-3$, roots of 1 st equation $: 1,-3$ and 2 nd equation : $1,-\frac{1}{3}$
So other lines: $\frac{x}{y}=-3$ and $\frac{x}{y}=-\frac{1}{3}$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2020
Let \(\exp (x)\) denote exponential function ex. If \(f(x)=\exp \left(x^{\frac{1}{x}}\right), x>0\) then the minimum value of \(f\) in the interval \([2,5]\) is

Options:
A \(\exp \left(e^{\frac{1}{c}}\right)\)
B \(\exp \left(2^{\frac{1}{2}}\right)\)
C \(\exp \left(5^{\frac{1}{5}}\right)\)
D \(\exp \left(3^{\frac{1}{3}}\right)\)

The correct answer is:
\(\exp \left(5^{\frac{1}{5}}\right)\)

Solution:
Given that,
$$
f(x)=e^{(x)^{\frac{1}{x}}, x>0}
$$
Taking log on both sides, we get
$$
\log +(x)=(x)^{\frac{1}{x}}=g(x)
$$
Here, $g(x)=x^{\frac{1}{x}}$
$\Rightarrow \quad \log g(x)=\frac{1}{x} \log x$
On differentiating w.r.t. $x$, we get
$$
\begin{aligned}
\frac{1}{g(x)} \cdot g^{\prime}(x) &=\frac{x \cdot \frac{1}{x}-\log x}{x^{2}} \\
&=\left(\frac{1-\log x}{2}\right)
\end{aligned}
$$
$\Rightarrow \quad g^{\prime}(x)=x^{\left(\frac{1}{x}-2\right)}(1-\log x)$
For maximum or minimum of $g(x)$ put
$$
g^{\prime}(x)=0
$$
$\Rightarrow \quad x^{\left(\frac{1}{x}-2\right)}(1-\log x)=0$
$$
\begin{array}{lr}
\Rightarrow & \log x=1=\log e \\
\Rightarrow & x=e
\end{array}
$$
and $\left.\quad g^{\prime \prime}(x)\right|_{x=e}>0$
So, $g(x)$ is minimum at $x=e$ $\therefore g(x)$ increases in $(0, e)$ and decreases in $(\theta, \infty)$ it will be minimum at either 2 or 5 $\therefore 2^{\frac{1}{2}}>5^{\frac{1}{5}} \Rightarrow$ Minimum value of $f(x)=e^{(5)^{\frac{1}{5}}}$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2013
Let $f:[1,3] \rightarrow R$ be a continuous function that is differentiable in (1,3) an $f^{\prime}(x)=|f(x)|^{2}+4$ for all $x \in(1,3).$ Then,

Options:
A $f(3)-f(1)=5$ is true
B $f(3)-f(1)=5$ is false
C $f(3)-f(1)=7$ is true
D $f(3)-f(1) < 0$ only at one point of (1,3)

The correct answer is:
$f(3)-f(1)=5$ is false

Solution:
Given that $f:[1,3] \rightarrow R$ be a continuous and differentiable in (1,3)
and $f^{\prime}(x)=|f(x)|^{2}+4$
By applying $L M V T,$ there exist at least one point $c \in(1,3)$ such that
$$
\begin{array}{c}
\frac{f(3)-f(1)}{3-1}=f^{\prime}(c)\left[\because f^{\prime}(c)=\frac{f(b)-f(a)}{b-a}\right] \\
\Rightarrow \quad \frac{f(3)-f(1)}{2}=f^{\prime}(c) \\
\Rightarrow \quad \frac{f(3)-f(1)}{2}=\left[f|c|^{2}+4\right] \\
\Rightarrow f^{\prime}(c)=|f(c)|^{2}+4 \\
\Rightarrow \quad f(3)-f(1)=2 \cdot|f(c)|^{2}+8
\end{array}
$$
$\Rightarrow \quad f(3)-f(1) \geq 8$

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2019
Let $f(\theta)=\left(1+\sin ^{2} \theta\right)\left(2-\sin ^{2} \theta\right) .$ Then, for all
values of $\theta$

Options:
A $f(\theta)>\frac{9}{4}$
B $f(\theta) < 2$
C $f(\theta)>\frac{11}{4}$
D $2 \leq f(\theta) \leq \frac{9}{4}$

The correct answer is:
$2 \leq f(\theta) \leq \frac{9}{4}$

Solution:
Given
$$
\begin{aligned}
f(\theta)=\left(1+\sin ^{2} \theta\right)\left(2-\sin ^{2} \theta\right) \\
&=2+2 \sin ^{2} \theta-\sin ^{2} \theta-\sin ^{4} \theta \\
&=-\sin ^{4} \theta+\sin ^{2} \theta+2 \\
&=-\left(\sin ^{4} \theta-\sin ^{2} \theta-2\right) \\
&=-\left\{\sin ^{4} \theta-\sin ^{2} \theta+\frac{1}{4}-\frac{9}{4}\right\} \\
&=+\frac{9}{4}-\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} \\
\because \quad-1 \leq \sin \theta \leq 1 \Rightarrow 0 \leq \sin ^{2} \theta \leq 1\\
\Rightarrow & 0 \geq-\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} \geq-\frac{1}{4} \\
\Rightarrow \quad \quad-\frac{1}{2} \leq \sin ^{2} \theta-\frac{1}{2} \leq \frac{1}{2}\\
\Rightarrow \quad 0 \leq\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} \leq \frac{1}{4}
\Rightarrow & \frac{9}{4} \geq \frac{9}{4}-\left(\sin ^{2} \theta-\frac{1}{2}\right)^{2} \geq \frac{9}{4}-\frac{1}{4} \\
\Rightarrow \quad 2 \leq f(\theta) \leq \frac{9}{4} & \quad[\text { from Eq. }(0)]
\end{aligned}
$$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2013
Let $f_{1}(x)=e^{x}, f_{2}(x)=e^{f_{1}(x)} \ldots \ldots$
$f_{n+1}(x)=c^{f n(x)}$ for all $n \geq 1 .$ Then for any fixed
$n \cdot \frac{d}{d x} f_{n}(x)$ is

Options:
A $f_{n}(x)$
B $f_{n}(x)f_{n-1}(x)$
C $f_{n}(x)f_{n-1}(x)...f_{1}(x)$
D $f_{n}(x)...f_{1}(x)e^{x}$

The correct answer is:
$f_{n}(x)f_{n-1}(x)...f_{1}(x)$

Solution:
We have, $f_{1}(x)=e^{x}$
$$
\begin{array}{r}
f_{2}(x)=e^{f_{1}(x)} \\
\cdots \cdot \cdots \cdots \\
\cdots f_{n+1}(x)=e^{f_{n}(x)}
\end{array}
$$
$f_{n}(x)=e^{f_{n-1}(x)}$
On taking log both sides, we get
$\quad \log \left\{f_{n}(x)\right\}=f_{n-1}(x) \log e$
$\Rightarrow \quad \frac{d}{d x} \log \left(f_{n}(x)\right)=\frac{d}{d x} f_{n-1}(x) \quad(\because \log e=1)$
$\Rightarrow \quad \frac{1}{f_{n}(x)} \frac{d}{d x} f_{n}(x)=f_{n-1}^{\prime}(x)$
$\Rightarrow \quad \frac{d}{d x} f_{n}(x)=f_{n}(x) f_{n-1}^{\prime}(x) \quad \ldots$ (i)
Now, $\quad f_{1}^{\prime}(x)=e^{x}$
$\begin{array}{l}\text { and } \quad f_{2}(x)=e^{f_{1}(x)} \\ \Rightarrow \log f_{2}(x)=f_{1}(x) \log e=f_{1}(x) \\ \Rightarrow \quad \frac{1}{f_{2}(x)} \cdot f_{2}^{\prime}(x)=f_{1}^{\prime}(x) \\ \Rightarrow \quad \begin{aligned} f_{2}^{\prime}(x) &=f_{2}(x) \cdot f_{1}^{\prime}(x) \\=f_{2}(x) \cdot e^{x} &\left(\because f_{1}^{\prime}(x)=e^{x}\right) \end{aligned} \\ =f_{2}(x) \cdot f_{1}(x) & {\left[\because e^{x}=f_{1}(x)\right] \ldots(\mathrm{ii})}\end{array}$
From Eq. (i). $\frac{d}{d x} f_{n}(x)=f_{n}(x) \cdot f_{n-1}(x) \ldots f_{1}(x)[\mathrm{using} \mathrm{Eq}$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2018
Let $f:[-2,2] \rightarrow R$ be a continuous function such that $f(x)$ assumes only irrational values. If $f(\sqrt{2})=\sqrt{2},$ then

Options:
A $f(0)=0$
B $f(\sqrt{2}-1)=\sqrt{2}-1$
C $f(\sqrt{2}-1)=\sqrt{2}+1$
D $f(\sqrt{2}-1)=\sqrt{2}$

The correct answer is:
$f(\sqrt{2}-1)=\sqrt{2}$

Solution:
If a function $f(x)$ assumes only irrational values which is also continuous, then $f(x)$ must be constant function.
$\Rightarrow \quad f(x)=\sqrt{2}$
$$
[\because f(\sqrt{2})=\sqrt{2} \mid
$$
$\therefore \quad f(\sqrt{2}-1)=\sqrt{2}$

Topic: Mathematics, Functions, WBJEE, WBJEE 2015
Let $f:[a, b] \rightarrow \mathbb{R}$ be continuous in $[a, b]$, differentiable in $(a, b)$ and $f(a)=0=f(b)$. Then

Options:
A there exists at least one point $c \in(a, b)$ for which $f^{\prime}(c)=f(c)$
B $f^{\prime}(x)=f(x)$ does not hold at any point of $(a, b)$
C at every point of $(a, b), f^{\prime}(x)>f(x)$
D at every point of $(a, b), f^{\prime}(x) < f(x)$

The correct answer is:
there exists at least one point $c \in(a, b)$ for which $f^{\prime}(c)=f(c)$

Solution:
$:$ Let $g(x)=e^{-x} f(x)$
$$
\mathrm{g}(\mathrm{a})=\mathrm{g}(\mathrm{b})=0
$$
By Rolle's theorem, for atleast one $c \in(a, b)$ where $\mathrm{g}^{\prime}(\mathrm{c})=0$
or $e^{-c} f^{\prime}(c)-f(c) e^{-c}=0$
or $e^{-c}\left(f^{\prime}(c)-f(c)\right)=0$
or $f^{\prime}(c)=f(c) \quad$ for atleast one $c \in(a, b)$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2022
Let $f:[a, b] \rightarrow R$ be differentiable on $[a, b]$ and $k \in R$. Let $f(a)=0=f(b)$
Also let $J(x)=f'(x)+k f(x) .$ Then

Options:
A $J(x)>0$ for all $x \in[a, b]$
B $J(x) < 0$ for all $x \in[a, b]$
C $J(x)=0$ has at least one root in $(a, b)$
D $J(x)=0$ through $(a, b)$

The correct answer is:
$J(x)=0$ has at least one root in $(a, b)$

Solution:
We have, $f:[a, b] \longrightarrow R$ be differentiable
on $[a, b]$ and $k \in R,$ also $f(a)=0=f(b)$ and $\quad J(x)=f(x)+k f(x)$
Let $g(x)=k x f(x)$ which is continuous in $[a, b]$ and differentiable in $(a, b)$ such that
$$
g(a)=0=g(b)
$$
Then, for every $c \in(a, b), g^{\prime}(c)=0$
(by Rolle's theorem) $\begin{array}{lr}\text { Now, } & g^{\prime}(x)=k f(x)+k x f^{\prime}(x) \\ \Rightarrow \quad & g(c)=k f(c)+k c f^{\prime}(c)\end{array}$
$\Rightarrow \quad k f(c)+k c f(c)=0$
$\Rightarrow \quad f(x)=0,$ for every $x=c \in(a, b)$
$\therefore J(x)=0$ has atleast one root in $(a, b)$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2018
Let $f:[a, b] \rightarrow R$ be such that $f$ is differentiable in $(a, b)$ is continuous at $x=a$ and $x=b$ and moreover $f(a)=0=f(b)$. Then

Options:
A there exists at least one point $c$ in $(a,b)$ such that $f^{\prime}=f$
B $f^{\prime}(x)=f(x)$ does not hold at any point in $(a, b)$
C at every point of $(a, b) f^{\prime}(x)>f(x)$
D at every point of $(a, b), f^{\prime}(x) < f(x)$

The correct answer is:
there exists at least one point $c$ in $(a,b)$ such that $f^{\prime}=f$

Solution:
Let $g(x)=e^{-x} f(x)$
such that $g(a)=0, g(b)=0$
and $g(x)$ is continuous and differentiable.
Then, for atleast one value of $c \in(a, b)$ such that $g(c)=0$
Now, $g(x)=e^{-x} f^{\prime}(x)+\left(-e^{-x}\right) f(x)$
$\Rightarrow \quad g(c)=e^{-c} f^{\prime}(c)+\left(-e^{-\ell}\right) f(c)=0$
$\Rightarrow e^{-c} f^{\prime}(c)=e^{-c} f(c) \Rightarrow f^{\prime}(c)=f(c)$

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2018
Let $f$ and $g$ be differentiable on the interval $I$ and let $a, b \in I, a < b$. Then,

Options:
A If $f(a)=0=f(b)$, the equation $f^{\prime}(x)+f(x) g^{\prime}(x)=0$ is solvable in $(a, b)$
B If $f(a)=0=f(b)$, the equation $f^{\prime}(x)+f(x) g^{\prime}(x)=0$ may not be solvable in $(a, b)$
C If $g(a)=0=g(b)$, the equation $g^{\prime}(x)+k g(x)=0$ is solvable in $(a, b), k \in \mathbb{R}$
D If $g(a)=0=g(b)$, the equation $g^{\prime}(x)+k g(x)=0$ may not be solvable in $(a, b), k \in \mathbb{R}$.

The correct answers are:
If $f(a)=0=f(b)$, the equation $f^{\prime}(x)+f(x) g^{\prime}(x)=0$ is solvable in $(a, b)$, If $g(a)=0=g(b)$, the equation $g^{\prime}(x)+k g(x)=0$ is solvable in $(a, b), k \in \mathbb{R}$

Solution:
From option (a)
We have,
$f(a)=f(b)=0$
$\Rightarrow f^{\prime}(a) \cdot f^{\prime}(b) < 0$
Again, let $h(x)=f^{\prime}(x)+f(x) g^{\prime}(x)$
$\Rightarrow h(a)=f^{\prime}(a)+fg^{\prime}(a)=f^{\prime}(a)$
and $h(b)=f^{\prime}(b)+f(b) g^{\prime}(b)=f^{\prime}(b)$
$\therefore \quad h(a) \cdot h(b)=f^{\prime}(a) \cdot f^{\prime}(b) < 0$
$\therefore h(x)=0$ has root between $(a, b)$
From option (c)
We have, $g(a)=g(b)=0$
$\Rightarrow \quad g^{\prime}(a)\cdot g^{\prime}(b) < 0$
Again, let $m(x)=g^{\prime}(x)+k g(x)$
$\Rightarrow m(a)=g^{\prime}(a)+k g(a)=g^{\prime}(a)$
$\Rightarrow \quad m(b)=g^{\prime}(b)+k g(b)=g^{\prime}(b)$
$\therefore \quad m(a) \cdot m(b)=g^{\prime}\cdot g^{\prime}(b) < 0$
$\therefore \quad m(x)=0$ has root between $(a, b)$ So, option (a) and (c) are correct.

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2019
Let $f$ and $g$ be periodic functions with the periods $T_{1}$ and $T_{2}$ respectively. Then $f+g$ is

Options:
A periodic with period $\mathrm{T}_{1}+\mathrm{T}_{2}$
B non-periodic
C periodic with the period $\mathrm{T}_{1}$
D periodic when $T_{1}=T_{2}$

The correct answers are:
periodic when $T_{1}=T_{2}$

Solution:
$\mathrm{f}+\mathrm{g}$ is a periodic function.

Topic: Mathematics, Functions, WBJEE, WBJEE 2021
Let $f$, be a continuous function in $[0,1]$, then $\lim_{n \rightarrow \infty} \sum_{j=0}^n \frac{1}{n} f\left(\frac{j}{n}\right)$ is

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

The correct answer is:
$\int_{0}^{1} f(x) d x$

Solution:
\(\lim _{n \rightarrow \infty} \sum_{i=0}^{n} \frac{1}{n} f\left(\frac{i}{n}\right)\)
Let \(1 / n \rightarrow d x\)

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2020
Let $f$ be a differentiable function with $\lim _{x \rightarrow \infty} f(x)=0 .$ If $y^{\prime}+y f^{\prime}(x)-f(x) f^{\prime}(x)=0, \lim _{x \rightarrow \infty} y(x)=0$, then $\left(\right.$ where $\left.y^{\prime}=\frac{d y}{d x}\right)$

Options:
A $y+1=e^{f(x)}+f(x)$
B $y-1=e^{f(x)}+f(x)$
C $y+1=e^{-f(x)}+f(x)$
D $y-1=e^{-f f] x}+f(x)$

The correct answer is:
$y+1=e^{-f(x)}+f(x)$

Solution:
Hint:
$\frac{d y}{d x}+f^{\prime}(x) y=f^{\prime}(x) f(x)$
$\Rightarrow y \times e^{f(x)}=\int f^{\prime}(x) f(x) e^{f(x)} d x$
$\Rightarrow y \times e^{f(x)}=e^{f(x)}(f(x)-1)+c \quad[$ Putting $f(x)=0 ; y=0, c=1]$
$\Rightarrow y \times e^{f(x)}=e^{f(x)}(f(x)-1)+1$
$\Rightarrow y=f(x)-1+e^{-f(x)}$
$\Rightarrow y+1=e^{-f(x)}+f(x)$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2020
Let $f$ be a non-constant continuous function for all $x \geq 0$. Let $f$ satisfy the relation $f(x) f(a-x)=1$ for some $a \in R^{+}$. Then, $I=\int_{0}^{a} \frac{d x}{1+f(x)}$ is equal to

Options:
A $a$
B $\frac{a}{4}$
C $\frac{a}{2}$
D $f(a)$

The correct answers are:
$\frac{a}{2}$

Solution:
$I=\int_{0}^{a} \frac{d x}{1+f(x)}$
$=\int_{0}^{a} \frac{1}{1+f(a-x)} d x$
$\quad\left[\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]$
$=\int_{0}^{a} \frac{d x}{1+\frac{1}{f(x)}} \quad[\because f(x) f(a-x)=1]$
$I=\int_{0}^{a} \frac{f(x)}{f(x)+1} d x$
On adding Eqs. (i) and (ii), we get $2 I=\int_{0}^{a} \frac{f(x)+1}{f(x)+1} d x$
$\Rightarrow \quad 2 I=\int_{0}^{a} 1 d x$
$\begin{array}{ll}\Rightarrow & 2 t=[x]_{0}^{a} \\ \Rightarrow & 2 I=a \\ \Rightarrow & I=\frac{a}{2}\end{array}$
72. $(b,$ d) We have, $x=1-2 x$
$\Rightarrow y=\frac{1-2 x}{x}$
$\Rightarrow \frac{d y}{d x}=\frac{-2 x-(1-2 x) \cdot 1}{x^{2}}$
$\quad=\frac{-2 x-1+2 x}{x^{2}}=\frac{-1}{x^{2}} < 0$
since, $a x+b y+c=0$ is tangent to the curve $x=1-2 x$
$\therefore \quad \frac{-a}{b} < 0 \Rightarrow \frac{a}{b}>0$
$\Rightarrow$ Either $a>a b>0$ or $a < 0, b < 0$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2017