datavorous/entrance-exam-dataset huggingface dataset:
Let $f$ be a non-negative function defined in $[0, \pi / 2], f^{\prime}$ exists and be continuous for all $x$ and $\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} d t=\int_0^x f(t) d t$ and $f(0)=0$. Then

Options:
A $f\left(\frac{1}{2}\right) < \frac{1}{2}$ and $f\left(\frac{1}{3}\right) > \frac{1}{3}$
B $f\left(\frac{1}{2}\right) > \frac{1}{2}$ and $f\left(\frac{1}{3}\right) < \frac{1}{3}$
C $f\left(\frac{4}{3}\right) < \frac{4}{3}$ and $f\left(\frac{2}{3}\right) < \frac{2}{3}$
D $\mathrm{f}\left(\frac{4}{3}\right) > \frac{4}{3}$ and $\mathrm{f}\left(\frac{2}{3}\right) > \frac{2}{3}$

The correct answer is:
$f\left(\frac{4}{3}\right) < \frac{4}{3}$ and $f\left(\frac{2}{3}\right) < \frac{2}{3}$

Solution:
$\int_0^x \sqrt{1-\left(f^{\prime}(t)\right)^2} d t=\int_0^x f(t) d t$
Using Leibnitz Rule,
$\begin{aligned}
&\sqrt{1-\left(f^{\prime}(x)\right)^2}=f(x) \\
&\therefore f^{\prime}(x)=\pm \sqrt{1-f^2(x)} \Rightarrow \int \frac{d f(x)}{\sqrt{1-f^2(x)}}=\pm \int d x \\
&\sin ^{-1}(f(x))=\pm x+c \\
&\therefore f(x)=\sin (\pm x+C) \\
&\because f(0)=0 \Rightarrow C=0 \\
&\therefore f(x)=\sin x \text { or }-\sin x
\end{aligned}$
But $f$ is non-negative on $[0, \pi / 2]$
$\begin{aligned}
&\therefore \mathrm{f}(\mathrm{x})=\operatorname{Sin} \mathrm{x} \\
&\because \mathrm{f}(\mathrm{x}) < \mathrm{x} \text { for all } \mathrm{x} > 0 \\
&\therefore \mathrm{f}\left(\frac{4}{3}\right) < \frac{4}{3} \text { and } \mathrm{f}\left(\frac{2}{3}\right) < \frac{2}{3}
\end{aligned}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2022
Let \(f\) be a non-negative function defined on \(\left[0, \frac{\pi}{2}\right]\). If \(\int_0^x\left(f^{\prime}(t)-\sin 2 t\right) d t=\int_x^0 f(t) \tan t d t, f(0)=1\), then \(\int_0^{\frac{\pi}{2}} f(x) d x\) is

Options:
A 3
B \(3-\frac{\pi}{2}\)
C \(3+\frac{\pi}{2}\)
D \(\frac{\pi}{2}\)

The correct answers are:
\(3-\frac{\pi}{2}\)

Solution:
\(\begin{aligned}
\text {Hint } & f^{\prime}(x)-\sin 2 x=-f(x) \tan x \\
& \Rightarrow f^{\prime}(x)+\tan x f(x)=\sin 2 x \\
& \Rightarrow \frac{d f(x)}{d x}+\tan x . f(x)=\sin 2 x
\end{aligned}\)
\(\text { If }=e^{f \tan x d x}=e^{\ln |\sec x|}=|\sec x|=\sec x \quad \text { As, } x \in\left[0, \frac{\pi}{2}\right]\)
Solution is,
\(\begin{aligned}
& f(x) \times \sec x=\int \sin 2 x \sec x d x=\int 2 \sin x \cos x \sec x d x=2 \int \sin x d x \\
& f(x) \sec x=-2 \cos x+C \\
& \text { Put } x=0 ; f(0)=1 \\
& f(0) \sec 0=-2 \cos 0+c \\
& \Rightarrow 1=-2+C \quad \therefore C=3
\end{aligned}\)
\(\therefore\) Required solution is;
\(\begin{aligned}
& f(x) \sec x=-2 \cos x+3 \\
& f(x)=-2 \cos ^2 x+3 \cos x
\end{aligned}\)
\(\begin{aligned}
& =-2\left(\frac{1+\cos 2 x}{2}\right)+3 \cos x=-1-\cos 2 x+3 \cos x \\
& f(x)=-1-\cos 2 x+3 \cos x
\end{aligned}\)
Now \(\int_0^{\frac{\pi}{2}} f(x) d x\)
\(\int_0^{\frac{\pi}{2}}(-1-\cos 2 x+3 \cos x) d x=\left[-x-\frac{\sin 2 x}{2}+3 \sin x\right]_0 ^{-\frac{\pi}{2}}=\left(-\frac{\pi}{2}+3\right)-0=3-\frac{\pi}{2}\)
\(\left[-x-\frac{\sin 2 x}{2}+\left.3 \sin x\right|_0 ^{\frac{\pi}{2}}\right.\)

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2023
Let \(f\) be a strictly decreasing function defined on \(\mathbb{R}\) such that \(f(x) > 0, \forall x \in \mathbb{R}\). Let \(\frac{x^2}{f\left(a^2+5 a+3\right)}+\frac{y^2}{f(a+15)}=1\) be an ellipse with major axis along the \(y\)-axis. The value of ' \(a\) ' can lie in the interval (s)

Options:
A \((-\infty,-6)\)
B \((-6,2)\)
C \((2, \infty)\)
D \((-\infty, \infty)\)

The correct answers are:
\((-\infty,-6)\), \((2, \infty)\)

Solution:
Hint : \(f \Rightarrow\) strictly decreasing function \(\forall x \in R\)
\(\begin{aligned}
& f(x) > 0 \Rightarrow f^{\prime}(x) < 0 \\
& \frac{x^2}{f\left(a^2+5 a+3\right)}+\frac{y^2}{f(a+15)}=1
\end{aligned}\)
As major axis is \(y\)-axis
\(\begin{aligned}
& f(a+15) > f\left(a^2+5 a+3\right) \\
& \Rightarrow a+15 < a^2+5 a+3 \quad (\because f \text{ is decreasing) }\\
& \Rightarrow a^2+4 a-12 > 0 \\
& \Rightarrow a < -6 \text { or } a > 2
\end{aligned}\)

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2023
Let $f$ be any continuously differentiable function on $[a, b]$ and twice differentiable on $(a, b)$ such that $f(a)=f^{\prime}(a)=0$ and $f(b)=0$ Then.

Options:
A $f^{\prime \prime}(a)=0$
B $f^{\prime}(x)=0$ for some $x \in(a, b)$
C $f^{\prime \prime}(x) \neq 0$ for some $x \in(a, b)$
D $f^{\prime \prime \prime}(x)=0$ for some $x \in(a, b)$

The correct answers are:
$f^{\prime}(x)=0$ for some $x \in(a, b)$, $f^{\prime \prime}(x) \neq 0$ for some $x \in(a, b)$

Solution:
We have, $f$ is continuous and differentiably function on $[a, b]$ Also, $f(a)=f(b)=0$
By Rolle's theorem, there exists $c \in(a, b)$ such the $f^{\prime}(c)=0$
Thus, there exists $x \in(a, b)$ such that $f^{\prime}(x)=0$ Let at $x=c \in(a, b), f^{\prime}(c)=0$
Now, $f$ is continuously differentiable on $|a, b|$ $\Rightarrow \quad f^{\prime}$ is continuous on $\left.\mid a, b\right]$ Also, $f$ is twice differentiable on $(a, b)$ $\therefore f^{\prime}$ is differentiable on $(a, b)$
and $f^{\prime}(a)=0=f^{\prime}(c)$
By Rolle's theorem, there exists $k \in(a, c)$ such that. $f^{\prime \prime}(k)=0$
Thus, there exists $x \in(a, c)$ such that $f^{\prime \prime}(x)=0$ So, there exists $x \in(a, b)$ such that $f^{\prime \prime}(x)=0$ Let us consider, $f(x)=(x-a)^{2}(x-b)$
where $f(a)=f(b)=f^{\prime}(a)=0$ but
$f^{\prime \prime}(a) \neq 0$ and $f^{m}(x) \neq 0$ for any $x \in(a, b)$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2015
Let $\mathrm{f}$ be derivable in $[0,1]$, then

Options:
A there exists $c \in(0,1)$ such that $\int_0^c f(x) d x=(1-c) f(c)$
B there does not exist any point $d \in(0,1)$ for which $\int_0^d f(x) d x=(1-d) f(d)$
C $\int_0^c f(x) d x$ does not exist, for any $c \in(0,1)$
D $\int_0^c f(x) d x$ is independent of $c, c \in(0,1)$

The correct answer is:
there exists $c \in(0,1)$ such that $\int_0^c f(x) d x=(1-c) f(c)$

Solution:
Let $g(x)=x \int_0^x f(t) d t-\int_0^x f(t) d t$
Now, $g(0)=0$
$g(1)=0$
By Rolle's Theorem
$g^{\prime}(x)=0$
for some $x \in(0,1)$
$g^{\prime}(x)=x f(x)+\int_0^x f(t) d t-f(x)=0$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2022
Let $f: D \rightarrow R$ where $D=[0,1] \cup[2,4]$ be defined by $f(x)=\left\{\begin{array}{ll}x, & \text { if } x \in[0,1] \\ 4-x, & \text { if } x \in[2,4]\end{array}\right.$. Then,

Options:
A Rolle's theorem is applicable to $\mathrm{f}$ in $\mathrm{D}$
B Rolle's theorem is not applicable to $\mathrm{f}$ in $\mathrm{D}$
C there exists $\xi \in \mathrm{D}$ for which $\mathrm{f}^{\prime}(\xi)=0$ but Rolle's theorem is not applicable
D $f$ is not continuous in $D$

The correct answer is:
Rolle's theorem is not applicable to $\mathrm{f}$ in $\mathrm{D}$

Solution:
$f(x)=\left\{\begin{array}{ll}x, & x \in[0,1] \\ 4-x, & x \in[2,4]\end{array}\right.$
$f(x)$ is increasing in $[0,1]$
and $f(x)$ is decreasing in $[2,4]$
$\therefore$ Rolle's theorem is not applicable to $\mathrm{f}$ in $\mathrm{D}$.

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2021
Let $f(n)=2^{n+1}, g(n)=1+(n+1)^{2 n}$ for all $n \in \mathbb{N}$. Then

Options:
A $f(n) > g(n)$
B $f(n) < g(n)$
C $f(n)$ and $g(n)$ are not comparable
D $f(n) > g(n)$ if $n$ be even and $f(n) < g(n)$ if $n$ be odd

The correct answer is:
$f(n) < g(n)$

Solution:
$g(n)-f(n)=1+(n+1) 2^n-2 \cdot 2^n=1+n \cdot 2^n-2^n=1+(n-1) 2^n > 0$
$\because \mathrm{n} \geq 1 \Rightarrow(\mathrm{n}-1) \geq 0$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2022
Let $f: N \rightarrow R$ be such that $f(1)=1$ and $f(1)+2 f(2)+3 f(3)+\ldots+n f(n)=n(n+1) f(n),$ for all $n \in N, n \geq 2,$ where $N$ is the set of natural numbers and $R$ is the set of real numbers. Then, the value of $f(500)$ is

Options:
A 1000
B 500
C $1 / 500$
D $1 / 1000$

The correct answer is:
$1 / 1000$

Solution:
We have, $f: N \rightarrow R$ such that $f(1)=1$
$$
\begin{array}{l}
\text { and } f(1)+2 f(2)+3 f(3)+\ldots+n f(n) \\
\quad=n(n+1) f(n), \forall n \geq 2 \\
\text { Clearly, } f(1)+2 f(2)=2(2+1) f(2) \\
\Rightarrow \quad f(1)=6 f(2)-2 f(2) \\
\Rightarrow \quad f(1)=4 f(2) \\
\Rightarrow \quad f(2)=\frac{f(1)}{4}=\frac{1}{4}
\end{array}
$$
Similarly.
$$
f(1)+2 f(2)+3 f(3)=3(3+1) f(3)
$$
$\Rightarrow \quad 1+\frac{1}{2}+3 f(3)=12 f(3)$
$\Rightarrow \quad 9 f(3)=\frac{3}{2} \Rightarrow f(3)=\frac{1}{6}$
and $f(1)+2 f(2)+3 f(3)+4 f(4)=4(5) f(4)$
$\Rightarrow \quad 1+\frac{1}{2}+\frac{1}{2}=16 f(4)$
$\Rightarrow \quad f(4)=\frac{2}{16}=\frac{1}{8}$
In general, $f(n)=\frac{1}{2 n}$
$\therefore \quad f(500)=\frac{1}{1000}$

Topic: Mathematics, Functions, WBJEE, WBJEE 2015
Let f:R-{3}→R-{1} be defined by fx=x-2x-3. Let g : R→R be given as g(x)=2x-3. Then, the sum of all the values of x for which f-1x+g-1x=132 is equal to

Options:
A 7
B 2
C 5
D 3

The correct answer is:
5

Solution:
fx=y=x-2x-3

∴x=3y-2y-1

∴f-1x=3x-2x-1

& g(x)=y=2 x-3

∴x=y+32

∴g-1x=x+32

∵f-1x+g-1x=132

∴ x2-5x+6=

∴ sum of roots x1+x2=5

Topic: Mathematics, Functions, WBJEE, WBJEE 2014
Let $f: R \rightarrow R$ be a continuous function which satisfies $f(x)=\int_{0}^{x} f(t) d t .$ Then. the value of $f\left(\log _{e} 5\right)$ is

Options:
A 0
B 2
C 5
D 3

The correct answer is:
0

Solution:
Given, $\quad f(x)=\int_{0}^{x} f(t) d t$
Using Leibnitz theorem, we get
$$
f^{\prime}(x)=f(x) \Rightarrow f(x)=k e^{x}
$$
On putting $x=0$, we get $f(0)=\int_{0}^{0} f(t) d t$
$\Rightarrow \quad k e^{0}=0$
$\left.\because \int_{a}^{a} f(x) d x=0\right]$
$\left.\Rightarrow \quad k=0 \quad \mid \because e^{0}=1\right]$
$\therefore \quad f(x)=0 \Rightarrow f\left(\log _{e} 5\right)=0$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2015
Let $f: R \rightarrow R$ be a twice continuously differentiable function such that $f(0)=f(1)=f^{\prime}(0)=0 .$ Then

Options:
A $f^{\prime \prime}(0)=0$
B $f^{\prime \prime}=0$ for some $C \in A$
C if c $\neq 0,$ then $f^{\prime \prime}\neq 0$
D $f^{\prime}(x)>0$ for all $x \neq 0$

The correct answer is:
$f^{\prime \prime}=0$ for some $C \in A$

Solution:
We have.
$$f: \mathbb{R} \rightarrow \mathbb{R} $$ be a twice continuousty differentiable function such that $f(0)=f(1)=f'(0)=0$
Now, for atleast one value of $\epsilon_{1} \in(0,1)$ $f'\left(c_{1}\right)=0 \quad$ (by Rolle's theorem)
Again. $\quad f'(0)=0=f(c_{1})$
$f'(c)=$ ofor some $c \in\left(0, c_{1}\right)$ (by Rolle's theorem)

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2018
Let $f:R\rightarrow R$ be defined as
$f(x)=\left\{\begin{array}{lr}0, & x \text { is irrational} \\ \sin |x|, & x \text { is rational }\end{array}\right.$ Then, which of the following is true?

Options:
A f is discontinuous for all x
B f is continuous for all x
C f is discontinuous at $x=k\pi$ where k is an integer
D f is continuous at $x=k\pi$ where k is an integer

The correct answer is:
f is continuous at $x=k\pi$ where k is an integer

Solution:
We have, $f(x)=\left\{\begin{array}{ll}0, & x \text { is irrational } \\ \sin |x|, x \text { is rational }\end{array}\right.$
If $f(x)$ is continuous, then $\sin |x|=0$ $\Rightarrow x=k \pi,$ where $k$ is an integer.

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2015
Let $f: R \rightarrow R$ be defined as $f(x)=\frac{x^{2}-x+4}{x^{2}+x+4}$. Then, range of the function $f(x)$ is

Options:
A $\left[\frac{3}{5}, \frac{5}{3}\right]$
B $\left(\frac{3}{5}, \frac{5}{3}\right)$
C $\left(-\infty, \frac{3}{5}\right) \cup\left(\frac{5}{3}, \infty\right)$
D $\left[-\frac{5}{3},-\frac{3}{5}\right]$

The correct answer is:
$\left[\frac{3}{5}, \frac{5}{3}\right]$

Solution:
Let $y=\frac{x^{2}-x+4}{x^{2}+x+4}$
$\Rightarrow \quad x^{2} y+x y+4 y=x^{2}-x+4$
$\Rightarrow(y-1) x^{2}+(y+1) x+4 y-4=0$
For $x$ to be real, discriminant of the above quadratic equation should be greater than equal to 0
$\\ \Rightarrow \quad(y+1)^{2}-4(y-1)(4 y-4) \geq 0$
$\Rightarrow \quad(y+1)^{2}-16(y-1)^{2} \geq$
$\Rightarrow(y+1+4 y-4)(y+1-4 y+4) \geq$
$(5 y-3)(5-3 y) \geq 0$
$\therefore \quad y \in\left[\frac{3}{5}, \frac{5}{3}\right]$

Topic: Mathematics, Functions, WBJEE, WBJEE 2015
Let $f: R \rightarrow R$ be differentiable at $x=0$. If f(0)=0 and f'(0)=2, then the value of $$
\left.\lim _{x \rightarrow 0} \frac{1}{x} \mid f(x)+f(2 x)+f(3 x)+\ldots+f(2015 x)\right] \text{ is}
$$

Options:
A 2015
B 0
C $2015 \times 2016$
D $2015 \times 2014$

The correct answer is:
$2015 \times 2016$

Solution:
Given, $f(0)=0$ and $f^{\prime}(0)=2$
$\therefore \lim _{x \rightarrow 0} \frac{1}{x}[f(x)+f(2 x)+f(3 x)+\ldots+f(2015 x)]$
$=\lim _{x \rightarrow 0} \frac{\left[\begin{array}{c}f^{\prime}(x)+2 f^{\prime}(2 x)+3 f^{\prime}(3 x) \\ +\ldots+2015 f^{\prime}(2015 x)\end{array}\right]}{1}$
$=\frac{2+2 \times 2+3 \times 2+\ldots+2015 \times 2}{1}$
$=2[1+2+3+\ldots+2015]$
$=\frac{2(2015)(2015+1)}{2}=\frac{2 \times 2015 \times 2016}{2}$
$=2015 \times 2016$

Topic: Mathematics, Limits, WBJEE, WBJEE 2015
Let $f: R \rightarrow R$ be given by $f(x)=\left|x^{2}-1\right|, x \in R$. then

Options:
A f has a local minimum at $x=\pm 1$ but no local maximum
B f has a local maximum at $x=0$ but no local minimum
C $\mathrm{f}$ has a local minima at $\mathrm{x}=\pm 1$ and a local maxima at $\mathrm{x}=0$
D f has neither a local maxima nor a local minima at any point

The correct answer is:
$\mathrm{f}$ has a local minima at $\mathrm{x}=\pm 1$ and a local maxima at $\mathrm{x}=0$

Solution:
$f$ has a local minima at $x=\pm 1$ and local maximum at $x=0$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2021
Let $f: R \rightarrow R$ be such that $f(0)=0$ and $\left|f^{\prime}(x)\right| \leq 5$ for all $x$. Then $f(1)$ is in

Options:
A $(5,6)$
B $[-5,5]$
C $(-\infty,-5) \cup(5, \infty)$
D $[-4,4]$

The correct answer is:
$[-5,5]$

Solution:
$\left|f^{\prime}(x)\right| \leq 5$
$\int_{0}^{1}-5 d x \leq \int_{0}^{1} f^{\prime}(x) d x \leq \int_{0}^{1} 5 d x$
$\Rightarrow-5 \leq f(1)-0 \leq 5$
$\Rightarrow-5 \leq f(1) \leq 5$
$\Rightarrow f(1) \in[-5,5]$

Topic: Mathematics, Functions, WBJEE, WBJEE 2021
Let $f: R \rightarrow R$ be such that $f(2 x-1)=f(x)$ for all $x \in R .$ If $f$ is continuous at $x=1$ an $f(1)=1,$ then

Options:
A $f(2)=1$
B $f(2)=2$
C $f$ is continuous only at $x=1$
D $f$ is continuous at all points

The correct answers are:
$f$ is continuous only at $x=1$

Solution:
Given, $f: R \rightarrow R$
and $\quad f(2 x-1)=f(x), x \in R$
Hence, $f$ is continuous at $x=1$ and $f(1)=1$

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2015
Let $f: R \rightarrow R$ be such that $f$ is injective and $f(x) f(y)=f(x+y) \quad$ for $\quad$ all $\quad x, y \in R, \quad$ if $f(x), f(y)$ and $f(z)$ are in $G P,$ then $x, y$ and $z$ are in

Options:
A AP always
B GP always
C AP depending on the values of $x, y$ and $z$
D GP depending on the values of $x, y$ and $z$

The correct answer is:
AP always

Solution:
Let the function, $f(x)=a^{kx}$ Which define in $f: R \rightarrow R$ and injective also.
Now, we have
$$
f(x) f(y)=f(x+y)
$$
$\Rightarrow$
$a^{k x} \cdot a^{k y}=a^{k(x+y)}$
$\Rightarrow$
$a^{k(x+y)}=a^{k(x+y)}$
$\because \quad f(x), f(y)$ and $f(z)$ are in GP
$\therefore$
$f(y)^{2}=f(x) \cdot f(z)$
$\Rightarrow$
$a^{2 k y}=a^{k x} \cdot a^{k z}$
$\Rightarrow$
$e^{2 k y}=e^{k(x+z)}$
On comparing, we get $2 k y=k(x+z) \Rightarrow 2 y=x+z$
$\Rightarrow x, y$ and $z$ are in $A P$.

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2013
Let $f: R \rightarrow R$ be such that $f$ is injective and $f(x) f(y)=f(x+y)$ for $\forall x, y \in R .$ If $f(x)$
$f(y), f(z)$ are in G.P., then $x, y, z$ are in

Options:
A AP always
B GP always
C AP depending on the value of $x, y, z$
D GP depending on the value of $x, y, z$

The correct answer is:
AP always

Solution:
Since, $f: R \rightarrow R$ is injective and $f(x) f(y)=f(x+y), \forall x, y \in R$
$\therefore \quad f(x)=a^{x}$
Again, $f(x), f(y), f(z)$ are in GP. $\Rightarrow \quad(f(y))^{2}=f(x) \cdot f(y)$
$\Rightarrow \quad a^{2 y}=a^{x} \cdot a^{z}$
$\Rightarrow \quad a^{2 y}=a^{x+y}$
$\Rightarrow \quad 2 y=x+z$
$\therefore x, y, z$ are in $\mathrm{AP}$

Topic: Mathematics, Functions, WBJEE, WBJEE 2017
Let $f: R \rightarrow R$ be twice continuously differentiable. Let $f(0)=f(1)=f^{\prime}(0)=0 .$ Then,

Options:
A $f^{\prime \prime}(x) \neq 0$ for all $x$
B $f^{\prime \prime}(c)=0$ for some $c \in R$
C $f^{\prime \prime}(x) \neq 0$ if $x \neq 0$
D $f^{\prime}(x)>0$ for all $x$

The correct answers are:
$f^{\prime \prime}(c)=0$ for some $c \in R$

Solution:
Let function $f(x)=x^{2}(x-1)$
$\Rightarrow \quad f^{\prime}(x)=3 x^{2}-2 x$
and $\quad f^{\prime \prime}(x)=6 x-2$
$\begin{array}{ll}\text { Now, } & f(0)=f(1)=f^{\prime}(0)=0\end{array}$
Then, according to question, at $x=\frac{1}{3}, \quad f^{\prime \prime}(x)=0$
(i.e. $c=1 / 3$ for some $C \in R$ )

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2017
Let $f(x)>0$ for all $x$ and $f^{\prime}(x)$ exists for all $x$. If $f$ is the inverse function of $h$ and $h^{\prime}(x)=\frac{1}{1+\log x} \cdot$ Then, $f^{\prime}(x)$ will be

Options:
A $1+\log (f(x))$
B $1+f(x)$
C $1-\log (f(x))$
D $\log f(x)$

The correct answer is:
$1+\log (f(x))$

Solution:
Given,
$$
h\{f(x)\}=x
$$
differentiating w.r.t. $x$, we get
$$
\begin{array}{r}
h^{\prime}\{f(x)\}, f^{\prime}(x)=1 \\
f^{\prime}(x)=\frac{1}{h^{\prime}\{f(x)\}}
\end{array}
$$
$\begin{aligned} \Rightarrow \quad f^{\prime}(x)=\frac{1}{\frac{1}{1+\log \{f(x)\}}} & \\ \Rightarrow \quad\left[\because h^{\prime}(x)=\frac{1}{1+\log x}\right] \\ \Rightarrow & f^{\prime}(x)=1+\log (f(x)) \end{aligned}$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2019
$\operatorname{Let} f(x)=\left\{\begin{array}{l}0, \text { if }-1 \leq x < 0 \\ 1, \text { if } x=0 \\ 2, \text { if } 0 < x \leq 1\end{array}\right.$ and let $F(x)=\int_{-1}^{x} f(t) d t,-1 \leq x \leq 1$, then

Options:
A $\mathrm{F}$ is continuous function in $[-1,1]$
B $F$ is discontinuous function in $[-1,1]$
C $\mathrm{F}^{\prime}(\mathrm{x})$ exists at $\mathrm{x}=0$
D $\mathrm{F}^{\prime}(\mathrm{x})$ does not exists at $\mathrm{x}=0$

The correct answers are:
$\mathrm{F}$ is continuous function in $[-1,1]$, $\mathrm{F}^{\prime}(\mathrm{x})$ does not exists at $\mathrm{x}=0$

Solution:
$f(x)=\left\{\begin{array}{ll}0 & -1 \leq x < 0 \\ 1 & x=0 \\ 2 & 0 < x \leq 1\end{array}\right.$

$F(x)=\left\{\begin{array}{ll}0 & -1 \leq x \leq 0 \\ 2 x & 0 < x \leq 1\end{array}\right.$

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2021
Let $f(x)=\left\{\begin{array}{ll}\int_{0}^{x}|1-t| d t, & x>0 \\ x-\frac{1}{2}, & x \leq 1\end{array} .\right.$ Then

Options:
A $f(x)$ is continuous at $x=1$
B $f(x)$ is not continuous at $x=1$
C $f(x)$ is differentiable at $x=1$
D $f(x)$ is not differentiable at $x=1$

The correct answers are:
$f(x)$ is continuous at $x=1$, $f(x)$ is not differentiable at $x=1$

Solution:
Given. $f(x)=\left\{\begin{aligned} \int_{0}^{x}|1-t| d t, & x>1 \\ x-\frac{1}{2}, & x \leq 1 \end{aligned}\right.$
Now, for $x>1$, $\int_{0}^{x}|1-t| d t$
$$
\begin{aligned}
&=\int_{0}^{1}(1-t) d t+\int_{1}^{x}(t-1) d t \\
&=\left[t-\frac{t^{2}}{2}\right]_{0}^{1}+\left[\frac{t^{2}}{2}-t\right]_{1}^{x} \\
&=\left[1-\frac{1}{2}-0\right]+\left[\frac{x^{2}}{2}-x-\frac{1}{2}+1\right] \\
&=\frac{1}{2}+\frac{x^{2}}{2}-x-\frac{1}{2}+1 \\
\Rightarrow \quad \frac{x^{2}}{2}-x+1
\end{aligned}
$$
$\therefore$ Eq. (i) becomes, $f(x)=\left\{\begin{array}{ll}\frac{x^{2}}{2}-x+1, & x>1 \\ x-\frac{1}{2}, & x \leq 1\end{array}\right.$
Continuity at $x=1$
and $\begin{aligned} \mathrm{LHL} &=\lim (f(x)\\ &=\lim _{x \rightarrow 1}\left(x-\frac{1}{2}\right)=1-\frac{1}{2} \\ \mathrm{RHL} &=\lim _{x \rightarrow 1^{+}} f(x) \\ &=\lim _{x \rightarrow 1}\left(\frac{x^{2}}{2}-x+1\right) \\ &=\frac{1}{2}-1+1=\frac{1}{2} \\ f(1) &=1-\frac{1}{2}=\frac{1}{2} \end{aligned}$
$\therefore \quad f(1)=\lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x)$
Hence, $f(x)$ is continuous at $x=1$
Differentiability at $x=1$
$\begin{aligned} L H D &=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} \\ &=\lim _{h \rightarrow 0} \frac{(1-h)-\frac{1}{2}-\left(1-\frac{1}{2}\right)}{-h}=\lim _{h \rightarrow 0} \frac{-h}{-h}=1 \end{aligned}$
$\begin{aligned} \mathrm{RHD} &=\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\frac{(1+h)^{2}}{2}-(1+h)+1-\left(1-\frac{1}{2}\right)}{h} \\ &=\lim _{h \rightarrow 0} \frac{\frac{1+h^{2}+2 h}{2}-1-h+1-\frac{1}{2}}{h} \\ &=\lim _{h \rightarrow 0} \frac{\frac{1}{2}+\frac{h^{2}}{2}+h-h-\frac{1}{2}}{h} \\ &=\lim _{h \rightarrow 0} \frac{h^{2}}{2 h}=0 \end{aligned}$
$\mathrm{LHD} \neq \mathrm{RHD}$
Hence, $f(x)$ is not differentiable at $x=1$.

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2014
Let $f(x)=1-\sqrt{\left(x^{2}\right)}$ where the square root is to be taken positive, then

Options:
A $f$ has no extrema at $x=0$
B $f$ has minima at $x=0$
C $f$ has maxima at $x=0$
D $\mathrm{f}^{\prime}$ exists at 0

The correct answer is:
$f$ has maxima at $x=0$

Solution:
Hint:
$f(x)=1-|x|, f$ has maxima at $x=0$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2020
Let $f(x)=\left\{\begin{array}{cc}-2 \sin x, & \text { if } x \leq-\frac{\pi}{2} \\ A \sin x+B, & \text { if }-\frac{\pi}{2} < x < \frac{\pi}{2} \text { . Then, } \\ \cos x, & \text { if } x \geq \frac{\pi}{2}\end{array}\right.$

Options:
A $f$ is discontinuous for all $A$ and $B$
B $f$ is continuous for all $A=-1$ and $B=1$
C $f$ is continuous for all $A=1$ and $B=-1$
D $f$ is continuous for all real values of $A B$

The correct answer is:
$f$ is continuous for all $A=-1$ and $B=1$

Solution:
At $\quad x=-\frac{\pi}{2}$
$$
\begin{array}{l}
\text { LHL }=-2 \\
\text { RHL }=-A+B \\
\text { For continuity, } \mathrm{LHL}=\mathrm{RHL}=f(-\pi / 2)
\end{array}
$$
$\Rightarrow \quad-A+B=2$
At $x=\pi / 2$
$$
\begin{aligned}
\mathrm{LHL} &=A+B \\
\mathrm{RHL} &=0 \\
\text { For continuity, } \mathrm{LHL} &=\mathrm{RHL}=f(\pi / 2)
\end{aligned}
$$
$\Rightarrow \quad A+B=0$
On solving Eqs. (i) and (ii), we get
$$
A=-1 \text { and } B=1
$$

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2018
Let $f(x)=2 x^{2}+5 x+1$. If we write $f(x)$ as $f(x)=a(x+1)(x-2)+b(x-2)(x-1)+c(x-1)(x+1)$
for real numbers $a, b, c,$ then

Options:
A there are infinite number of choices for $a, b, c$
B only one choice for a but infinite number of choices for $b$ and $c$
C exactly one choice for each of $a, b, c$
D more than one but finite number of choices for $a, b, c$

The correct answer is:
exactly one choice for each of $a, b, c$

Solution:
$$
\begin{array}{l}
\text { Given, } f(x)=2 x^{2}+5 x+1 \\
\text { Also, } f(x)=a(x+1)(x-2)+b(x-2)(x-1) \\
+c(x-1)(x+1) \\
\qquad \begin{array}{r}
=a\left(x^{2}-x-2\right)+b\left(x^{2}-3 x+2\right) \\
+c\left(x^{2}-1\right) \\
f(x)=(a+b+c) x^{2}+(-a-3 b) x \\
+(-2 a+2 b-c)
\end{array}
\end{array}
$$
On equating the coefficients of $x^{2}$, $x$ and constant term we get
$$
\begin{array}{l}
a+b+c=2,-a-3 b=5 \\
\text { and }-2 a+2 b-c=1
\end{array}
$$
On solving above equations, we get $a=-\frac{35}{4}, b=\frac{5}{4}$ and $c=\frac{38}{4}$
Hence, exactly one choice for each of b and c

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2014
Let $f(x)=\frac{1}{3} x \sin x-(1-\cos x) .$ The smallest positive interger $k$ such that $\lim _{x \rightarrow 0} \frac{f(x)}{x^{k}} \neq 0$ is

Options:
A 4
B 3
C 2
D 1

The correct answers are:
2

Solution:
Hint:
It $\left.\frac{x \sin x-3(1-\cos x)}{3 x^{k}}=\frac{1}{3} \operatorname{lt}_{x \rightarrow 0}\left(\frac{\sin x / 2}{x / 2}\right)_{x \rightarrow 0} \mid t_{x \rightarrow 0} \frac{2 x \cos x / 2-6 \sin x / 2}{2 x^{k-1}}\right)$
$k-1=1 \Rightarrow k=2$

Topic: Mathematics, Limits, WBJEE, WBJEE 2020