datavorous/entrance-exam-dataset huggingface dataset:
Let $f(x)=3 x^{10}-7
x^{8}+5 x^{6}-21 x^{3}+3 x^{2}-7$
Then $\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{h^{3}+3 h}$

Options:
A does not exist
B is $\frac{50}{3}$
C is $\frac{53}{3}$
D is $\frac{22}{3}$

The correct answer is:
is $\frac{53}{3}$

Solution:
We have, $f(x)=3 x^{10}-7 x^{8}+5 x^{6}-21 x^{3}+3 x^{2}-7$
$\therefore \quad f(1-h)=3(1-h)^{10}-7(1-h)^{8}$
$+5(1-h)^{6}-21\left(1-h^{3}+3(1-h)^{2}-7\right.$
$=3\left(1-10 h+45 h^{2}-120 h^{3}+\ldots \ldots+h^{10}\right)$
$-7\left(1-8 h+28 h^{2}-56 h^{3}+\ldots . .+h^{8}\right)$
$+5\left(1-6 h+15 h^{2}-20 h^{3}+\ldots . .+h^{6}\right)$
$-21\left(1-3 h+3 h^{2}-h^{3}\right)$
$+3\left(1-2 h+h^{2}\right)-7$
$\Rightarrow f(1-h)=-24+53 h+h^{2}(-46)+h^{3}(-47)+\ldots$
and $\quad f(1)=-24$
$\therefore \quad \lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{h^{3}+3 h}$
$=\lim _{h \rightarrow 0} \frac{-24+53 h+h^{2}(-46)+h^{3}(-47)+\ldots-(-24)}{h\left(h^{2}+3\right)}$
$=\lim _{h \rightarrow 0} \frac{53 h+h^{2}(-46)+h^{3}(-47)+\ldots}{h\left(h^{2}+3\right)}$
$=\lim _{h \rightarrow 0} \frac{53+h(-46)+h^{2}(-47)+\ldots}{h^{2}+3}=\frac{53}{3}$

Topic: Mathematics, Limits, WBJEE, WBJEE 2018
Let $f(x)=a_0+a_1|x|+a_2|x|^2+a_3|x|^3$, where $a_0, a_1, a_2, a_3$ are real constants. Then $f(x)$ is differentiable at $x=0$

Options:
A whatever be $a_0, a_1, a_2, a_3$
B for no values of $a_0, a_1, a_2, a_3$
C only if $a_1=0$
D only if $a_1=0, a_3=0$

The correct answer is:
only if $a_1=0$

Solution:
$f^{\prime}\left(0^{+}\right)=\lim _{x \rightarrow 0}\left(a_1+2 a_1 x+3 a_3 x^2\right)=a_1$
$f^{\prime}\left(0^{-}\right)=\lim _{x \rightarrow 0}\left(-a_1+2 a_1 x-3 a_3 x^2\right)=-a_1$
$f^{\prime}\left(0^{+}\right)=f^{\prime}\left(0^{-}\right) \quad \Rightarrow a_1=0$

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2022
Let $f(x)=a x^{2}+b x+c, g(x)=p x^{2}+q x+r$
$\begin{array}{lll}\text { such that } & f(1)=g(1), f(2)=g(2) & \text { and }\end{array}$
$f(3)-g(3)=2 .$ Then, $f(4)-g(4)$ is

Options:
A 4
B 5
C 6
D 7

The correct answer is:
6

Solution:
Given, $f(x)=e x^{2}+b x+c, g(x)=p x^{2}+q x+r$
since, $f(1)=g(1)$
$\Rightarrow \quad a+b+c=p+q+r$
$f(2)=8(2)$
$\Rightarrow \quad 4 a+2 b+c=4 p+2 q+r$
Subtracting Eq.
(tii) from Eq. (i), we get $3 a+b=3 p+q$
$f(3)-g(3)=2$
$\Rightarrow(9 a+3 b+c)-(9 p+3 q+r)=2$
$\Rightarrow \quad 3(3 a+b)+c-3(3 p+q)-r=2$
$\Rightarrow$
$c-r=2 \quad \ldots \text { (iv) }$
$(3 a+b=3 p+q)$
From Eq. (i), $(a-p)+(b-q)+(c-r)=0$
$\Rightarrow \quad(a-p)+(b-q)+2=0$
From Eq. (ii), $4(a-p)+2(b-q)+c-r=0$
$\Rightarrow \quad 2(a-p)+(b-q)+1=0$
Subtracting Eq.(v) from Eq. (vi), we get $(a-p)-1=0$
$a-p=1$
$\therefore$ From Eq. $(v), \quad b-q=-3$
Now, $f(4)-g(4)=(16 a+4 b+c)-(16 p+4 q+r)$
$=16(a-p)+4(b-q)+(c-r) \ldots(v$
Substituting the values of $(a-p),(b-q),(c-r)$ from above in Eq. (vii). we get $\begin{aligned} f(4)-g(4) &=16 \times 1+4(-3)+2 \\ &=16-12+2=6 \end{aligned}$

Topic: Mathematics, Linear Programming, WBJEE, WBJEE 2012
Let $f(x)$ be a continuous periodic function with period $T$. Let $I=\int_{a}^{a+T} f(x) d x$. Then

Options:
A $\mathrm{I}$ is linear function in ' $\mathrm{a}$ '
B I does not depend on ' $a$ '
C $0 < \mathrm{I} < \mathrm{a}^{2}+1$ where $\mathrm{I}$ depends on ' $\mathrm{a}$ '
D $\mathrm{I}$ is quadratic function in ' $\mathrm{a}$ '

The correct answer is:
I does not depend on ' $a$ '

Solution:
$I=\int_{a}^{a+T} f(x) d x=\int_{0}^{T} f(x) d x$
I does not depend on a.

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2021
Let $f(x)$ be a derivable function, $f^{\prime}(x)>f(x)$ and $f(0)=0 .$ Then

Options:
A $f(x)>0$ for all $x>0$
B $f(x) < 0$ for all $x>0$
C no sign of $f(x)$ can be ascertained
D $f(x)$ is a constant function

The correct answer is:
$f(x)>0$ for all $x>0$

Solution:
Let $g(x)=e^{-x} f(x)$
$\therefore$ $g^{\prime}(x)=e^{-x} f^{\prime}(x)-e^{-x} f(x)$
$\quad=e^{-x}\left(f^{\prime}(x)-f(x)\right)$
Since, $f^{\prime}(x)>f(x),$ so $f^{\prime}(x)-f(x)>0$
$\therefore e^{-x}\left(f^{\prime}(x)-f(x)\right)>0$
$\Rightarrow \quad g^{\prime}(x)>0$
$\therefore g(x)$ is an increasing function.
Now, $g(x)>g(0) \forall x>0$
$\begin{array}{lr}\Rightarrow e^{-x} f(x)>e^{0} f(0) & \\ \Rightarrow e^{-x} f(x)>0 & {[\because f(0)=0]}\end{array}$
$\Rightarrow \quad f(x)>0, \forall x>0$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2019
Let $f(x)$ be a differentiable function and $f^{\prime}(4)=5 .$ Then, $\lim _{x \rightarrow 2} \frac{f(4) - f\left(x^{2}\right)}{x-2}$ equals

Options:
A 0
B 5
C 20
D -20

The correct answer is:
-20

Solution:
Given, $f^{\prime}(4)=5$
$$
\begin{array}{ll}
\text { Now, } \lim _{x \rightarrow 2} \frac{f(4)-f\left(x^{2}\right)}{x-2} & {\left[\frac{0}{0} \text { form }\right]} \\
=\lim _{x \rightarrow 2} \frac{0-f^{\prime}\left(x^{2}\right) \cdot 2 x}{1} \\
=\frac{-f^{\prime}(4) \cdot 2 \times 2}{1} \\
=-(5) \times 4=-20
\end{array}
$$

Topic: Mathematics, Limits, WBJEE, WBJEE 2014
Let $f(x)$ be a differentiable function in $[2,7]$. If $f(2)=3$ and $f^{\prime}(x) \leq 5$ for all $x$ in $(2,7),$ then the maximum posaible value of $f(x)$ at $x=7$ is

Options:
A 7
B 15
C 28
D 14

The correct answer is:
28

Solution:
Given, $\quad f^{\prime}(x) \leq 5$
$\frac{f(7)-f(2)}{x-2} \leq 5$
(by Lagrange mean value theorem] $\begin{array}{ll}\Rightarrow & \frac{f(7)-3}{7-2} \leq 5 \\ \Rightarrow & f(7) \leq 5(7-2)+3 \\ \therefore & f(7) \leq 28\end{array}$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2014
Let $f(x)=\cos \left(\frac{\pi}{x}\right), x \neq 0,$ then assuming $k$ as an integer,

Options:
A $f(x)$ increases in the interval $\left(\frac{1}{2 k+1}, \frac{1}{2 k}\right)$
B $f(x)$ decreases in the interval $\left(\frac{1}{2 k+1}, \frac{1}{2 k}\right)$
C $f(x)$ decreases in the interval $\left(\frac{1}{2 k+2}, \frac{1}{2 k+1}\right)$
D $f(x)$ increases in the interval $\left(\frac{1}{2 k+2}, \frac{1}{2 k+1}\right)$

The correct answers are:
$f(x)$ increases in the interval $\left(\frac{1}{2 k+1}, \frac{1}{2 k}\right)$, $f(x)$ decreases in the interval $\left(\frac{1}{2 k+2}, \frac{1}{2 k+1}\right)$

Solution:
$f(x)=\cos \left(\frac{\pi}{x}\right)$
$\Rightarrow \quad f'(x)=-\sin \left(\frac{\pi}{x}\right)\left(\frac{-\pi}{x^{2}}\right)=\frac{\pi}{x^{2}} \sin \frac{\pi}{x}$
For increasing function, $f'(x)>0$ $\Rightarrow \sin \left(\frac{\pi}{x}\right)>0 \Rightarrow 2 k \pi < \frac{\pi}{x} < (2 k+1) \pi$
$\Rightarrow \quad \frac{1}{2 k}>x>\frac{1}{2 k+1}$
For decreasing function, $f(x) < 0$ $\Rightarrow \quad \sin \left(\frac{\pi}{x}\right) < 0$
$\Rightarrow \frac{\pi}{x} \in[(2 k+1) \pi,(2 k+2) \pi] \Rightarrow x \in\left(\frac{1}{2 k+2}, \frac{1}{2 k+1}\right)$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2018
Let $f(x)$ denotes the fractional part of a real number $x$. Then, the value of $\int_{0}^{\sqrt{3}} f\left(x^{2}\right) d x$ is

Options:
A $2 \sqrt{3}-\sqrt{2}-1$
B 0
C $\sqrt{2}-\sqrt{3}+1$
D $\sqrt{3}-\sqrt{2}+1$

The correct answer is:
$\sqrt{2}-\sqrt{3}+1$

Solution:
Let $I=\int_{0}^{\sqrt{3}} f\left(x^{2}\right) d x=\int_{0}^{\sqrt{3}}\left\{x^{2}\right\} d x$
$\left.=\int_{0}^{\sqrt{3}}\left(x^{2}-\mid x^{2}\right)\right) d x$
$=\int_{0}^{\sqrt{3}} x^{2} d x-\int_{0}^{\sqrt{3}}\left|x^{2}\right| d x$
$=\left[\frac{x^{3}}{3}\right]_{0}^{\sqrt{3}}-1 \int_{0}^{1}\left|x^{2}\right| d x+\int_{1}^{\sqrt{2}}\left|x^{2}\right| d x+\int_{\sqrt{2}}\left|x^{2}\right| d x \mid$
$=\sqrt{3}-\left[\int_{0}^{1} 0 d x+\int_{1}^{\sqrt{2}} 1 d x+\int_{\sqrt{2}}^{\sqrt{3}} 2 d x\right]$
$=\sqrt{3}-\left\{0+|x|_{i}^{\sqrt{2}}+|2 x| \sqrt{2}\right\}$
$=\sqrt{3}-\sqrt{2}+1-2 \sqrt{3}+2 \sqrt{2}$
$=\sqrt{2}-\sqrt{3}+1$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2015
Let $F(x)=e^{x}, G(x)=e^{-x}$ and $H(x)=G(F(x)),$ where $x$ is a real variable. Then, $\frac{d H}{d x}$ at $x=0$ is

Options:
A 1
B -1
C $-\frac{1}{e}$
D $-e$

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

Solution:
We have, $F(x)=e^{x}, G(x)=e^{-x}$ $\therefore \quad
H(x)=G(F(x))$ $\begin{aligned} &=G\left(e^{x}\right) \\ &=e^{-e^{x}} \\ \frac{d H}{d x} &=e^{-e^{x}} \cdot\left(-e^{x}\right) \\ &=-e^{x} e^{-e^{x}} \end{aligned}$
$\left.\therefore \quad \frac{d H}{d x}\right|_{x=0}=-e^{0} \cdot e^{-e^{0}}$
$=-1 . e^{-1}$
$=-e^{-1}$
$=\frac{-1}{e}$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2017
Let $f(x)=\max \{x+|x|, x-[x]\}, \quad$ where $\quad \mid x]$ denotes the greateat integer $\leq x$. Then, the value of $\int_{-3}^{3} f(x) d x$ is

Options:
A 0
B $51 / 2$
C $21 / 2$
D 1

The correct answer is:
$21 / 2$

Solution:
Given, $f(x)=\max \{x+|x|, x-[x]\}$
$$
\begin{aligned}
=\left\{\begin{array}{ll}
2 x, & x \geq 0 \\
x-[x], & x \leq 0
\end{array}\right.\\
\begin{aligned}
\int_{-3}^{3} f(x) d x &=\int_{-3}^{0} x-[x] d x+\int_{0}^{3} 2 x d x \\
&=3 \int_{-1}^{0}(1+x) d x+2 \int_{0}^{3} x d x
\end{aligned}
\end{aligned}
$$
[$\because$ $x-[x]$ is a periodic function at $x=1]$
$$
\begin{array}{l}
=3\left[x+\frac{x^{2}}{2}\right]_{-1}^{0}+2\left[\frac{x^{2}}{2}\right]_{0}^{3} \\
=3\left[0-0-\left(-1+\frac{1}{2}\right)\right]+3^{2}-0 \\
=3\left[\frac{1}{2}\right]+9=\frac{3}{2}+9 \\
=\frac{21}{2}
\end{array}
$$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2014
Let $f(x)=\int_{\sin x}^{\cos x} e^{-t^2} d t$. Then $f^{\prime}\left(\frac{\pi}{4}\right)$ equals

Options:
A $\sqrt{1 / e}$
B $-\sqrt{2 / e}$
C $\sqrt{2 / e}$
D $-\sqrt{1 / e}$

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

Solution:
$f^{\prime}(x)=-e^{-\cos ^2 x} \sin x-e^{-\sin ^2 x} \cos x$
$\therefore f^{\prime}\left(\frac{\pi}{4}\right)=-e^{-\frac{1}{2}} \frac{1}{\sqrt{2}}-e^{-\frac{1}{2}} \frac{1}{\sqrt{2}}=-\frac{2}{\sqrt{2 e}}=-\sqrt{\frac{2}{e}}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2022
Let $f(x)=\left\{\begin{array}{ccc}\frac{x^{p}}{(\sin x)^{q}} & , & \text { if } 0 < x \leq \frac{\pi}{2} \\ 0 & , & \text { if } x=0\end{array}\right.$
$(p, q \in R) .$ Then, Lagrange's mean value theorem is applicable to $f(x)$ in closed interval $[0, x]$

Options:
A for all $p, q$
B only when $p>q$
C only when $p < q$
D for no value of $p, q$

The correct answer is:
only when $p>q$

Solution:
Since, Lagrange's mean value theorem is applicable on
$\begin{array}{ll}\therefore & \lim _{x \rightarrow 0} \frac{x^{P}}{(\sin x)^{q}}=f(0) \\ \Rightarrow \quad & \lim _{x \rightarrow 6} \frac{x^{p}}{(\sin x)^{q}}=0\end{array}$
Above equation holds only when $p>q$.

Topic: Mathematics, Limits, WBJEE, WBJEE 2017
Let $f(x)=\sin x+\cos a x$ be periodic function. Then

Options:
A 'a' is any real number
B 'a' is any irrational number
C 'a' is rational number
D $a=0$

The correct answer is:
'a' is rational number

Solution:
Hint:
LCM of rational multiple of same irrational is defined

Topic: Mathematics, Functions, WBJEE, WBJEE 2020
Let $f(x)=\frac{\sqrt{x+3}}{x+1}$ then the value of $\underset{x \rightarrow-3-0}{\operatorname{Lt}} f(x)$ is

Options:
A 0
B does not exist
C $\frac{1}{2}$
D $-\frac{1}{2}$

The correct answer is:
does not exist

Solution:
Hints : Because on left hand side of 3 function is not defined.

Topic: Mathematics, Limits, WBJEE, WBJEE 2009
Let $f(x)=(x-2)^{17}(x+5)^{24}$. Then

Options:
A $f$ does not have a critical point at $x=2$
B $f$ has a minimum at $x=2$
C $f$ has neither a maximum nor a minimum at $x=2$
D f has a maximum at $x=2$

The correct answer is:
$f$ has neither a maximum nor a minimum at $x=2$

Solution:
$\begin{aligned} & f(x)=(x-2)^{17}(x+5)^{24} \\ \Rightarrow & f^{\prime}(x)=17(x-2)^{16}(x+5)^{24}+24(x-2)^{17}(x+5)^{23}=(x-2)^{16}(x+5)^{23}(17 x+85+24 x-48)=(x-2)^{16}(x+5)^{23}(41 x+37) \end{aligned}$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2022
Let $f(x)=x^2+x \sin x-\cos x$. Then

Options:
A $f(x)=0$ has at least one real root
B $f(x)=0$ has no real root
C $f(x)=0$ has at least one positive root
D $f(x)=0$ has at least one negative root

The correct answers are:
$f(x)=0$ has at least one real root, $f(x)=0$ has at least one positive root, $f(x)=0$ has at least one negative root

Solution:
$f^{\prime}(x)=2 x+x \cos x+\sin x+\sin x=2(x+\sin x)+x \cos x$
$\Rightarrow f^{\prime}(x) > 0 \quad \forall x > 0, \quad f^{\prime}(x) < 0 \quad \forall x < 0 \Rightarrow f(0)=-1$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2022
Let $f(x)=x^{4}-4 x^{3}+4 x^{2}+c, c \in R .$ Then,

Options:
A $f(x)$ has infinitely many zeros in (1,2) for all $c$
B $f(x)$ has exactly one zero in (1,2) if $-1 < c < 0$
C $f(x)$ has double zeros in (1,2) if $-1 < c < 0$
D whatever be the value of $c, f(x)$ has no zero in (1,2)

The correct answer is:
$f(x)$ has exactly one zero in (1,2) if $-1 < c < 0$

Solution:
Given,
$$
\begin{array}{ll}
f(x) & =x^{4}-4 x^{3}+4 x^{2}+c \\
f(1) & =1+c \\
\text { and } \quad f(2) & =2^{4}-4(2)^{3}+4(2)^{2}+c
\end{array}
$$
$$
\begin{array}{l}
\begin{aligned}
=& 16-32+16+c \\
&=c
\end{aligned} \\
\text { Now, } f(1) \cdot f(2)=c(1+c) \\
\text { Let } f(1) \cdot f(2) < 0 \\
\Rightarrow \quad c(1+c) < 0 \Rightarrow c \in(-1,0)
\end{array}
$$
For $c \in(-1,0), f(x)$ has exactly one zero is (1,2)

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2019
Let $f(x)=x+1 / 2$. Then, the number of real values of $x$ for which the three unequal terms $f(x), f(2 x), f(4 x)$ are in $\mathrm{HP}$ is

Options:
A 1
B 0
C 3
D 2

The correct answer is:
1

Solution:
Given, $f(x)=x+\frac{1}{2}=\frac{2 x+1}{2}$
$\therefore$
$f(2 x)=2 x+\frac{1}{2}$
$\Rightarrow$
$f(2 x)=\frac{4 x+1}{2}$
and $f(4 x)=4 x+\frac{1}{2} \Rightarrow f(4 x)=\frac{8 x+1}{2}$
since, $f(x), f(2 x)$ and $f(4 x)$ are in $\mathrm{HP}$. $\frac{1}{f(x)} \cdot \frac{1}{f(2 x)}$ and $\frac{1}{f(4 x)}$ are in AP.
$\Rightarrow \quad \frac{1}{f(2 x)}=\frac{\frac{1}{f(x)}+\frac{1}{f(4 x)}}{2}$
$\Rightarrow \quad \frac{2}{4 x+1}=\frac{\frac{2}{2 x+1}+\frac{2}{8 x+1}}{2}$
$\Rightarrow \quad-\frac{2}{4 x+1}=\frac{10 x+2}{(2 x+1)(8 x+1)}$
$\Rightarrow(2 x+1)(8 x+1)=(5 x+1)(4 x+1)$
$\Rightarrow 16 x^{2}+10 x+1=20 x^{2}+9 x+1$
$\Rightarrow \quad 4 x^{2}-x=0 \Rightarrow x(4 x-1)=0$
$\Rightarrow \quad x=\frac{1}{4} \quad[\because x \neq 0$
Hence, one real value of $x$ for which the thrce unequal terms are in HP.

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2014
Let $f(x)=x^{13}+x^{11}+x^{9}+x^{7}+x^{5}+x^{3}+x+12$. Then

Options:
A $f(x)$ has 13 non-zero real roots
B $\mathrm{f}(\mathrm{x})$ has exactly one real root
C $\mathrm{f}(\mathrm{x})$ has exactly one pair of imaginary roots
D $f(x)$ has no real root

The correct answer is:
$\mathrm{f}(\mathrm{x})$ has exactly one real root

Solution:
Hint:
$f^{\prime}(x)>0 \quad \forall x \in R$
$f(x)=0$ has exactly one real root

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2020
Let $f(x)=x^{13}+x^{11}+x^{9}+x^{7}+x^{5}$
$+x^{3}+x+19 .$ Then, $f(x)=0$ has

Options:
A 13 real roots
B only one positive and only two negative real roots
C not more than one real root
D has two positive and one negative real root

The correct answer is:
not more than one real root

Solution:
We have, $f(x)=x^{13}+x^{11}+x^{9}+x^{7}+x^{5}+x^{3}+x+19$
$$
\begin{array}{r}
\Rightarrow f^{\prime}(x)=13 x^{12}+11 x^{10}+9 x^{4} \\
+7 x^{6}+5 x^{4}+3 x^{2}+1
\end{array}
$$
$f^{\prime}(x)$ has no real root.
$\therefore f(x)=0$ has not more than one real root.

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2017
Let $f: X \rightarrow X$ be such that $f[f(x)]=x,$ for all
$x \in X$ and $X \subseteq R,$ then

Options:
A $f$ is one-to-one
B $f$ is onto
C $f$ is one-to-one but not onto
D $f$ is onto but not one-to-one

The correct answers are:
$f$ is one-to-one, $f$ is onto

Solution:
Given, $f(f(x)]=x$
Now, $\quad f^{-1}(x)=f(x)$
Le. $f(x)$ is bijective.
Hence, $f(x)$ has to be one-one and onto.

Topic: Mathematics, Functions, WBJEE, WBJEE 2016
Let $f: X \rightarrow Y$ and $A, B$ are non-void subsets of $Y$, then (where the symbols have their usual interpretation)

Options:
A $f^{-1}(A)-f^{-1}(B) \supset f^{-1}(A-B)$ but the opposite does not hold.
B $f^{-1}(A)-f^{-1}(B) \subset f^{-1}(A-B)$ but the opposite does not hold.
C $f^{-1}(A-B)=f^{-1}(A)-f^{-1}(B)$
D $f^{-1}(A-B)=f^{-1}(A) \cup f^{-1}(B)$

The correct answer is:
$f^{-1}(A-B)=f^{-1}(A)-f^{-1}(B)$

Solution:
Let A={1,2,3}
and B={2,3,4}

here, $f^{-1}(3)=1$
$f^{-1}(4)=2$ and $f^{-1}(2)=3$
It is clear that option (C) $f^{-1}(A-B)=f^{-1}(A)$
$-f^{-1}(B)$ is correct.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2019
Let \(f:[1,3] \rightarrow \mathbb{R}\) be continuous and be derivable in \((1,3)\) and \(f^{\prime}(x)=[f(x)]^2+4 \forall x \in(1,3)\). Then

Options:
A \(f(3)-f(1)=5\) holds
B \(f(3)-f(1)=5\) does not hold
C \(f(3)-f(1)=3\) holds
D \(f(3)-f(1)=4\) holds

The correct answer is:
\(f(3)-f(1)=5\) does not hold

Solution:
\(\begin{aligned}
& \text { Hint : } f^{\prime}(c)=(f(c))^2+4=\frac{f(3)-f(1)}{2} \\
& \Rightarrow f(3)-f(1)=2(f(c))^2+8
\end{aligned}\)

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2023
Let for all $x>0, f(x)=\lim _{n \rightarrow \infty} n\left(x^{1 / n}-1\right),$ then

Options:
A $f(x)+f\left(\frac{1}{x}\right)=1$
B $f(x y)=f(x)+f(y)$
C $f(x y)=x f(y)+y f(x)$
D $f(x y)=x f(x)+y f(y)$

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

Solution:
We have, $f(x)=\lim _{n \rightarrow \infty} n\left(x^{1 / n}-1\right)$
$=\lim _{n \rightarrow \infty} \frac{x^{1 / n}-1}{1 / n}$
Let $\quad \frac{1}{n}=y$
$\begin{aligned} \therefore \quad f(x)=& \lim _{y \rightarrow 0} \frac{x^{y}-1}{y} \\ &=\log x \end{aligned}$
$\therefore \quad f(x y)=\log (x, y)$
$=\log x+\log y$
$=f(x)+f(y)$

Topic: Mathematics, Limits, WBJEE, WBJEE 2017
Let \(f(x)=\left[x^2\right] \sin \pi x, x > 0\). Then

Options:
A \(f\) is discontinuous everywhere.
B \(f\) is continuous everywhere.
C \(f\) is continuous at only those points which are perfect squares.
D None of these

The correct answer is:
None of these

Solution:
Hint: \(\underset{\downarrow \atop \large \text{discontinuous}}{\left[x^2\right]} \sin \pi x\)
at all points where \(x^2\) is integer
If \(x^2\) is integer and \(x\) is also integer then \(f(x)\) will be continuous, but if \(x^2\) is integer and \(x\) is not integer then \(f(x)\) will be discontinuous.

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2023
Let \(f(x)=x^3 e^{-3 x}, x > 0\). Then the maximum value of \(f(x)\) is

Options:
A \(\mathrm{e}^{-3}\)
B \(3 \mathrm{e}^{-3}\)
C \(27 \mathrm{e}^{-9}\)
D \(\infty\)

The correct answer is:
\(\mathrm{e}^{-3}\)

Solution:
\(\begin{aligned}
& \text { Hints: } f(x)=x^3 \cdot e^{-3 x} \\
& =f^{\prime}(x)=3 x^2 e^{-3 x}+x^3 e^{-3 x}(-3) \\
& =x^2 3 e^{-3 x}[1-x]=0, x=1 \\
& \text { Maximum at } x=1 \\
& f(1)=e^{-3}
\end{aligned}\)

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2011
Let \(f(x)=x^m, m\) being a non-negative integer. The value of \(m\) so that the equality \(f^{\prime}(a+b)=f^{\prime}(a)+f^{\prime}(b)\) is valid for all \(a\), \(b > 0\) is

Options:
A 0
B 1
C 2
D 3

The correct answers are:
0, 2

Solution:
Hint : Let \(f(x)=x^m, m \geq 0, m \in I\)
\(\begin{aligned}
& f^{\prime}(a+b)=f^{\prime}(a)+f^{\prime}(b), a, b > 0 \\
& \Rightarrow m(a+b)^{m-1}=m a^{m-1}+m \cdot b^{m-1} \\
& (a+b)^{m-1}=a^{m-1}+b^{m-1} \quad f^{\prime}(x)=m n^{m-1} \\
& \text { for } m=0 \Rightarrow 0=0 \\
& (a+b)^{-1}=a^{-1}+b^{-1}
\end{aligned}\)
Not satisfied
for \(m=1\)
\(\begin{aligned}
& (a+b)^0=a^{1-1}+b^{1-1} \\
& 1=2
\end{aligned}\)
Not satisfied
for \(m=2\)
\(\begin{aligned}
& (a+b)^{2-1}=a^{2-1}+b^{2-1} \\
& a+b=a+b \text { (satisfied) } \\
& \text { for } m=3 \\
& (a+b)^{3-1}=a^{3-1}+b^{3-1}
\end{aligned}\)
Not satisfied

Topic: Mathematics, Functions, WBJEE, WBJEE 2023
Let $g(x)=\int_{x}^{2 x} \frac{f(t)}{t} d t$ where $x>0$ and $f$ be continuous function and $f(2 x)=f(x)$, then

Options:
A $g(x)$ is strictly increasing function
B $\mathrm{g}(\mathrm{x})$ is strictly decreasing function
C $\mathrm{g}(\mathrm{x})$ is constant function
D $\mathrm{g}(\mathrm{x})$ is not derivable function

The correct answer is:
$\mathrm{g}(\mathrm{x})$ is constant function

Solution:
$g(x)=\int_{x}^{2 x} \frac{f(t)}{t} d t$
$g^{\prime}(x)=\frac{f(2 x)}{2 x} \cdot 2^{\prime}-\frac{f(x)}{x} \cdot 1=\frac{f(2 x)-f(x)}{x}=\frac{f(x)-f(x)}{x}[\because f(2 x)=f(x)]$
$g^{\prime}(x)=0$
$g(x)=$ constant.

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2021
Let $I=\int_{0}^{1} \frac{x^{3} \cos 3 x}{2+x^{2}} d x$. Then

Options:
A $-\frac{1}{2} < I < \frac{1}{2}$
B $-\frac{1}{3} < I < \frac{1}{3}$
C $-1 < I < 1$
D $-\frac{3}{2} < I < \frac{3}{2}$

The correct answers are:
$-\frac{1}{3} < I < \frac{1}{3}$

Solution:
$I=\int_{0}^{1} \frac{x^{3} \cos 3 x}{2+x^{2}} d x$
Here, $\quad-1 < \cos 3 x < 1$
$\Rightarrow \quad-x^{3} < x^{3} \cos 3 x < x^{3}$
$\Rightarrow \quad \frac{-x^{3}}{x^{2}} < \frac{-x^{3}}{x} < \frac{-x^{3}}{2+x^{2}} < \frac{x^{3} \cos 3 x}{2+x^{2}}$
$ < \frac{x^{3}}{2+x^{2}} < \frac{x^{3}}{x} < \frac{x^{3}}{x^{2}}$
$\Rightarrow \quad \int_{0}^{1}-x^{2} d x < I < \int_{0}^{1} x^{2} d x$
$\Rightarrow$
$\left(\frac{-x^{3}}{3}\right)_{0}^{1} < I < \left(\frac{x^{3}}{3}\right)_{0}^{1}$
$\frac{-1}{3} < I < \frac{1}{3}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2018
Let $I=\int_{0}^{{100 \pi }} \sqrt{(1-\cos 2 x)} d x$, then

Options:
A $I=0$
B $I=200 \sqrt{2}$
C $I=\pi \sqrt{2}$
D $I=100$

The correct answer is:
$I=200 \sqrt{2}$

Solution:
$I=\int_{0}^{1004} \sqrt{1-\cos 2 x} d x$
$=\int_{0}^{100 \pi} \sqrt{2 \sin ^{2} x} d x$
$=\sqrt{2} \int_{0}^{100 \pi} \sin x \mid d x$
$=\sqrt{2} \times 100 \int_{0}^{n} \sin x|d x|$
$[\sin x \mid$ has period of $\pi]$
$=100 \sqrt{2} \int_{0}^{\pi} sin x d x$
$=100 \sqrt{2}[-\cos x]_{0}^{\pi}$
$=100 \sqrt{2}\{(-\cos \pi)-(-\cos 0)]$
$=100 \sqrt{2}[-(-1)-(-1)]$
$=100 \sqrt{2} \times 2$
$=200 \sqrt{2}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2017
Let $I_{1}=\int_{0}^{n}[x] d x$ and $I_{2}=\int_{0}^{n}\{x\} d x,$ where $[x]$
and $\{x\}$ are integral and fractional parts of $x$ and $n \in N-\{1\} .$ Then, $I_{1} / I_{2}$ is equal to

Options:
A $\frac{1}{n-1}$
B $\frac{1}{n}$
C $n$
D $n-1$

The correct answer is:
$n-1$

Solution:
We have, $I_{1}=\int_{0}^{n}[x] d x$
$=\int_{0}^{1}[x] d x+\int_{1}^{2}[x] d x$
$\quad+\int_{2}^{3}[x] d x+\ldots+\int_{n-1}^{n}[x] d x$
$=\int_{0}^{1} 0 d x+\int_{2}^{3} 1 d x+\int_{2}^{3} 2 d x$
$\quad+\ldots+\int_{n-1}^{n}(n-1) d x$
$=0+[x]_{1}^{2}+2[x]_{2}^{3}+\ldots+(n-1)[x]_{n-1}^{n}$
$=(2-1)+2(3-2)+\ldots+$
$=\frac{n(n-1)}{2}$
$\begin{aligned} \text { Now, } I_{2} &=\int_{0}^{n}\{x\} d x=\int_{0}^{n} x-[x] d x \\ &=\int_{0}^{n} x d x-\int_{0}^{n}[x] d x \\ &=\left[\frac{x^{2}}{2}\right]_{0}^{n}-I_{1} \\ &=\frac{n^{2}}{2}-\frac{n(n-1)}{2} \\ &=\frac{n^{2}-n^{2}+n}{2} \\ &=\frac{n}{2} \\ \therefore \quad \frac{I_{1}}{I_{2}}=\frac{\frac{n(n-1)}{2}}{\frac{n}{2}} \\=&(n-1) \end{aligned}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2017
Let $I=\int_{\pi / 4}^{\pi / 3} \frac{\sin x}{x} d x$. Then

Options:
A $\frac{1}{2} \leq 1 \leq 1$
B $4 \leq 1 \leq 2 \sqrt{30}$
C $\frac{\sqrt{3}}{8} \leq 1 \leq \frac{\sqrt{2}}{6}$
D $1 \leq 1 \leq \frac{2 \sqrt{3}}{\sqrt{2}}$

The correct answer is:
$\frac{\sqrt{3}}{8} \leq 1 \leq \frac{\sqrt{2}}{6}$

Solution:
We have,
$$
I=\int_{\pi / 4}^{\pi / 3} \frac{\sin x}{x} d x
$$
since, $\frac{\sin x}{x}$ is a decreasing function.
$$
\begin{array}{c}
\frac{\pi}{12} \times \frac{\sin \pi / 3}{\pi / 3} \leq 1 \leq \frac{\pi}{12} \times \frac{\sin \pi / 4}{\pi / 4} \\
\frac{\sqrt{3}}{8} \leq I \leq \frac{\sqrt{2}}{6}
\end{array}
$$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2018
Let $\mathrm{I}=\int_{\pi / 4}^{\pi / 3} \frac{\sin \mathrm{x}}{\mathrm{x}} \mathrm{dx}$. Then

Options:
A $\frac{\sqrt{3}}{8} \leq \mathrm{I} \leq \frac{\sqrt{2}}{6}$
B $\frac{\sqrt{3}}{2 \pi} \leq \mathrm{I} \leq \frac{2 \sqrt{3}}{\pi}$
C $\frac{\sqrt{3}}{9} \leq \mathrm{I} \leq \frac{\sqrt{2}}{16}$
D $\pi \leq \mathrm{I} \leq \frac{4 \pi}{3}$

The correct answer is:
$\frac{\sqrt{3}}{8} \leq \mathrm{I} \leq \frac{\sqrt{2}}{6}$

Solution:
$f(x)=\frac{\sin x}{x}$ (decreasing function)
$\Rightarrow f\left(\frac{\pi}{3}\right) < f(x) < f\left(\frac{\pi}{4}\right)$
$\Rightarrow \frac{\sqrt{3}}{2} \times \frac{3}{\pi} \int_{\pi / 4}^{\pi / 3} d x < I < \frac{1}{\sqrt{2}} \times \frac{4}{\pi} \times \int_{\pi / 4}^{\pi / 3} d x$
$\Rightarrow \frac{\sqrt{3}}{8} < I < \frac{\sqrt{2}}{6}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2021