datavorous/entrance-exam-dataset huggingface dataset:
If $e^{\sin x}-e^{-\sin x}-4=0,$ then the number ofreal values of $x$ is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
$$
\text { Given that, } e^{\sin x}-e^{-\sin x}-4=0
$$
$$
\begin{aligned}
&\begin{array}{l}
\text { Let } e^{\sin x}=t \\
\therefore \quad t-t^{-1}-4=0 \\
\Rightarrow \quad t-\frac{1}{t}-4=0 \\
\Rightarrow \quad t^{2}-1-4 t=0 \\
\quad t=\frac{4 \pm 2 \sqrt{5}}{2}=2 \pm \sqrt{5} \\
\Rightarrow \quad e^{\sin x}=2 \pm \sqrt{5} \\
\Rightarrow \quad \quad \quad e^{\sin x}=2+\sqrt{5} \text { and } e^{\sin x}=2-\sqrt{5} \\
\Rightarrow \quad e^{\sin x}=2 + \sqrt{5} \\
\text { Also, }-1 \leq \sin x \leq 1 \\
\therefore \frac{1}{e} \leq e^{\sin x} \leq e \\
\therefore e^{\sin x} \neq 2+\sqrt{5}
\end{array}\\
&\text { Hence, there is no value of } x \text { . }
\end{aligned}
$$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2019
If electrolysis of aqueous CuSO $_{4}$ solution is carried out using Cu-electrodes, the reaction taking place at the anode is

Options:
A $\mathrm{H}^{+}+\mathrm{e}^{-} \rightarrow \mathrm{H}$
B $\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s})$
C $\mathrm{SO}_{4}^{2-}(\mathrm{aq})-2 \mathrm{e}^{-} \rightarrow \mathrm{SO}_{4}$
D $\mathrm{Cu}(\mathrm{s})-2 e^{-} \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})$

The correct answer is:
$\mathrm{Cu}(\mathrm{s})-2 e^{-} \rightarrow \mathrm{Cu}^{2+}(\mathrm{aq})$

Solution:
Electrolysis of copper sulphate solution using Cu-electrodes,
Ions present : $\mathrm{Cu}^{2+}, \mathrm{H}^{+}, \mathrm{SO}_{4}^{2-}, \mathrm{OH}^{-}$
At cathode $\mathrm{Cu}^{2+}(a q)+2 e^{-} \rightarrow \mathrm{Cu}(s)$
At anode $\mathrm{Cu}(s)-2 e^{-} \rightarrow \mathrm{Cu}^{2+}(a q)$
So, the option (d) is correct answer.

Topic: Chemistry, Electrochemistry, WBJEE, WBJEE 2019
If $f:[0, \pi / 2) \rightarrow R$ is defined as
$f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$ Then, the range of $f$ is

Options:
A $(2, \infty)$
B $(-\infty,-2]$
C $[2, \infty)$
D $(-\infty, 2]$

The correct answer is:
$(2, \infty)$

Solution:
We have, $f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$
$=1\left(1+\tan ^{2} \theta\right)-\tan \theta$
$(-\tan \theta+\tan \theta)+1\left(\tan ^{2} \theta+1\right)$
$=2\left(1+\tan ^{2} \theta\right)-0=2 \sec ^{2} \theta$
$\therefore$ Range of $f=(2, \infty)$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2015
If $f^{\prime \prime}(0)=k, k \neq 0,$ then the value of $\lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^{2}}$ is

Options:
A $k$
B $2 k$
C $3 k$
D $4 k$

The correct answer is:
$3 k$

Solution:
$\lim _{x \rightarrow 0} \frac{2 f(x)-3 f(2 x)+f(4 x)}{x^{2}}$
$=\lim _{x \rightarrow 0} \frac{2 f^{\prime}(x)-3 f^{\prime}(2 x) \cdot 2+f^{\prime}(4 x) \cdot 4}{2 x}$
$=\lim _{x \rightarrow 0} \frac{f^{\prime}(x)-3 f^{\prime}(2 x)+2 f^{\prime}(4 x)}{x}$
$=\lim _{x \rightarrow 0} \frac{f^{\prime \prime}(x)-3 f''{(2 x)} \cdot 2+2 f^{\prime \prime}(4 x) \cdot 4}{1}$
$=\lim _{x \rightarrow 0} f^{\prime \prime}(x)-6 f''{(2 x)}+8 f''(4 x)$
$=f^{\prime \prime}(0)-6 f^{\prime \prime}(0)+8 f''{(0)}$
$=k-6 k+8 k$
$=3 k$

Topic: Mathematics, Limits, WBJEE, WBJEE 2017
If $\mathrm{f}(5)=7$ and $\mathrm{f}^{\prime}(5)=7$ then $\underset{x \rightarrow 5}{ } \frac{x f(5)-5 f(x)}{x-5}$ is given by

Options:
A 35
B -35
C 28
D -28

The correct answer is:
-28

Solution:
Hints: $\operatorname{Lt}_{x \rightarrow 5} \frac{x f(5)-\operatorname{tf}(x)}{x-5}=\operatorname{Lt}_{x \rightarrow 5} \frac{f(5)-5 f^{\prime}(x)}{1}=f(5)-5 f^{\prime}(5)=7-5 \times 7=-28$

Topic: Mathematics, Limits, WBJEE, WBJEE 2010
If $f$ is a real-valued differentiable function such that $f(x) f^{\prime}(x) < 0$ for all real $x,$ then

Options:
A $f(x)$ must be an increasing function
B $f(x)$ must be a decreasing function
C $|f(x)|$ must be an increasing function
D $|f(x)|$ must be a decreasing function

The correct answer is:
$|f(x)|$ must be a decreasing function

Solution:
Given, $f(x) f^{\prime}(x) < 0$ $\Rightarrow f(x)$ and $f^{\prime}(x)$ must be of opposite sign.
(i) Let $f(x)=e^{-x}$
$\therefore \quad f^{\prime}(x)=-e^{-x}$
$\Rightarrow f(x)>0$ and $f^{\prime}(x) < 0, \forall x \in R$
(ii) Let $f(x)=-e^{-x}$
$\therefore \quad f^{\prime}(x)=e^{-x}$
$\Rightarrow f(x) < 0$ and $f^{\prime}(x)>0, \forall x \in R$
But $|f(x)|=\left|\pm e^{-x}\right|=e^{-x}$ in both cases
$\therefore \frac{d}{d x}|f(x)|=-e^{-x} < 0$ in both case, $\forall x \in R$
$\Rightarrow|f(x)|$ must be a decreasing function

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2012
If $f: R \rightarrow R$ be defined by $f(x)=e^{x}$ and $g: R \rightarrow R$ be defined by $g(x)=x^{2}$. The mapping gof : $R \rightarrow R$ be defined by (gof) $(x)$ $=g f(x) \mid \forall x \in R .$ Then.

Options:
A gof is bijective but $f$ is not injective
B gof is injective and $g$ is injective
C gof is injective but $g$ is not bijective
D gof is surjective and $g$ is surjective

The correct answer is:
gof is injective but $g$ is not bijective

Solution:
We have, $f: R \rightarrow R$, defined by $f(x)=e^{x}$ and $g: R \rightarrow R$ defined by $g(x)=x^{2}$
Now, We have
$$
\begin{aligned}
(g v f)(x) &=g(f(x)) \\
&=g\left(e^{x}\right) \\
&=\left(e^{x}\right)^{2} \\
&=e^{2 x}, \forall x \in R
\end{aligned}
$$
$\Rightarrow$ gof is injective and $g$ is neither injective nor surjective. $\Rightarrow g o f$ is injective but $g(x)$ is not bijective.

Topic: Mathematics, Functions, WBJEE, WBJEE 2018
If $\mathrm{f}: \mathrm{S} \rightarrow \mathbb{R}$ where $S$ is the set of all non-singular matrices of order 2 over $\mathbb{R}$ and $f\left[\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\right]=a d-b c$, then

Options:
A $\mathrm{f}$ is bijective mapping
B $\mathrm{f}$ is one-one but not onto
C $\mathrm{f}$ is onto but not one-one
D $\mathrm{f}$ is neither one-one nor onto

The correct answer is:
$\mathrm{f}$ is neither one-one nor onto

Solution:
Hint :
$\mathrm{f}\left[\left(\begin{array}{ll}2 & 0 \\ 0 & 2\end{array}\right)\right]=4=\mathrm{f}\left[\left(\begin{array}{ll}4 & 0 \\ 0 & 1\end{array}\right)\right]$
$\Rightarrow$ not one-one
As $0 \in \mathbb{R}$ but S does not contain any singular matrix so, $f$ is not onto

Topic: Mathematics, Functions, WBJEE, WBJEE 2020
If $F(x)=\int_{0}^{x} \frac{\cos t}{\left(1+t^{2}\right)} d t, 0 \leq x \leq 2 \pi .$ Then

Options:
A $F$ is increasing in $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$ and decreasing in $\left(0, \frac{\pi}{2}\right)$ and $\left(\frac{3 \pi}{2}, 2 \pi\right)$
B $F$ is increasing in $(0, \pi)$ and decreasing in $(\pi, 2 \pi)$
C $F$ is increasing $(\pi, 2 \pi)$ and decreasing in $(0, \pi)$
D $F$ is increasing in $\left(0, \frac{\pi}{2}\right)$ and $\left(\frac{3 \pi}{2}, 2 \pi\right)$ and decreasing in $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$

The correct answer is:
$F$ is increasing in $\left(0, \frac{\pi}{2}\right)$ and $\left(\frac{3 \pi}{2}, 2 \pi\right)$ and decreasing in $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$

Solution:
Given function is
$$
F(x)=\int_{0}^{x} \frac{\cos t}{\left(1+t^{2}\right)} d t, 0 \leq x \leq 2 \pi
$$
On differentiation w.r.t. $x$. (apply Leibnitz rule) $F(x)=\frac{\cos x}{1+x^{2}} \times 1=\frac{\cos x}{1+x^{2}},$ where $\left(1+x^{2}\right)>0$
Here, $\cos x>0 \Rightarrow x \in\left(0, \frac{\pi}{2}\right) \cup\left(\frac{3 \pi}{2}, 2 \pi\right)$
and $\cos x < 0 \Rightarrow x \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$
So, $F$ is increasing in $\left(0, \frac{\pi}{2}\right)$ and $\left(\frac{3 \pi}{2}, 2 \pi\right)$ and decreasing in $\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2013
$$
\text { If } f(x)=\begin{array}{|ccc|}
1 & x & x+1\\
2 x & x(x-1) & (x+1)x\\
3 x(x-1) & x(x-1)(x-2) & (x+1)x(x-1)
\end{array}
$$ Then, $f(100)$ is equal to

Options:
A 0
B 1
C 100
D 10

The correct answer is:
0

Solution:
We have, $f(x)= \begin{array}{|ccc|}1 & x &x+1 \\ 2 x & x(x-1) & (x+1)x\\ 3 x(x-1) & x(x-1)(x-2) &(x+1)x(x-1)\end{array}$
Taking common $x,(x-1)$ from $R_{2}$ and $R_{3}$ respectively, we ge $f(x)=x(x-1)\left|\begin{array}{ccc}1 & x & x+1 \\ 2 & x-1 & x+1 \\ 3 x & x(x-2) & (x+1) x\end{array}\right|$
Taking commen $(x+1)$ from $C_{2}$, we get
$f(x)=x(x-1)(x+1)\left|\begin{array}{ccc}1 & x & 1 \\ 2 & x-1 & 1 \\ 3 x & x(x-2) & x\end{array}\right|$
Taking common $x$ from $R_{3}$, we get $f(x)=x^{2}(x-1)(x+1)\left|\begin{array}{ccc}1 & x & 1 \\ 2 & x-1 & 1 \\ 3 & x-2 & 1\end{array}\right|$
Applying $R_{1} \rightarrow R_{1}-R_{2}, R_{2}$ we get
$f(x)=x^2(x-1)(x+1)\left|\begin{array}{ccc}-1 & 1 & 0 \\ -1 & 1 & 0 \\ 3 & (x-2) & 1\end{array}\right|$
$\Rightarrow f(100)=0$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2015
If $f(x)=\int_{-1}^{x}|t|$ dt, then for any $x \geq 0, f(x)$ is equal to

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

The correct answer is:
$\frac{1}{2}\left(1+x^{2}\right)$

Solution:
We have,
$\begin{aligned} f(x) &=\int_{-1}^{x} t\left|d t=\int_{-1}^{0}\right| t\left|d t+\int_{0}^{x}\right| t \mid d t \\ &=-\int_{-1}^{0} t d t+\int_{0}^{x} t d t \\ &=\left[-\frac{t^{2}}{2}\right]_{-1}^{0}+\left[\frac{t^{2}}{2}\right]_{0}^{x} \\ &=-\frac{1}{2}\left(0^{2}-(-1)^{2}\right)+\frac{1}{2}\left(x^{2}-0^{2}\right) \\ &=-\frac{1}{2}(-1)+\frac{1}{2}\left(x^{2}\right)=\frac{1}{2}\left(1+x^{2}\right) \end{aligned}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2017
If $f(x)=2^{100} x+1, g(x)=3^{100} x+1$, then the set of real numbers $x$ such that $f[g(x)\}=x$ is

Options:
A emply
B a singleton
C a finite set with more than one element
D infinite

The correct answer is:
a singleton

Solution:
Given, $\quad f(x)=2^{100} \cdot x+1$
$g(x)=3^{300} \cdot x+1$
Now, $f\{g(x)\}=x$
$\Rightarrow \quad f\left(3^{100}. x+1\right)=x$
$\Rightarrow \quad 2^{100}\left\{3^{100} \cdot x+1\right\}+1=x$
$\Rightarrow \quad 6^{100} \cdot x+2^{100}+1=x$
$\Rightarrow \quad x\left(1-6^{100}\right)=\left(1+2^{100}\right)$
$\Rightarrow \quad x=\frac{1+2^{100}}{1-6^{100}}$
Hence, fog $(x)=x$ represent a singleton set.

Topic: Mathematics, Functions, WBJEE, WBJEE 2013
If $f(x)=\left\{\begin{array}{ll}2 x^{2}+1, & x \leq 1 \\ 4 x^{3}-1, & x>1\end{array},\right.$ then $\int_{0}^{2} f(x) d x$ is

Options:
A $47 / 3$
B $50 / 3$
C $1 / 3$
D $47 / 2$

The correct answer is:
$47 / 3$

Solution:
Given, $f(x)=\left\{\begin{array}{ll}2 x^{2}+1, & x \leq 1 \\ 4 x^{3}-1, & x>1\end{array}\right.$
$\begin{aligned} \therefore \int_{0}^{2} f(x) d x &=\int_{0}^{1} f(x) d x+\int_{1}^{2} f(x) d x \\ &=\int_{0}^{1}\left(2 x^{2}+1\right) d x+\int_{1}^{2}\left(4 x^{3}-1\right) d x \\ &=\left[\frac{2 x^{3}}{3}+x\right]_{0}^{1}+\left[\frac{4 x^{4}}{4}-x\right]_{1}^{2} \\ &=\frac{2}{3}(1)^{3}+1-(0+0) \\ &+\left[(2)^{4}-2-\left\{(1)^{4}-1\right)\right.\end{aligned}$
$$
\begin{array}{l}
=\frac{2}{3}+1+[16-2-0] \\
=\frac{2}{3}+15=\frac{2+45}{3} \\
=\frac{47}{3} \mathrm{sq} \text { units }
\end{array}
$$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2014
If $f(x)$ and $g(x)$ are twice differentiable. functions on (0,3)
satisfying $f^{\prime \prime}(x)=g^{\prime \prime}(x), f^{\prime}(1)=4, g^{\prime}(1)=6, \quad f(2)=3$
$g(2)=9,$ then $f(1)-g(1)$ is

Options:
A 4
B -4
C 0
D -2

The correct answer is:
-4

Solution:
According to question, $f^{\prime \prime}(x)=g^{\prime \prime}(x)$
Integrating w.r.t. $x$, we get $\begin{array}{ll} & f^{\prime}(x)=g^{\prime}(x)+c_{1} \\ \text { Put } & x=1 \Rightarrow f^{\prime}(1)=g^{\prime}(1)+C_{1} \\ \Rightarrow & 4=6+C_{1} \\ \therefore & C_{1}=-2 \\ \therefore \quad & f^{\prime}(x)=g^{\prime}(x)-2\end{array}$
Again, integrating w.r.t. $x$, we get $f(x)=g(x)-2 x+C_{2}$
$\Rightarrow \quad 3=9-4+C_{2} \Rightarrow C_{2}=-2$
$\therefore \quad f(x)=g(x)-2 x-2$
Put $x=1,$ we get $f(1)-g(1)=-2(1)-2=-4$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2012
If $f(x)=e^{x}(x-2)^{2},$ then

Options:
A $f$ is increasing in $(-\infty, 0)$ and $(2, \infty)$ and decreasing $\operatorname{lin}(0,2)$
B $t$ is increasing in $(-\infty, 0)$ and decreasing in $(0, \infty)$
C $f$ is increasing in $(2, \infty)$ and decreasing in $(-\infty, 0)$
D $f$ is increasing in (0,2) and decreasing in $(-\infty, 0)$ and $(2, \infty)$

The correct answer is:
$f$ is increasing in $(-\infty, 0)$ and $(2, \infty)$ and decreasing $\operatorname{lin}(0,2)$

Solution:
Given function is, $f(x)=e^{x}(x-2)^{2}$
$$
\begin{aligned}
\Rightarrow f^{\prime}(x) &=e^{x}(x-2)^{2}+2(x-2) e^{x} \\
&=e^{x}(x-2)(x-2+2)=x(x-2) e^{x}
\end{aligned}
$$
Now, sign scheme of $f^{\prime}(x)$ is

So, $t$ is increasing in $(-\infty, 0)$ and $(2, \infty)$ and decreasing in (0,2)

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2013
If $f(x)=f(a-x)$ then $\int_0^a x f(x) d x$ is equal to

Options:
A $\int_0^a f(x) d x$
B $\frac{a^2}{2} \int_0^{\mathrm{a}} \mathrm{f}(\mathrm{x}) \mathrm{dx}$
C $\frac{a}{2} \int_0^a f(x) d x$
D $-\frac{a}{2} \int_0^{\mathrm{a}} f(x) d x$

The correct answer is:
$\frac{a}{2} \int_0^a f(x) d x$

Solution:
Hints : $f(x)=f(a-x), I=\int_0^a x f(x) d x=\int_0^a(a-x) f(a-x) d x$
$$
\begin{aligned}
& =\int_0^a(a-x) f(x) d x=a \int_0^a f(x) d x-I \\
& \therefore 2 I=a \int_0^a f(x) d x \Rightarrow I=\frac{a}{2} \int_0^a f(x) d x
\end{aligned}
$$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2009
If $f(x)$ is a function such that $f^{\prime}(x)=(x-1)^{2}(4-x),$ then

Options:
A $f(0)=0$
B $f(x)$ is increasing in (0,3)
C $x=4$ is a critical point of $f(x)$
D $f(x)$ is decreasing in (3,5)

The correct answers are:
$f(x)$ is increasing in (0,3), $x=4$ is a critical point of $f(x)$

Solution:
Given
$$
f^{\prime}(x)=(x-1)^{2}(4-x)
$$
The sign scheme of $f^{\prime}(x)$

Clearly. $f(x)$ is increasing, for $x \in(-\infty, 4)$ and decreasing, for $n \in(4, \infty)$
Since, $f^{\prime}(x)=0$ at $x=4$
$\mathrm{SO}, \mathrm{x}=4$ is a critical point.

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2016
If $f(x)$ is an odd differentiable function defined on $(-\infty, \infty)$ such that $f^{\prime}(3)=2,$ then $f^{\prime}(-3)$ is equal to

Options:
A 0
B 1
C 2
D 4

The correct answer is:
2

Solution:
Given that $f(x)$ is an odd differentiable function.
Then, $f(-x)=-f(x)$
$\begin{array}{lr}\Rightarrow & -f^{\prime}(-x)=-f^{\prime}(x) \\ \Rightarrow & f^{\prime}(-x)=f^{\prime}(x)\end{array}$
Put $x=3$ in Eq. (i), we get
$\therefore$
$$
\begin{array}{l}
f^{\prime}(-3)=f^{\prime}(3) \\
f^{\prime}(-3)=2
\end{array}
$$

Topic: Mathematics, Functions, WBJEE, WBJEE 2016
If $f(x)=\log _{5} \log _{3} x,$ then $f^{\prime}(e)$ is equal to

Options:
A $e \log _{e} 5$
B $e \log _{4} 3$
C $\frac{1}{e \log _{e} 5}$
D $\frac{1}{e \log _{e} 3}$

The correct answer is:
$\frac{1}{e \log _{e} 5}$

Solution:
We have,
$\therefore$
$\begin{aligned} f(x) &=\log _{5} \log _{3} x \\ &=\frac{\log \log _{3} x}{\log 5} \\ &=\frac{\log \left(\frac{\log x}{\log 3}\right)}{\log 5} \\ &=\frac{\log \log x-\log \log 3}{\log 5} \\ f^{\prime}(x)=\frac{1}{\log 5} \cdot \frac{1}{\log x} \cdot \frac{1}{x} \end{aligned}$
$\begin{aligned} f^{\prime}(e) &=\frac{1}{e \log 5 \cdot \log e} \\ &=\frac{1}{e \log 5} \quad[\because \log e=1] \\ &=\frac{1}{e \log _{n} 5} \end{aligned}$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2017
If $f(x)=\sin x+2 \cos ^{2} x, \frac{\pi}{4} \leq x \leq \frac{3 \pi}{4} .$ Then, $f$ attains its

Options:
A minimum at $x=\frac{\pi}{4}$
B maximum at $x=\frac{\pi}{2}$
C minimum $x=\frac{\pi}{2}$
D maximum at $x=\sin ^{-1}\left(\frac{1}{4}\right)$

The correct answer is:
minimum $x=\frac{\pi}{2}$

Solution:
Given, $f(x)=\sin x+2 \cos ^{2} x, x \in\left[\frac{\pi}{4}, \frac{3 \pi}{4}\right]$
$\therefore \quad f^{\prime}(x)=\cos x-4 \cos x \cdot \sin x$
and $f^{\prime}(x)=-\sin x-4 \cos 2 x$
For maximum or minimum of $f(x)$
Put $\quad f^{\prime}(x)=0$
$\Rightarrow \quad \cos x-4 \cos x \cdot \sin x=0$
$\Rightarrow \quad \cos x(1-4 \sin x)=0$
$\Rightarrow \quad \cos x=0=\cos \frac{\pi}{2}$ and $\sin x \neq \frac{1}{4}$
$\because x \in\left[\frac{\pi}{4}, \frac{3 \pi}{4}\right] \Rightarrow x=\frac{\pi}{2}$
Now, $\quad f^{''}\left(\frac{\pi}{2}\right)=-\sin \frac{\pi}{2}-4 \cos \pi$
$$
=-1+4=3>0(\mathrm{min})
$$
So, $f(x)$ is minimum at $x=\frac{\pi}{2}$
and its minimum value is $f\left(\frac{\pi}{2}\right)=\sin \frac{\pi}{2}+2 \cos ^{2} \frac{\pi}{2}=1-2 \times 0=1$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2013
If $\int f(x) \sin x \cos x d x=\frac{1}{2\left(b^{2}-a^{2}\right)} \log f(x)+c$
where $c$ is the constant of integration, then $f(x)$ is equal to

Options:
A $\frac{2}{\left(b^{2}-a^{2}\right) \sin 2 x}$
B $\frac{2}{a b \sin 2 x}$
C $\frac{2}{\left(b^{2}-a^{2}\right) \cos 2 x}$
D $\frac{2}{a b \cos 2 x}$

The correct answer is:
$\frac{2}{\left(b^{2}-a^{2}\right) \cos 2 x}$

Solution:
We have. $\int f(x) \sin x \cos x d x=\frac{1}{2 b^{2}-a^{2}} \log (f(x))+c$
$\Rightarrow f(x) \sin x \cos x=\frac{1}{\left.2 b^{2}-a^{2}\right)} \cdot \frac{1}{f(x)} \cdot f'(x)$
$\Rightarrow \quad f(x) \sin 2 x=\frac{1}{b^{2}-a^{2}} \cdot \frac{f'(x)}{f(x)}$
$\Rightarrow \quad \sin 2 x=\frac{1}{b^{2}-a^{2}} \frac{f(x)}{(f(x))^{2}}$
$\Rightarrow \quad \int \sin 2 x d x=\frac{1}{b^{2}-a^{2}} \int \frac{f(x)}{(f(x))^{2}} d x$
$\Rightarrow \quad \frac{-\cos 2 x}{2}=\frac{1}{b^{2}-a^{2}} \cdot\left(\frac{-1}{f(x)}\right)$
$\Rightarrow \quad \quad \frac{\cos 2 x\left(b^{2}-a^{2}\right)}{2}=\frac{1}{f(x)}$
$\Rightarrow \quad f(x)=\frac{2}{\left(b^{2}-a^{2}\right) \cos 2 x}$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2018
If $f(x)=\tan ^{-1}\left[\frac{\log \left(\frac{e}{x^{2}}\right)}{\log \left(e x^{2}\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right]$ then the value of $f^{\prime \prime}(x)$ is equal to

Options:
A $x^{2}$
B $x$
C 1
D 0

The correct answer is:
0

Solution:
Given, $f(x)=\tan ^{-1}\left[\frac{\log \left(\frac{e}{x^{2}}\right)}{\log \left(e x^{2}\right)}\right]+\tan ^{-1}\left[\frac{3+2 \log x}{1-6 \log x}\right]$
$=\tan ^{-1}\left[\frac{\log e-2 \log x}{\log e+2 \log x}\right]+\tan ^{-1}\left[\frac{3+2 \log _{x} x}{1-6 \log x}\right]$
Let $2 \log x=\tan \theta,$ we get
$=\tan ^{-1}\left[\frac{1-\tan \theta}{1+\tan \theta}\right]+\tan ^{-1}\left[\frac{3+\tan \theta}{1-3 \tan \theta}\right]$
$\left.=\tan ^{-1}\left[\tan \left(\frac{\pi}{4}-\theta\right)\right]+[\tan ^{-1} (\tan ((\tan ^{-1} 3)+\theta))\right]$
$=\frac{\pi}{4}-\theta+\tan ^{-1} 3+\theta=\frac{\pi}{4}+\tan ^{-1} 3$
$\therefore \quad f(x)=$ constant
$\therefore f^{\prime}(x)=0$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2016
If $f(x)=\left\{\begin{array}{ll}x^{3}-3 x+2, & x < 2 \\ x^{3}-6 x^{2}+9 x+2, & x \geq 2\end{array}\right.$ then

Options:
A $\lim _{x \rightarrow 2} f(x)$ does not exist
B $f$ is not continuous at $x=2$
C $f$ is continuous but not differentiable at $x=2$
D $t$ is continuous and differentiable at $x=2$

The correct answer is:
$f$ is continuous but not differentiable at $x=2$

Solution:
Given. $f(x)=\left\{\begin{aligned} x^{3}-3 x+2, & x < 2 \\ x^{3}-6 x^{2}+9 x+2, & x \geq 2 \end{aligned}\right.$
LHL
$$
\begin{array}{l}
=f(2-0)=\lim _{h \rightarrow 0}(2-h)^{3}-3(2-h)+2 \\
=(2)^{3}-6+2=8-6+2=4 \\
\begin{aligned}
LH L=f(2+0) &=\lim _{h \rightarrow 0}(2+h)^{3}-6(2+h)^{2} \\
&=(2)^{3}-6(2)^{2}+9(2)+2 \\
&=8-24+18+2=4
\end{aligned}
\end{array}
$$
$\because \mathrm{LHL}=\mathrm{RHL}$
$\therefore \lim _{x \rightarrow 2} f(x)$ exist
and $\begin{aligned} f(2)=(2)^{3} &-6(2)^{2}+9(2)+2 \\ &=8-24+18+2=4 \end{aligned}$
$\therefore \quad \mathrm{LHL}=\mathrm{RHL}=f(2)$
So, $f(x)$ is continuous at $x=2$
Now, $\quad f^{\prime}(x)=\left\{\begin{array}{cc}3 x^{2}-3, & x < 2 \\ 3 x^{2}-12 x+9, & x \geq 2\end{array}\right.$
$\begin{array}{ll}\therefore \quad & L f^{\prime}(2)=3(2)^{2}-3=12-3=9 \\ \text { and } \quad & R f^{\prime}(2)=3(2)^{2}-12(2)+9 \\ & =12-24+9=-3\end{array}$
$\because \quad L f^{\prime}(2) \neq R f^{\prime}(2)$
$\therefore f(x)$ is not differentiable at $x=2$
Hence. $f$ is continuous but not differentiable at $x=2$

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2013
If $f(x)=x^{2 / 3}, x \geq 0 .$ Then, the area of the region enclosed by the curve $y=f(x)$ and the three lines $y=x, x=1$ and $x=8$ is

Options:
A $\frac{63}{2}$
B $\frac{93}{5}$
C $\frac{105}{7}$
D $\frac{129}{10}$

The correct answer is:
$\frac{129}{10}$

Solution:
Given, $f(x)=x^{2 / 3}, x \geq 0$ and line $y=x$

$\therefore$ Required area $A=\int_{x=1}^{8}\left(x-x^{2 / 3}\right) d x$
$=\left[\frac{x^{2}}{2}-\frac{3}{5} x^{5 / 3}\right]_{1}^{9}=\left(32-\frac{3}{5} \times 32\right)-\left(\frac{1}{2}-\frac{3}{5}\right)$
$=32 \times \frac{2}{5}-\frac{(5-6)}{10}=\frac{64}{5}+\frac{1}{10}$
$=\frac{128+1}{10}=\frac{129}{10}$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2013
If $f(x)=x^{n}, n$ being a non-negative integer, then the values of $n$ for which
$f^{\prime}(\alpha+\beta)=f^{\prime}(\alpha)+f^{\prime}(\beta)$ for all $\alpha, \beta>0$ is

Options:
A 1
B 2
C 0
D 5

The correct answers are:
2, 0

Solution:
We have,
$\begin{aligned} f(x) &=x^{n} \\ f^{\prime}(x) &=n x^{n-1} \end{aligned}$
Now, $f^{\prime}(\alpha+\beta)=f^{\prime}(\alpha)+f^{\prime}(\beta)$
$\Rightarrow \quad n(\alpha+\beta)^{n-1}=n \alpha^{n-1}+n \beta^{n-1}$
$\Rightarrow(a+\beta)^{n-1}=a^{n-1}+\beta^{n-1}$
From options we see that $n=2$, satisfy the above equation.
$\therefore$
$$
n=2
$$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2017
If $f(x)=x\left(\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}\right), x>1$. Then

Options:
A $f(x) \leq 1$
B $1 < f(x) \leq 2$
C $2 < f(x) \leq 3$
D $f(x)>3$

The correct answer is:
$f(x)>3$

Solution:
Given function is
$$
\begin{aligned}
f(x) &=x\left\{\frac{1}{x-1}+\frac{1}{x}+\frac{1}{x+1}\right\}, x>1 \\
&=x\left\{\frac{2 x}{x^{2}-1}+\frac{1}{x}\right\}=\frac{2 x^{2}}{x^{2}-1}+1 \\
&=\left\{\frac{2}{\left(1-\frac{1}{x^{2}}\right)}+1\right\}>3
\end{aligned}
$$
$\therefore \quad f(x)>3$

Topic: Mathematics, Functions, WBJEE, WBJEE 2013
If four distinct points $(2 k, 3 k),(2,0),(0,3)$ (0,0) lie on a circle, then

Options:
A $k < 0$
B $0 < k < 1$
C $k=1$
D $k>1$

The correct answer is:
$k=1$

Solution:
Since, join of (2,0) and (0,3) subtends $90^{\circ}$ at (0,0)
$\Rightarrow$ It is a diameter.

$\therefore$ Equation is $(x-2)(x-0)+(y-0)(y-3)=0$
$x^{2}+y^{2}-2 x-3 y=0$
$(2 k, 3 k)$ lies on it $\Rightarrow \quad 4 k^{2}+9 k^{2}-4 k-9 k=0$
$\Rightarrow$
. $13 k^{2}=13 k$
$\Rightarrow$
$k=1$
since, $k \neq 0$ otherwise $(2 k, 3 k)$ will be (0,0)

Topic: Mathematics, Circle, WBJEE, WBJEE 2012
If \(\frac{1}{6} \sin \theta, \cos \theta, \tan \theta\) are in G.P., then the solution set of \(\theta\) is

Options:
A \(2 \mathrm{n} \pi \pm \frac{\pi}{6}\)
B \(2 \mathrm{n} \pi \pm \frac{\pi}{3}\)
C \(\mathrm{n} \pi+(-1)^{\mathrm{n}} \frac{\pi}{3}\)
D \(\mathrm{n} \pi+\frac{\pi}{3}\)

The correct answer is:
\(2 \mathrm{n} \pi \pm \frac{\pi}{3}\)

Solution:
\(\begin{aligned}
& \text {Hint : } 6 \cos ^3 \theta+\cos ^2 \theta-1=0 \\
& \Rightarrow \quad(2 \cos \theta-1)\left(3 \cos ^2 \theta+2 \cos \theta+1\right)=0 \\
& \Rightarrow \quad \cos \theta=\frac{1}{2}\left(3 \cos ^2 \theta+2 \cos \theta+1 \neq 0\right)
\end{aligned}\)

Topic: Mathematics, Trigonometric Equations, WBJEE, WBJEE 2023
If from a point $\mathrm{P}(\mathrm{a}, \mathrm{b}, \mathrm{c})$, perpendiculars $\mathrm{PA}$ and $\mathrm{PB}$ are drawn to $\mathrm{YZ}$ and $\mathrm{ZX}$ planes respectively, then the equation of the plane $\mathrm{OAB}$ is

Options:
A $b c x+$ cay $+a b z=0$
B bcx $+$ cay $-a b z=0$
C $\mathrm{bcx}-\mathrm{cay}+\mathrm{abz}=0$
D bcx - cay - abz = 0

The correct answer is:
bcx $+$ cay $-a b z=0$

Solution:
$\mathrm{A}(\mathrm{o}, \mathrm{b}, \mathrm{c}) \mathrm{B}(\mathrm{a}, \mathrm{o}, \mathrm{c}), \mathrm{O}(\mathrm{o}, \mathrm{o}, \mathrm{o})$ Perpendicular vector is
$\overrightarrow{\mathrm{OA}} \times \overrightarrow{\mathrm{OB}}$
$\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{j} & \hat{k} \\ 0 & b & c \\ a & 0 & c\end{array}\right|=b c \hat{i}+a c \hat{j}-a b \hat{k}$
$\therefore$ Equation of plane is $b c x+a c y-a b z=0$

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2021