datavorous/entrance-exam-dataset huggingface dataset:
The value of $\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}$ is

Options:
A 0
B 2
C 3
D 1

The correct answer is:
2

Solution:
Hints: $\frac{\cot 54^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\cot 70^{\circ}}=\frac{\tan 36^{\circ}}{\tan 36^{\circ}}+\frac{\tan 20^{\circ}}{\tan 20^{\circ}}=1+1=2$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2010
The value of $\tan \frac{\pi}{5}+2 \tan \frac{2 \pi}{5}+4 \cot \frac{4 \pi}{5}$ is

Options:
A $\cot \frac{\pi}{5}$
B $\cot \frac{2 \pi}{5}$
C $\cot \frac{4 \pi}{5}$
D $\cot \frac{3 \pi}{5}$

The correct answer is:
$\cot \frac{\pi}{5}$

Solution:
$\tan \frac{\pi}{5}+2 \tan \frac{2 \pi}{5}+4 \cot \frac{4 \pi}{5}$
$=\frac{\sin \frac{\pi}{5}}{\cos \frac{\pi}{5}}+2 \frac{\sin \frac{2 \pi}{5}}{\cos \frac{2 \pi}{5}}+4 \frac{\cos \frac{4 \pi}{5}}{\sin \frac{4 \pi}{5}}$
$=\frac{\sin \frac{\pi}{5}}{\cos \frac{\pi}{5}}+2 \frac{\sin \frac{2 \pi}{5}}{\cos \frac{2 \pi}{5}}+\frac{4\left(\cos ^{2} \frac{2 \pi}{5}-\sin ^{2} \frac{2 \pi}{5}\right)}{2 \sin \frac{2 \pi}{5} \cos \frac{2 \pi}{5}}$
$=\frac{\sin \frac{\pi}{5}}{\cos \frac{\pi}{5}}+2 \frac{\sin \frac{2 \pi}{5}}{\cos \frac{2 \pi}{5}}+\frac{2\left(\cos ^{2} \frac{2 \pi}{5}-\sin ^{2} \frac{2 \pi}{5}\right)}{\sin \frac{2 \pi}{5} \cos \frac{2 \pi}{5}}$
$=\frac{\sin \frac{\pi}{5}}{\cos \frac{\pi}{5}}+2\left[\frac{\sin ^{2} \frac{2 \pi}{5}+\cos ^{2} \frac{2 \pi}{5}-\sin ^{2} \frac{2 \pi}{5}}{\cos \frac{2 \pi}{5} \sin \frac{2 \pi}{5}}\right]$
$=\frac{\sin \frac{\pi}{5}}{\cos \frac{\pi}{5}}+\frac{2 \cos \frac{2 \pi}{5}}{\sin \frac{2 \pi}{5}}$
$=\frac{\sin \frac{\pi}{5}}{\cos \frac{\pi}{5}}+\frac{2\left(\cos ^{2} \frac{\pi}{5}-\sin ^{2} \frac{\pi}{5}\right)}{2 \sin \frac{\pi}{5} \cos \frac{\pi}{5}}$
$=\frac{\sin ^{2} \frac{\pi}{5}+\cos ^{2} \frac{\pi}{5}-\frac{\sin ^{2} \pi}{5}}{\sin \frac{\pi}{5} \cos \frac{\pi}{5}}$
$=\frac{\cos ^{2} \frac{\pi}{5}}{\sin \frac{\pi}{5} \cos \frac{\pi}{5}}$
$=\cot \frac{\pi}{5}$

Topic: Mathematics, Trigonometric Equations, WBJEE, WBJEE 2014
The value of the infinite series $\frac{1^{2}+2^{2}}{3 !}$ $+\frac{1^{2}+2^{2}+3^{2}}{4 !}+\frac{1^{2}+2^{2}+3^{2}+4^{2}}{5 !}+\ldots$

Options:
A $e$
B $5 e$
C $\frac{5 e}{6}-\frac{1}{2}$
D $\frac{5 e}{6}$

The correct answer is:
$\frac{5 e}{6}-\frac{1}{2}$

Solution:
Given infinite series is $\frac{1^{2}+2^{2}}{3 !}+\frac{1^{2}+2^{2}+3^{2}}{4 !}$
$\begin{aligned} \frac{1^{2}+2^{2}}{3 !} &+\frac{1^{2}+2^{2}+3^{2}}{4 !} \\ &+\frac{1^{2}+2^{2}+3^{2}+4^{2}}{5 !}+\ldots \end{aligned}$
nth term, $t_{n}=\frac{1^{2}+2^{2}+3^{2}+4^{2}+\ldots+(r+1)^{2}}{(r+2) !}$
Now, $S_{n}=\Sigma t_{n}=\sum_{r=1}^{n} \frac{1^{2}+2^{2}+3^{2}+\ldots+(r+1)^{2}}{(r+2) !}$
$=\sum_{r=1}^{n} \frac{(r+1)(r+2)(2 n+3)}{6(r+2) !}$
$\left[\because \Sigma n^{2}=\frac{n(n+1)(2 n+1)}{6}\right]$
$=\sum_{r=1}^{n} \frac{(2 r+3)}{6 \cdot r !}=\frac{1}{6} \sum_{r=1}^{n}\left\{\frac{2}{(r-1) !}+\frac{3}{r !}\right\}$
$=\frac{1}{6}\left\{\left(\frac{2}{1}+\frac{3}{1 !}\right)+\left(\frac{2}{1 !}+\frac{3}{2 !}\right)+\left(\frac{2}{2 !}+\frac{3}{3 !}\right)+\ldots\right\}$
$=\frac{1}{6}\left\{\left(\frac{2}{1 !}+\frac{2}{1 !}+\frac{2}{2 !}+\ldots\right)+\left(\frac{3}{1 !}+\frac{3}{2 !}+\frac{3}{3 !}+\ldots\right)\right\}$
$=\frac{1}{6}\left\{2\left(1+\frac{1}{1 !}+\frac{1}{2 !}+\ldots\right)+3\left(\frac{1}{1 !}+\frac{1}{2 !}+\ldots\right)\right\}$
$=\frac{1}{6}\{2 e+3(e-1)\}$
$=\frac{1}{6}\{2 e+3 e-3\}$
$=\frac{1}{6}(5 e-3)=\frac{5 e}{6}-\frac{1}{2}$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2013
The value of the integral $\int_{0}^{1} e^{x^{2}} d x$

Options:
A is less than 1
B is greater than 1
C is less than or equal to 1
D lies in the closed interval $[1, e]$

The correct answer is:
lies in the closed interval $[1, e]$

Solution:
We know that, $0 \leq x \leq 1$ $\Rightarrow \quad x^{2} \leq 1$
$\begin{array}{ll}\Rightarrow & e^{x^{2}} \leq e \\ \Rightarrow & \int_{0}^{1} e^{x^{2}} d x \leq \int_{0}^{1} e d x \\ \Rightarrow & \int_{0}^{1} e^{x^{2}} d x \leq e \\ \therefore \quad \int_{0}^{-} e^{x^{2}} \in[1, e]\end{array}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2017
The value of the integral $\int_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{-101}} d x$ is equal to

Options:
A 1
B $\frac{\pi}{6}$
C $\frac{\pi}{8}$
D $\frac{\pi}{4}$

The correct answer is:
$\frac{\pi}{4}$

Solution:
Let $I=\int_{0}^{\pi / 2} \frac{1}{1+(\tan x)^{-101}} d x$
$=\int_{0}^{\pi / 2} \frac{d x}{1+\left\{\tan \left(\frac{\pi}{2}-x\right)\right\}^{101}}$
$=\int_{0}^{\pi / 2} \frac{d x}{1+(\cot x)^{101}}$
$=\int_{0}^{\pi / 2} \frac{\tan x^{10 t}}{\tan x^{101}+1} d x$
$=\int_{0}^{\pi / 2} \frac{1+\tan x^{101}-1}{1+\tan x^{101}}=[x]_{0}^{\pi / 2}-I$
$\Rightarrow \quad 1=\frac{\pi}{2}-I \Rightarrow 2 I=\frac{\pi}{2} \Rightarrow I=\frac{\pi}{4}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2012
The value of the integral $\int_0^{\pi / 2} \sin ^5 x d x$ is

Options:
A $\frac{4}{15}$
B $\frac{8}{5}$
C $\frac{8}{15}$
D $\frac{4}{5}$

The correct answer is:
$\frac{8}{15}$

Solution:
Hints : $I=\int_0^{\frac{\pi}{2}} \sin ^4 \mathrm{x} d \mathrm{dx} \quad \cos \mathrm{x}=\mathrm{f}, \sin \mathrm{d} \mathrm{x}=\mathrm{dt}$
$=-\int_1^0\left(1-t^2\right) d t=\int_0^2\left(t^4-2 t^2+1\right) d t$
$=\frac{1}{5}\left(\mathrm{t}^5\right)_0^1-\frac{2}{3}\left(\mathrm{t}^3\right)_0^1+(\mathrm{t})_0^1==\frac{1}{5}-\frac{2}{5}+1=\frac{3-10+15}{15}=\frac{8}{15}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2010
The value of the integral $\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$
is equal to

Options:
A $\log _{e} 2$
B $\log _{e} 3$
C $\frac{1}{4} \log _{e} 2$
D $\frac{1}{4} \log _{e} 3$

The correct answer is:
$\frac{1}{4} \log _{e} 3$

Solution:
Let $I=\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+\sin 2 x} d x$
$=\int_{0}^{\pi / 4} \frac{\sin x+\cos x}{3+2 \sin x \cos x} d x$
$=\int_{0}^{\pi / 4}-\frac{\sin x+\cos x}{(\sin x-\cos x)^{2}-4} d x$
Put $\sin x-\cos x=t$
$\Rightarrow \quad(\cos x+\sin x) d x=d t$
when $x=0 \Rightarrow t=-1$
and $x=\frac{\pi}{4} \Rightarrow t=0$
$\therefore$ Eq. (i) becomes, $I=-\int_{-1}^{0} \frac{d t}{t^{2}-4}=-\left.\frac{1}{4}\left[\log \mid \frac{t-2}{t+2}\right]\right|_{-1} ^{0}$
$=-\frac{1}{4}(\log 1-\log 3)$
$=-\frac{1}{4}(0-\log 3)=\frac{1}{4} \log 3$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2012
The value of the integral $\int_{-1}^{1}\left\{\frac{x^{2013}}{e^{|x|}\left(x^{2}+\cos x\right)}+\frac{1}{e^{|x|}}\right\} d x$ is equal to

Options:
A 0
B $1-e^{-1}$
C $2 e^{-1}$
D $2\left(1-e^{-1}\right)$

The correct answer is:
$2\left(1-e^{-1}\right)$

Solution:
Let $\quad I=\int_{-1}^{1}\left\{\frac{x^{2013}}{e^{\mid x\left|(x^{2}+\cos x\right)}}+\frac{1}{e^{|x|}}\right\} d x$
$\Rightarrow \quad I=\int_{-1}^{1} \frac{x^{2013}}{e^{|x|+\left(x^{2}+\cos x\right)}} d x+\int_{-1}^{1} \frac{1}{e^{|x|}} d x$
Here, $\frac{x^{2013}}{e^{|x|}\left(x^{2}+\cos x\right)}$ is an odd function
and $\frac{1}{e^{\mid x|}}$ is an even function. $\left\{\because \int_{-a}^{a} f(x) d x=\left\{\begin{array}{cc}2 \int_{0}^{9}(x) d x ; & f(x) \text { is even } \\ 0, & f(x) \text { is odd }\end{array}\right\}\right.$
$\left.\therefore \quad 1=0+2 \int_{0}^{1} e^{-x} d x=-2 (e^{-x})\right\}_{0}^{1}=-2\left(e^{-1}\right)$
$=2\left(1-e^{-1}\right)$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2013
The value of the integral $\int_{-1}^{1}\left\{\frac{x^{2015}}{e^{\mid x \mid}\left(x^{2}+\cos x\right)}+\frac{1}{e^{\mid{x} \mid}}\right\} d x$ is equal to

Options:
A 0
B $1-e^{-1}$
C $2 e^{-1}$
D $2\left(1-e^{-1}\right)$

The correct answer is:
$2\left(1-e^{-1}\right)$

Solution:
Let $I=\int_{-1}^{1}\left\{\frac{x^{2015}}{\left(e^{\mid x\mid} \left(x^{2}+\cos x\right)\right.}+\frac{1}{e^{|x|}}\right\} d x$
$\begin{aligned}=\int_{-1}^{1} \frac{x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)} & d x+\int_{-1}^{1} \frac{1}{e^{|x|}} d x \\ \text { Let } f(x)=\frac{x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)} \text { and } g(x)=\frac{1}{e^{|x|}} \\ \text { Again, } f(-x)=\frac{(-x)^{2015}}{e^{|-x|}\left((-x)^{2}+\cos (-x)\right)} \\ &=\frac{-x^{2015}}{e^{|x|}\left(x^{2}+\cos x\right)}=-f(x) \\ \text { and } \quad g(-x)=\frac{1}{e^{|-x|}}=\frac{1}{e^{|x|}}=g(x) \end{aligned}$
$\because f(x)$ is odd function and $g(x)$ is even function $\int_{-1}^{1} f(x) d x=0$ and $\int_{-1}^{1} g(x) d x=2 \int_{0}^{1} g(x) d x$
$\begin{aligned} I &=2 \int_{0}^{1} \frac{1}{e^{\mid x \mid}} d x \\ &=2 \int_{0}^{1} \frac{1}{e^{x}} d x=2 \int_{0}^{1} e^{-x} d x \\ &=-2\left[e^{-x}\right]_{0}^{1} \\ &=-2\left[e^{-1}-e^{0}\right] \\ &=-2\left(e^{-1}-1\right) \\ &=2\left(1-e^{-1}\right) \end{aligned}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2019
The value of the integral $\int_{1}^{2} e^{x}\left(\log _{e} x+\frac{x+1}{x}\right) d x$ is

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

The correct answer is:
$e^{2}\left(1+\log _{e} 2\right)-e$

Solution:
Let $I=\int_{1}^{2} e^{x}\left(\log _{e} x+\frac{x+1}{x}\right) d x$
$\Rightarrow \quad I=\int_{1}^{2}\left(e^{x} \cdot \log _{e} x+e^{x}+\frac{e^{x}}{x}\right) d x$
$\Rightarrow \quad I=\int_{1}^{2} e^{x} \log _{e} x d x+\int_{1}^{2} e^{x} d x+\int_{1}^{2} \frac{e^{x}}{x} d x$
$-\int_{1}^{2} e^{x} \log _{e} x d x$
$\Rightarrow \quad I=\left(e^{2}-e^{1}\right)+\left(e^{2} \log _{e} 2-0\right)$
$=e^{2}\left(1+\log _{e} 2\right)-e$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2013
The value of the integral $\int_{1}^{5}[|x-3|+|1-x|] d x$ is equal to

Options:
A 4
B 8
C 12
D 16

The correct answer is:
12

Solution:
$\int_{1}^{5}[|x-3|+|1-x|] d x$
$=\int_{1}^{5}|x-3| d x+\int_{1}^{5}|1-x| d x$
$=\int_{1}^{3}|x-3| d x+\int_{3}^{5}|x-3| d x+\int_{1}^{5}|1-x| d x$
$=\int_{1}^{3}-(x-3) d x+\int_{3}^{5}(x-3) d x$
$+\int_{1}^{5}-(1-x) d x$
$=\int_{1}^{3}(3-x) d x+\int_{3}^{5}(x-3) d x+\int_{1}^{5}(x-1) d x$
$=\left[3 x-\frac{x^{2}}{2}\right]^{3}+\left[\frac{x^{2}}{2}-3 x\right]_{3}^{5}+\left[\frac{x^{2}}{2}-x\right]_{0}^{5}$
$=\left(3 \times 3-\frac{9}{2}\right)-\left(3 \times 1-\frac{1}{2}\right)+\left(\frac{5 \times 5}{2}-3 \times 5\right)$
$\begin{aligned}-\left(\frac{3 \times 3}{2}-3 \times 3\right)+\left(\frac{5 \times 5}{2}-5\right)-\left(\frac{1}{2}-1\right) \\=\left(9-\frac{9}{2}\right)-\left(3-\frac{1}{2}\right)+\left(\frac{25}{2}-15\right)-\left(\frac{9}{2}-9\right) \\ &+\left(\frac{25}{2}-5\right)-\left(-\frac{1}{2}\right) \\=& \frac{9}{2}-\frac{5}{2}-\frac{5}{2}+\frac{9}{2}+\frac{15}{2}+\frac{1}{2} \\=& \frac{9-5-5+9+15+1}{2}=\frac{24}{2}=12 \end{aligned}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2012
The value of the integral $\int_{-2}^{2}(1+2 \sin x) e^{|x|} d x$ is
equal to

Options:
A 0
B $e^{2}-1$
C $2\left(e^{2}-1\right)$
D 1

The correct answer is:
$2\left(e^{2}-1\right)$

Solution:
$\int_{-2}^{2}(1+2 \sin x) e^{|x|} d x$
$=\int_{-2}^{2} e^{|x|} d x+2 \int_{-2}^{2} \sin x e^{|x|} d x$
(Since, first function is even and second function is odd). $=2 \int_{0}^{2} e^{|x|} d x+2 \text{ instead of ,} 0=2\left[e^{|x|}\right]_{0}^{2}=2\left(e^{2}-1\right)$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2012
The value of the integral $\int_{\pi / 6}^{\pi / 2}\left(\frac{1+\sin 2 x+\cos 2 x}{\sin x+\cos x}\right) d x$ is equal to

Options:
A 16
B 8
C 4
D 1

The correct answer is:
1

Solution:
Let $I=\int_{\pi / 6}^{\pi / 2}\left(\frac{1+\sin 2 x+\cos 2 x}{\sin x+\cos x}\right) d x$
$=\int_{\pi / 6}^{\pi / 2}\left(\frac{1+2 \sin x \cos x+2 \cos ^{2} x-1}{(\sin x+\cos x)}\right) d x$
$=\int_{\pi / 6}^{\pi / 2} \frac{2 \cos x(\sin x+\cos x)}{(\sin x+\cos x)} d x$
$=\int_{\pi / 6}^{\pi / 2} 2 \cos x d x=2[\sin x]_{\pi / 6}^{\pi / 2}$
$=2\left(\sin \frac{\pi}{2}-\sin \frac{\pi}{6}\right)=2\left(1-\frac{1}{2}\right)=2 \times \frac{1}{2}=1$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2012
The value of the integral $\int_{-1 / 2}^{1 / 2}\left\{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2\right\}^{1 / 2} d x$ is equal to

Options:
A $\log _{e}\left(\frac{4}{3}\right)$
B $4 \log _{e}\left(\frac{3}{4}\right)$
C $4 \log _{e}\left(\frac{4}{3}\right)$
D $\log _{e}\left(\frac{3}{4}\right)$

The correct answer is:
$4 \log _{e}\left(\frac{4}{3}\right)$

Solution:
$\int_{-1 / 2}^{1 / 2} \sqrt{\left(\frac{x+1}{x-1}\right)^{2}+\left(\frac{x-1}{x+1}\right)^{2}-2} d x=\int_{-1 / 2}^{y / 2}\left|\frac{x+1}{x-1}-\frac{x-1}{x+1}\right| d x$
$=\int_{-1 / 2}^{0} \frac{-4 x}{-(x-1)(x+1)} d x+\int_{0}^{1 / 2} \frac{4 x}{-(x-1)(x+1)} d x \int_{-\frac{1}{2}}^{0} \frac{4 x}{x^{2}-1}-\int_{0}^{\frac{1}{2}} \frac{4 x}{x^{2}-1} d x=4 \ln 4 / 3$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2021
The value of the integral $\int \frac{d x}{\left(e^x+e^{-x}\right)^2}$ is

Options:
A $\frac{1}{2}\left(\mathrm{e}^{2 \mathrm{x}}+1\right)+\mathrm{C}$
B $\frac{1}{2}\left(e^{-2 x}+1\right)+C$
C $-\frac{1}{2}\left(e^{2 x}+1\right)^{-1}+C$
D $\frac{1}{4}\left(e^{2 x}-1\right)+C$

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

Solution:
Hints: $\int \frac{e^{2 x} d x}{\left(e^{2 x}+1\right)^2} \quad e^x=t ; e^x d x=d t$
$$
=\frac{1}{2} \int \frac{2 \mathrm{tdt}}{\left(\mathrm{t}^2+1\right)^2}=\frac{1}{2}\left\{-\frac{1}{\left(t^2+1\right)}\right\}+c=-\frac{1}{2\left(\mathrm{e}^{2 x}+1\right)}+c
$$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2010
The value of the integral $I=\int_{1 / 2014}^{2014} \frac{\tan ^{-1} x}{x} d x$ is

Options:
A $\frac{\pi}{4} \log 2014$
B $\frac{\pi}{2} \log 2014$
C $\pi \log 2014$
D $\frac{1}{2} \log 2014$

The correct answer is:
$\frac{\pi}{2} \log 2014$

Solution:
We have,
$$
I=\int_{1 / 2014}^{2014} \frac{\tan ^{-1} x}{x} d x
$$
Let
$$
x=\frac{1}{t}
$$
$\Rightarrow$
$$
d x=\frac{-1}{t^{2}} d t
$$
Now,
$$
\begin{aligned}
I &=\int_{2014}^{1 / 2014} \frac{\tan ^{-1}(1 / 7)}{1 / t}\left(\frac{-1}{t^{2}} d t\right) \\
&=\int_{1 / 2014}^{2014} \frac{\cot ^{-1} t}{t} d t \\
&=\int_{1 / 2014}^{2014} \frac{\cot ^{-1} x}{x} d x
\end{aligned}
$$
On adding Eqs. (i) and (ii), we get
$$
\begin{aligned}
2 I &=\int_{1 / 2014}^{2014} \frac{\pi / 2}{x} d x=\frac{\pi}{2}(\log x)_{1 / 2014}^{2014} \\
&=\frac{\pi}{2}(\log 2014-\log 1 / 2014)
\end{aligned}
$$
$\Rightarrow$
$$
\begin{aligned}
I &=\frac{\pi}{4}\left(\log 2014-\log \frac{1}{2014}\right) \\
&=\frac{\pi}{4}(\log 2014+\log 2014) \\
&=\frac{\pi}{4}(2 \log 2014)=\frac{\pi}{2} \log 2014
\end{aligned}
$$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2018
The value of the integration $\int_{-\pi / 4}^{\pi / 4}\left(\lambda|\sin x|+\frac{\mu \sin x}{1+\cos x}+\gamma\right) d x$

Options:
A is independent of $\lambda$ only
B is independent of $\mu$ only
C is independent of $\gamma$ only
D depends on $\lambda, \mu$ and $\gamma$

The correct answer is:
is independent of $\mu$ only

Solution:
Let $I=\int_{-\frac{\pi}{4}}^{\pi / 4}\left(\lambda|\sin x|+\frac{\mu \sin x}{1+\cos x}+\gamma\right)$
$\begin{aligned} &=\int_{-\pi / 4}^{\pi / 4}(\lambda|\sin x|+\gamma) d x+\mu \int_{-\frac{\pi}{4}}^{\pi / 4} \frac{\sin x}{1+\cos x} d x \\ \text { Let } f(x)=\frac{\sin x}{1+\cos x} & \\ \Rightarrow f(-x)=\frac{\sin (-x)}{1+\cos (-x)} &=\frac{-\sin x}{1+\cos x}=-f(x) \end{aligned}$
$f(x)$ is an odd function.
$\int_{-\pi}^{\pi / 4}\left(\frac{\sin x}{1+\cos x}\right) d x=0$
$\left.I=\int_{-\frac{\pi}{6}}^{\pi / 4} \lambda|\sin x|+\gamma\right) d x$
$\therefore I$ is independent of $\mu$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2019
The value of the limit $\lim _{x \rightarrow 1} \frac{\sin \left(e^{x-1}-1\right)}{\log x}$ is

Options:
A 0
B $e$
C $\frac{1}{e}$
D 1

The correct answer is:
1

Solution:
$$
\text { Hints } \begin{aligned}
&: \operatorname{Lt}_{h \rightarrow 0} \frac{\sin \left(e^h-1\right)}{\log (1+h)} \quad \text { Put } x=1+h \\
&=\operatorname{Lt}_{h \rightarrow 0} \frac{\sin \left(e^h-1\right)}{\left(e^h-1\right)} \cdot \frac{\left(e^h-1\right)}{\log (1+h)} \\
&=\operatorname{Lt}_{h \rightarrow 0} \frac{\sin \left(e^h-1\right)}{\left(e^h-1\right)} \cdot \frac{\left(e^h-1\right)}{h} \cdot \frac{h}{\log (1+h)} \\
&=1.1 .1 \\
&=1
\end{aligned}
$$

Topic: Mathematics, Limits, WBJEE, WBJEE 2009
The value of the sum $\left({ }^{n} C_{1}\right)^{2}+\left({ }^{n} C_{2}\right)^{2}+\left({ }^{n} C_{3}\right)^{2}+\ldots+\left({ }^{n} C_{n}\right)^{2}$ is

Options:
A $\left({ }^{2 n} C_{n}\right)^{2}$
B ${ }^{2 n} C_{n}$
C ${ }^{2 n} \mathrm{C}_{n}+1$
D ${ }^{2 n} C_{n}-1$

The correct answer is:
${ }^{2 n} C_{n}-1$

Solution:
We know that
$$
\begin{array}{r}
(1+x)^{n}={ }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2} \\
\quad+\ldots+{ }^{n} C_{n} x^{n} \\
\text { and }(x+1)^{n}={ }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1} \\
\quad+{ }^{n} C_{2} x^{n-2}+\ldots+{ }^{n} C_{n}
\end{array}
$$
On multiplying Eqs. (i) and (ii), we get
$$
\begin{array}{r}
(1+x)^{2 n}=\left({ }^{n} C_{0}+{ }^{n} C_{1} x+{ }^{n} C_{2} x^{2}+\ldots+{ }^{n} C_{n} x^{n}\right) \\
\quad \times\left({ }^{n} C_{0} x^{n}+{ }^{n} C_{1} x^{n-1}+{ }^{n} C_{2} x^{n-2}+\ldots+{ }^{n} C_{n}\right)
\end{array}
$$
Coefficient of $x^{n}$ in RHS
$$
=\left({ }^{n} C_{0}\right)^{2}+\left({ }^{n} C_{1}\right)^{2}+\ldots+\left({ }^{n} C_{n}\right)^{2}
$$
and coefficient of $x^{n}$ in $\mathrm{LHS}={ }^{2 n} \mathrm{C}_{n}$
$$
\begin{array}{c}
\therefore\left({ }^{n} C_{0}\right)^{2}+\left({ }^{n} C_{1}\right)^{2}+\ldots+\left({ }^{n} C_{n}\right)^{2}=\frac{2 n !}{n ! n !} \\
\Rightarrow\left({ }^{n} C_{1}\right)^{2}+\ldots+\left({ }^{n} C_{n}\right)^{2}=\frac{(2 n) !}{n ! n !}-1 \\
\quad={ }^{2 n} C_{n}-1
\end{array}
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2014
The value of $\int \frac{(x-2)}{\left\{(x-2)^{2}(x+3)^{7}\right\rangle^{1 / 3}} d x$ is

Options:
A $\frac{3}{20}\left(\frac{x-2}{x+3}\right)^{4 / 3}+C$
B $\frac{3}{20}\left(\frac{x-2}{x+3}\right)^{3 / 4}+C$
C $\frac{5}{12}\left(\frac{x-2}{x+3}\right)^{4 / 3}+C$
D $\frac{3}{20}\left(\frac{x-2}{x+3}\right)^{5 / 3}+C$

The correct answer is:
$\frac{3}{20}\left(\frac{x-2}{x+3}\right)^{4 / 3}+C$

Solution:
Let $I=\int \frac{(x-2)}{\left\{(x-2)^{2}(x+3)^{7}\right\}^{1 / 3}} d x$
$$
\begin{array}{l}
=\int \frac{d x}{(x-2)^{-1 / 3}(x+3)^{7 / 3}} \\
=\int \frac{d x}{(x-2)^{2}\left(\frac{x+3}{x-2}\right)^{7 / 3}} \\
\text { Put } \frac{x+3}{x-2}=t \Rightarrow \frac{-5 d x}{(x-2)^{2}}=d t
\end{array}
$$
$\therefore \quad I=-\frac{1}{5} \int \frac{d t}{t^{7 / 3}}=-\frac{1}{5} \int t^{-7 / 3} d t$
$=-\frac{1}{5}\left[\frac{t^{-7 / 3+1}}{-\frac{7}{3}+1}\right]+C=-\frac{1}{5}\left[\frac{t^{-4 / 3}}{-\frac{4}{3}}\right]+C$
$=\frac{3}{20 t^{4 / 3}}+C=\frac{3}{20}\left(\frac{x-2}{x+3}\right)^{4 / 3}+C$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2015
The value of $|z|^{2}+|z-3|^{2}+|z-i|^{2}$ is minimum when $z$ equals

Options:
A $2-\frac{2}{3} i$
B $45+3 i$
C $1+\frac{i}{3}$
D $1-\frac{i}{3}$

The correct answer is:
$1+\frac{i}{3}$

Solution:
Let $z=x+i y$
$\therefore|z|^{2}+|z-3|^{2}+|z-i|^{2}$
$$
\begin{aligned}
=|x+i y|^{2}+\mid &\mid x-3)+\left.i y\right|^{2} \\
&+|x+i(y-1)|^{2}
\end{aligned}
$$
$=x^{2}+y^{2}+(x-3)^{2}+y^{2}+x^{2}+(y-1)^{2}$
$=x^{2}+y^{2}+x^{2}-6 x+9+y^{2}+x^{2}+y^{2}$
$=3 x^{2}+3 y^{2}-6 x-2 y+10$
$=3\left(x^{2}-2 x+1\right)+3\left(y^{2}-\frac{2}{3} y+\frac{1}{9}\right)$
$\quad=3(x-1)^{2}+3\left(y-\frac{1}{3}\right)^{2}+\frac{20}{3}$
It is minimum, when $x-1=0$ and $y-\frac{1}{3}=0$
$\therefore \quad x=1 \quad$ and $\quad y=\frac{1}{3}$
$\therefore \quad z=1+\frac{1}{3} i$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2014
The values of $a, b, c$ for which the function $f(x)=\left\{\begin{array}{l}\frac{\sin (a+1) x+\sin x}{x}, x < 0 \\ c, x=0 \\ \frac{\left(x+b x^2\right)^{1 / 2}-\sqrt{x}}{b x^{1 / 2}}, x>0\end{array}\right.$ is continuous at $x=0$, are

Options:
A $a=\frac{3}{2}, b=-\frac{3}{2}, c=\frac{1}{2}$
B $\mathrm{a}=-\frac{3}{2}, \mathrm{c}=\frac{3}{2}, \mathrm{~b}$ is arbitrary non-zero real number
C $\mathrm{a}=-\frac{5}{2}, \mathrm{~b}=-\frac{3}{2}, \mathrm{c}=\frac{3}{2}$
D $a=-2, b \in \mathbb{R}-\{0\}, c=0$

The correct answer is:
$a=-2, b \in \mathbb{R}-\{0\}, c=0$

Solution:
$\quad f\left(0^{+}\right)=\underset{x \rightarrow 0^{+}}{L t} \frac{(1+b x)^{1 / 2}-1}{b}=0$
$f(0)=c$
$f\left(0^{-}\right)=a+2$
$\Rightarrow c=0$ and $a=-2, b \in R-\{0\}$

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2022
The values of $\Delta H$ and $\Delta S$ of a certain reaction are $-400 \mathrm{k} \mathrm{J} \mathrm{mol}^{-1}$ and $-20 \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}$ respectively. The temperature below which the reaction is spontaneous, is

Options:
A $100 \mathrm{K}$
B $20^{\circ} \mathrm{C}$
C $20 \mathrm{K}$
D $120^{\circ} \mathrm{C}$

The correct answer is:
$20 \mathrm{K}$

Solution:
Given, $\quad \Delta H=-400 \mathrm{kJ} \mathrm{mol}^{-1}$
$$
\Delta S=-20 \mathrm{kJ} \mathrm{mol}^{-1} \mathrm{K}^{-1}
$$
Gibbs Helmholtz equation is
$$
\Delta G=\Delta H-T \Delta S
$$
For a reaction to be spontaneous, $\Delta G$ must be $\therefore \quad 0 \geq \Delta H-T \Delta S$
$$
\begin{array}{r}
T \Delta S \leq \Delta H \\
T \leq \frac{\Delta H}{\Delta S} \leq \frac{-400}{-20} \leq 20 \mathrm{K}
\end{array}
$$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2014
The variance of first 20 natural numbers is

Options:
A $133 / 4$
B $279 / 12$
C 133/2
D 399/4

The correct answer is:
$133 / 4$

Solution:
Variance of $n$ natural numbers $=\frac{n^{2}-1}{12}=\frac{(20)^{2}-1}{12} \quad$ $[\because n=20]$,
$=\frac{400-1}{12}=\frac{399}{12}=\frac{133}{4}$

Topic: Mathematics, Statistics, WBJEE, WBJEE 2015
The variation of electric field along the $Z$-axis due to a uniformly charged circular ring of radius ' $a$ ' in $X Y$ plane is shown in the figure. The value of coordinate $M$ will be

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

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

Solution:
For $E_{z}$ to be maximum,
$$
\frac{d E_{z}}{d z}=0 \quad \therefore z=a / \sqrt{2} \quad \text { or } \frac{z}{a}=\frac{1}{\sqrt{2}}
$$

Topic: Physics, Electrostatics, WBJEE, WBJEE 2021
The vectors $\mathrm{A}$ and $\mathrm{B}$ are such that $|\mathbf{A}+\mathbf{B}|=|\mathbf{A}-\mathbf{B}| .$ The angle between the two vectors will be

Options:
A $0^{\circ}$
B $60^{\circ}$
C $90^{\circ}$
D $45^{\circ}$

The correct answer is:
$90^{\circ}$

Solution:
Given, $|A+B|=|A-B|$
$\therefore \quad(A+B)^{2}=(A-B)^{2}$
$A^{2}+B^{2}+2 A B \cos \theta=A^{2}+B^{2}-2 A B \cos \theta$
$2 A B \cos \theta+2$ AB $\cos \theta=0$
$4 A B \cos \theta=0$
Here, $\quad A \neq 0, B \neq 0$
$\therefore \quad \cos \theta=0$
$\theta=90^{\circ}$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2016
The vectors are given by $\dot{\mathbf{A}}=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+2 \hat{\mathbf{k}}$ and $\mathbf{B}=3 \hat{\mathbf{i}}+6 \hat{\mathbf{j}}+2 \mathbf{k}$. Another vector $\mathbf{C}$ has the same magnitude as $\mathbf{B}$ but has the same direction as A. Then which of the following vectors represents C?

Options:
A $\frac{7}{3}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$
B $\frac{3}{7}(i-2 \hat{j}+2 \hat{k})$
C $\frac{7}{9}(i-2 \hat{j}+2 \hat{k})$
D $\frac{9}{7}(\vec{i}+2 \hat{j}+2 \hat{k})$

The correct answer is:
$\frac{7}{3}(\hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}})$

Solution:
(a) Given, $A=i+2 j+2 \hat{k}$ and $B=3 \hat{i}+6 j+2 \hat{k}$
So, $\quad C=\frac{\hat{i}+2 \hat{j}+2 \hat{k}}{\sqrt{1+4+4}} \times \sqrt{3^{2}+6^{2}+2^{2}}$
$=\frac{\hat{i}+2 \hat{j}+2 \hat{k}}{3} \times \sqrt{49}=\frac{7}{3}(\hat{i}+2 \hat{j}+2 \hat{\mathbf{k}})$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2013
The velocity of a car travelling on a straight road is $3.6 \mathrm{kmh}^{-1}$ at an instant of time. Now travelling with uniform acceleration for $10 \mathrm{s}$, the velocity becomes exactly double. If the wheel radius of the car is $25 \mathrm{cm}$, then which of the following is the closest to the number of revolutions that the wheel makes during this 10 s?

Options:
A 84
B 95
C 126
D 135

The correct answer is:
95

Solution:
We have, $\theta=2 \pi n=\frac{\left(\frac{v_{f}^{2}}{r^{2}}-\frac{v_{1}^{2}}{r^{2}}\right)}{\left(2 \frac{a}{r}\right)}$
$$
n=\frac{v_{f}^{2}-v_{i}^{2}}{(2 a r)(2 \pi)}=95
$$

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2013