datavorous/entrance-exam-dataset huggingface dataset:
The value of $\lambda$ for which the curve $(7 x+5)^{2}+(7 y+3)^{2}=\lambda^{2}(4 x+3 y-24)^{2}$ representa a parabola is

Options:
A $\pm \frac{6}{5}$
B $\pm \frac{7}{5}$
C $\pm \frac{1}{5}$
D $\pm \frac{2}{5}$

The correct answer is:
$\pm \frac{7}{5}$

Solution:
Given curve is
$$
\begin{array}{l}
(7 x+5)^{2}+(7 y+3)^{2}=\lambda^{2}(4 x+3 y-24)^{2} \\
\Rightarrow 49 x^{2}+25+70 x+49 y^{2}+9+42 y \\
=\lambda^{2}\left(16 x^{2}+9 y^{2}+576+24 x y-144 y-192 x\right) \\
\Rightarrow \left(49-16 \lambda^{2}\right) x^{2}+\left(49-9 \lambda^{2}\right) y^{2}+\left(70+192 \lambda^{2}\right) x \\
+\left(42+144 \lambda^{2}\right) y-24 \lambda^{2} x y+\left(25-576 \lambda^{2}\right)=0
\end{array}
$$
On comparing with $a x^{2}+2 h x y+b y^{2}+2 g x+2 f y+c=0,$ we get
$a=49-16 \lambda^{2}, b=\left(49-9 \lambda^{2}\right), h=-12 \lambda^{2}$
Condition for parabola is ab $=h^{2}$.
$$
\begin{array}{l}
\therefore \quad\left(49-16 \lambda^{2}\right)\left(49-9 \lambda^{2}\right)=\left(-12 \lambda^{2}\right)^{2} \\
\Rightarrow \quad 2401-441 \lambda^{2}-784 \lambda^{2}+144 \lambda^{4}=144 \lambda^{4} \\
\Rightarrow \quad 2401-1225 \lambda^{2}=0 \\
\Rightarrow \quad \lambda^{2}=\frac{2401}{1225}
\end{array}
$$
$\Rightarrow \quad \lambda^{2}=\frac{(49)^{2}}{(35)^{2}}$
$\Rightarrow \quad \lambda=\pm \frac{49}{35}$
$\Rightarrow \quad \lambda=\pm \frac{7}{5}$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2014
The value of $\lambda$ for which the straight line $\frac{x-\lambda}{3}=\frac{y-1}{2+\lambda}=\frac{z-3}{-1}$ may lie on the plane $x-2 y=0,$ is

Options:
A 2
B 0
C $-\frac{1}{2}$
D there is no such $\lambda$

The correct answer is:
there is no such $\lambda$

Solution:
If a line $\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}$ lies
on a plane $a_{2} x+b_{2} y+c_{2} z=d,$ then
(i) $a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0$
(ii) $a_{2} x_{1}+b_{2} y_{1}+c_{2} z_{1}=d$
According to the question. $3(1)+(2+\lambda)(-2)+(-1)(0)=0$
$\Rightarrow \quad 3-4-2 \lambda=0$
$\Rightarrow \quad \lambda=-1 / 2$
Also, $\quad \lambda-2=0$
$\Rightarrow \quad \lambda=2$
So. there is no such value of $\lambda$

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2015
The value of ' $\lambda$ ' for which the two vectors $\vec{a}=5 \hat{i}+\lambda \hat{j}+\hat{k}$ and $\vec{b}=\hat{i}-2 \hat{j}+\hat{k}$ are perpendicular to each other is

Options:
A 2
B -2
C 3
D -3

The correct answer is:
3

Solution:
Hints : For $\vec{a} \perp \vec{b}$
$\begin{aligned}
& \vec{a} \cdot \vec{b}=0 \\
& \text { i.e., } 5-2 \lambda+1=0 \\
& \lambda=3
\end{aligned}$

Topic: Physics, Mathematics in Physics, WBJEE, WBJEE 2010
The value of $\Delta H$ for cooling 2 mole of an ideal monoatomic gas from $225^{\circ} \mathrm{C}$ to $125^{\circ} \mathrm{C}$ at constant pressure will be [given $\left.C_{p}=\frac{5}{2} R\right]$

Options:
A $250 \mathrm{R}$
B $-500 R$
C $500 \mathrm{R}$
D $-250 R$

The correct answer is:
$-500 R$

Solution:
$\Delta H=-n C_{p} \Delta T=-2 \times \frac{5}{2} R(225-125)$
$$
=-5 R(100)=-500 R
$$

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2014
The value of $I=\int_{0}^{\frac{\pi}{4}}\left(\tan ^{n+1} x\right) d x$
$+\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \tan ^{n+1}\left(\frac{x}{2}\right) d x$ is

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

The correct answer is:
$\frac{1}{n}$

Solution:
.Given. $I=\int_{0}^{\pi / 4}\left(\tan ^{n+1} x\right) d x+\frac{1}{2} \int_{0}^{\pi / 2} \tan ^{n-1}\left(\frac{x}{2}\right) d x$
In second integral, putt $=\frac{x}{2} \Rightarrow d x=2 d t$
$\Rightarrow \quad$ Also, when $x=0$ then $t=0$
When $x=\pi / 2,$ then $t=\pi / 4$
Then, $\begin{aligned} I &=\int_{0}^{\pi / 4}\left(\tan ^{n+1} x\right) d x+\int_{0}^{\pi / 4} \tan ^{n-1} t d t \\ & I=\int_{0}^{\pi / 4} \tan ^{n+1} x \cdot d x+\int_{0}^{\pi / 4} \tan ^{n-1} x d x \\ & \quad\left\{\because \int_{a}^{b} f(x) d x=\int_{a}^{b} f(y) d y\right\} \end{aligned}$
$\Rightarrow \quad I=\int_{0}^{\pi / 4}\left(\tan ^{n+1} x+\tan ^{n-1} x\right) d x$
$\Rightarrow \quad I=\int_{0}^{\pi / 4} \tan ^{n-1} x \cdot\left(\tan ^{2} x+1\right) d x$
$\Rightarrow \quad I=\int_{0}^{\pi / 4} \tan ^{n-1} x\left(\sec ^{2} x\right) d x$
Put $\quad t=\tan x$
$\Rightarrow \quad d t=\sec ^{2} x d x$
Also, when $x=0,$ then $t=0$
when $x=\pi / 4,$ then $t=1$ $I=\int_{0}^{1} t^{n-1} d t=\left[\frac{t^{n}}{n}\right]_{0}^{1}=\frac{1}{n}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2013
The value of $I=\int_{-\pi / 2}^{\pi / 2}|\sin x| d x$ is

Options:
A 0
B 2
C -2
D $-2 < $ I $ < 2$

The correct answer is:
2

Solution:
Hints: $I=2 \int_0^{\frac{\pi}{2}} \sin x d x=2(1)=2$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2010
The value of
$I=\int_{\pi / 2}^{5 \pi / 2} \frac{e^{\tan ^{-1}}(\sin x)}{e^{\tan ^{-1}(\sin x)}+e^{\tan ^{-1}(\cos x)}} d x,$ is

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

The correct answer is:
$\pi$

Solution:
Let
$I=\int_{\pi / 2}^{5 \pi / 2} \frac{e^{\tan ^{-1}}(\sin x)}{e^{\tan ^{-1}(\sin x)}+e^{\tan ^{-1}(\cos x)}} d x$
$=\int_{0}^{5 \pi / 2} \frac{e^{\tan ^{-1}(\sin x)}}{e^{\tan ^{-1}(\sin x)}+e^{\tan ^{-1}(\cos x)}} d x$
$-\int_{0}^{\pi / 2} \frac{e^{\tan ^{-1}}(\sin x)}{e^{\tan ^{-1}(\sin x)}+e^{\tan ^{-1}(\cos x)}} d x \ldots(\mathrm{i}$
$I=\int_{0}^{5 \pi / 2} \frac{e^{\tan ^{-1}(\cos x)}}{e^{\tan ^{-1}(\cos x)}+e^{\tan ^{-1}(\sin x)}} d x$
$\quad-\int_{0}^{\pi / 2} \frac{e^{\tan ^{-1}}(\cos x)}{e^{\tan ^{-1}(\cos x)}+e^{\tan ^{-1}(\sin x)}} d x \ldots$ (ii)
$\quad\left[\right.$ using, $\left.\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right]$
On adding Eqs. (i) and (ii), we get
$$
\begin{array}{l}
2 I=\int_{0}^{5 \pi / 2} d x-\int_{0}^{\pi / 2} d x \\
\Rightarrow \quad 2 I=(x)_{0}^{5 \pi / 2}-(x)_{0}^{\mathrm{v} / 2} \\
\Rightarrow \quad 2 I=\frac{5 \pi}{2}-\frac{\pi}{2}=2 \pi \quad \Rightarrow \quad I=\pi
\end{array}
$$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2018
The value of \(\int_0^\pi \sin ^{50} x \cos ^{49} x d x\) is

Options:
A 0
B \(\pi / 4\)
C \(\pi / 2\)
D 1

The correct answer is:
0

Solution:
Hints: \(I=\int_0^\pi \sin ^{50} x \cdot \cos ^{49} x d x \int_0^4 f(x)=\int_0^4 f(a-x)\)
\(\begin{aligned}
& I=\int_0^\pi \sin ^{50} x\left(-\cos ^{49}(x)\right)=-\int_0^\pi \sin ^{50} x \cdot \cos ^{49} x \\
& =I=-I \\
& I=0
\end{aligned}\)

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2011
The value of \(\int_0^{1 / 2} \frac{d x}{\sqrt{1-x^{2 n}}}\) is \((n \in \mathbb{N})\)

Options:
A less than or equal to \(\frac{\pi}{6}\)
B greater than or equal to 1
C less than \(\frac{1}{2}\)
D greater than \(\frac{\pi}{6}\)

The correct answer is:
less than or equal to \(\frac{\pi}{6}\)

Solution:
Hint : For 0 to \(\frac{1}{2}\)
\(\begin{aligned} & x^2 \geq x^{2 n} \\ & \Rightarrow-x^2 \leq-x^{2 n} \\ & \Rightarrow 1-x^2 \leq 1-x^{2 n} \\ & \Rightarrow \frac{1}{\sqrt{1-x^2}} \geq \frac{1}{\sqrt{1-x^{2 n}}} \\ & \Rightarrow \int_0^{1 / 2} \frac{1}{\sqrt{1-x^2}} d x \geq \int_0^{1 / 2} \frac{1}{\sqrt{1-x^{2 n}}} d x \\ & \Rightarrow\left[\sin ^{-1} x\right]_0^{1 / 2} \geq \mathrm{I} \\ & \Rightarrow \frac{\pi}{6} \geq I \\ & \end{aligned}\)

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2023
The value of \(\int_{-2}^2(x \cos x+\sin x+1) d x\) is

Options:
A 2
B 0
C -2
D 4

The correct answer is:
4

Solution:
Hints: \(\int_{-2}^2(x \cos x+\sin x+1) d x=\int_{-2}^2 d x=4\)

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2011
The value of integral $\int_{-1}^1 \frac{|x+2|}{x+2} d x$ is

Options:
A 1
B 2
C 0
D -1

The correct answer is:
2

Solution:
Hints : $I=\int_{-1}^1 \frac{|x+2|}{x+2} d x \quad, \quad x+2=v \Rightarrow d x=d v$
$$
\therefore \mathrm{I}=\int_1^3 \frac{|v|}{V} d v=\int_1^3 \frac{v}{V} d v=\int_1^3 d v=2
$$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2009
The value of integral $\int_{\frac{\pi}{6}}^{\frac{\pi}{3}}\frac{(sinx-xcosx)}{x(x+sinx)}dx$ is

Options:
A $\log _{e}\left\{\frac{2(\pi+3)}{(2 \pi+3 \sqrt{3})}\right\}$
B $\log _{e}\left\{\frac{\pi+3}{2(2 \pi+3 \sqrt{3})}\right\}$
C $\log _{e}\left\{\frac{2 \pi+3 \sqrt{3}}{2(\pi+3)}\right\}$
D $\log _{e}\left\{\frac{2(2 \pi+3 \sqrt{3})}{\pi+3}\right\}$

The correct answer is:
$\log _{e}\left\{\frac{2(\pi+3)}{(2 \pi+3 \sqrt{3})}\right\}$

Solution:
Let $l=\int_{\pi / 6}^{\pi / 3} \frac{(\sin x-x \cos x)}{x(x+\sin x)} d x$
$\Rightarrow \quad I=\int_{\pi / 6}^{\pi / 3} \frac{(x+\sin x)-x(1+\cos x)}{x(x+\sin x)} d x$
$\Rightarrow \quad I=\int_{\pi / 6}^{\pi / 3}\left(\frac{1}{x}-\frac{1+\cos x}{x+\sin x}\right) d x$
$\Rightarrow \quad 1=[\log \times]_{\pi / 6}^{\pi / 3}-\int_{\pi / 6}^{\pi / 3} \frac{1+\cos x}{x+\sin x} \cdot d x$
$\left\{\begin{array}{l}\text { put } t=x+\sin x \\ \text { ot }=(1+\cos x) d x\end{array}\right.$ in lind term
$\Rightarrow \quad I=\left(\log \frac{\pi}{3}-\log \frac{\pi}{6}\right)-\int\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right) \frac{d t}{t}$
$\Rightarrow \quad I=\log 2-\left[\log t \mid\left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)\right.$
$\Rightarrow \quad 1=\log 2-\left[\log \left(\frac{\pi}{3}+\frac{\sqrt{3}}{2}\right)-\log \left(\frac{\pi}{6}+\frac{1}{2}\right)\right]$
$\Rightarrow \quad I=\log 2-\log \left(\frac{2 \pi+3 \sqrt{3}}{\pi+3}\right)$
$\left(\because \log m-\log n=\log \frac{m}{n}\right)$
$I=\log \left(\frac{2(\pi+3)}{2 \pi+3 \sqrt{3}}\right)$
$=\log \left(\frac{2 \pi+6}{2 \pi+3 \sqrt{3}}\right)$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2013
The value of $K$ in order that $f(x)=\sin x-\cos x-K x+5$ decreases for all positive real values of $x$ is given by

Options:
A $K < 1$
B $K \geq 1$
C $K>\sqrt{2}$
D $K < \sqrt{2}$

The correct answer is:
$K>\sqrt{2}$

Solution:
We have,
$\begin{aligned} f(x) &=\sin x-\cos x-K x+5 \\ \Rightarrow f^{\prime}(x) &=\cos x+\sin x-K \end{aligned}$
For decreasing, $f^{\prime}(x) < 0$ $\Rightarrow \cos x+\sin x-K < 0$
$\begin{array}{ll}\Rightarrow & K>\cos x+\sin x \\ \Rightarrow & \quad K>\sqrt{2}\left[\frac{1}{\sqrt{2}} \cos x+\frac{1}{\sqrt{2}} \sin x\right] \\ \Rightarrow & K>\sqrt{2} \sin \left(\frac{\pi}{4}+x\right) \\ \therefore \quad & K>\sqrt{2}\end{array}$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2017
The value of $\left(\frac{1}{\log _3 12}+\frac{1}{\log _4 12}\right)$ is

Options:
A 0
B $\frac{1}{2}$
C 1
D 2

The correct answer is:
1

Solution:
Hints : $\log _{12} 3+\log _{12} 4=\log _{12} 12=1$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2009
The value of $\frac{\log _3 5 \times \log _{25} 27 \times \log _{49} 7}{\log _{81} 3}$ is

Options:
A 1
B 6
C $\frac{2}{3}$
D 3

The correct answer is:
3

Solution:
Hints: $\frac{\left(\frac{\log 5}{\log 3} \times \frac{3 \log 3}{2 \log 5} \times \frac{\log 7}{2 \log 7}\right)}{\left(\frac{\log 3}{4 \log 3}\right)}=3$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2010
The value of $\lim _{n \rightarrow \infty} \frac{(n !)^{n}}{\frac{1}{n}}$ is

Options:
A 1
B $\frac{1}{e^{2}}$
C $\frac{1}{2 e} \quad$
D $\frac{1}{e}$

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

Solution:
$\lim _{x \rightarrow \infty} \frac{(n !)^{1 / n}}{n}=\lim _{n \rightarrow-}\left(\frac{n !}{n^{n}}\right)^{1 / n}$
We have, $\frac{n !}{n^{n}}=\frac{1 \cdot 2 \cdot 3}{n \cdot n \cdot n} \frac{n}{n}$
$\therefore \quad\left\{\frac{n !}{n^{n}}\right\}^{1 / n}=\left\{\frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{r}{n} \cdots \frac{n}{n}\right\}^{1 / n}$
$\Rightarrow \lim _{n \rightarrow \infty}\left\{\frac{n !}{n^{n}}\right\}^{1 / n}=\lim _{n \rightarrow \infty}\left\{\frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{r}{n} \cdots \frac{n}{n}\right\}^{1 / n}$
Let $\quad A=\lim _{n \rightarrow \infty}\left\{\frac{n !}{n^{n}}\right\}^{1 / n}$
Then, $A=\lim _{n \rightarrow-}\left\{\frac{1}{n} \cdot \frac{2}{n} \cdot \frac{3}{n} \cdots \frac{r}{n} \cdots \frac{n}{n}\right\}^{1 / n}$
$\begin{aligned} \Rightarrow \quad \log A &=\lim _{n \rightarrow \infty} \frac{1}{n} \Sigma \log \left(\frac{r}{n}\right)=\int_{0}^{1} \log x d x \\ &=\left[x \log x-\int \frac{1}{x} \cdot x d x\right]_{0}^{1} \end{aligned}$
Integrating by parts $=[x \log x-x]_{0}^{1}=-1$
$\Rightarrow \quad A=e^{-1}=\frac{1}{e}$

Topic: Mathematics, Limits, WBJEE, WBJEE 2012
The value of
$\lim _{n \rightarrow \infty} \frac{1}{n}\left\{\sec ^{2} \frac{\pi}{4 n}+\sec ^{2} \frac{2 \pi}{4 n}+\ldots+\sec ^{2} \frac{n \pi}{4 n}\right] is$

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

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

Solution:
We have, $\lim _{n \rightarrow \infty} \frac{1}{n}\left\{\sec ^{2} \frac{\pi}{4 n}+\sec ^{2} \frac{2 \pi}{4 n}+\ldots+\sec ^{2} \frac{n \pi}{4 n}\right\}$
$\quad=\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \frac{1}{n} \sec ^{2}\left(\frac{r \pi}{4 n}\right)=\int_{0}^{1} \sec ^{2}\left(\frac{\pi x}{4}\right)$
$=\frac{4}{\pi}\left[\tan \left(\frac{\pi x}{4}\right)\right]_{0}^{1}$
$=\frac{4}{\pi} \times 1=\frac{4}{\pi}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2018
The value of $\lim _{n \rightarrow\infty}\left[\frac{n}{n^{2}+1^{2}}+\frac{n}{n^{2}+2^{2}}+\ldots+\frac{1}{2 n}\right]$ is

Options:
A $\frac{n \pi}{4}$
B $\frac{\pi}{4}$
C $\frac{\pi}{4 n}$
D $\frac{\pi}{2 n}$

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

Solution:
$\lim _{n \rightarrow-}\left[\frac{n}{n^{2}+1^{2}}+\frac{n}{n^{2}+2^{2}}+\ldots+\frac{1}{2 n}\right]$
$=\lim _{n \rightarrow-}\left[\frac{n}{n^{2}+1^{2}}+\frac{n}{n^{2}+2^{2}}+\ldots+\frac{n}{n^{2}+n^{2}}\right]$
$=\lim _{n \rightarrow-} \sum_{r=1}^{-\infty} \frac{n}{n^{2}+r^{2}}$
$=\lim _{n \rightarrow-\infty} \frac{1}{n} \sum_{r=1}^{-\infty} \frac{1}{1+\left(\frac{r}{n}\right)^{2}}$
$\quad=\int_{0}^{1} \frac{1}{1+x^{2}} d x$
$\quad=\left[\tan ^{-1} x\right]_{0}^{1}$
$\quad=\tan ^{-1} 1-\tan ^{-1} 0$
$\quad=\frac{\pi}{4}-0=\frac{\pi}{4}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2017
The value of $\lim _{n \rightarrow \infty}\left[\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\ldots \ldots . . \frac{n}{n^2+n^2}\right]$ is

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

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

Solution:
$$
\text { Hints : } \lim _{n \rightarrow \infty}\left[\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\ldots+\frac{n}{n^2+n^2}\right]
$$
$$
=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{n}{n^2+r^2}=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{1}{1+\left(\frac{r}{n}\right)^2}=\int_0^1 \frac{d x}{1+x^2}=\left[\tan ^{-1} x\right]_0^1=\frac{\pi}{4}
$$

Topic: Mathematics, Limits, WBJEE, WBJEE 2009
The value of $\lim _{n \rightarrow \infty}\left\{\frac{\sqrt{n+1}+\sqrt{n+2}+\ldots+\sqrt{2 n-1}}{n^{3 / 2}}\right\}$ is

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

The correct answer is:
$\frac{2}{3}(2 \sqrt{2}-1)$

Solution:
$\because \lim _{n \rightarrow \infty \left[\frac{\sqrt{n+1}+\sqrt{n+2}+\ldots+\sqrt{2 n-1}}{n^{\frac{3}{2}}}\right]}$
$=\lim _{n \rightarrow \infty 1}\left[\sqrt{1+\frac{1}{n}}+\sqrt{1+\frac{2}{n}}+\ldots+\sqrt{1+\frac{n-1}{n}}\right] \frac{1}{n}$
$=\lim _{n \rightarrow \infty 1} \sum_{r=1}^{n-1} \frac{1}{n} \sqrt{1+\frac{r}{n}}$
$=\int_{0}^{1} \sqrt{1+x} d x=\left[\frac{2}{3}(1+x)^{3 / 2}\right]_{0}^{1}=\frac{2}{3}(2 \sqrt{2}-1)$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2016
The value of
\(\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^3}{r^4+n^4} \text { is }\)

Options:
A \(\frac{1}{2} \log _{\mathrm{e}}(1 / 2)\)
B \(\frac{1}{4} \log _e(1 / 2)\)
C \(\frac{1}{4} \log _{\mathrm{e}} 2\)
D \(\frac{1}{2} \log _{\mathrm{e}} 2\)

The correct answer is:
\(\frac{1}{4} \log _{\mathrm{e}} 2\)

Solution:
Hints: \(\operatorname{Lt}_{n \rightarrow \infty} \cdot \sum \frac{n^3\left(\frac{r}{n}\right)^3}{n^4\left[\left(\frac{r}{n}\right)^4+1\right]}\)
\(\begin{aligned}
& =\frac{1}{4} \cdot \int_0^1 \frac{x^3}{1+x^4} d x=\frac{1}{4}\left[\log \left(1+x^4\right)\right]_0^1 \\
& =\frac{1}{4}(\log 2-\log 1)=\frac{1}{4} \log 2
\end{aligned}\)

Topic: Mathematics, Limits, WBJEE, WBJEE 2011
The value of \(\lim _{n \rightarrow \infty}\left[\left(\frac{1}{2 \cdot 3}+\frac{1}{2^2 \cdot 3}\right)+\left(\frac{1}{2^2 \cdot 3^2}+\frac{1}{2^3 \cdot 3^2}\right)+\ldots+\left(\frac{1}{2^n \cdot 3^n}+\frac{1}{2^{n+1} \cdot 3^n}\right)\right]\) is

Options:
A \(\frac{3}{8}\)
B \(\frac{3}{10}\)
C \(\frac{3}{14}\)
D \(\frac{3}{16}\)

The correct answer is:
\(\frac{3}{10}\)

Solution:
\(\begin{aligned}
\text{Hint : } & \lim _{n \rightarrow \infty} \sum_{x=1}^n\left(\frac{1}{2^x \times 3^x}+\frac{1}{2^{x+1} \times 3^x}\right) \\
& =\lim _{n \rightarrow \infty} \sum_{x=1}^n \frac{1}{6^x}\left(1+\frac{1}{2}\right) \\
& =\frac{3}{2} \times \underbrace{\lim _{n \rightarrow \infty}\left(\frac{1}{6}+\frac{1}{6^2}+\ldots \cdots \cdot \frac{1}{6^n}\right)}_{\text {infinite GP }}=\frac{3}{2} \times \frac{\frac{1}{6}}{1-\frac{1}{6}}=\frac{3}{10}
\end{aligned}\)

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2023
The value of $\lim _{x \rightarrow 0^{+}} \frac{x}{p}\left[\frac{q}{x}\right]$ is

Options:
A $\frac{[q]}{p}$
B 0
C 1
D $\infty$

The correct answer is:
$\frac{[q]}{p}$

Solution:
$$
\begin{aligned}
&\text { Given, } \lim _{x \rightarrow 0^{+}} \frac{x}{p}\left[\frac{q}{x}\right]\\
&\begin{array}{l}
=\lim _{x \rightarrow 0^{+}} \frac{x}{p}\left(\frac{q}{x}-\{\frac{q}{x}\right\}\right) \\
=\frac{[q]}{p}-\frac{x}{p}\left\{\frac{q}{x}\right\} \\
=\frac{[q]}{p}-0=\frac{[q]}{p}
\end{array}
\end{aligned}
$$

Topic: Mathematics, Limits, WBJEE, WBJEE 2019
The value of $\lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$ is equal to

Options:
A $e^{\sin ^{2} y}$
B $e^{2 \sin y}$
C $e^{\mid \sin y \mid}$
D $e^{\operatorname{cosec}^{2} y}$

The correct answer is:
$e^{\sin ^{2} y}$

Solution:
We have
$\lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t-\int_{x+y}^{a} e^{\sin ^{2} t} d t\right]$
$=\lim _{x \rightarrow 0} \frac{1}{x}\left[\int_{y}^{a} e^{\sin ^{2} t} d t+\int_{a}^{x+y} e^{\sin ^{2} t} d t\right]$
$=\lim _{x \rightarrow 0} \frac{1}{x} \int_{y}^{x+y} e^{\sin ^{2} t} d t$
$=\lim _{x \rightarrow 0} \frac{\int_y^{x+y}e^{sin^{2t}}dt}{x}$
$=\lim _{x \rightarrow 0} \frac{e^{\sin ^{2}(x+y)}(1+0)-0}{1}=e^{\sin ^{2} y}$

Topic: Mathematics, Limits, WBJEE, WBJEE 2019
The value of $\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \cos \left(t^{2}\right) d t}{x \sin x}$ is

Options:
A 1
B -1
C 2
D $\log _{e} 2$

The correct answer is:
1

Solution:
$\lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \cos \left(t^{2}\right) d t}{x \sin x} \quad\left[\frac{0}{0}\text{ form}\right]$
$=\lim _{x \rightarrow 0} \frac{\cos \left(x^{4}\right) \times 2 x}{\sin x+x \cos x} \quad$ [L' Hospital's rule]
$=\lim _{x \rightarrow 0} \frac{2\left[\cos x^{4}-x \sin \left(x^{4}\right) \times 4 x^{3}\right]}{\cos x+\cos x-x \sin x}$
[L' Hospital's rule]
$=\frac{2[\cos 0-0]}{\cos 0+\cos 0-0}=\frac{2}{1+1}=1$

Topic: Mathematics, Limits, WBJEE, WBJEE 2014
The value of $\lim _{x \rightarrow 2} \int_{2}^{x} \frac{3 t^{2}}{(x-2)} d t$ is

Options:
A 10
B 12
C 8
D 16

The correct answer is:
12

Solution:
$\lim _{x \rightarrow 2} \int_{2}^{x} \frac{3 t^{2}}{(x-2)}$ dt $=\frac{\lim _{x \rightarrow 2} \int_{2}^{x} 3 t^{2} d t}{\lim _{x \rightarrow 2}(x-2)}$
$=\frac{\lim _{x \rightarrow 2} 3 x^{2}}{1} \quad$ [using L' Hospital's nule]
$=3 \times(2)^{2}=12$

Topic: Mathematics, Limits, WBJEE, WBJEE 2015
The value of \(\lim _{x \rightarrow 1} \frac{x+x^2+\ldots+x^n-n}{x-1}\) is

Options:
A \(\mathrm{n}\)
B \(\frac{\mathrm{n}+1}{2}\)
C \(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)
D \(\frac{\mathrm{n}(\mathrm{n}-1)}{2}\)

The correct answer is:
\(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

Solution:
Hints: \(\operatorname{Lt}_{x \rightarrow 1} \frac{(x-1)+\left(x^2-1\right)+\left(x^3-1\right) \ldots \ldots \ldots \ldots .\left(x^4-1\right)}{x-1}\)
\(=1+2+3 \ldots \ldots \ldots \ldots \ldots \ldots+n=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

Topic: Mathematics, Limits, WBJEE, WBJEE 2011
The value of $\operatorname{Lt}_{x \rightarrow 0}\left(\frac{1+5 x^2}{1+3 x^2}\right)^{\frac{1}{x^2}}$ is

Options:
A $\mathrm{e}^2$
B $\mathrm{e}$
C $\frac{1}{\mathrm{e}}$
D $\frac{1}{\mathrm{e}^2}$

The correct answer is:
$\mathrm{e}^2$

Solution:
Hints: $\operatorname{Lim}_{x \rightarrow 0}\left(\frac{1+5 x^2}{1+3 x^2}\right)^{\frac{1}{x^2}}=\mathrm{e}^{\operatorname{Lim}_{x \rightarrow 0} \frac{1}{x^2}\left(\frac{1+5 x^2}{1+3 x^2}-1\right)}=\mathrm{e}^{\operatorname{Lim}_{x \rightarrow 0} \frac{2 x^2}{x^2\left(1+3 x^2\right)}}=\mathrm{e}^2$

Topic: Mathematics, Limits, WBJEE, WBJEE 2010
The value of $\operatorname{Lt}_{x \rightarrow 0} \frac{\sin ^2 x+\cos x-1}{x^2}$ is

Options:
A 1
B $\frac{1}{2}$
C $-\frac{1}{2}$
D 0

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

Solution:
Hints: $\operatorname{Lim}_{x \rightarrow 0} \frac{\sin ^2 x+\cos x-1}{x^2}=\operatorname{Lim}_{x \rightarrow 0} \frac{\cos x-\cos ^2 x}{x^2}=\operatorname{Lim}_{x \rightarrow 0}\left(\frac{1-\cos x}{x^2}\right) \cos x$
$$
=\operatorname{Lim}_{x \rightarrow 0} \frac{2 \sin ^2 \frac{x}{2}}{\left(\frac{x}{2}\right)^2 \times 4}=\frac{1}{2}
$$

Topic: Mathematics, Limits, WBJEE, WBJEE 2010
The value of $\sum_{\mathrm{n}=1}^{10} \int_{-2 n-1}^{-2 \pi} \sin ^{27} x d x+\sum_{n=1}^{10} \int_{2 n}^{2 n+1} \sin ^{27} x d x$ is equal to

Options:
A 27
B 54
C $-54$
D 0

The correct answer is:
0

Solution:
Hint:
$-\sum_{\mathrm{n}=1}^{10} \int_{2 n}^{2 n+1} \sin ^{27} x d x+\sum_{n=1}^{10} \int_{2 n}^{2 n+1} \sin ^{27} x d x$
$=0$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2020
The value of $\sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right), i=\sqrt{-1}$ is

Options:
A $i$
B $i-1$
C 1
D 0

The correct answer is:
$i-1$

Solution:
We have.
$\begin{aligned} & \sum_{n=1}^{13}\left(i^{n}+i^{n+1}\right) \\=& \sum_{n=1}^{13} i^{n}+\sum_{n=1}^{13} i^{n+1} \\=& i \left(\frac{1-i^{13}}{1-i}\right)+i^{2}\left(\frac{1-i^{13}}{1-i}\right) \\=& i \left(\frac{1-i}{1-i}\right)-\left(\frac{1-i}{1-i}\right) \\=& i-1 \end{aligned}$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2016
The value of $\mathrm{n}$ for which $\frac{\mathrm{x}^{\mathrm{n}+1}+\mathrm{y}^{\mathrm{n}+1}}{\mathrm{x}^{\mathrm{n}}+\mathrm{y}^{\mathrm{n}}}$ is the geometric mean of $\mathrm{x}$ and $\mathrm{y}$ is

Options:
A $\mathrm{n}=-\frac{1}{2}$
B $\mathrm{n}=\frac{1}{2}$
C $\mathrm{n}=1$
D $\mathrm{n}=-1$

The correct answer is:
$\mathrm{n}=-\frac{1}{2}$

Solution:
Hints: $\frac{x^{n+1}+y^{n+1}}{x^n+y^n}=\sqrt{x y} \Rightarrow x^{n+1}+y^{n+1}=\sqrt{x y}\left(x^n+y^n\right)$
$$
\mathrm{x}^{\mathrm{n}+\frac{1}{2}}\left(\mathrm{x}^{\frac{1}{2}}-\mathrm{y}^{\frac{1}{2}}\right)=\mathrm{y}^{\mathrm{n}+\frac{1}{2}}\left(\mathrm{x}^{\frac{1}{2}}-\mathrm{y}^{\frac{1}{2}}\right),\left(\frac{\mathrm{x}}{\mathrm{y}}\right)^{\mathrm{n}+\frac{1}{2}}=1 \quad \mathrm{n}=-\frac{1}{2}
$$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2010
The value of $\sum_{r=2}^{\infty} \frac{1+2+\quad+(r-1)}{r !}$

Options:
A $e$
B $2 e$
C $\frac{e}{2}$
D $\frac{3 e}{2}$

The correct answer is:
$\frac{e}{2}$

Solution:
$\sum_{r=2}^{\infty} \frac{(r-1) r}{2 r !}=\sum_{r=2}^{\infty} \frac{1}{2(r-2) !}$
$=\frac{1}{2}\left[\frac{1}{0 !}+\frac{1}{1 !}+\frac{1}{2 !}+\quad \infty\right]=\frac{1}{2} e$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2012
The value of $\frac{\sin 55^0-\cos 55^{\circ}}{\sin 10^0}$ is

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

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

Solution:
Hints: $\frac{\sin 55-\sin 35}{\sin 10}=\frac{2 \cos 45 \cdot \sin 10}{\sin 10}=\sqrt{2}$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2010
The value of $\lambda$ such that the system of equations $2x-y-2 y=2 ; x-2 y+z=-4$,$x+y+\lambda z=4,$ has no solution, is

Options:
A 3
B 1
C 0
D -3

The correct answer is:
-3

Solution:
If the given system of equations has no solution. then $\left|\begin{array}{ccc}2 & -1 & -2 \\ 1 & -2 & 1 \\ 1 & 1 & \lambda\end{array}\right|=0$
$\Rightarrow 2(-2 \lambda-1)+1(\lambda-1)-2(1+2)=0$
$\Rightarrow \quad-4 \lambda-2+\lambda-1-6=0$
$\Rightarrow \quad-3 \lambda-9=0$
$\Rightarrow \quad \lambda=-3$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2015