datavorous/entrance-exam-dataset huggingface dataset:
Let $I=\int_{10}^{19} \frac{\sin x}{1+x^{6}} d x$. Then,

Options:
A $|I| < 10^{-9}$
B $|I| < 10^{-7}$
C $|I| < 10^{-5}$
D $|I|>10^{-7}$

The correct answer is:
$|I| < 10^{-7}$

Solution:
For $x>10,$ we have
$|\sin x| < 1$ and $1+x^{8}>10^{8}$
$\Rightarrow \quad \frac{1}{1+x^{8}} \leq 10^{-8}$
$\therefore \quad\left|\int_{10}^{19} \frac{\sin x}{1+x^{8}} d x\right| \leq \int_{10}^{19} \frac{|\sin x|}{1+x^{8}} d x$
$\quad \leq \int_{10}^{19} 10^{-8}=9 \times 10^{-8} < 10^{-7}$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2017
Let $I$ denote the $3 \times 3$ identity matrix and $P$ be a matrix obtained by rearranging the columns of
$I$. Then,

Options:
A there are six distinct choices for $P$ and $\operatorname{det}(P)=1$
B there are six distinct choices for $P$ and $\operatorname{det}(P)=\pm 1$
C there are more than one choices for $P$ and some of them are not invertible
D there are more than one choices for $P$ and $P^{-1}=1$ in each choice

The correct answer is:
there are six distinct choices for $P$ and $\operatorname{det}(P)=\pm 1$

Solution:
Given, $I=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Then, $\quad \operatorname{det}(I)=1$
If we take $I$ as
$$
A_{1}=\left[\begin{array}{lll}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array}\right]
$$
Then, $\quad \operatorname{det}\left(I_{1}\right)=-1$
Similarly, there are four other possibilities,
$$
\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array}\right] \cdot\left[\begin{array}{lll}
0 & 0 & 1 \\
1 & 0 & 0 \\
0 & 1 & 0
\end{array}\right]
$$
$$
\left[\begin{array}{lll}
0 & 1 & 0 \\
0 & 0 & 1 \\
1 & 0 & 0
\end{array}\right] \cdot\left[\begin{array}{lll}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0
\end{array}\right]
$$
who will give a determinant either -1 or 1 . Hence, there are six distinct choices for $P$ and
$\operatorname{det}(P)=\pm 1$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2014
Let $\vec{\alpha}=\hat{i}+\hat{j}+\hat{k}, \vec{\beta}=\hat{i}-\hat{j}-\hat{k}$ and
$\vec{\gamma}=-\hat{i}+\hat{j}-\hat{k}$ be three vectors. A vector $\vec{\delta},$ in the plane of $\vec{\alpha}$ and $\vec{\beta}$, whose projection on $\vec{\gamma}$ is $\frac{1}{\sqrt{3}},$ is given by

Options:
A $-\hat{i}-3 \hat{j}-3 \hat{k}$
B $\hat{i}-3 \hat{j}-3 \hat{k}$
C $-\hat{i}+3 \hat{j}+3 \hat{k}$
D $\hat{i}+3 \hat{j}-3 \hat{k}$

The correct answer is:
$-\hat{i}+3 \hat{j}+3 \hat{k}$

Solution:
We have,
$\overrightarrow{8}=\vec{a}+\lambda \vec{\beta}$
$=(\hat{i}+\hat{j}+\hat{k})+\lambda(\hat{i}-\hat{j}-\hat{k})$
$\Rightarrow \quad \vec{\delta}=(1+\lambda) \hat{i}+(1-\lambda) \hat{j}+(1-\lambda) \hat{k}$
Given, $\quad \frac{\vec{\delta} \cdot \vec{\gamma}}{|\vec{\gamma}|}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{(1+\lambda)(-1)+(1-\lambda)(1)+(1-\lambda)(-1)}{\sqrt{3}}=\frac{1}{\sqrt{3}}$
$\Rightarrow \quad \lambda=-2$
$\therefore$
$\vec{\delta}=-\hat{i}+3 \hat{j}+3 \hat{k}$

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2018
Let $I_{n}=\int_{0}^{1} x^{n} \tan ^{-1} x d x .$ If $a_{n} I_{n+2}+b_{n} I_{n}=c_{n}$ for all $n \geq 1,$ then

Options:
A $a_{1}, a_{2}, a_{3}$ are in GP
B $b_{1}, b_{2}, b_{3}$ are in AP
C $c_{1}, c_{2}, c_{3}$ are in HP
D $a_{1}, a_{2}, a_{3}$ are in $A P$

The correct answers are:
$b_{1}, b_{2}, b_{3}$ are in AP, $a_{1}, a_{2}, a_{3}$ are in $A P$

Solution:
We have,
$I_{n}=\int_{0}^{1} x^{n} \tan ^{-1} x d x$
$=\left[\tan ^{-1} x \cdot \frac{x^{n+1}}{n+1}\right]_{0}^{1}-\int_{0}^{1} \frac{x^{n+1}}{n+1} \cdot \frac{1}{1+x^{2}} d x$
$=\frac{\pi}{4(n+1)}-\frac{1}{n+1} \int_{0}^{1} \frac{x^{2} \cdot x^{n-1}}{1+x^{2}} d x$
$=\frac{\pi}{4(n+1)}-\frac{1}{n+1} \int_{0}^{1} \frac{\left(1+x^{2}\right) x^{n-1}-x^{n-1}}{1+x^{2}} d x$
$=\frac{\pi}{4(n+1)}-\frac{1}{n+1} \int_{0}^{1} x^{n-1} d x+\frac{1}{n+1} \int_{0}^{1} \frac{x^{n-1}}{1+x^{2}} d x$
$=\frac{\pi}{4(n+1)}-\frac{1}{n+1}\left[\frac{x^{n}}{n}\right]_{0}^{1}+\frac{1}{n+1}$
$\left[x^{n-1} \cdot \tan ^{-1} x\right]_{0}^{1}-\int_{0}^{1} \tan ^{-1} x \cdot(n-1) x^{n-2} d x$
$\begin{aligned}=\frac{\pi}{4(n+1)}-\frac{1}{n(n+1)}+\frac{\pi}{4(n+1)}-\frac{n-1}{n+1} & \int_{0}^{1} x^{n-2} \tan ^{-1} x d x \end{aligned}$
$\Rightarrow \quad I_{n}=\frac{\pi}{2(n+1)}-\frac{1}{n(n+1)}-\frac{n-1}{n+2} I_{n-2}$
$\Rightarrow \quad I_{n}+\frac{n-1}{n+2} I_{n-2}=\frac{\pi}{2(n+1)}-\frac{1}{n(n+1)}$
Put $n=n+2,$ we get $I_{n+2}+\frac{n+1}{n+3} I_{n}=\frac{\pi}{2(n+3)}-\frac{1}{(n+2)(n+3)}$
$\Rightarrow(n+3) I_{n+2}+(n+1) I_{n}=\frac{\pi}{2}-\frac{1}{n+2}$
$\therefore a_{n}=n+3, b_{n}=n+1, c_{n}=\frac{\pi}{2}-\frac{1}{n+2}$
$\therefore a_{n}$ and $b_{n}$ are in $\mathrm{AP}$.

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2019
Let $\mathrm{I}(\mathrm{n})=\mathrm{n}^{\mathrm{n}}, \mathrm{J}(\mathrm{n})=1.3 .5 \ldots . .(2 \mathrm{n}-1)$ for all $(\mathrm{n}>1), \mathrm{n} \in \mathrm{N}$, then

Options:
A $\mathrm{I}(\mathrm{n})>\mathrm{J}(\mathrm{n})$
B $\mathrm{I}(\mathrm{n}) < \mathrm{J}(\mathrm{n})$
C $\mathrm{I}(\mathrm{n})=\mathrm{J}(\mathrm{n})$
D $\mathrm{I}(\mathrm{n})=\frac{1}{2} \mathrm{~J}(\mathrm{n})$

The correct answer is:
$\mathrm{I}(\mathrm{n})>\mathrm{J}(\mathrm{n})$

Solution:
Hint:
$\mathrm{AM} \geq \mathrm{GM}$
$\frac{1+3+5+7+\ldots+(2 n-1)}{\mathrm{n}}>(\mathrm{J}(\mathrm{n}))^{\frac{1}{n}}, \quad \frac{\mathrm{n}^{2}}{\mathrm{n}}>(\mathrm{J}(\mathrm{n}))^{\frac{1}{n}}, \quad \mathrm{n}^{\mathrm{n}}>\mathrm{J}(\mathrm{n}), \mathrm{I}(\mathrm{n})>\mathrm{J}(\mathrm{n})$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2020
Let $L$ be the length and d be the diameter of cross-section of a wire. Wires of the same material with different $L$ and $d$ are subjected to the same tension along the length of the wire. In which of the following cases, the extension of wire will be the maximum?

Options:
A $L=200 \mathrm{cm}, \mathrm{d}=0.5 \mathrm{mm}$
B $L=300 \mathrm{cm}, \mathrm{d}=1.0 \mathrm{mm}$
C $L=50 \mathrm{cm}, \quad d=0.05 \mathrm{mm}$
D $L=100 \mathrm{cm}, \mathrm{d}=0.2 \mathrm{mm}$

The correct answer is:
$L=50 \mathrm{cm}, \quad d=0.05 \mathrm{mm}$

Solution:
We know
$$
\Delta l=\frac{F \times l}{A \times Y}=\frac{4 L}{\pi d^{2}}
$$
Here $\frac{4}{\pi}$ is a constant.
Hence, $\Delta l \propto \frac{L}{d^{2}}$

Topic: Physics, Mechanical Properties of Solids, WBJEE, WBJEE 2012
Let $\lim _{c \rightarrow 0} \int_c^x \frac{b t \cos 4 t-a \sin 4 t}{t^2} d t=\frac{a \sin 4 x}{x}-1,(0 < x < \pi / 4)$. Then a and $b$ are given by

Options:
A $a=2, b=2$
B $a=1 / 4, b=1$
C $a=-1, b=4$
D $a=2, b=4$

The correct answer is:
$a=1 / 4, b=1$

Solution:
Let $g(x)=\lim _{c \rightarrow 0} \int_E^x \frac{b t \cos 4 t-a \sin 4 t}{t^2} d t=\frac{a \sin 4 x}{x}-1$
$\lim _{x \rightarrow 0} g(x)=0=4 a-1 \quad \Rightarrow a=1 / 4
$$
$$
g^{\prime}(x)=\frac{b x \cos 4 x-a \sin 4 x}{x^2}$
Comparing, $b=4 a=1$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2022
Let \(\mathrm{a}, \mathrm{b}, \mathrm{c}\) be three real numbers such that \(\mathrm{a}+2 \mathrm{~b}+4 \mathrm{c}=0\). Then the equation \(\mathrm{ax}^2+\mathrm{bx}+\mathrm{c}=0\)

Options:
A has both the roots complex
B hat its roots lying within \(-1 < x < 0\)
C has one of roots equal to \(\frac{1}{2}\)
D has its roots lying within \(2 < x < 6\)

The correct answer is:
has one of roots equal to \(\frac{1}{2}\)

Solution:
Hints: \(\frac{1}{4} \mathrm{a}+\frac{1}{2} \mathrm{~b}+\mathrm{c}=0\)
\(\begin{aligned}
& \left(\frac{1}{2}\right)^2 a+\left(\frac{1}{2}\right) b+c=0 \\
& \therefore x=\frac{1}{2}
\end{aligned}\)

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2011
Let \(\mathrm{f}(\mathrm{x})=\tan ^{-1} \mathrm{x}\). Then \(\mathrm{f}^{\prime}(\mathrm{x})+\mathrm{f}^{\prime \prime}(\mathrm{x})\) is \(=0\), when \(\mathrm{x}\) is equal to

Options:
A 0
B +1
C 1
D \(-i\)

The correct answer is:
+1

Solution:
Hints: \(f(x)=\tan ^{-1} x\)
\(\begin{aligned}
& f^{\prime}(x)=\frac{1}{1+x^2} \\
& f^{\prime \prime}(x)=\frac{-1}{\left(1+x^2\right)} \cdot 2 x, \frac{1}{1+x^2}=\frac{2 x}{\left(1+x^2\right)^2} \\
& 1+x^2=2 x,(x-1)^2=0 \\
& x=1
\end{aligned}\)

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2011
Let $n \geq 2$ be an integer. $A=\left[\begin{array}{ccc}\cos (2 \pi / n) & \sin (2 \pi / n) & 0 \\ -\sin (2 \pi / n) & \cos (2 \pi / n) & 0 \\ 0 & 0 & 1\end{array}\right]$ and $I$ is the identity matrix of order 3 . Then,

Options:
A $A^{n}=I$ and $A^{n-1} \neq I$
B $A^{m} \neq I$ for any positive integer $m$
C $A$ is not invertible
D $A^{m}=O$ for a positive integer $m$

The correct answer is:
$A^{n}=I$ and $A^{n-1} \neq I$

Solution:
Given,
$A=\left[\begin{array}{ccc}\cos \left(\frac{2 \pi}{n}\right) & \sin \left(\frac{2 \pi}{n}\right) & 0 \\ -\sin \left(\frac{2 \pi}{n}\right) & \cos \left(\frac{2 \pi}{n}\right) & 0 \\ 0 & 0 & 1\end{array}\right]$
Now, $A \times A=\left[\begin{array}{ccc}\cos \left(\frac{2 \pi}{n}\right) & \sin \left(\frac{2 \pi}{n}\right) & 0 \\ -\sin \left(\frac{2 \pi}{n}\right) & \cos \left(\frac{2 \pi}{n}\right) & 0 \\ 0 & 0 & 1\end{array}\right] \times$$\left[\begin{array}{ccc}\cos \left(\frac{2 \pi}{n}\right) & \sin \left(\frac{2 \pi}{n}\right) & 0 \\ -\sin \left(\frac{2 \pi}{n}\right) & \cos \left(\frac{2 \pi}{n}\right) & 0 \\ 0 & 0 & 1\end{array}\right]$
$\left[\begin{array}{ccc}\cos ^{2}\left(\frac{2 \pi}{n}\right)-\sin ^{2}\left(\frac{2 \pi}{n}\right) & \cos \left(\frac{2 \pi}{n}\right) \sin \left(\frac{2 \pi}{n}\right)+\sin \left(\frac{2 \pi}{n}\right) \cos \left(\frac{2 \pi}{n}\right) & 0\\ -\sin \left(\frac{2 \pi}{n}\right) \cos \left(\frac{2 \pi}{n}\right)-\sin \left(\frac{2 \pi}{n}\right) \cos \left(\frac{2 \pi}{n}\right) & -\sin ^{2}\left(\frac{2 \pi}{n}\right)+\cos ^{2}\left(\frac{2 \pi}{n}\right) & 0\\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{cccc}\cos \left(2 \times \frac{2 \pi}{n}\right) & 2 \sin \left(\frac{2 \pi}{n}\right) \cos \left(\frac{2 \pi}{n}\right) & 0 \\ -2 \sin \left(\frac{2 \pi}{n}\right) \cos \left(\frac{2 \pi}{n}\right) & \cos \left(2 \times \frac{2 \pi}{n}\right) & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{ccc}
\cos \left(2 \times \frac{2 \pi}{n}\right) & \sin \left(2 \times \frac{2 \pi}{n}\right) & 0 \\
-\sin \left(2 \times \frac{2 \pi}{n}\right) & \cos \left(2 \times \frac{2 \pi}{n}\right) & 0 \\
0 & 0 & 1
\end{array}\right]$
Similarly,
$A^{n}=\left[\begin{array}{cccc}\cos \left(2^{n-1} \times \frac{2 \pi}{n}\right) & \sin \left(2^{n-1} \times \frac{2 \pi}{n}\right) & 0 \\ -\sin \left(2^{n-1} \times \frac{2 \pi}{n}\right) & \cos \left(2^{n-1} \times \frac{2 \pi}{n}\right) & 0 \\ 0 & 0 & 1\end{array}\right]$
$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=I$
and
$\begin{aligned}
A^{n-1}=&\left[\begin{array}{cccc}
\cos \left(2^{n-2} \times \frac{2 \pi}{n}\right) & \sin \left(2^{n-2} \times \frac{2 \pi}{n}\right) & 0 \\
-\sin \left(2^{n-2} \times \frac{2 \pi}{n}\right) & \cos \left(2^{n-2} \times \frac{2 \pi}{n}\right) & 0 \\
0 & 0 & 1
\end{array}\right] \\
\neq I
\end{aligned}$
$\therefore A^{n-1} \neq I$ for any positive integer $m$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2014
Let $n$ be a positive even integer. If the ratio of the largest coefficient and the 2nd largest coefficient in the expansion of $(1+x)^{n}$ is $11: 10 .$ Then, the number of terms in the expansion of $(1+x)^{n}$ is

Options:
A 20
B 21
C 10
D 11

The correct answer is:
21

Solution:
Let the number of terms, $n=2 \mathrm{m}$
Now, by condition
$\frac{\text { Largest coefficient in }(1+x)^{\circ}}{\text { Second largest coefficient in }(1+x)^{n}}=\frac{11}{10}$
(given)
$\Rightarrow$
$$
\begin{aligned}
\frac{2 m}{2 m} C_{m} &=\frac{11}{10} \\
\frac{(m-1) !(m+1) !}{m ! m !} &=\frac{11}{10}
\end{aligned}
$$
$\Rightarrow \quad \frac{1}{m} \frac{m(m-1) !(m+1) m !}{m ! m !}=\frac{11}{10}$
$\Rightarrow \quad \frac{m+1}{m} \frac{m+m !}{m ! m !}=\frac{11}{10}$
$\Rightarrow \quad 10 m+10=11 m \Rightarrow m=10$
$\therefore \quad n=20$
Hence, total number of term
$$
=n+1=20+1=21
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2013
Let \(O\) be the vertex, \(Q\) be any point on the parabola \(x^2=8 y\). If the point \(P\) divides the line segment \(O Q\) internally in the ratio \(1: 3\), then the locus of \(P\) is

Options:
A \(x^2=y\)
B \(y^2=x\)
C \(y^2=2 x\)
D \(x^2=2 y\)

The correct answer is:
\(x^2=2 y\)

Solution:
Hint :
\(\begin{aligned} & O P: P Q=1: 3 \\ & \Rightarrow \mathrm{h}=\mathrm{t}, \mathrm{k}=\frac{\mathrm{t}^2}{2} \\ & \Rightarrow \mathrm{k}=\frac{\mathrm{h}^2}{2}\end{aligned}\)

Topic: Mathematics, Parabola, WBJEE, WBJEE 2023
Let $P(2,-3), Q(-2,1)$ be the vertices of the $\Delta P Q R .$ If the centroid of $\Delta P Q R$ lies on the line $2 x+3 y=1,$ then the locus of $R$ is

Options:
A $2 x+3 y=9$
B $2 x-3 y=7$
C $3 x+2 y=5$
D $3 x-2 y=5$

The correct answer is:
$2 x+3 y=9$

Solution:
Let the vertices of $R$ be $(h, k)$

$\therefore$ Centroid of triangle is given by
$$
\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)
$$
Le, $\left(\frac{2+h-2}{3}, \frac{-3+1+k}{3}\right)$
Le, $\left(\frac{h}{3}, \frac{-2+k}{3}\right)$
since, the point $\left(\frac{h}{3}, \frac{-2+k}{3}\right)$ lies on the line $2 x+3 y=1,$ therefore the point $\left(\frac{h}{3}, \frac{-2+k}{3}\right)$ must be satisfy the equation of the given line.
Le., $2\left(\frac{h}{3}\right)+3\left(\frac{-2+k}{3}\right)=1$
$\Rightarrow \quad \frac{2 h-6+3 k}{3}=1$
$\Rightarrow \quad 2 h-6+3 k=3$
$\Rightarrow h+3 k=9$
Replace $h=x$ and $k=y,$ we get the required locus of $R$. ie, $\quad 2 x+3 y=9$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2012
Let $\mathrm{P}(3 \sec \theta, 2 \tan \theta) \mathrm{Q}(3 \sec \phi, 2 \tan \phi)$ be two points on $\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{4}=1$ such that $\theta+\phi=\frac{\pi}{2} .0 < \theta, \phi < \frac{\pi}{2}$. Then the ordinate of the point of intersection of the normals at $P$ and $Q$ is

Options:
A $\frac{13}{2}$
B $-\frac{13}{2}$
C $\frac{5}{2}$
D $-\frac{5}{2}$

The correct answer is:
$-\frac{13}{2}$

Solution:
$\begin{aligned} & \frac{a^2 x}{x_1}+\frac{b^2 y}{y_1}=a^2+b^2, P(3 \sec \theta, 2 \tan \theta) \\ \therefore & \frac{9 x}{3 \sec \theta}+\frac{4 y}{2 \tan \theta}=9+4=13 \\ \Rightarrow & 3 x \cos \theta+2 y \cot \theta=13 \end{aligned}$
Similarly, $3 x \cos \phi+2 y \cot \phi=13$
$3 x \cos \left(\frac{\pi}{2}-\theta\right)+2 y \cot \left(\frac{\pi}{2}-\theta\right)=13$
$\Rightarrow 3 x \sin \theta+2 y \tan \theta=13 \ldots \ldots(2)$
$\therefore \frac{13-2 \mathrm{y} \cot \theta}{3 \cos \theta}=\frac{13-2 \mathrm{y} \tan \theta}{3 \sin \theta}$
$\Rightarrow 13 \sin \theta-2 y \cos \theta=13 \cos \theta-2 y \sin \theta$
$\Rightarrow 13 \tan \theta-2 y=13-2 y \tan \theta$
$\Rightarrow 13(\tan \theta-1)=-2 y(\tan \theta-1)$
$\therefore \mathrm{y}=-\frac{13}{2}$
$\therefore$ Ordinate $=-\frac{13}{2}$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2022
Let $P(4,3)$ be a point on the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=\mathrm{1}$. If the normal at $P$ intersects the $X$ -axis at (16,0) , then the eccentricity of the hyperbola is

Options:
A $\frac{\sqrt{5}}{2}$
B 2
C $\sqrt{2}$
D $\sqrt{3}$

The correct answer is:
2

Solution:
Given hyperbola equation
$H: \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$
$P(4,3) $ lie on $H$
$\Rightarrow$ $\frac{16}{a^{2}}-\frac{9}{b^{2}}=1$
Normal equation at $P(4,3)$ for $H$
$$
a^{2} y_{1}\left(x-x_{1}\right)+b^{2} x_{1}\left(y-y_{1}\right)=0
$$
$\Rightarrow \quad 3 a^{2}(x-4)+4 b^{2}(y-3)=0$
Normal cuts the $X$ -axis at (16,0) $\Rightarrow \quad 3 a^{2}(16-4)+4 b^{2}(0-3)=0$
$\Rightarrow \quad 3 a^{2} 12+4 b^{2}(-3)=0$
$\Rightarrow$ $36 a^{2}-12 b^{2}=0$
$\Rightarrow$ $3 a^{2}-b^{2}=0$
$\Rightarrow$ $3 a^{2}=b^{2}$
$\therefore$ Eccentricity of the hyperbola is $e=\frac{\sqrt{a^{2}+b^{2}}}{a}$
$=\frac{\sqrt{a^{2}+3 a^{2}}}{a}=\frac{\sqrt{4 a^{2}}}{a}=\frac{2 a}{a}=2$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2019
Let $P\left(a t^{2}, 2 a t\right), Q,$ R(ar $^{2}, 2ar)$ be three points on a parabola $y^{2}=4 a x$. If $P Q$ is the focal chord and $P K, Q R$ are parallel where the co-ordinates of $K$ is $(2 a, 0)$ then the value of $r$ is

Options:
A $\frac{t}{1-t^{2}}$
B $\frac{1-t^{2}}{t}$
C $\frac{t^{2}+1}{t}$
D $\frac{t^{2}-1}{t}$

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

Solution:
Here, coordinate of $Q$ will be $\left(\frac{a}{t^{2}}, \frac{-2 a}{t}\right)$
Slope of $Q R=\frac{2}{r-\frac{1}{t}}$
Slope of $P K=\frac{2 a t}{a t^{2}-2 a}=\frac{2 t}{t^{2}-2}$
since, slope of $Q R=$ slope of $P K$
$\therefore \quad \frac{2}{r-\frac{1}{t}}=\frac{2 t}{t^{2}-2}$
$\Rightarrow \quad r=\frac{t^{2}-1}{t}$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2018
Let $P$ and $Q$ be the points on the parabola $y^{2}=4 x$ so that the line segment $P Q$ subtends right angle at the vertex. If $P Q$ intersects the axis of the parabola at $R,$ then the distance of the vertex from $R$ is

Options:
A 1
B 2
C 4
D 6

The correct answer is:
4

Solution:
According to question, it is given that slope of $P X$ and $Q X$ are perpendicular to each other, i.e., slope of $P X \times$ slope of $X Q=-1$ $\Rightarrow \quad \frac{2 t}{t^{2}} \times \frac{2 m}{m^{2}}=-1$
$\Rightarrow \quad t_{m}=-4$
$P Q: \frac{x-t^{2}}{t^{2}-m^{2}}=\frac{y-2 t}{2 t-2 m}$

Let $P Q$ meets axis of parabola i.e, $x$ -axis at $R(\alpha, 0),$ then
$\frac{\alpha-t^{2}}{(t+m)(t-m)}=\frac{0-2 t}{2(t-m)}$
$\Rightarrow \quad \alpha-t^{2}=-t^{2}-t m=-t^{2}+4$
$\Rightarrow \quad \alpha=4 \Rightarrow A R=4$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2012
Let $P$ and $T$ be the subsets of $x$, $y$ -plane defined by $P=\left\{(x, y): x>0, y>0\right.$ and $\left.x^{2}+y^{2}=1\right\}$, $T=\left\{(x, y): x>0, y>0\right.$ and $\left.x^{8}+y^{8} < 1\right\}$. Then, $P \cap T$ is

Options:
A the void set $\phi$
B $P$
C $T$
D $P-T^{C}$

The correct answer is:
$P$

Solution:
From the graph, it is clear that there is common region which satisfies $x^{2}+y^{2}=1$ and $x^{8}+y^{8} < 1$
$P \cap T=P$

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2019
Let $P$ be a point on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$ and the line through $P$ parallel to the $Y$ -axis mects the circle $x^{2}+y^{2}=9$ at $Q$ where $P, Q$ are on the same side of the $X$ -axis. If $R$ is a point on $P Q$ such that $\frac{P R}{R Q}=\frac{1}{2},$ then the locus of $R$ is

Options:
A $\frac{x^{2}}{9}+\frac{9 y^{2}}{49}=1$
B $\frac{x^{2}}{49}+\frac{y^{2}}{9}=1$
C $\frac{x^{2}}{9}+\frac{y^{2}}{49}=1$
D $\frac{9 x^{2}}{49}+\frac{y^{2}}{9}=1$

The correct answer is:
$\frac{x^{2}}{9}+\frac{9 y^{2}}{49}=1$

Solution:
Since, point $P$ on the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$
$\therefore$
$P(3 \cos \theta, 2 \sin \theta)$
Now, equation of line parallel of $Y$ -axis is
$$
x=3 \cos \theta
$$
and above line meets circle at $Q$
$\therefore$
$Q(3 \cos \theta, 3 \sin \theta)$
Given, $\frac{P R}{R Q}=\frac{1}{2}$
$\therefore h=\frac{3 \cos \theta+6 \cos \theta}{3}, k=\frac{3 \sin \theta+4 \sin \theta}{3}$
$\Rightarrow \quad h=3 \cos \theta, \quad k=\frac{7}{3} \sin \theta$
$\Rightarrow \quad \cos \theta=h / 3, \sin \theta=\frac{3 k}{7}$
Now, $\cos ^{2} \theta+\sin ^{2} \theta=h^{2} / 9+\frac{9 k^{2}}{49}=1$
Hence, locus of a point is $\frac{x^{2}}{9}+\frac{9 y^{2}}{49}=1$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2018
Let $P$ be a variable point on a circle $C$ and $Q$ be a fixed point outside $C$. If $R$ is the midpoint of the line segment $P Q$, then locus of $R$ is

Options:
A a circle
B a circle and a pair of straight lines
C a rectangular hyperbola
D a pair of sraight lines

The correct answers are:
a circle

Solution:
Let circle be $\left(x-x_{1}\right)=r \cos \theta$
$y-y_{1}=r \sin \theta$
$h=\frac{a+x_{1}+r \cos \theta}{2}$
$k=\frac{b+y_{1}+r \sin \theta}{2}$
$\Rightarrow\left(h-\frac{\left(a+x_{1}\right)}{2}\right)^{2}+\left(k-\frac{\left(b+y_{1}\right)}{2}\right)^{2}=\frac{r^{2}}{4}$

Topic: Mathematics, Circle, WBJEE, WBJEE 2021
Let $P$ be the foot of the perpendicular from focus $S$ of hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ on the line $b x-a y=0$ and let $C$ be the centre of the hyperbola. Then, the area of the rectangle whose sides are equal to that of $S P$ and $C P$ is

Options:
A $2 a b$
B $a b$
C $\frac{\left(a^{2}+b^{2}\right)}{2}$
D $\frac{a}{b}$

The correct answer is:
$a b$

Solution:
Given, equation of hyperbola is $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

From figure,
$\begin{aligned} S P &=\mid \frac{a b e}{\sqrt{b^{2}+a^{2}}} \\ &=\left|\frac{a b e}{a e}\right|=b \end{aligned}$
and $C S=a e$
Again, $\Delta S P C$ is right angled triangle at $P$.
$$
\begin{aligned}
\therefore \quad C P &=\sqrt{C S^{2}-S P^{2}} \\
&=\sqrt{a^{2} e^{2}-b^{2}} \\
&=\sqrt{a^{2}\left(1+\frac{b^{2}}{a^{2}}\right)-b^{2}} \\
&=\sqrt{a^{2}+b^{2}-b^{2}}=a
\end{aligned}
$$
$\therefore$ Area of rectangle $=C P \times S P$ $=a b$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2017
Let $P$ be the mid-point of a chord joining the vertex of the parabola $y^{2}=8 x$ to another point on it. Then, the locus of $P$ is

Options:
A $y^{2}=2 x$
B $y^{2}=4 x$
C $\frac{x^{2}}{4}+y^{2}=1$
D $x^{2}+\frac{y^{2}}{4}=1$

The correct answer is:
$y^{2}=4 x$

Solution:
Let the chord be MN. Let other end of chord joining from vertex to it lying on $y^{2}=8 x$ be $N\left(2 t^{2}, 4 t\right)$

$\therefore$ Mid-point of $M N=\left(t^{2}, 2 t\right)$ $\therefore \quad x=t^{2}, y=2 t \Rightarrow x=\left(\frac{y}{2}\right)^{2}$
$\Rightarrow \quad y^{2}=4 x$
is the required locus of $P$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2012
Let $P$ be the point $(2,0)$ and $Q$ be a variable point on $(y-6)^2=2(x-4)$. Then the locus of mid-point of $P Q$ is

Options:
A $y^2+x+6 y+12=0$
B $y^2-x+6 y+12=0$
C $y^2+x-6 y+12=0$
D $y^2-x-6 y+12=0$

The correct answer is:
$y^2-x-6 y+12=0$

Solution:
$$
\begin{gathered}
\mathrm{P}(2,0) \\
\mathrm{Q}\left(4+\frac{1}{2} \mathrm{t}^2, 6+\mathrm{t}\right)
\end{gathered}
$$
Let midpoint of $\mathrm{PQ}$ be R(h, k)
$$
\begin{aligned}
&\mathrm{h}=3+\frac{1}{4} \mathrm{t}^2, \mathrm{k}=3+\frac{\mathrm{t}}{2} \\
&\Rightarrow(\mathrm{k}-3)^2=(\mathrm{h}-3) \\
&\Rightarrow \mathrm{y}^2-6 \mathrm{y}-\mathrm{x}+12=0
\end{aligned}
$$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2022
Let $P$ be the set of all non-singular matrices of order 3 over $R$ and $Q$ be the set of all orthogonal matrices of order 3 over $R .$ Then,

Options:
A $P$ is proper subset of $Q$
B $Q$ is proper subset of $P$
C Neither $P$ is proper subset of $Q$ nor $Q$ is proper subset of $P$
D $P \cap Q=\phi$, the void set

The correct answer is:
$Q$ is proper subset of $P$

Solution:
All the orthogonal matrix are non-singular matrix. $\therefore Q$ is proper subset of $P$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2017
Let $p, q$ and $r$ be the altitudes of a triangle with area $S$ and perimeter $2 t .$ Then, the value of $\frac{1}{p}+\frac{1}{q}+\frac{1}{r}$ is

Options:
A $\frac{s}{t}$
B $\frac{t}{S}$
C $\frac{s}{2 t}$
D $\frac{2 S}{t}$

The correct answer is:
$\frac{t}{S}$

Solution:
According to question, $S=\frac{1}{2} a \cdot p$

$$
\begin{array}{c}
=\frac{1}{2} b \cdot q=\frac{1}{2} c \cdot r \\
\therefore \quad \frac{1}{p}=\frac{a}{2 S}, \frac{1}{q}=\frac{b}{2 S}, \frac{1}{r}=\frac{c}{2 S} \\
\therefore \quad \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=\frac{1}{2 S}(a+b+c)=\frac{2 t}{2 S}=\frac{t}{S}
\end{array}
$$

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2012
Let $p, q$ and $r$ be the side4s opposite to the angles $P, Q$ and $R,$ respectively in a $\Delta P Q R$. Then, $2 p r \sin \left(\frac{p-Q+R}{2}\right)$ equals

Options:
A $p^{2}+q^{2}+r^{2}$
B $p^{2}+r^{2}-q^{2}$
C $q^{2}+r^{2}-p^{2}$
D $p^{2}+q^{2}-r^{2}$

The correct answer is:
$p^{2}+r^{2}-q^{2}$

Solution:
In $\Delta P Q R, P+Q+R=180^{\circ}$
$\begin{aligned} \therefore 2 p r \sin \left(\frac{P-Q+R}{2}\right) &=2 p r \sin \frac{180^{\circ}-Q-Q}{2} \\ &=2 p r \sin \left(90^{\circ}-Q\right) \\ &=2 p r \cos Q \\=&\left(\quad \ln \Delta P Q R \cos Q=\frac{p^{2}+r^{2}-q^{2}}{2 p r}\right) \\=& 2 p r\left(\frac{p^{2}+r^{2}-q^{2}}{2 p r}\right)=p^{2}+r^{2}-q^{2} \end{aligned}$

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2012