datavorous/entrance-exam-dataset huggingface dataset:
If $M$ is a $3 \times 3$ matrix such that $(0~1~2) M=(1~0~0),(3~4~5) M=(0~1~0)$, then $(6~7~8) M$ is equal to

Options:
A $(\left.\begin{array}{llll}2 & 1 & -2\end{array}\right)$
B $(\left.\begin{array}{llll}0 & 0 & 1\end{array}\right)$
C $(\left.\begin{array}{llll}-1 & 2 & 0\end{array}\right)$
D $(\left.\begin{array}{llll}9 & 10 & 8\end{array}\right)$

The correct answer is:
$(\left.\begin{array}{llll}-1 & 2 & 0\end{array}\right)$

Solution:
$\left(\begin{array}{lll}0 & 1 & 2\end{array}\right) \mathrm{m}=\left(\begin{array}{lll}1 & 0 & 0\end{array}\right)$ ....(i)
$\left(\begin{array}{lll}3 & 4 & 5\end{array}\right) \mathrm{m}=\left(\begin{array}{lll}0 & 1 & 0\end{array}\right)$
$\left(\begin{array}{lll}6 & 8 & 10\end{array}\right) \mathrm{m}=\left(\begin{array}{lll}0 & 2 & 0\end{array}\right)$ ...(ii)
(ii) $-$ (i)
$\left(\begin{array}{lll}6 & 8 & 10\end{array}\right) \mathrm{m}-\left(\begin{array}{lll}0 & 1 & 2\end{array}\right) \mathrm{m}$ = $\left(\begin{array}{lll}0 & 2 & 0\end{array}\right)-\left(\begin{array}{lll}1 & 0 & 0\end{array}\right)$
$\Rightarrow$ $\left(\begin{array}{lll}6 & 7 & 8\end{array}\right) \mathrm{m}=\left(\begin{array}{lll}-1 & 2 & 0\end{array}\right)$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2021
If $M$ is any square matrix of order 3 over $R$. and if $M^{\prime}$ be the transpose of $M$, then $adj\left(M^{\prime}\right)-(\text { adj } M)^{\prime}$ is equal to

Options:
A $M$
B $M{'}$
C null matrix
D identity matrix

The correct answer is:
null matrix

Solution:
$$
\begin{aligned}
&\begin{aligned}
\text { Since, } \operatorname{adj}\left(M^{\prime}\right)=&(\operatorname{adj} M)^{\prime} \\
=& \operatorname{adj}\left(M^{\prime}\right)-(\operatorname{adj} M)^{\prime} \\
&\left[\because \operatorname{adj}\left(A^{\prime}\right)=(\operatorname{adj} A)^{\prime}\right]
\end{aligned}\\
&=0 \text { , a null matrix. }
\end{aligned}
$$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2019
If \(\mathrm{A}\) and \(\mathrm{B}\) are two matrices such that \(\mathrm{A}+\mathrm{B}\) and \(\mathrm{AB}\) are both defined, then

Options:
A A and \(B\) can be any matrices
B A, B are square matrices not necessarily of the same order
C A, B are square matrices of the same order
D Number of columns of \(A=\) number of rows of \(B\)

The correct answer is:
A, B are square matrices of the same order

Solution:
Hints : Addition is defined if order of \(A\) is equal to order of \(B\)
A B
\(\mathrm{nxm} \mathrm{nxm}\) is defined if \(\mathrm{m}=\mathrm{n}\)
\(\Rightarrow A, B\) are square matrices of same order

Topic: Mathematics, Matrices, WBJEE, WBJEE 2011
If \(\mathrm{g}\) is the acceleration due to gravity on the surface of the earth, the gain in potential energy of an object of mass \(\mathrm{m}\) raised from the earth's surface to a height equal to the radius \(R\) of the earth is

Options:
A \(\frac{\mathrm{mg} \mathrm{R}}{4}\)
B \(\frac{m g R}{2}\)
C \(m g R\)
D \(2 \mathrm{mg} \mathrm{R}\)

The correct answer is:
\(\frac{m g R}{2}\)

Solution:
Hints: \(\Delta \mathrm{U}=\frac{\mathrm{mgh}}{1+\frac{\mathrm{h}}{\mathrm{R}}}=\frac{\mathrm{mgR}}{1+\frac{\mathrm{R}}{\mathrm{R}}}=\frac{\mathrm{mgR}}{2}\)

Topic: Physics, Gravitation, WBJEE, WBJEE 2011
If \(\mathrm{n} > 1\) is an integer and \(\mathrm{x} \neq 0\), then \((1+\mathrm{x})^{\mathrm{n}}-\mathrm{nx}-1\) is divisible by

Options:
A \(\mathrm{nx}^3\)
B \(\mathrm{n}^3 \mathrm{x}\)
C \(\mathrm{x}\)
D \(\mathrm{nx}\)

The correct answer is:
\(\mathrm{x}\)

Solution:
\(\begin{array}{r}\text {Hints: }
(1+\mathrm{x})^n={ }^n C_0+{ }^n C_1 x+{ }^n C_2 x^2+{ }^n C_3 \mathrm{x} 3+\ldots \ldots . \\
=1+\mathrm{nx}+\mathrm{x}^2\left({ }^n C_2+{ }^n C_3 \mathrm{x}+\ldots \ldots . .\right) \\
(1+\mathrm{x})^{\mathrm{n}}-\mathrm{nx}-1=\mathrm{x}^2\left({ }^n C_2+{ }^n C_3 \mathrm{x}+\ldots \ldots .\right)
\end{array}\)

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2011
If momentum is increased by $20 \%$, then kinetic energy increases by

Options:
A $48 \%$
B $44 \%$
C $40 \%$
D $36 \%$

The correct answer is:
$44 \%$

Solution:
Hints: $\mathrm{K}=\frac{\mathrm{P}^2}{2 m}$
Here $\mathrm{P}^{\prime}=1.2 \mathrm{P}$
Hence, $K^{\prime}=\frac{(1.2 \mathrm{P})^2}{2 m}$
$$
\mathrm{K}^{\prime}=1.44 \frac{\mathrm{P}^2}{2 m}
$$
$\mathrm{K}^{\prime}=1.44 \mathrm{~K}$ or Percentage increase in $\mathrm{K}=44 \%$

Topic: Physics, Work Power Energy, WBJEE, WBJEE 2010
If ${ }^{\mathrm{n}-1} \mathrm{C}_3+{ }^{\mathrm{n}-1} \mathrm{C}_4>{ }^{\mathrm{n}} \mathrm{C}_3$, then $\mathrm{n}$ is just greater than integer

Options:
A 5
B 6
C 4
D 7

The correct answer is:
7

Solution:
$$
\begin{aligned}
& \text { Hints: }:^{n-1} C_3+{ }^{n-1} C_4>{ }^n C_3 \\
& \qquad \Rightarrow^n C_4>^n C_3 \Rightarrow \frac{n !}{4 !(n-4) !}>\frac{n !}{3 !(n-3) !} \Rightarrow \frac{1}{4}>\frac{1}{(n-3)} \Rightarrow n-3>4 \Rightarrow n>7
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2010
If ${ }^{n} C_{r-1}=36,{ }^{n} C_{r}=84$ and ${ }^{n} C_{r+1}=126$ then the value of ${ }^{n} C_{8}$ is

Options:
A 10
B 7
C 9
D 8

The correct answer is:
9

Solution:
Given,
$$
\begin{aligned}
{ }^{n} C_{r-1} &=36 \\
{ }^{n} C_{r} &=84 \\
{ }^{n} C_{r+1} &=126
\end{aligned}
$$
On dividing
$$
\frac{r}{n-r+1}=\frac{36}{84}
$$
$\Rightarrow \quad 84 r=36 n-36 r+36$
$\Rightarrow \quad 120 r=36 n+36$
Also,
$$
\begin{array}{r}
\frac{r+1}{n-r}=\frac{64}{126} \\
\Rightarrow \quad 126 r+126=84 n-84 r
\end{array}
$$
$\Rightarrow$
$$
210r=84 n-126
$$
On solving Eqs. (iv) and (v). we get $n=9$ and $r=3$
\begin{array}{l}
\hline
\end{array}
$$
{ }^{n} C_{8}={ }^{9} C_{8}=9
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2016
If \({ }^n C_4,{ }^n C_5\) and \({ }^n C_6\) are in A.P., then \(n\) is

Options:
A 7 or 14
B 7
C 14
D 14 or 21

The correct answer is:
7 or 14

Solution:
Hints: \({ }^n \mathrm{C}_4,{ }^n \mathrm{C}_5,{ }^n \mathrm{C}_6\) are in \(\mathrm{AP}\)
\(\begin{aligned}
& \text { 2. }{ }^n \mathrm{C}_5={ }^n \mathrm{C}_4+{ }^n \mathrm{C}_6 \\
& \frac{2}{5(n-5)}=\frac{1}{(n-4)}+\frac{1}{30} \\
&
\end{aligned}\)
by solving \(n=14\) or 7

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2011
If $n$ denotes a positive integer, $h$ the Planck constant, $q$ the charge and $B$ the magnetic field, then the quantity $\left[\frac{n h}{2 \pi q B}\right]$ has the

Options:
A area
B length
C speed
D acceleration

The correct answer is:
area

Solution:
$\left[\frac{n h}{2 \pi q B}\right]$
where $n$ and $2 \pi$ dimensionless quantity
So, $[h]=[m v r]$
$\begin{aligned}
\mid q B] &=\left[\frac{q v B}{v}\right]=\left[\frac{F}{v}\right] \\
\therefore \quad\left[\frac{m v r}{F / v}\right] &=\frac{[m v r][v]}{[F]} \\
&=\left[\frac{\left[m v^{2} r\right]}{r}\right]=\left[r^{2}\right\}=\text { Area }
\end{aligned}$

Topic: Physics, Units and Dimensions, WBJEE, WBJEE 2014
If $n$ is even posirive integer, then the condition that the greatest term in the expansion of $(a+x)^{n}$ may also have the greatest coefficient, is

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

The correct answer is:
$\frac{n}{n+2} < x < \frac{n+2}{n}$

Solution:
For greatest term of $(1+x)^{n},$ we have
$\quad \frac{n}{2} < \frac{n+1}{1+x} < \frac{n}{2}+1$
$\Rightarrow \quad \frac{n}{2} < \frac{n+1}{1+x}$ and $\frac{n+1}{1+x} < \frac{n}{2}+1$
$\Rightarrow \quad 1+x < \frac{n+1}{n / 2}$ and $\frac{n+1}{\frac{n}{2}+1} < 1+x$
$\Rightarrow \quad x < \frac{n+1-n / 2}{n / 2}$
and $\quad \frac{n+1-(n / 2-1)}{\frac{n+2}{2}} < x$
$\Rightarrow \quad x < \frac{n+2}{n}$ and $\frac{n}{n+2} < x$
$\Rightarrow \quad \frac{n}{n+2} < x < \frac{n+2}{n}$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2018
If $n_t$ number of radioatoms are present at time $t$, the following expression will be a constant :

Options:
A $\mathrm{n}_{\mathrm{t}} / \mathrm{t}$
B $\operatorname{In~}_{n_t / t}$
C $\mathrm{d} \operatorname{In} \mathrm{n} / \mathrm{dt}$
D $\mathrm{tn}_{\mathrm{t}}$

The correct answer is:
$\mathrm{d} \operatorname{In} \mathrm{n} / \mathrm{dt}$

Solution:
$\frac{d N}{d t}=\lambda \mathrm{N} \Rightarrow-\frac{d \ln N}{d t}=\lambda
Hence C

Topic: Chemistry, Chemical Kinetics, WBJEE, WBJEE 2009
If \(\omega \neq 1\) is a cube root of unity, then the sum of the series \(S=1+2 \omega+3 \omega^2+\ldots \ldots \ldots . .+3 n \omega^{3 n-1}\) is

Options:
A \(\frac{3 n}{\omega-1}\)
B \(3 n(\omega-1)\)
C \(\frac{\omega-1}{3 n}\)
D 0

The correct answer is:
\(\frac{3 n}{\omega-1}\)

Solution:
\(\begin{aligned}
\text {Hints : } & s=1+2 \omega+3 \omega^2+\ldots \ldots \ldots .+3 n \omega^{3 n-1} \\
& s \omega=\omega+2 \omega^2+\ldots \ldots \ldots \ldots .+(3 n-1) \omega^{3 n}+3 n \omega^{3 n} \\
& s(1-\omega)=1+\omega+\omega^2+\ldots \ldots \ldots .+\omega^{3 n-1}-3 n \omega^{3 \mathrm{~h}} \\
& =0-3 n \\
& s=\frac{-3 n}{1-\omega}=\frac{3 n}{\omega-1}
\end{aligned}\)

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2011
If one end of a diameter of the circle $3 x^{2}+3 y^{2}-9 x+6 y+5=0$ is $(1,2),$ then the other end is

Options:
A (2,1)
B (2,4)
C (2,-4)
D (-4,2)

The correct answer is:
(2,-4)

Solution:
Given equation of circle is
$$
\begin{array}{r}
3 x^{2}+3 y^{2}-9 x+6 y+5=0 \\
\Rightarrow \quad x^{2}+y^{2}-3 x+2 y+\frac{5}{3}=0
\end{array}
Centre
=\left(\frac{3}{2},-1\right)
$$
and radius $=\sqrt{\frac{9}{4}+1-\frac{5}{3}}$
$$
=\sqrt{\frac{19}{12}}=\frac{1}{2} \sqrt{\frac{19}{3}}
$$
We know that, centre of the circle is the mid-point of the diameter.
Lives one and of point of diameter in (1,2) Let the other end point of diameter is $(h, k)$
Then, $\quad\left(\frac{3}{2},-1\right)=\left(\frac{1+h}{2}, \frac{2+k}{2}\right)$
$\Rightarrow \quad \frac{1+h}{2}=\frac{3}{2}$
$\Rightarrow 1+h=3$
$\Rightarrow h=2$
and $\frac{2+k}{2}=-1$
$\Rightarrow 2+k=-2$
$\Rightarrow k=-4$
So, the other end point is (2,-4)

Topic: Mathematics, Circle, WBJEE, WBJEE 2013
If one of the diameter of the circle, given by the equation $x^{2}+y^{2}+4 x+6 y-12=0,$ is a chord of a circle $S,$ whose centre is $(2,-3$ ), the radius of $S$ is

Options:
A $\sqrt{41}$ unit
B $3 \sqrt{5}$ unit
C $5 \sqrt{2}$ unit
D $2 \sqrt{5}$ unit

The correct answer is:
$\sqrt{41}$ unit

Solution:
Given, equation of circle is $x^{2}+y^{2}+4 x+6 y-12=0$
Centre $C(-2,-3)$ and radius $=\sqrt{(-2)^{2}+(-3)^{2}+12}=5$
$$
x^{2}+y^{2}+4 x+6 y-12=0
$$
$$
\begin{aligned}
C_{1} C_{2} &=\sqrt{\left(2+{2})^{2}+(-3+3)^{2}\right.} \\
&=\sqrt{(4)^{2}+(0)^{2}}=4
\end{aligned}
$$
$\therefore$ Radius of circle, $S$
$$
=\sqrt{(4)^{2}+({5})^{2}}=\sqrt{16+25}=\sqrt{41} \text { unit }
$$

Topic: Mathematics, Circle, WBJEE, WBJEE 2018
If one of the diameters of the curve $x^{2}+y^{2}-4 x-6 y+9=0$ is a chord of a circle with centre $(1,1),$ the radius of this circle is

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

The correct answer is:
3

Solution:
Given circle
$x^{2}+y^{2}-4 x-6 y+9=0$
$\therefore$ Centre $=\left(-\frac{1}{2} \times(-4),-\frac{1}{2} \times(-6)\right)$
$=(2.3)$
$\begin{aligned} \text { Radius } &=\sqrt{(-2)^{2}+(-3)^{2}-9} \\ &=\sqrt{4+9-9} \\ &=2 \end{aligned}$

$$
\begin{aligned}
\therefore \quad A B &=\sqrt{(2-1)^{2}+(3-1)^{2}}=\sqrt{1+4}=\sqrt{5} \\
\therefore \quad A C &=\sqrt{A B^{2}+B C^{2}}=\sqrt{(\sqrt{5})^{2}+(2)^{2}} \\
&=\sqrt{5+4}=\sqrt{9}=3
\end{aligned}
$$
Radius of required circle $=3$

Topic: Mathematics, Circle, WBJEE, WBJEE 2017
If one root of \(x^2+p x-q^2=0, p\) and \(q\) are real, be less than 2 and other be greater than 2, then

Options:
A \(4+2 p+q^2 > 0\)
B \(4+2 p+q^2 < 0\)
C \(4+2 p-q^2 > 0\)
D \(4+2 p-q^2 < 0\)

The correct answer is:
\(4+2 p-q^2 < 0\)

Solution:
Hint :
\(\alpha < 2\) and \(\beta > 2\)
Let, \(\mathrm{p}(\mathrm{x})=\mathrm{x}^2+\mathrm{px}-\mathrm{q}^2\)
\(\begin{aligned}
& \therefore \mathrm{p}(2) < 0 \\
& \Rightarrow 2^2+2 p-\mathrm{q}^2 < 0 \\
& \Rightarrow 4+2 p-\mathrm{q}^2 < 0
\end{aligned}\)

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2023
If $P(0,0), Q(1,0)$ and $R\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$ are three given points, then the centre of the circle for which the lines $P Q, Q R$ and $R P$ are the tangents is

Options:
A $\left(\frac{1}{2}, \frac{1}{4}\right)$
B $\left(\frac{1}{2}, \frac{\sqrt{3}}{4}\right)$
C $\left(\frac{1}{2}, \frac{1}{2 \sqrt{3}}\right)$
D $\left(\frac{1}{2}, \frac{-1}{\sqrt{3}}\right)$

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

Solution:
$\begin{aligned}
&\text { } I=\left(\frac{a x_{1}+b x_{2}+c x_{3}}{a+b+c}, \frac{a y_{1}+b y_{2}+c y_{3}}{a+b+c}\right)\\
&a=\sqrt{\frac{1}{4}+\frac{3}{4}}=1, b=1, c=1\\
&\text { Centre of the circle. }\\
&=\left(\frac{1 \times 0+1 \times 1+1 \times \frac{1}{2}}{1+1+1}, \frac{1 \times 0+1 \times 0+1 \times \sqrt{3} / 2}{1+1+1}\right)\\
&=\left(\frac{0+1+\frac{1}{2}}{3} \frac{0+0+\frac{\sqrt{3}}{2}}{3}\right)\\
&=\left(\frac{3}{\frac{2}{3}}, \frac{\frac{\sqrt{3}}{2}}{3}\right)
\end{aligned}$
$=\left(\frac{3}{2} \times \frac{1}{3}, \frac{\sqrt{3}}{2} \times \frac{1}{3}\right)$
$=\left(\frac{1}{2}, \frac{1}{2 \sqrt{3}}\right)$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2019
If $P=\left[\begin{array}{lll}1 & 2 & 1 \\ 1 & 3 & 1\end{array}\right] Q=P P^{T}$, then the value of the
determinant of $Q$ is

Options:
A 2
B -2
C 1
D 0

The correct answer is:
2

Solution:
Given, $P=\left[\begin{array}{lll}1 & 2 & 1 \\ 1 & 3 & 1\end{array}\right]$
$$
\begin{array}{l}
\therefore \quad Q=P P^{T}=\left[\begin{array}{lll}
1 & 2 & 1 \\
1 & 3 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 1 \\
2 & 3 \\
1 & 1
\end{array}\right] \\
=\left[\begin{array}{ll}
1 \times 1+2 \times 2+1 \times 1 & 1 \times 1+2 \times 3+1 \times 1 \\
1 \times 1+3 \times 2+1 \times 1 & 1 \times 1+3 \times 3+1 \times 1
\end{array}\right] \\
=\left[\begin{array}{ll}
1+4+1 & 1+6+1 \\
1+6+1 & 1+9+1
\end{array}\right]=\left[\begin{array}{lr}
6 & 8 \\
8 & 11
\end{array}\right] \\
\therefore \text { The determintant of } Q=\left|\begin{array}{lr}
6 & 8 \\
8 & 11
\end{array}\right| \\
=66-64=2
\end{array}
$$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2012
If $\mathrm{p}=\left[\begin{array}{lll}1 & \alpha & 3 \\ 1 & 3 & 3 \\ 2 & 4 & 4\end{array}\right]$ is the adjoint of the $3 \times 3$ matrix $\mathrm{A}$ and $\operatorname{det} \mathrm{A}=4$, then $\alpha$ is equal to

Options:
A 4
B 11
C 5
D 0

The correct answer is:
11

Solution:
$|\mathrm{p}|=2 \alpha-6=(\operatorname{det} \mathrm{A})^2=16$
$\Rightarrow \alpha=11$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2022
If $P=1+\frac{1}{2 \times 2}+\frac{1}{3 \times 2^{2}}+\ldots$
and $Q=\frac{1}{1 \times 2}+\frac{1}{3 \times 4}+\frac{1}{5 \times 6}+\ldots,$ then

Options:
A $P=Q$
B $2 P=Q$
C $P=2 Q$
D $P=4 Q$

The correct answer is:
$P=2 Q$

Solution:
Given,
$$
P=1+\frac{1}{2 \times 2}+\frac{1}{3 \times 2^{2}}+\ldots
$$
and $\quad Q=\frac{1}{1 \times 2}+\frac{1}{3 \times 4}+\frac{1}{5 \times 6}+\ldots$
Now, $\quad \frac{P}{2}=\frac{(1 / 2)^{1}}{1}+\frac{(1 / 2)^{2}}{2}+\frac{(1 / 2)^{3}}{3}+\ldots$
$\Rightarrow \quad-p / 2=-\left(\frac{1 / 2}{1}\right)^{1}-\frac{(1 / 2)^{2}}{3}-\frac{(1 / 2)^{3}}{3} \ldots$
$\Rightarrow \quad-P / 2=\log _{e}\left(1-\frac{1}{2}\right)$
$\Rightarrow \quad-\frac{P}{2}=\log _{\theta} \frac{1}{2}$
$\Rightarrow \quad P=2 \log _{e} 2$
Now, $Q=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\left(\frac{1}{5}-\frac{1}{6}\right)+\ldots$
$\Rightarrow \quad Q=\log _{e} 2$
Now, from Eqs. (i) and (ii), we get
$$
P=2 Q
$$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2013
If $P_1 P_2$ and $P_3 P_4$ are two focal chords of the parabola $y^2=4 a x$ then the chords $P_1 P_3$ and $P_2 P_4$ intersect on the

Options:
A directrix of the parabola
B axis of the parabola
C latus-rectum of the parabola
D $y$-axis

The correct answer is:
directrix of the parabola

Solution:
Let $P_i\left(a_i{ }^2, 2 a t_i\right), i=1,2,3,4$
then, $t_1 t_2=t_3 t_4=-1$
equation of $P_1 P_3$ : $\left(t_1+t_3\right) y=2 x+2 a t_1 t_3$ ...(1)
$P_2 P_4:\left(t_2+t_4\right) y=2 x+2 a t_2 t_4$ ...(2)
Putting $x=-a$ in equation (1) gives $y=\frac{2 a t_1 t_3-2 a}{t_1+t_3}$ Putting $x=-a$ in equation (2) gives
$y=\frac{2 \mathrm{at}_2 \mathrm{t}_4-2 \mathrm{a}}{\mathrm{t}_2+\mathrm{t}_4}=\frac{\frac{2 \mathrm{a}}{\mathrm{t}_4 \mathrm{t}_3}-2 \mathrm{a}}{-\frac{1}{\mathrm{t}_1}-\frac{1}{\mathrm{t}_3}}=\frac{2 \mathrm{a}\left(1-\mathrm{t}_1 \mathrm{t}_3\right)}{-\left(\mathrm{t}_1+\mathrm{t}_3\right)}=\frac{2 \mathrm{a}\left(\mathrm{t}_1 \mathrm{t}_3-1\right)}{\left(\mathrm{t}_1+\mathrm{t}_3\right)}$
$\therefore$ These lines meet at $\left(-\mathrm{a}, \frac{2 \mathrm{a}\left(\mathrm{t}_1 \mathrm{t}_3-1\right)}{\mathrm{t}_1+\mathrm{t}_3}\right)$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2022
If $P=\left(\begin{array}{rrr}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right),$ then $P^{5}$ is equal to

Options:
A $P$
B $2 P$
C $-P$
D $-2 P$

The correct answer is:
$P$

Solution:
Given, $P=\left(\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right)$
$\begin{aligned} P^{2} &=P \cdot P=\left(\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right)\left(\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right) \\ &=\left(\begin{array}{ccc}4+2-4 & -4-6+8 & -8 & -8+12 \\ -2-3+4 & 2+9-8 & 4+12-12 \\ 2+2-3 & -2-6+6 & -4-8+9\end{array}\right) \\ &=\left(\begin{array}{ccc}2 & -2 & -4 \\ -1 & 3 & 4 \\ 1 & -2 & -3\end{array}\right)=P \\ \therefore P^{4}=P^{2}=P \\ \Rightarrow P^{5}=P^{2}=P \end{aligned}$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2013
If $\mathrm{P}=\frac{1}{2} \sin ^2 \theta+\frac{1}{3} \cos ^2 \theta$ then

Options:
A $\frac{1}{3} \leq \mathrm{P} \leq \frac{1}{2}$
B $\mathrm{P} \geq \frac{1}{2}$
C $\quad 2 \leq \mathrm{P} \leq 3$
D $-\frac{\sqrt{13}}{6} \leq \mathrm{P} \leq \frac{\sqrt{13}}{6}$

The correct answer is:
$\frac{1}{3} \leq \mathrm{P} \leq \frac{1}{2}$

Solution:
Hints: $P=\frac{1}{2} \sin ^2 \theta+\frac{1}{3} \cos ^2 \theta=\frac{1}{2} \sin ^2 \theta+\frac{1}{3}\left(1-\sin ^2 \theta\right)=\frac{1}{3}+\frac{1}{6} \sin ^2 \theta$
$$
\begin{aligned}
& 0 \leq \sin ^2 \theta \leq 1 \Rightarrow \frac{1}{3} \leq \frac{1}{3}+\frac{1}{6} \sin ^2 \theta \leq \frac{1}{3}+\frac{1}{6} \\
& \Rightarrow \frac{1}{3} \leq \mathrm{P} \leq \frac{1}{2}
\end{aligned}
$$

Topic: Mathematics, Functions, WBJEE, WBJEE 2009
If $p^{\circ}$ and $p$ are the vapour pressure of the pure solvent and solution and $n_{1}$ and $n_{2}$ are the moles of solute and solvent respectively in the solution then the correct relation between $p$ and $p^{\circ}$ is

Options:
A $p^{\circ}=\rho\left[\frac{n_{1}}{n_{1}+n_{2}}\right]$
B $D^{\circ}=p\left[\frac{n_{2}}{n_{1}+n_{2}}\right]$
C $p=p^{\circ}\left[\frac{n_{2}}{n_{1}+n_{2}}\right]$
D $p=p^{\circ}\left[\frac{n_{1}}{n_{1}+n_{2}}\right]$

The correct answer is:
$p=p^{\circ}\left[\frac{n_{2}}{n_{1}+n_{2}}\right]$

Solution:
Given,
$$
\begin{array}{l}
p^{\circ}= \text {vapour pressure of pure solvent}\\
p=\text { vapour pressure of solution } \\
n_{1}=\text { moles of solute } \\
n_{2}=\text { moles of solvent }
\end{array}
$$
Thus, according to Raoult's law.
$$
\frac{p^{\circ}-p}{p^{\circ}}=x_{1}=\frac{n_{1}}{n_{1}+n_{2}}
$$
$\left(x_{1}=\right.$ mole fraction of solute)
or, $\quad 1-\frac{p}{p^{\circ}}=\frac{n_{1}}{n_{1}+n_{2}}$
$$
\frac{p}{p^{\circ}}=1-\frac{n_{1}}{n_{1}+n_{2}}
$$
$\Rightarrow$
$$
\begin{array}{c}
\frac{n_{1}+n_{2}-n_{1}}{n_{1}+n_{2}} \\
\frac{p}{p_{\circ}}=\frac{n_{2}}{n_{1}+n_{2}}
\end{array}
$$
$\Rightarrow p=p^{\circ} \left( \frac{n_{2}}{n_{1}+n_{2}} \right)$

Topic: Chemistry, Solutions, WBJEE, WBJEE 2016
If $P$ be a point on the parabola $y^{2}=4 a x$ with focus $F$. Let $Q$ denote the foot of the perpendicular from $P$ onto the directrix. Then, $\frac{\tan \angle P Q F}{\tan \angle P F Q}$ is

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

The correct answer is:
1

Solution:
Equation of parabola is
$$
y^{2}=4 a x
$$

Let the parametric coordinate of point $P$ on the parabola is $(a, 2 a)$
Now. $Q F=2 \sqrt{2} a$
$$
P Q=2 a \text { and } P F=2 a
$$
we observe that, $O F^{2}=P Q^{2}+P F^{2}$
$\Rightarrow \quad 8 a^{2}=4 a^{2}+4 a^{2}=8 a^{2}$
So, $\Delta Q P F$ form a right angle isoceles triangle. In which, $\angle P Q F=\angle P F Q$
$\Rightarrow$
$\tan \angle P Q F=\tan \angle P F Q$
$\Rightarrow$
$\frac{\tan \angle P Q F}{\tan \angle P F Q}=1$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2013
If $p=\left(\begin{array}{cc}\cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \\ \sin \frac{\pi}{4} & \cos \frac{\pi}{4}\end{array}\right)$ and $X=\left(\begin{array}{c}\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right)$. Then,
$P^{3} X$ is equal to

Options:
A $\left(\begin{array}{l}0 \\ 1\end{array}\right)$
B $\left(\begin{array}{c}-\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}}\end{array}\right)$
C $\left(\begin{array}{r}-1 \\ 0\end{array}\right)$
D $\left(\begin{array}{c}-\frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}}\end{array}\right)$

The correct answer is:
$\left(\begin{array}{r}-1 \\ 0\end{array}\right)$

Solution:
Given, $P=\left(\begin{array}{cc}\cos \frac{\pi}{4} & -\sin \frac{\pi}{4} \\ \sin \frac{\pi}{4} & \cos \frac{\pi}{4}\end{array}\right)=\left(\begin{array}{cc}\frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}}\end{array}\right)$
$\Rightarrow \quad P=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right)$
$\begin{aligned} \text { Now, } P^{2} &=P \cdot P=\frac{1}{2}\left(\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right)\left(\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right) \\ &=\frac{1}{2}\left(\begin{array}{cc}1-1 & -1 & -1 \\ 1+1 & -1+1\end{array}\right) \\ &=\frac{1}{2}\left(\begin{array}{cc}0 & -2 \\ 2 & 0\end{array}\right)=\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right) \end{aligned}$
$\begin{aligned} P^{3} &=P \cdot P^{2}=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}1 & -1 \\ 1 & 1\end{array}\right) \cdot\left(\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right) \\ &=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}0-1 & -1-0 \\ 0+1 & -1+0\end{array}\right) \\ &=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}-1 & -1 \\ 1 & -1\end{array}\right) \end{aligned}$
Also, given $X=\left(\begin{array}{c}1 / \sqrt{2} \\ \frac{1}{\sqrt{2}}\end{array}\right)=\frac{1}{\sqrt{2}}\left(\begin{array}{l}1 \\ 1\end{array}\right)$
$\begin{aligned} P^{3} X &=\frac{1}{\sqrt{2}}\left(\begin{array}{cc}-1 & -1 \\ 1 & -1\end{array}\right) \cdot \frac{1}{\sqrt{2}}\left(\begin{array}{l}1 \\ 1\end{array}\right) \\ &=\frac{1}{2}\left(\begin{array}{c}-1-1 \\ 1-1\end{array}\right)=\frac{1}{2}\left(\begin{array}{c}-2 \\ 0\end{array}\right)=\left(\begin{array}{c}-1 \\ 0\end{array}\right) \end{aligned}$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2013
If $P, Q$ and $R$ are angles of an isosceles triangle and $\angle P=\frac{\pi}{2},$ then the value of $\left(\cos \frac{P}{3}-i \sin \frac{P}{3}\right)^{3}+(\cos Q+i \sin Q) \\ (\cos R-i \sin R)+(\cos P-i \sin P) \\ (\cos Q-i \sin Q)(\cos R-i \sin R) \text { is }$

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

The correct answer is:
$-i$

Solution:
Given that, $P, Q$ and $R$ are angles of an isosceles triangle and $\angle P=\frac{\pi}{2}$
$\therefore \quad Q=R=\frac{\pi}{4}\left(\because P+Q+R=180^{\circ}\right)$
Now, $\left(\cos \frac{P}{3}-i \sin \frac{P}{3}\right)^{3}+(\cos Q+i \sin Q)$
$(\cos R-i \sin R)$
$+(\cos P-i \sin P)(\cos Q-i \sin Q)$
$\quad(\cos R-i \sin R)$
$=\left(e^{-i \frac{P}{3}}\right)^{3}+e^{i Q} \cdot e^{-i R}+e^{-i \rho} e^{-i Q} \cdot e^{-i R}$
$\left(\because \cos \theta+i \sin \theta=e^{i \theta}\right)$
$=e^{-i P}+e^{\gamma Q-R)}+e^{-i(P+Q+R)}$
$=e^{-i \pi / 2}+e^{(0)}+e^{-i \pi} \quad\left(Q=R=\frac{\pi}{2} P=\frac{\pi}{2}\right)$
$=-i+1+(-1)=-i$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2013