datavorous/entrance-exam-dataset huggingface dataset:
If the coefficient of $x^{8}$ in $\left(a x^{2}+\frac{1}{b x}\right)^{13}$ is equal to the coefficient of $x^{-8}$ in $\left(a x-\frac{1}{b x^{2}}\right)^{13},$ then $a$ and $b$ will satisfy the relation

Options:
A $a b+1=0$
B $a b=1$
C $a=1-b$
D $a+b=-1$

The correct answer is:
$a b+1=0$

Solution:
The general term in $\left(a x^{2}+\frac{1}{b x}\right)^{13}$ is
$T_{r+1}={ }^{13} C_{r}\left(a x^{2}\right)^{13-r}\left(\frac{1}{b x}\right)^{r}$
$$
={ }^{13} C_{r} a^{13-r} \times b^{-r}(x)^{26-3 r}
$$
For coefficient of $x^{8}$. put $26-3 r=8$
$\Rightarrow \quad 3 r=18 \Rightarrow r=6$
$\therefore$
$T_{7}={ }^{13} C_{6} a^{13-6} b^{-6} x^{8}$
$={ }^{13} C_{6} a^{7} b^{-6} x^{8}$
Now, the general term in $\left(a x-\frac{1}{b x^{2}}\right)^{13}$ is
$$
\begin{aligned}
T_{r+1}^{'} &={ }^{13} C_{r}(a x)^{13-r}\left(-\frac{1}{b x^{2}}\right)^{r} \\
&={ }^{13} C_{r}^{*} a^{13-r} \times b^{-r} \times(x)^{13-3 r}(-1)^{\prime}
\end{aligned}
$$
For coefficient of $x^{-8}$, put $13-3 r=-8$
According to the given condition, Coefficient of $x^{8}$ in $\left(a x^{2}+\frac{1}{b x}\right)^{13}$
$$
\begin{array}{l}
=\text { Coefficient of } x^{-8} \text {in }\left(a x-\frac{1}{b x^{2}}\right)^{13} \\
\therefore{ }^{13} C_{6} a^{7} b^{-6}=-{ }^{13} C_{7} a^{5} b^{-7} \\
\Rightarrow{ }^{13} C_{7} \frac{a^{7}}{b^{6}}=-{ }^{13} C_{7} \frac{a^{6}}{b^{7}} \\
\Rightarrow \quad \frac{a^{7}}{a^{6}}=-\frac{b^{6}}{b^{7}} \Rightarrow \quad a=-\frac{1}{b} \Rightarrow a b+1=0
\end{array}
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2014
If the co-efficients of $x^2$ and $x^3$ in the expansion of $(3+a x)^9$ be same, then the value of ' $a$ ' is

Options:
A $\frac{3}{7}$
B $\frac{7}{3}$
C $\frac{7}{9}$
D $\frac{9}{7}$

The correct answer is:
$\frac{9}{7}$

Solution:
$$
\begin{aligned}
& \text { Hints : }(3+a x)^9={ }^9 \mathrm{C}_0 3^9+{ }^9 \mathrm{C}_1 3^8(a x)+{ }^9 \mathrm{C}_2 37(a x)^2+{ }^9 \mathrm{C}_3 3{ }^6(a x)^3 \\
& { }^9 \mathrm{C}_2 3^7 a^2={ }^9 \mathrm{C}_3 3{ }^6 \mathrm{a}^3 \\
& \frac{9}{7}=a
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, WBJEE, WBJEE 2009
If the coordinates of one end of a diameter of the circle \(x^2+y^2+4 x-8 y+5=0\), is \((2,1)\), the coordinates of the other end is

Options:
A \((-6,-7)\)
B \((6,7)\)
C \((-6,7)\)
D \((7,-6)\)

The correct answer is:
\((-6,7)\)

Solution:
Hints: \(x^2+y^2+9 x-8 y+5=0\)
Centre circle \((-2,4)\)
\(\begin{aligned}
& \frac{\mathrm{h}+2}{2}=-2 \\
& \mathrm{~h}=-4-2=-6 \\
& \frac{\mathrm{k}+1}{2}=4 \Rightarrow \mathrm{k}=7 \\
& (\mathrm{~h}, \mathrm{k}) \rightarrow(-6,7)
\end{aligned}\)

Topic: Mathematics, Circle, WBJEE, WBJEE 2011
If the displacement, velocity and acceleration of a particle at time, $\mathrm{t}$ be $\mathrm{x}, \mathrm{v}$ and $\mathrm{f}$ respectively, then which one is true?

Options:
A $\mathrm{f}=\mathrm{v}^3 \frac{\mathrm{d}^2 t}{d \mathrm{x}^2}$
B $f=-v^3 \frac{d^2 t}{d x^2}$
C $\mathrm{f}=\mathrm{v}^2 \frac{\mathrm{d}^2 \mathrm{t}}{\mathrm{dx}^2}$
D $f=-v^2 \frac{d^2 t}{d x^2}$

The correct answer is:
$f=-v^3 \frac{d^2 t}{d x^2}$

Solution:
$$
\begin{aligned}
& \text { Hints: } \frac{d^2 t}{d x^2}=\frac{d\left(\frac{d t}{d x}\right)}{d x}=\frac{d\left(\frac{1}{v}\right)}{d x}=-\frac{1}{v^2} \frac{d v}{d t} \times \frac{1}{v} \\
& \Rightarrow f=-v^3 f \frac{d^2 t}{d x^2}
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2010
If the distance between the foci of an ellipse is equal to the length of the latusrectum, then its eccentricity is

Options:
A $\frac{1}{4}(\sqrt{5}-1)$
B $\frac{1}{2}(\sqrt{5}+1)$
C $\frac{1}{2}(\sqrt{5}-1)$
D $\frac{1}{4}(\sqrt{5}+1)$

The correct answer is:
$\frac{1}{2}(\sqrt{5}-1)$

Solution:
Given,
In ellipse the distance between the foci
= Length of the latusrectum
$\begin{array}{ll}\Rightarrow \quad & \text { 2ae }=\frac{2 b^{2}}{a} \\ \Rightarrow \quad & \quad a^{2} e=b^{2} \Rightarrow e=\frac{b^{2}}{a^{2}}\end{array}$
$\because \quad e^{2}=1-\frac{b^{2}}{a^{2}}$
$\Rightarrow \quad e^{2}=1-e \quad$ [from Eq.
$\Rightarrow e^{2}+e-1=0$
$\Rightarrow \quad e=\frac{-1 \pm \sqrt{1+4}}{2}$
(by quadratic formula) $\Rightarrow \quad e=\frac{-1 \pm \sqrt{5}}{2}$
$\Rightarrow$
$$
e=\frac{\sqrt{5}-1}{2} \quad(\because 1>e>0)
$$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2013
If the distance between the plane \(a x-2 y+z=k\) and the plane containing the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\) is \(\sqrt{6}\), then \(|k|\) is

Options:
A 36
B 12
C 6
D \(2 \sqrt{3}\)

The correct answer is:
6

Solution:
Hint: Equation of plane
\(\begin{aligned}
& \left|\begin{array}{ccc}
x-1 & y-2 & z-3 \\
2 & 3 & 4 \\
3 & 4 & 5
\end{array}\right|=0 \\
& \Rightarrow x-2 y+z=0 \quad ---(1)
\end{aligned}\)
& given \(\alpha x-2 y+z=k\) ---(2)
(1) // (2) \(\Rightarrow \alpha=1, k \neq 0\)
\(\therefore \frac{|k|}{\sqrt{6}}=\sqrt{6} \Rightarrow|k|=6\)

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2023
If the Earth were to suddenly contract to \(\frac{1}{n}\) th of its present radius without any change in its mass, the duration of the new day will be nearly

Options:
A \(24 / \mathrm{n} \mathrm{hr}\).
B \(24 \mathrm{n} \mathrm{hr}\).
C \(24 / \mathrm{n}^2 \mathrm{hr}\).
D \(24 n^2 h r\).

The correct answer is:
\(24 / \mathrm{n}^2 \mathrm{hr}\).

Solution:
Hints: \(\mathrm{I}_1 \omega_1=\mathrm{I}_2 \omega_2\)
\(\begin{aligned}
& \frac{2}{5} \mathrm{MR}^2\left(\frac{2 \pi}{\mathrm{T}_1}\right)=\frac{2}{5} \mathrm{M} \cdot \frac{\mathrm{R}^2}{\mathrm{n}^2}\left(\frac{2 \pi}{\mathrm{T}_2}\right) \\
& \mathrm{T}_2=\frac{\mathrm{T}_1}{\mathrm{n}^2}=\frac{24}{\mathrm{n}^2}
\end{aligned}\)

Topic: Physics, Rotational Motion, WBJEE, WBJEE 2011
If the electron in a hydrogen atom jumps from an orbit with level $\mathrm{n}_1=2$ to an orbit with level $\mathrm{n}_2=1$ the emitted radiation has a wavelength given by

Options:
A $\lambda=5 / 3 \mathrm{R}$
B $\lambda=4 / 3 \mathrm{R}$
C $\lambda=\mathrm{R} / 4$
D $\lambda=3 \mathrm{R} / 4$

The correct answer is:
$\lambda=4 / 3 \mathrm{R}$

Solution:
$\frac{1}{\lambda}=R\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)=R\left(\frac{1}{1^2}-\frac{1}{2^2}\right)=\frac{3 R}{4}$
$\Rightarrow \lambda=\frac{4}{3 R}$

Topic: Physics, Atomic Physics, WBJEE, WBJEE 2009
If the equation of one tangent to the circle with centre at $(2,-1)$ from the origin is $3 x+y=0$, then the equation of the other tangent through the origin is

Options:
A $3 x-y=0$
B $x+3 y=0$
C $x-3 y=0$
D $x+2 y=0$

The correct answer is:
$x-3 y=0$

Solution:
$$
\begin{aligned}
&\because \mathrm{CA}=\mathrm{CB} \\
&\Rightarrow \frac{5}{\sqrt{10}}=\left|\frac{2 \mathrm{~m}+1}{\sqrt{1+\mathrm{m}^2}}\right|
\end{aligned}
$$
squaring
$$
\frac{25}{10}=\frac{4 m^2+4 m+1}{1+m^2}
$$
$$
\Rightarrow 5+5 m^2=8 m^2+8 m+2 \Rightarrow 3 m^2+8 m-3=0 \Rightarrow m=\frac{1}{3},-3
$$
$\therefore$ Equation of tangent $\mathrm{OB}$ is $y=\frac{x}{3}$

Topic: Mathematics, Circle, WBJEE, WBJEE 2022
If the equation $x^{2}-ca+d=0$ has roots equal to the fourth powers of the roots of $x^{2}+a x+b=0,$ where $a^{2}>4 b,$ then the roots of $x^{2}-4 b x+2 b^{2}-c=0$ will be

Options:
A both real
B both negative
C both positive
D one positive and one negative

The correct answers are:
both real, one positive and one negative

Solution:
Let $\alpha$ and $\beta$ are the roots of $x^{2}+a x+b=0$ and the roots of $x^{2}-cx+d=0$ are $\alpha^{4}$ and $\beta^{4}$
Now, $\alpha+\beta=-a, \alpha \beta=b$ ...(i)
and $\alpha^{4}+\beta^{4}-c, \alpha^{4} \beta^{4}=d$ ...(ii)
From Eqs. (i) and (ii),
$b^{4}=d$ and $\alpha^{4}+\beta^{4}=c$
$\left(\alpha^{2}+\beta^{2}\right)^{2}-2(\alpha \beta)^{2}=c$
$\left.\Rightarrow \quad\left[(\alpha+\beta)^{2}-2 \alpha \beta\right]^{2}-2 (\alpha\beta\right)^{2}=c$
$\Rightarrow \quad\left(a^{2}-2 b\right)^{2}-2 b^{2}=c$
$\Rightarrow \quad 2 b^{2}+c=\left(a^{2}-2 b\right)^{2}$
$\Rightarrow \quad 2 b^{2}+c=a^{4}+4 b^{2}-4 a^{2} b$
$\Rightarrow \quad 2 b^{2}-c=4 a^{2} b-a^{4}$
$\Rightarrow \quad 2 b^{2}-c=a^{2}\left(4 b-a^{2}\right)$
and for equation $x^{2}-4 b x+2 b^{2}-c=0$
$$
D=(4 b)^{2}-4(1)\left(2 b^{2}-c\right)
$$
$=16 b^{2}-8 b^{2}+4 c=8 b^{2}+4 c$
$=4\left(2 b^{2}+c\right)=4\left(a^{2}-2 b\right)^{2}>0=$ real root
Now, $f(0)=2 b^{2}-c=a^{2}\left(4 b-a^{2}\right) < 0 \quad\left(\because a^{2}>4 b\right)$
= roots are opposite in sign

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2018
If the equation $x^{2}+y^{2}-10 x+21=0$ has real roots $x=\alpha$ and $y=\beta,$ then

Options:
A $3 \leq x \leq 7$
B $3 \leq y \leq 7$
C $-2 \leq y \leq 2$
D $-2 \leq x \leq 2$

The correct answers are:
$3 \leq x \leq 7$, $-2 \leq y \leq 2$

Solution:
Given, $\quad x^{2}+y^{2}-10 x+21=0$
$\Rightarrow$
$x^{2}-10 x+y^{2}+21=0$
Now, for real roots of $x, D \geq 0$
$\Rightarrow$
$$
\begin{array}{r}
100-4\left(y^{2}+21\right) \geq 0 \\
y^{2} \leq 4 \Rightarrow-2 \leq y \leq 2
\end{array}
$$
Also.
$$
y^{2}=-x^{2}+10 x-21
$$
Now, for real roots of $y$.
$$
\begin{array}{r}
-4\left(x^{2}-10 x+21\right) \geq 0 \\
-x^{2}+10 x-21 \geq 0
\end{array}
$$
$\Rightarrow \quad(x-7)(x-3) \leq 0 \therefore 3 \leq x \leq 7$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2016
If the equilibrium constants of the following equilibria
\(\mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \rightleftharpoons \mathrm{SO}_3 \text { and } 2 \mathrm{SO}_3 \rightleftharpoons 2 \mathrm{SO}_2+\mathrm{O}_2\)
are given by \(\mathrm{K}_1\) and \(\mathrm{K}_2\) respectively, which of the following relations is correct

Options:
A \(\mathrm{K}_2=\left(\frac{1}{\mathrm{~K}_1}\right)^2\)
B \(K_1=\left(\frac{1}{K_2}\right)^3\)
C \(K_2=\left(\frac{1}{K_1}\right)\)
D \(K_2=\left(K_1\right)^2\)

The correct answer is:
\(\mathrm{K}_2=\left(\frac{1}{\mathrm{~K}_1}\right)^2\)

Solution:
Hints: \(\mathrm{K}_1=\frac{\left[\mathrm{SO}_3\right]}{\left[\mathrm{SO}_2\right]\left[\mathrm{O}_2\right]^{1 / 2}}\)
\(\mathrm{K}_2=\frac{\left[\mathrm{SO}_2\right]^2\left[\mathrm{O}_2\right]}{\left[\mathrm{SO}_3\right]^2}\)
Thus \(\mathrm{K}_2=\left(\frac{1}{\mathrm{~K}_1}\right)^2\)

Topic: Chemistry, Chemical Equilibrium, WBJEE, WBJEE 2011
If the first and $(2 n-1)$ th terms of an $\mathrm{AP}$, GP and $\mathrm{HP}$ are equal and their $n$ th terms are respectively $a, b, c,$ then always

Options:
A $a=b=c$
B $a \geq b \geq c$
C $a+c=b$
D $a c-b^{2}=0$

The correct answers are:
$a \geq b \geq c$, $a c-b^{2}=0$

Solution:
$\because a, b$ and $c$ are respectively the $A M,$ GM and HM of first and $(2 n-1)$ th terms. and also, we know $A M \geq G M \geq H M$ and $A M \cdot H M=G M^{2}$
$\therefore \quad a \geq b \geq c$ and $a \cdot c=b^{2}$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2016
If the first ionization energy of $\mathrm{H}$ atom is $13.6 \mathrm{eV},$ then the second ionization energy of He atom is

Options:
A $27.2 \mathrm{eV}$
B $40.8 \mathrm{eV}$
C $54.4 \mathrm{eV}$
D $108.8 \mathrm{eV}$

The correct answer is:
$54.4 \mathrm{eV}$

Solution:
Second ionisation energy $=13.6 \times \frac{2^{2}}{1^{2}}$
$$
=54.4 \mathrm{eV}
$$

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2012
If the focal length of the eye piece of a telescope is doubled, its magnifying power \((\mathrm{m})\) will be

Options:
A \(2 \mathrm{~m}\)
B \(3 \mathrm{~m}\)
C \(\frac{\mathrm{m}}{2}\)
D \(4 \mathrm{~m}\)

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

Solution:
Hints: \(\mathrm{m}=\frac{-\mathrm{f}_0}{\mathrm{f}_{\mathrm{e}}}\)
\(\mathrm{m}^{\prime}=\frac{\mathrm{m}}{2}\)

Topic: Physics, Ray Optics, WBJEE, WBJEE 2011
If the following three linear equations have a non-trivial solution, then
$$
\begin{array}{l}
x+4 a y+a z=0 \\
x+3 b y+b z=0 \\
x+2 c y+c z=0
\end{array}
$$

Options:
A $a, b, c$ are in $A P$
B $a, b, c$ are in $G$
C $a, b, c$ are in $\mathrm{HP}$
D $a+b+c=0$

The correct answer is:
$a, b, c$ are in $\mathrm{HP}$

Solution:
For non-trivial solution, We have, $\left|\begin{array}{ccc}1 & 4 a & a \\ 1 & 3 b & b \\ 1 & 2 c & c\end{array}\right|=0$
$\Rightarrow \quad 1(3 b c-2 b c)-(4 a c-2 a c)+1(4 a b-3 a b)=0$
$\Rightarrow b c-2 a c+a b=0$
$\Rightarrow b c+a b=2 a c$
$\Rightarrow b(a+c)=2 a c$
$\Rightarrow b=\frac{2 a c}{a+c}$
$\Rightarrow a, b, c$ are in $\mathrm{HP}$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2018
If the four points with position vectors $-2\hat i+\hat j + \hat k, \hat i+\hat j+\hat k,\hat j- \hat k$ and $\lambda \hat j + \hat k$ are coplanar, then $\lambda$ is equal to

Options:
A 1
B 2
C -1
D 0

The correct answer is:
1

Solution:
If four points $\left(x_{1}, y_{1}, z_{1}\right),\left(x_{2}, y_{2}, z_{2}\right),\left(x_{3}, y_{3}, z_{3}\right)$ and
$\left(x_{4}, y_{4}, z_{4}\right)$ are coplanar, then $\left|\begin{array}{lll}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1} \\ x_{4}-x_{1} & y_{4}-y_{1} & z_{4}-z_{1}\end{array}\right|=0$
Now, $\quad\left|\begin{array}{ccc}3 & 0 & 0 \\ 2 & 0 & -2 \\ 2 & \lambda-1 & 0\end{array}\right|=0$
$\Rightarrow \quad 3(0+2 \lambda-2)=0$
$\lambda=1$

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2015
If the function $f: R \rightarrow R$ is defined by $f(x)=\left(x^{2}+1\right)^{35}, \forall x \in R,$ then $f$ is

Options:
A one-one but not onto
B onto but not one-one
C neither one-one nor onto
D both one-one and onto

The correct answer is:
neither one-one nor onto

Solution:
Given, $f(x)=\left(x^{2}+1\right)^{35}, \forall x \in R$
Since, $f(x)$ is even function. Hence, the function is not one-one. And $f(x)>0, y \times \epsilon R$
Hence, the function is not onto.
Given, the function is neither one-one nor onto.

Topic: Mathematics, Functions, WBJEE, WBJEE 2016
If the function $f(x)=2 x^{3}-9 a x^{2}+12 a^{2} x+1[a>0]$ attains its maximum and minimum at $p$ and $q$ respectively such that $\mathrm{p}^{2}=\mathrm{q}$, then $\mathrm{a}$ is equal to

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

The correct answer is:
2

Solution:
Hint:
$f^{\prime}(x)=6 x^{2}-18 a x+12 a^{2} \Rightarrow f^{\prime \prime}(x)=12 x-18 a \Rightarrow f^{\prime}(x)=0 \Rightarrow x=a, 2 a$
$f^{\prime}(a) < 0 ; p=a$ (maximum)
$f^{\prime}(2 a)>0 ; q=2 a$ (minimum)
$a^{2}=2 a ; a(a-2)=0, \quad a=2$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2020
If the function
\(f(x)= \begin{cases}\frac{x^2-(A+2) x+A}{x-2} & \text { for } x \neq 2 \\ 2 & \text { for } x=2\end{cases}\)
is continuous at \(x=2\), then

Options:
A \(\mathrm{A}=0\)
B \(\mathrm{A}=1\)
C \(\mathrm{A}=-1\)
D \(\mathrm{A}=2\)

The correct answer is:
\(\mathrm{A}=0\)

Solution:
Hints: \(\frac{4-(A+2) 2+A}{0}=\frac{-A}{0}\)
Put \(\mathrm{A}=0\).

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2011
If the given four electronic configurations.
(i) $n=4, l=1$
(ii) $n=4, l=0$
(iii) $n=3, l=2$
(iv) $n=3, l=1$
are arranged in order of increasing energy, then the order will be

Options:
A (iv) $ < $ (ii) $ < $ (iii) $ < $ (i)
B (ii) $ < ($ iv $) < (i) < ($ iii $)$
C (i) $ < $ (iii) $ < $ (ii) $ < $ (iv)
D (iii) $ < $ (i) $ < $ (iv) $ < $ (ii)

The correct answer is:
(iv) $ < $ (ii) $ < $ (iii) $ < $ (i)

Solution:
According to $(n+l)$ rule;
(i) More be the sum of $(n+l)$ value, more be the energy.
(ii) For same value of sum, more be the value of $n$, more be the energy.
For
(i) $n+l \rightarrow 4+1=5$
(ii) $n+l \rightarrow 4+0=4$
(iii) $n+l \rightarrow 3+2=5$
(iv) $n+l \rightarrow 3+1=4$
Hence, order of energy will be
(iv) $ < $ (ii) $ < $ (iii) $ < $ (i)

Topic: Chemistry, Structure of Atom, WBJEE, WBJEE 2017
If the kinetic energies of an electron, an alpha particle and a proton having same de-Broglie wavelength are $\varepsilon_1, \varepsilon_2$ and $\varepsilon_3$ respectively, then

Options:
A $\varepsilon_1>\varepsilon_3>\varepsilon_2$
B $\varepsilon_1=\varepsilon_2=\varepsilon_2$
C $\varepsilon_1 < \varepsilon_3 < \varepsilon_2$
D $\varepsilon_1>\varepsilon_2>\varepsilon_3$

The correct answer is:
$\varepsilon_1>\varepsilon_3>\varepsilon_2$

Solution:
$\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}, \sqrt{\mathrm{mk}}=\mathrm{const}$
$\mathrm{k} \propto \frac{1}{\mathrm{~m}}, \varepsilon_1>\varepsilon_3>\varepsilon_2$

Topic: Physics, Dual Nature of Matter, WBJEE, WBJEE 2022
If the kinetic energy of a body changes by $20 \%$ then its momentum would change by -

Options:
A $20 \%$
B $24 \%$
C $40 \%$
D None of the above

The correct answer is:
None of the above

Solution:
$\frac{\frac{p_f^2}{2 m}-\frac{p_i^2}{2 m}}{\frac{p_i^2}{2 m}} \times 100=20$
$\Rightarrow \frac{p_f}{p_i}=\sqrt{1.2}=1.095 \Rightarrow \frac{p_f-p_i}{p_i}=0.095$
Therefore $\%$ increase $=9.5 \%$

Topic: Physics, Center of Mass Momentum and Collision, WBJEE, WBJEE 2009
If the letters of the word 'PROBABILITY' are written down at random in a row, then probability that two B's are together, is

Options:
A $\frac{2}{11}$
B $\frac{10}{11}$
C $\frac{3}{11}$
D $\frac{5}{11}$

The correct answer is:
$\frac{2}{11}$

Solution:
Total number of ways in which the letters of the word 'PROBABILITY can be amanged in a row $=\frac{11!}{2!2 !}$
Number of ways in which two B's are togother $=\frac{10!}{2!}$
$\therefore$ Required probability
$=\frac{\frac{10}{2 !}}{\frac{11!}{2!2!}}=\frac{10 ! \times 2!}{11!}=\frac{2}{11}$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2015
If the line $a x+b y+c=0, a b \neq 0,$ is a tangent
to the curve $x y=1-2 x,$ then

Options:
A $a>0, b < 0$
B $a>0, b>0$
C $a < 0, b>0$
D $a < 0, b < 0$

The correct answers are:
$a>0, b>0$, $a < 0, b < 0$

Solution:
We have, $xy=1-2 x$
$\Rightarrow y=\frac{1-2 x}{x}$
$\Rightarrow \frac{d y}{d x}=\frac{-2 x-(1-2 x) \cdot 1}{x^{2}}$
$\quad=\frac{-2 x-1+2 x}{x^{2}}=\frac{-1}{x^{2}} < 0$
since, $a x+b y+c=0$ is tangent to the curve $xy=1-2 x$
$\therefore \quad \frac{-a}{b} < 0 \Rightarrow \frac{a}{b}>0$
$\Rightarrow$ Either $a>b>0$ or $a < 0, b < 0$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2017
If the line $a x+b y+c=0$ is a tangent to the curve $x y=4$, then

Options:
A a $ < 0$, b $>0$
B a $\leq 0$, b $>0$
C a $ < 0$, b $ < 0$
D a $\leq 0$, b $ < 0$

The correct answer is:
a $ < 0$, b $ < 0$

Solution:
$$
\begin{aligned}
& \text { Hints: Slope of line }=-\frac{a}{b} \\
& y=\frac{4}{x}=1, \frac{d y}{d x}=-\frac{4}{x^2},-\frac{a}{b}=-\frac{4}{x^2} \Rightarrow \frac{a}{b}=\frac{4}{x^2}>0 \\
& a < 0, b < 0
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2010
If the line $\mathrm{y}=\mathrm{x}$ is a tangent to the parabola $\mathrm{y}=\mathrm{ax}^{2}+\mathrm{bx}+\mathrm{c}$ at the point $(1,1)$ and the curve passes through $(-1,0)$, then

Options:
A $a=b=-1, c=3$
B $a=b=\frac{1}{2}, c=0$
C $a=c=\frac{1}{4}, b=\frac{1}{2}$
D $a=0, b=c=\frac{1}{2}$

The correct answer is:
$a=c=\frac{1}{4}, b=\frac{1}{2}$

Solution:
Hint:
$\frac{d y}{d x}=2 a x+b$
at $(1,1) \frac{d y}{d x}=2 a+b=1$ ...(1)
Now $(1,1)$ and $(-1,0)$ satisfies the curve
$a-b+c=0$ ...(2)
$a+b+c=1$ ...(3)
$\mathrm{b}=\frac{1}{2} ; \mathrm{a}=\frac{1}{4} ; \mathrm{c}=\frac{1}{4}$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2020
If the lines joining the focii of the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) where \(a > b\), and an extremity of its minor axis is inclined at an angle \(60^{\circ}\), then the eccentricity of the ellipse is

Options:
A \(\frac{\sqrt{3}}{2}\)
B \(\frac{1}{2}\)
C \(\frac{\sqrt{7}}{3}\)
D \(\frac{1}{\sqrt{3}}\)

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

Solution:
Hint: \(\frac{b}{a e}=\tan 60^{\circ} \Rightarrow \frac{b}{a}=e \sqrt{3}\)
\(\begin{aligned}
& \because e^2=1-\frac{b^2}{a^2}=1-3 e^2 \\
& \Rightarrow e=+\frac{1}{2} \quad\left(e \neq-\frac{1}{2}\right)
\end{aligned}\)

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2023
If the matrices $A=\left[\begin{array}{lll}2 & 1 & 3 \\ 4 & 1 & 0\end{array}\right]$ and $B=\left[\begin{array}{rr}1 & -1 \\ 0 & 2 \\ 5 & 0\end{array}\right]$, then $\mathrm{AB}$ will be

Options:
A $\left[\begin{array}{cr}17 & 0 \\ 4 & -2\end{array}\right]$
B $\left[\begin{array}{ll}4 & 0 \\ 0 & 4\end{array}\right]$
C $\left[\begin{array}{cr}17 & 4 \\ 0 & -2\end{array}\right]$
D $\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]$

The correct answer is:
$\left[\begin{array}{cr}17 & 0 \\ 4 & -2\end{array}\right]$

Solution:
$$
\text { Hints : } \mathrm{AB}=\left[\begin{array}{lll}
2 & 1 & 3 \\
4 & 1 & 0
\end{array}\right]\left[\begin{array}{rr}
1 & -1 \\
0 & 2 \\
5 & 0
\end{array}\right]=\left[\begin{array}{lr}
17 & 0 \\
4 & -2
\end{array}\right]
$$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2010
If the matrix $A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2\end{array}\right],$ then
$A^{n}=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ b & 0 & a\end{array}\right], n \in N,$ where

Options:
A $a=2 n, b=2^{n}$
B $a=2^{n}, b=2 n$
C $a=2^{n}, b=n 2^{n-1}$
D $a=2^{n}, b=n 2^{n}$

The correct answer is:
$a=2^{n}, b=n 2^{n}$

Solution:
We have. $A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2\end{array}\right]$
$\therefore \quad A^{2}=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2\end{array}\right]\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 2 & 0 & 2\end{array}\right]=\left[\begin{array}{lll}4 & 0 & 0 \\ 0 & 4 & 0 \\ 8 & 0 & 4\end{array}\right]$
$=\left[\begin{array}{ccc}2^{2} & 0 & 0 \\ 0 & 2^{2} & 0 \\ 2 \cdot 2^{2} & 0 & 2^{2}\end{array}\right]$
Now, $\quad A^{n}=\left[\begin{array}{ccc}2^{n} & 0 & 0 \\ 0 & 2^{n} & 0 \\ n-2^{n} & 0 & 2^{n}\end{array}\right]$
$A^{n}=\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ b & 0 & a\end{array}\right]$
$\therefore \quad\left[\begin{array}{lll}a & 0 & 0 \\ 0 & a & 0 \\ b & 0 & a\end{array}\right]=\left[\begin{array}{ccc}2^{n} & 0 & 0 \\ 0 & 2^{n} & 0 \\ n \cdot 2^{n} & 0 & 2^{n}\end{array}\right]$
Hence, $\quad a=2^{n}$
and $\quad b=n 2^{n}$

Topic: Mathematics, Matrices, WBJEE, WBJEE 2016
If the molecular wt. of \(\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3\) and \(\mathrm{I}_2\) are \(\mathrm{M}_1\) and \(\mathrm{M}_2\) respectively, then what will be the equivalent wt. of \(\mathrm{Na}_2 \mathrm{~S}_2 \mathrm{O}_3\) and \(\mathrm{I}_2\) in the following reaction?
\(2 \mathrm{~S}_2 \mathrm{O}_3^{2-}+\mathrm{I}_2 \longrightarrow \mathrm{S}_4 \mathrm{O}_6^{2-}+2 \mathrm{I}^{-}\)

Options:
A \(\mathrm{M}_1, \mathrm{M}_2\)
B \(\mathrm{M}_1, \mathrm{M}_2 / 2\)
C \(2 \mathrm{M}_1, \mathrm{M}_2\)
D \(\mathrm{M}_1, 2 \mathrm{M}_2\)

The correct answer is:
\(\mathrm{M}_1, \mathrm{M}_2 / 2\)

Solution:
Hints:
\(\begin{aligned}
& \text { n.f. }\left(\mathrm{S}_2 \mathrm{O}_3^{2-}\right)=1, \text { Equivalent mass }=\frac{\mathrm{M}_1}{1}=\mathrm{M}_1 \\
& \mathrm{n} \mathrm{f} .\left(\mathrm{I}_2\right)=2, \quad \text { Equivalent mass }=\frac{\mathrm{M}_2}{2}
\end{aligned}\)

Topic: Chemistry, Redox Reactions, WBJEE, WBJEE 2011
If the \(n\) terms \(a_1, a_2, \ldots \ldots \ldots . ., a_n\) are in A.P. with increment \(r\), then the difference between the mean of their squares & the square of their mean is

Options:
A (A) \(\frac{r^2\left\{(n-1)^2-1\right\}}{12}\)
B (B) \(\frac{r^2}{12}\)
C (C) \(\frac{r^2\left(n^2-1\right)}{12}\)
D (D) \(\frac{\mathrm{n}^2-1}{12}\)

The correct answer is:
(C) \(\frac{r^2\left(n^2-1\right)}{12}\)

Solution:
Hint: \(\frac{a_1^2+a_2^2+\ldots+a_n^2}{n}-\left(\frac{a_1+a_2+\ldots+a_n}{n}\right)^2\)
\(=\frac{a_1^2+\left(a_1+r\right)^2+\ldots+\left\{a_1+(n-1) r\right\}^2}{n}-\left(\frac{n a_1+r \cdot \frac{n(n-1)}{2}}{n}\right)^2\)
\(\begin{aligned}
& =\frac{n a_1^2+r^2 \cdot \frac{(n-1) \cdot n(2 n-1)}{6}+a_1 r(n-1) \cdot n}{n}-a_1^2-a_1 r(n-1)-\frac{r^2(n-1)^2}{4} \\
& =a_1^{2}+r^2 \frac{(n-1)(2 n-1)}{6}+a_1 r(n-1)-a_1^2-a_1 r(n-1)-\frac{r^2(n-1)^2}{4}
\end{aligned}\)
\(\begin{aligned} & =\frac{r^2(n-1)}{2}\left(\frac{2 n-1}{3}-\frac{n-1}{2}\right) \\ & =\frac{r^2(n-1)}{2} \cdot\left(\frac{4 n-2-3 n+3}{6}\right) \\ & =\frac{r^2(n-1)(n+1)}{12} \\ & =\frac{r^2\left(n^2-1\right)}{12}\end{aligned}\)

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2023
If the normal to the curve $y=f(x)$ at the point $(3,4)$ make an angle $3 \pi / 4$ with the positive $x$-axis, then $f(3)$ is

Options:
A 1
B -1
C $-\frac{3}{4}$
D $\frac{3}{4}$

The correct answer is:
1

Solution:
Hints: $\frac{d y}{d x}=f^{\prime}(x)$, Slope of normal $=-\frac{1}{f^{\prime}(x)},-\frac{1}{f^{\prime}(3)}=\tan \frac{3 \pi}{4}=-1$ $f^{\prime}(3)=1$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2010