datavorous/entrance-exam-dataset huggingface dataset:
If the parabola $x^{2}=a y$ makes an intercept of length $\sqrt{40}$ units on the line $y-2 x=1$, then $a$ is equal to

Options:
A 1
B -2
C -1
D 2

The correct answers are:
1, -2

Solution:
Gven, equation of the parabola
$x^{2}=a y$
ie $y=\frac{x^{2}}{a}$
and the equation of line
$y-2 x=1$
$y=2 x+1$
$\frac{x^{2}}{a}=2 x+1$
$x^{2}=2 a x+a$
$x^{2}-2 a x-a=0$
$\begin{aligned} x &=\frac{2 a \pm \sqrt{4 a^{2}+4 a}}{2} \\ &=\frac{2\left(a \pm \sqrt{a^{2}+a}\right)}{2}=a \pm \sqrt{a^{2}+a} \end{aligned}$
Points of intersection of the parabola and the line are
$$
\left(a+\sqrt{a^{2}+a}, \left(2\left(a+\sqrt{a^{2}+a}\right)+1\right)\right)
$$
and $\quad\left(a-\sqrt{a^{2}+a}, \left(2 ( a-\sqrt{a^{2}+a})+1\right)\right)$
$\because$ Distance between the points $=\sqrt{40}$ $\therefore \quad \sqrt{40}=\sqrt{\left(2 \sqrt{a^{2}+a}\right)^{2}+\left(4 \sqrt{a^{2}+a}\right)^{2}}$
$\Rightarrow \quad 40=4\left(a^{2}+a\right)+16\left(a^{2}+a\right)$
$\Rightarrow \quad 2=a^{2}+a \Rightarrow a=1-2$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2016
If the point $(2 \cos \theta, 2 \sin \theta)$ for $0 \in(0,2 \pi)$ lies in the region between the lines $x+y=2$ and $x-y=2$ containing the origin, then $\theta$ lies in

Options:
A $\left(\theta, \frac{\pi}{2}\right) \cup\left(\frac{3 \pi}{2}, 2 \pi\right)$
B $[0, n]$
C $\left(\frac{\pi}{2} \cdot \frac{3 \pi}{2}\right)$
D $\left[\frac{\pi}{4}, \frac{\pi}{2}\right]$

The correct answer is:
$\left(\frac{\pi}{2} \cdot \frac{3 \pi}{2}\right)$

Solution:
Given that $(2 \cos \theta, 2 \sin \theta)$ will lie on the circle $x^{2}+y^{2}=4$ (from the given figure). Since, point lies on the region containing origin.

So, point will be on the shaded region.
$\therefore$
$\theta \in\left(\frac{\pi}{2}, \frac{3 \pi}{2}\right)$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2015
If the point of intersection of the lines $2 a x+4 a y+c=0$ and $7 b x+3 b y-d=0$ lies in the 4 th quadrant and is equidistant from the two axes, where $a, b, c$ and $d$ are non-zero numbers, then $a d: b c$ equals to

Options:
A $2: 3$
B $2: 1$
C $1: 1$
D $3: 2$

The correct answer is:
$2: 1$

Solution:
Let coordinate of the point be $(\alpha,-\alpha)$
since, $(\alpha,-\alpha)$ lie on $2 a x+4 a y+c=0$
and $\quad 7 b x+3 b y-d=0$
$\therefore \quad 2 a \alpha-4 a \alpha+c=0$
$\Rightarrow \quad-2 a \alpha+c=0$
$\Rightarrow \quad \alpha=\frac{c}{2 a}$
Also, $7 b \alpha-3 b \alpha-d=0$
$\Rightarrow \quad 4 b \alpha-d=0$
$\Rightarrow \quad \alpha=\frac{d}{4 b}$
From Eqs. (i) and (ii),
$\begin{aligned} \frac{c}{2 a} &=\frac{d}{4 b} \\
\Rightarrow 2 a d &=4 b c \end{aligned}$
$\Rightarrow \quad \frac{a d}{b c}=\frac{4}{2}$
$\Rightarrow \quad \frac{a d}{b c}=\frac{2}{1}$
$a d: b c=2: 1$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2019
If the polynomial $f(x)=\left|\begin{array}{ccc}(1+x)^{a} & (2+x)^{b} & 1 \\ 1 & (1+x)^{a} & (2+x)^{b} \\ (2+x)^{b} & 1 & (1+x)^{a}\end{array}\right|,$ then the
constant term of $f(x)$ is $[a$ and $b$ are positive integers $]$

Options:
A $2-3 \cdot 2^{b}+2^{3 b}$
B $2+3 \cdot 2^{b}+2^{3 b}$
C $2+3 \cdot 2^{b}-2^{3 b}$
D $2-3 \cdot 2^{b}-2^{3 b}$

The correct answer is:
$2-3 \cdot 2^{b}+2^{3 b}$

Solution:
Given, $f(x)=\left|\begin{array}{ccc}(1+x)^{a} & (2+x)^{b} & 1 \\ 1 & (1+x)^{4} & (2+x)^{b} \\ (2+x)^{b} & 1 & (1+x)^{a}\end{array}\right|$
For constant term put $x=0$ $f(0)=\left|\begin{array}{ccc}1 & 2^{b} & 1 \\ 1 & 1 & 2 \\ 2^{b} & 1 & 1\end{array}\right|$
$=1\left(1-2^{b}\right)-2^{b}\left(1-2^{2 b}\right)+1\left(1-2^{b}\right)$
$=1-2^{b}-2^{b}+2^{3 b}+1-2^{b}$
$=2-3 \cdot 2^{b}+2^{3 b}$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2018
If the potential energy of a hydrogen atom in the first excited state is assumed to be zero, then the total energy of \(n\) \(=\infty\) state is,

Options:
A \(3.4 \mathrm{eV}\)
B \(6.8 \mathrm{eV}\)
C 0
D \(\infty\)

The correct answer is:
\(6.8 \mathrm{eV}\)

Solution:
Hint : Potential energy in 1 st emited stage \(\left(\mathrm{U}_2\right)=-6.8 \mathrm{eV}\). When \(U_2\) is assumed to be zero then potential energy in \((n=\infty)\)
will be \(6.8 \mathrm{eV}\)
so, total energy for \((n=\infty)=6.8 \mathrm{eV}\)

Topic: Physics, Atomic Physics, WBJEE, WBJEE 2023
If the pressure, temperature and density of an ideal gas are denoted by $p, T$ and $\rho$, respectively, the velocity of sound in the gas is

Options:
A proportional to $\sqrt{p},$ when $T$ is constant.
B proportional to $\sqrt{T}$
C proportional to $\sqrt{p}$, when $\rho$ is constant.
D proportional to $T$.

The correct answers are:
proportional to $\sqrt{T}$, proportional to $\sqrt{p}$, when $\rho$ is constant.

Solution:
$v=\sqrt{\frac{\gamma R T}{M}}=\sqrt{\frac{\gamma P}{\rho}}$
(i) Velocity of sound is proportional to $\sqrt{T}$
$\therefore$
$v \propto \sqrt{T}$
(ii) When $\rho$ is constant, the velocity of sound is proportional to root of pressure $P$
$\therefore$
$$
v \propto \sqrt{P}
$$

Topic: Physics, Waves and Sound, WBJEE, WBJEE 2017
If the radius of a spherical balloon increases by $0.1 \%$, then its volume increases approximately by

Options:
A $0.2 \%$
B $0.3 \%$
C $0.4 \%$
D $0.05 \%$

The correct answer is:
$0.3 \%$

Solution:
Let $V$ be the volume of spherical ballon of
radius $r$
Then, $V=\frac{4}{3} \pi r^{3}$
$\Rightarrow \quad \log V=\log \left(\frac{4 \pi}{3}\right)+3 \log r \Rightarrow \frac{1}{V} \frac{d V}{d r}=0+\frac{3}{r}$
$\Rightarrow \quad \frac{1}{V} \frac{d V}{d r}=\frac{3}{r} \Rightarrow \frac{1}{V} \frac{\Delta V}{\Delta r}=\frac{3}{r}$
$\Rightarrow \quad \frac{\Delta V}{V}=3 \frac{\Delta r}{r}$
$\Rightarrow \frac{\Delta V}{V} \times 100=3 \frac{\Delta r}{r} \times 100$
$\Rightarrow \frac{\Delta V}{V} \times 100=3 \times 01=0.3$
$\therefore$ Percentage increase in volume is $0.3 \%$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2019
If the rate of increase of the radius of a circle is $5 \mathrm{~cm} / \mathrm{sec}$, then the rate of increase of its area, when the radius is $20 \mathrm{~cm}$, will be

Options:
A $10 \pi$
B $20 \pi$
C $200 \pi$
D $400 \pi$

The correct answer is:
$200 \pi$

Solution:
Hints : $\mathrm{A}=\pi \mathrm{r}^2 \quad \frac{d r}{d t}=5$
$$
\begin{aligned}
& \frac{\mathrm{dA}}{\mathrm{dt}}=2 \pi r \frac{\mathrm{dr}}{\mathrm{dt}}=2 \pi 20(5) \\
& =200 \pi
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2009
If the ratio of the roots of the equation \(\mathrm{px}^2+\mathrm{qx}+\mathrm{r}=0\) is \(a: b\), then \(\frac{\mathrm{ab}}{(\mathrm{a}+\mathrm{b})^2}=\)

Options:
A \(\frac{\mathrm{p}^2}{\mathrm{qr}}\)
B \(\frac{\mathrm{pr}}{\mathrm{q}^2}\)
C \(\frac{q^2}{\mathrm{pr}}\)
D \(\frac{\mathrm{pq}}{\mathrm{r}^2}\)

The correct answer is:
\(\frac{\mathrm{pr}}{\mathrm{q}^2}\)

Solution:
Hints: Let roots are \(a \alpha\) and \(b \alpha\)
\(\Rightarrow(\mathrm{a}+\mathrm{b}) \alpha=\frac{-\mathrm{q}}{\mathrm{p}}\)
\(\begin{aligned} & a b \alpha^2=\frac{r}{p} \\ & \frac{a b \alpha^2}{(a+b)^2 \alpha^2}=\frac{r}{p} \cdot \frac{p^2}{q^2} \\ & \frac{a b}{(a+b)^2}=\frac{1 p}{q^2}\end{aligned}\)

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2011
If the rms velocity of hydrogen gas at a certain temperature is $c,$ then the ras velocity of oxygen gas at the same temperature is

Options:
A $\frac{c}{8}$
B $\frac{c}{10}$
C $\frac{c}{4}$
D $\frac{c}{2}$

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

Solution:
$v_{\ln m}=\sqrt{\frac{3 R T}{M}}$
$R=$ gas constant
$T=$ absolute temperature
$M=$ mass of gas
$v_{\text {rms }}$ of $\mathrm{H}_{2}=\sqrt{\frac{3 R T}{M_{\mathrm{H}_{2}}}}$
$v_{\text {rms }}$ of $\mathrm{O}_{2}=\sqrt{\frac{3 R T}{M_{\mathrm{O}_{2}}}}$
$\frac{v_{\mathrm{rms}} \cdot \mathrm{H}_{2}}{v_{\mathrm{rma}} \cdot \mathrm{O}_{2}}=\sqrt{\frac{M_{O_{2}}}{M_{\mathrm{H}_{1}}}}$
Given $\quad v_{\operatorname{rms}} H_{2}=c$
$\Rightarrow \quad \frac{c}{v_{\operatorname{rms}} \mathrm{O}_{2}}=\sqrt{\frac{M_{0}}{M_{H_{1}}}}=\sqrt{\frac{32}{2}}=\sqrt{16}=4$
$\therefore \quad v_{rms}=\frac{C}{4}$

Topic: Physics, Kinetic Theory of Gases, WBJEE, WBJEE 2016
If the solution of the differential equation $x \frac{d y}{d x}+y=x e^{x}$ be $x y=e^{x} \phi(x)+C,$ then $\phi(x)$ is equal to

Options:
A $x+1$
B $x-1$
C $1-x$
D $x$

The correct answer is:
$x-1$

Solution:
Given, $\quad x \frac{d y}{d x}+y=x e^{x}$
$\Rightarrow \quad \frac{d y}{d x}+\frac{y}{x}=e^{x}$
On comparing Eq. by $\frac{d y}{d x}+P y=Q$, we get
$\therefore$
$$
\begin{array}{l}
P=\frac{1}{x} \text { and } Q=e^{x} \\
IF=e^{\int \frac{1}{x} d x}
\end{array}
$$
$$
=e^{log^{x}}=x
$$
Hence, solution of differential equation,
$$
\begin{array}{ll}
\quad y \cdot x=\int x e^{x} d x+C \\
\Rightarrow \quad x y=x e^{x}-\int e^{x} d x+C \\
\Rightarrow \quad x y=x e^{x}-e^{x}+C
\end{array}
$$
$\Rightarrow \quad x y=e^{x}(x-1)+C$
$\because xy=e^{x} \phi(x)+c$ [given]
$$
\phi(x)=(x-1)
$$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2016
If the straight line $(a-1) x-b y+4=0$ is normal to the hyperbola $x y=1$, then which of the following does not hold?

Options:
A $a>1, b>0$
B $a>1, b < 0$
C $a < 1, b < 0$
D $a < 1, b>0$

The correct answers are:
$a>1, b>0$, $a < 1, b < 0$

Solution:
Given equation of straight line is
$$
(a-1) x-b y+4=0
$$
Slope $=-\frac{\text { Coefficient of } x}{\text { Coefficient of } y}$ $=\frac{-(a-1)}{-b}=\frac{a-1}{b}$
Equation of hyperbola is $x y=1$
$\Rightarrow \quad y=\frac{1}{x}$
$\Rightarrow \quad \frac{d y}{d x}=-\frac{1}{x^{2}}$
Slope of normal $=x^{2}>0$
$\therefore \quad \frac{a-1}{b}>0$
$\Rightarrow \quad a-1>0 \quad$ and $\quad b>0$
or $\quad a-1 < 0$ and $b < 0$ $\Rightarrow \quad a>1, b>0$ or $a < 1, b < 0$

Topic: Mathematics, Hyperbola, WBJEE, WBJEE 2015
If the straight line \(y=m x\) lies outside of the circle \(x^2+y^2-20 y+90=0\), then the value of \(m\) will satisfy

Options:
A \(m < 3\)
B \(|m| < 3\)
C \(\mathrm{m} > 3\)
D \(|\mathrm{m}| > 3\)

The correct answer is:
\(|m| < 3\)

Solution:
Hints: \(x^2+m^2 x^2-20 m x+90\)
\(\begin{aligned}
& x^2\left(1+m^2\right)-20 m x+90=0 \\
& D < 0 \\
& 400 m^2-4 \times 90\left(1+m^2\right) < 0 \\
& 40 m^2 < 360 \\
& m^2 < 9 ;|m| < 3
\end{aligned}\)

Topic: Mathematics, Circle, WBJEE, WBJEE 2011
If the sum of distances from a point $\mathrm{P}$ on two mutually perpendicular straight lines is 1 unit, then the locus of $\mathrm{P}$ is

Options:
A a parabola
B a circle
C an ellipse
D a straight line

The correct answer is:
a straight line

Solution:
Hints: $|x|+|y|=1$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2010
If the sum of the distances of a point from two perpendicular lines in a plane is 1 unit, then its locus is

Options:
A a square
B a circle
C a straight line
D two intersecting lines

The correct answer is:
a square

Solution:
$$
|x|+|y|=1
$$
Locus is square

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2022
If the sum of two unit vectors is a unit vector, then the magnitude of their difference is

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

The correct answer is:
$\sqrt{3}$ units

Solution:
$\begin{aligned} &\text { Let a and } \mathbf{b} \text { are two unit vectors. }\\ &\text { Then, } \quad|\mathbf{a}+\mathbf{b}|=1\\ &\Rightarrow \quad|\mathbf{a}+\mathbf{b}|^{2}=1 \end{aligned}$
$\Rightarrow \quad(a+b) \cdot(a+b)=1$
$\Rightarrow|a|^{2}+|b|^{2}+2 a \cdot b=1$
$\Rightarrow \quad 1+1+2 \mathbf{a} \cdot \mathbf{b}=1$
$\Rightarrow \quad 2 \mathbf{a} \cdot \mathbf{b}=-1$
Again, $|\hat{\mathbf{a}}-\mathbf{b}|^{2}=(\mathbf{a}-\hat{\mathbf{b}}) \cdot(\mathbf{a}-\hat{\mathbf{b}})$
$=|\mathbf{a}|^{2}+|\mathbf{b}|^{2}-2 \mathbf{a} \cdot \mathbf{b}$
$=1+1-(-1)$
$=1+1+1=3$
$|\hat{a}-\hat{b}|=\sqrt{3}$

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2017
If the tangent at the point $P$ with co-ordinates $(h, k)$ on the curve $y^{2}=2 x^{3}$ is perpendicular to the straight line $4 x=3 y$, then

Options:
A $(\mathrm{h}, \mathrm{k})=(0,0)$ only
B $(\mathrm{h}, \mathrm{k})=\left(\frac{1}{8},-\frac{1}{16}\right)$ only
C $(\mathrm{h}, \mathrm{k})=(0,0)$ or $\left(\frac{1}{8},-\frac{1}{16}\right)$
D no such point P exists

The correct answer is:
$(\mathrm{h}, \mathrm{k})=\left(\frac{1}{8},-\frac{1}{16}\right)$ only

Solution:
$2 y \frac{d y}{d x}=6 x^{2} \Rightarrow \frac{d y}{d x}=\frac{3 x^{2}}{y}=-\frac{3}{4} \Rightarrow y=-4 x^{2}$ and $y^{2}=2 x^{3}$
$\Rightarrow 16 x^{4}=2 x^{3} \Rightarrow x=\frac{1}{8} \quad(x \neq 0)$
$y=\frac{-1}{16} \quad(y \neq 0)$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2021
If the tangent to the curve $y^{2}=x^{3}$ at $\left(m^{2}, m^{3}\right)$ is also a normal to the curve at $\left(M^{2}, M^{3}\right)$, then the value of $m M$ is

Options:
A $-\frac{1}{9}$
B $-\frac{2}{9}$
C $-\frac{1}{3}$
D $-\frac{4}{9}$

The correct answer is:
$-\frac{4}{9}$

Solution:
Hint:
$2 y y_{1}=3 x^{2}$
$y_{1}=\frac{3 x^{2}}{2 y} \Rightarrow\left(y_{1}\right)_{m^{2}, m^{3}}=\frac{3 \times m^{4}}{2 \times m^{3}}=\frac{3 m}{2}$
Again; slope of normal $=-\frac{2}{3 M}, m M=-\frac{4}{9}$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2020
If the tangent to $y^{2}=4 a x$ at the point $\left(a t^{2}, 2 a t\right)$ where $|t|>1$ is a normal to $x^{2}-y^{2}=a^{2}$ at the point $(a \sec \theta, a \tan \theta),$ then

Options:
A $t=-\operatorname{cosec} \theta$
B $t=-\sec \theta$
C $t=2 \tan \theta$
D $t=2 \cot \theta$

The correct answers are:
$t=-\operatorname{cosec} \theta$, $t=2 \tan \theta$

Solution:
$\left(a,\right.$Equation of tangent to $y^{2}=4 a x$ at $\left(a t^{2}, 2 a t\right)$ will be
$$
x-y t=-a t^{2}
$$
Also, equation of normal to $x^{2}-y^{2}=a^{2}$ at $(a \sec \theta, a \tan \theta)$ will be
$\frac{x}{a \sec \theta}+\frac{y}{a \tan \theta}=2$
$\Rightarrow x+y \operatorname{cosec} \theta=2 a \sec \theta$
since, Eqs. (i) and (ii) are identical.
$\therefore t=-\operatorname{cosec} \theta$ or $t=2 \tan \theta$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2017
If the temperature of the Sun gets doubled, the rate of energy received on the Earth will increase by a factor of

Options:
A 2
B 4
C 8
D 16

The correct answer is:
16

Solution:
Suppose temperature of $\mathrm{Sun}=T$
When it is doubled, $T^{\prime}=2 T$
From law of radiation (Stefan's law)
$$
E \propto T^{4}=\sigma T^{4}
$$
$\sigma=$ Stefan's constant.
$$
\begin{array}{l}
E_{1}=T^{4} \\
E_{2} \approx(2 \Pi)^{4} \\
\frac{E_{1}}{E_{2}}=\frac{T^{4}}{2^{4} \cdot T^{4}} \\
E_{2}=16 E_{1}
\end{array}
$$

Topic: Physics, Thermal Properties of Matter, WBJEE, WBJEE 2016
If the three points $(3 q, 0),(0,3 p)$ and $(1,1)$ are collinear then which one is true ?

Options:
A $\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=1$
B $\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=1$
C $\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=3$
D $\frac{1}{\mathrm{p}}+\frac{3}{\mathrm{q}}=1$

The correct answer is:
$\frac{1}{\mathrm{p}}+\frac{1}{\mathrm{q}}=3$

Solution:
Hints: $\mathrm{A}(3 \mathrm{q}, 0) \mathrm{B}(0,3 \mathrm{p}) \mathrm{C}(11)$
Slope $=1 \mathrm{AC}=5 \log \mathrm{BC}$
$$
\begin{aligned}
& \frac{1-0}{1-3 q}=\frac{1-3 p}{1-0}=3, \frac{1}{1-3 q}=\frac{1-3 p}{1} \\
& 1=(1-3 p)(1-3 q), 1=1-3 q-3 p+9 p q \\
& \Rightarrow 3 p+3 q=9 p q, \frac{1}{q}+\frac{1}{p}=3
\end{aligned}
$$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2010
If the three points \(A(1,6), B(3,-4)\) and \(C(x, y)\) are collinear then the equation satisfying by \(x\) and \(y\) is

Options:
A \(5 x+y-11=0\)
B \(5 x+13 y+5=0\)
C \(5 x-13 y+5=0\)
D \(13 x-5 y+5=0\)

The correct answer is:
\(5 x+y-11=0\)

Solution:
\(\begin{aligned}
& \text {Hints: }\left|\begin{array}{ccc}
1 & 1 & 6 \\
1 & 3 & -4 \\
1 & x & y
\end{array}\right|=0 \\
& \Rightarrow 1(3 y+4 x)-(y-6 x)+1(-4-18)=0 \Rightarrow 3 y+4 x-y+6 x-12=0 \\
& \Rightarrow 2 y+10 x-22=0 \\
& y+5 x=11
\end{aligned}\)

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2011
If the total number of m-element subsets of the set $\mathrm{A}=\left\{\mathrm{a}{1}, \mathrm{a}{2}, \ldots . \mathrm{a}{\mathrm{n}}\right\}$ is $\mathrm{k}$ times the number of $\mathrm{m}$ element subsets containing $\mathrm{a}{4}$, then $\mathrm{n}$ is

Options:
A $(m-1) k$
B $\mathrm{mk}$
C $(m+1) k$
D $(m+2) k$

The correct answer is:
$\mathrm{mk}$

Solution:
Hint:
$n_{c_{m}}=k^{n-1} c_{m-1} \Rightarrow n=m k$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2020
If the transformation $z=\log \tan \frac{x}{2}$ reduces the differential equation $\frac{d^2 y}{d x^2}+\cot x \frac{d y}{d x}+4 y \operatorname{cosec}{ }^2 x=0$ into the form $\frac{d^2 y}{d z^2}+k y=0$ then $k$ is equal to

Options:
A $-4$
B 4
C 2
D $-2$

The correct answer is:
4

Solution:
$\frac{d^2 y}{d z^2}=\frac{d}{d z}\left(\frac{d y}{d z}\right)=\frac{d}{d z}\left(\frac{\frac{d y}{d x}}{\frac{d z}{d x}}\right)$
$=\frac{d}{d z}\left(\frac{\frac{d y}{d x}}{\frac{1}{\sin x}}\right) \quad\left(\because \frac{d z}{d x}=\frac{1}{\sin x}\right)$
$=\frac{d}{d z}\left(\sin x \cdot \frac{d y}{d x}\right)=\frac{\frac{d}{d x}\left(\sin x \cdot \frac{d y}{d x}\right)}{\frac{d z}{d x}}$
$=\sin x\left[\cos x \cdot \frac{d y}{d x}+\sin x \frac{d^2 y}{d x^2}\right]$
$\because \frac{d^2 y}{d z^2}+k y=0$
$\therefore \sin x \cdot \cos x \cdot \frac{d y}{d x}+\sin ^2 x \frac{d^2 y}{d x^2}+k y=0$
Dividing the equation by $\sin ^2 x$, we get $\frac{d^2 y}{d x^2}+\cot x \cdot \frac{d y}{d x}+k \cdot \operatorname{cosec}^2 x \cdot y=0$
Comparing with the equation
$\frac{d^2 y}{d x^2}+\cot x \cdot \frac{d y}{d x}+4 \operatorname{cosec}^2 x \cdot y=0$
We get $k=4$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2022
If the vectors $\vec{\alpha}=\hat{\mathbf{i}}+a \hat{\mathbf{j}}+\mathbf{a}^{2} \hat{\mathbf{k}}, \vec{\beta}=\hat{\mathbf{i}}+b \hat{\mathbf{j}}+\mathbf{b}^{2} \hat{\mathbf{k}}$, and $\vec{\gamma}=\hat{\mathbf{i}}+c \hat{\mathbf{j}}+c^{2} \hat{k}$ are three non-coplanar vectors and $\left|\begin{array}{lll}\mathbf{a} & \mathbf{a}^{2} & 1+\mathrm{a}^{3} \\ \mathbf{b} & \mathbf{b}^{2} & 1+\mathrm{b}^{3} \\ \mathrm{c} & \mathrm{c}^{2} & 1+\mathrm{c}^{3}\end{array}\right|=0$,
then the value of abc is

Options:
A 1
B 0
C $-1$
D 2

The correct answer is:
$-1$

Solution:
Hint:
$\left|\begin{array}{lll}\mathrm{a} & \mathrm{a}^{2} & 1 \\ \mathrm{~b} & \mathrm{~b}^{2} & 1 \\ \mathrm{c} & \mathrm{c}^{2} & 1\end{array}\right|(1+\mathrm{abc})=0$
$a b c=-1[\because \vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are non-coplanar vector $]$

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2020
If the velocity of light in vacuum is $3 \times 10^{8} \mathrm{ms}^{-1}$, the time taken (in nanosecond) to travel through a glass plate of thickness $10 \mathrm{cm}$ and refractive index 1.5 is

Options:
A 0.5
B 1
C 2
D 3

The correct answer is:
0.5

Solution:
Given $v=3 \times 10^{8} \mathrm{ms}^{-1}, d=10 \mathrm{m}$ and $\mu=1.5$
Here, $t=\frac{d}{c / \mu}$
$$
t=\frac{10 \times 10^{-2}}{2 \times 10^{8}}
$$
$\left(\therefore \frac{c}{\mu}=\frac{3 \times 10^{8}}{1.5}=2 \times 10^{8} \mathrm{ms}^{-1}\right)$
$t=0.5 \times 10^{-9} \mathrm{s}$

Topic: Physics, Ray Optics, WBJEE, WBJEE 2012
If the vertex of the conic $y^{2}-4 y=4 x-4 a$ always lies between the straight lines ${x}+y=3$ and $2 x+2 y-1=0,$ then

Options:
A $2 < a < 4$
B $-\frac{1}{2} < a < 2$
C $0 < a < 2$
D $-\frac{1}{2} < a < \frac{3}{2}$

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

Solution:
Equation of the vertex of conic
$$
y^{2}-4 y=4 x-4 a
$$
$\Rightarrow \quad y^{2}-4 y+4=4 x-4 a+4$
$\Rightarrow \quad(y-2)^{2}=4[x-(a-1)]$
$\therefore$
Vertex $=(a-12)$

From figure,
clearly. $-\frac{3}{2} < a-1 < 1$
$\therefore$
$-\frac{1}{2} < a < 2$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2016
If the vertex of the conic $y^{2}-4 y=4 x-4 a$ always lies between the straight lines $x+y=3$ and $2 x+2 y-1=0 .$ then

Options:
A $2 < a < 4$
B $-\frac{1}{2} < a < 2$
C $0 < a < 2$
D $-\frac{1}{2} < a < \frac{3}{2}$

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

Solution:
We have, $y^{2}-4 y=4 x-4 a$ $\Rightarrow \quad(y-2)^{2}-4=4 x-4 a$
$\Rightarrow \quad(y-2)^{2}=4 x-4 a+4$
$\Rightarrow \quad(y-2)^{2}=4|x-(a-1)|$
Hence, vertex is $(a-1,2)$
Since, vertex lies between the lines $x+y=3$ and $2 x+2 y-1=0,$ then
$(a-1+2-3)(2 a-2+4-1) < 0$
$\Rightarrow(a-2)(2 a+1) < 0 \Rightarrow a \in\left(-\frac{1}{2}, 2\right)$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2015
If the vertices of a square are \(z_1, z_2, z_3\) and \(z_4\) taken in the anti-clockwise order, then \(z_3=\)

Options:
A \(-i \mathrm{z}_1-(1+i) z_2\)
B \(\mathrm{z}_1-(1+\mathrm{i}) \mathrm{z}_2\)
C \(z_1+(1+i) z_2\)
D \(-i \mathrm{z}_1+(1+\mathrm{i}) \mathrm{z}_2\)

The correct answer is:
\(-i \mathrm{z}_1+(1+\mathrm{i}) \mathrm{z}_2\)

Solution:
Hint : In \(\triangle A B C \frac{z_1-z_2}{z_3-z_2}=\left|\frac{z_1-z_2}{z_3-z_2}\right|\mathrm{i}^{\pi / 2}=\left|\frac{z_1-z_2}{z_3-z_2}\right| \cdot i=\frac{A B}{B C} \cdot i=i\)
\(\Rightarrow \mathrm{z}_1-\mathrm{z}_2=\mathrm{i}\left(\mathrm{z}_3-\mathrm{z}_2\right) \Rightarrow-i \mathrm{z}_1+\mathrm{i} \mathrm{z}_2=\mathrm{z}_3-\mathrm{z}_2 \Rightarrow \mathrm{z}_3=-\mathrm{i} \mathrm{z}_1+(\mathrm{i}+1) \mathrm{z}_2=-\mathrm{i} \mathrm{z}_1+(1+\mathrm{i}) \mathrm{z}_2\)

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2023
If the volume of the parallelopiped with \(\vec{a} \times \vec{b}, \vec{b} \times \vec{c}\) and \(\vec{c} \times \vec{a}\) as coterminous edges is 9 cu. units., then the volume of the parallelopiped with \((\vec{a} \times \vec{b}) \times(\vec{b} \times \vec{c}),(\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})\) and \((\vec{c} \times \vec{a}) \times(\vec{a} \times \vec{b})\) as coterminous edges is

Options:
A 9 cu. units
B 729 cu. units
C 81 cu. units
D 243 cu. units

The correct answer is:
81 cu. units

Solution:
Hint : Volume of Parallelopiped whose coterminous edges are \(\vec{a}, \vec{b}\) and \(\vec{c}=[\vec{a} \vec{b} \vec{c}]\)
So, Given that
\(\therefore[\overrightarrow{\mathrm{a}} \overrightarrow{\mathrm{b}} \overrightarrow{\mathrm{c}}]=3
\quad \ldots \ldots (II)\)
Now volume of parallelopiped whose coterminous edges are
\(\begin{aligned} & (\vec{a} \times \vec{b}) \times(\vec{b} \times \vec{c}),(\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a}),(\vec{c} \times \vec{a}) \times(\vec{a} \times \vec{b}) \\ & {[(\vec{a} \times \vec{b}) \times(\vec{b} \times \vec{c})(\vec{b} \times \vec{c}) \times(\vec{c} \times \vec{a})(\vec{c} \times \vec{a}) \times(\vec{a} \times \vec{b})]} \\ & {[\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}]^2}\end{aligned}\)
\(\begin{aligned}
& =(9)^2 \quad \text{(from (1))}\\
& =81
\end{aligned}\)

Topic: Mathematics, Vector Algebra, WBJEE, WBJEE 2023
If \(\theta+\phi=\frac{\pi}{4}\), then \((1+\tan \theta)(1+\tan \phi)\) is equal to

Options:
A 1
B 2
C \(5 / 2\)
D \(1 / 3\)

The correct answer is:
2

Solution:
\(\begin{aligned}
& \text { Hints: }(1+\tan \theta)\left(1+\frac{(1-\tan \theta)}{1+\tan \theta}\right) \\
& =(1+\tan \theta) \frac{2}{1+\tan \theta}=2
\end{aligned}\)

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2011
If three positive real numbers $a, b, c$ are in A.P. and $a b c=4$ then minimum possible value of $b$ is

Options:
A $2^{3 / 2}$
B $2^{2 / 3}$
C $2^{1 / 3}$
D $2^{5 / 2}$

The correct answer is:
$2^{2 / 3}$

Solution:
$$
\text { Hints } \begin{aligned}
& (b-d) b(b+d)=4 \\
& \left(b^2-d^2\right) b=4 \\
& b^3=4+d^2 b \\
& b^3 \geq 4 \Rightarrow b \geq(2)^{2 / 3}
\end{aligned}
$$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2009
If $u(x)$ and $v(x)$ are two independent solutions of the differential equation $\frac{d^{2} y}{d x^{2}}+b \frac{d y}{d x}+c y=0$ then additional solution(s) of the given differential equation is(are)

Options:
A $y=5 u(x)+8 v(x)$
B $y=c_{1}\{u(x)-v(x)\}+c_{2} v(x), c_{1}$ and $c_{2}$ are arbitrary constants
C $y=c_{1}u(x) v(x)+c_{2} u(x) / v(x) . \quad c_{1}$ and $c_{2}$ are arbitrary constants
D $y=u(x) \vee(x)$

The correct answers are:
$y=5 u(x)+8 v(x)$, $y=c_{1}\{u(x)-v(x)\}+c_{2} v(x), c_{1}$ and $c_{2}$ are arbitrary constants

Solution:
We know that $u(x)$ and $v(x)$ are two independent solutions of the given differential equation, then their linear combination is also the solution of the given equation.
Here, we see that $y=5 u(x)+8 v(x)$ is a linear combination and $y=c_{1}\{u(x)-v(x)\}+c_{2} v(x)$ is also a linear combination of two independent solutions.

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2014