datavorous/entrance-exam-dataset huggingface dataset:
Tangent is drawn at any point $\mathrm{P}(\mathrm{x}, \mathrm{y})$ on a curve, which passes through $(1,1)$. The tangent cuts X-axis and Y-axis at A and B respectively. If AP:BP = 3:1, then

Options:
A the differential equation of the curve is $3 \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0$
B the differential equation of the curve is $3 \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}-\mathrm{y}=0$
C the curve passes through $\left(\frac{1}{8}, 2\right)$
D the normal at $(1,1)$ is $x+3 y=4$

The correct answers are:
the differential equation of the curve is $3 \mathrm{x} \frac{\mathrm{dy}}{\mathrm{dx}}+\mathrm{y}=0$, the curve passes through $\left(\frac{1}{8}, 2\right)$

Solution:
Hint:

$\frac{P A}{P B}=\frac{3}{1}$
Equation of tangent $\mathrm{AB}$ is $\mathrm{Y}-\mathrm{y}=\frac{\mathrm{d} \mathrm{y}}{\mathrm{dx}}(\mathrm{X}-\mathrm{x})$
$\mathrm{A}\left(\frac{\mathrm{xy}^{\prime}-\mathrm{y}}{\mathrm{y}^{\prime}}, 0\right)$ and $\mathrm{B}\left(0, \mathrm{y}-\mathrm{xy}^{\prime}\right)$
Using section formula : $\mathrm{y}=\frac{1 \times 0+3 \times\left(\mathrm{y}-\mathrm{xy}^{\prime}\right)}{4} \Rightarrow 4 \mathrm{y}=3 \mathrm{y}-3 \mathrm{xy}^{\prime} \Rightarrow 3 \mathrm{xy}^{\prime}=-\mathrm{y} \Rightarrow \frac{3 \mathrm{xdy}}{\mathrm{dx}}+\mathrm{y}=0$
$\frac{3 d y}{y}+\frac{d x}{x}=0$
$\mathrm{xy}^{3}=1$
$\frac{\mathrm{dy}}{\mathrm{dx}_{(1,1)}}=\frac{-1}{3}$
Slope of Normal $=3$
Equation of Normal $\Rightarrow y-1=3(x-1) \Rightarrow y-3 x+2=0$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2020
Tangents are drawn to the ellipse $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$ at the ends of both latusrectum. The area of the quadrilateral, so formed is

Options:
A 27 sq units
B $\frac{13}{2}$ sq units
C $\frac{15}{4} \mathrm{sq}$ units
D 45 sq units

The correct answer is:
27 sq units

Solution:
We have, $\frac{x^{2}}{9}+\frac{y^{2}}{5}=1$
$\Rightarrow \quad a=3$ and $b=\sqrt{5}$
$\begin{aligned} \therefore \quad e &=\sqrt{1-\frac{b^{2}}{a^{2}}} \\ &=\sqrt{1-\frac{5}{9}}=\frac{2}{3} \end{aligned}$
$\therefore$ Foci $=(\pm a e, 0)$
$=\left(\pm 3 \times \frac{2}{3}, 0\right)=(\pm 2,0)$
$\therefore$ Ends of latusrectum $=\left(\pm 2, \pm \frac{5}{3}\right)$

Equation of tangent at the end of
latusrectum $P$ is
$\frac{x \times 2}{9}+\frac{y \times 5 / 3}{5}=1$
$\frac{2}{9} x+\frac{1}{3} y=1$
$2 x+3 y=9$
$O A=\frac{9}{2}$ and $O B=3$
$\therefore$ Area of quadrilateral $A B C D$
$$
=4 \times \text { Area of } \Delta O A B
$$
$=4 \times \frac{1}{2} \times O A \times O B$
$=4 \times \frac{1}{2} \times \frac{9}{2} \times 3$
$=27 \mathrm{sq}$ units.

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2017
Tautomerism is exhibited by

Options:
A $\left(\mathrm{Me}_{3} \mathrm{CCO}\right)_{3} \mathrm{CH}$
B
C
D

The correct answers are:
$\left(\mathrm{Me}_{3} \mathrm{CCO}\right)_{3} \mathrm{CH}$, ,

Solution:
Availability of acidic $\alpha-H$ - atoms at $\left(^{*}\right)$ marked positions, enable the compounds to show ketoenol tautomerism.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2013
Temperature of an ideal gas, initially at $27^{\circ} \mathrm{C}$, is raised by $6^{\circ} \mathrm{C}$. The rms velocity of the gas molecules will

Options:
A increase by nearly $2 \%$
B decrease by nearly $2 \%$
C increase by nearly $1 \%$
D decrease by nearly $1 \%$

The correct answer is:
increase by nearly $1 \%$

Solution:
Initial temperature of ideal gas,
$$
T_{1}=273+27=300 \mathrm{K}
$$
when temperature of gas is raised by $6^{\circ} \mathrm{C},$ the final temperature of gas
$$
\begin{aligned}
T_{2} &=273+6+27 \\
&=306 \mathrm{K}
\end{aligned}
$$
Let initial velocities are $v_{r m_{1}},$ and $v_{r m_{2}}$
$v_{\mathrm{rms}} \propto \sqrt{T}$
$\begin{aligned} \frac{v_{\operatorname{mas}_{1}}}{v_{\operatorname{mas}_{2}}} &=\sqrt{\frac{T_{1}}{T_{2}}} \\ v_{\mathrm{rms}_{2}} &=\sqrt{\frac{T_{2}}{T_{1}} \times v_{\mathrm{ms}_{1}}} \\ &=\sqrt{\frac{306}{300}} \times v_{\mathrm{rms}} \\ &=1 \cdot 00 \times v_{\mathrm{rms}_{1}} \end{aligned}$
So, it will increase by $1 \%$

Topic: Physics, Kinetic Theory of Gases, WBJEE, WBJEE 2017
The 4 th higher homologue of ethane is

Options:
A butane
B pentane
C hexane
D heptane

The correct answer is:
hexane

Solution:
The molecular formula of ethane is $\mathrm{C}_{2} \mathrm{H}_{\mathrm{6}}$ To get its fourth homologue, we have to add
$$
\begin{array}{r}
4 \times \mathrm{CH}_{2}=\mathrm{C}_{4} \mathrm{H}_{8} \text { to it. } \\
\mathrm{C}_{2} \mathrm{H}_{6}+\mathrm{C}_4 \mathrm{H}_{8}=\mathrm{C}_6 \mathrm{H}_{14}
\end{array}
$$
$\mathrm{C}_{6} \mathrm{H}_{14}$ is the molecular formula of hexane. Thus, hexane is the fourth homologue of ethane.

Topic: Chemistry, General Organic Chemistry, WBJEE, WBJEE 2014
The above conversion can be carried out by,

Options:
A $\mathrm{Zn}-\mathrm{Hg} /$ Conc. $\mathrm{HCl}$
B i. $\mathrm{H}_2 \mathrm{NNH}_2$ ii. $\mathrm{NaOH}$ in ethylene glycol, $\Delta$
C i. $\mathrm{HSCH}_2 \mathrm{CH}_2 \mathrm{SH} / \mathrm{H}^{\oplus}$ ii. $\mathrm{H}_2 / \mathrm{Ni}$
D Bromine water

The correct answers are:
$\mathrm{Zn}-\mathrm{Hg} /$ Conc. $\mathrm{HCl}$, i. $\mathrm{H}_2 \mathrm{NNH}_2$ ii. $\mathrm{NaOH}$ in ethylene glycol, $\Delta$, i. $\mathrm{HSCH}_2 \mathrm{CH}_2 \mathrm{SH} / \mathrm{H}^{\oplus}$ ii. $\mathrm{H}_2 / \mathrm{Ni}$

Solution:
(A) $\mathrm{Zn}-\mathrm{Hg} /$ Conc. $\mathrm{HCl}-$ Clemmensen Reduction
(B) $\mathrm{NH}_2-\mathrm{NH}_2, \mathrm{NaOH}$ in ethylene glycol, $\Delta-$ Wolf-Kishner reduction
(C) i. $\mathrm{HSCH}_2 \mathrm{CH}_2 \mathrm{SH} / \mathrm{H}^{+}$ii. $\mathrm{H}_2 / \mathrm{Ni}-$ Mozingo method.
These are known to reduce $-\mathrm{CO}-$ to $-\mathrm{CH}_2-$

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2022
The acceleration ' $a$ ' (in $\mathrm{ms}^{-2}$ ) of a body, starting from rest varies with time $t$ (in $s$ ) following the equation $a=3 t+4$ The velocity of the body at time $t=2 \mathrm{~s}$ will be

Options:
A $10 \mathrm{~ms}^{-1}$
B $18 \mathrm{~ms}^{-1}$
C $14 \mathrm{~ms}^{-1}$
D $26 \mathrm{~ms}^{-1}$

The correct answer is:
$14 \mathrm{~ms}^{-1}$

Solution:
$a=3 t+4$
$\begin{aligned} & \frac{d V}{d t}=3 t+4 \\ & \int_0^V d V=\int_0^t(3 t+4) d t \\ & V=\frac{3 t^2}{2}+4 t=\frac{12}{2}+8=14 \mathrm{~m} / \mathrm{s}\end{aligned}$

Topic: Physics, Motion In One Dimension, WBJEE, WBJEE 2009
The acceleration of a particle starting from rest moving in a straight line with uniform acceleration is \(8 \mathrm{~m} / \mathrm{sec}^2\). The time taken by the particle to move the second metre is

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

The correct answer is:
\(\frac{\sqrt{2}-1}{2} \mathrm{sec}\)

Solution:
Hint:
\(\begin{array}{l|l}
\mathrm{S}=\mathrm{ut}+\frac{1}{2} \mathrm{at}^2 & \mathrm{~S}=\mathrm{uT}+\frac{1}{2} \mathrm{aT}^2 \\
1=\frac{1}{2} \cdot 8 \cdot \mathrm{t}^2 & 2=\frac{1}{2} 8 \cdot \mathrm{T}^2 \\
\frac{1}{4}=\mathrm{t}^2 & \mathrm{~T}^2=\frac{1}{2} \\
\mathrm{t}=\frac{1}{2} & \mathrm{~T}=\frac{1}{\sqrt{2}}
\end{array}\)
\(\text {Time }=\frac{1}{\sqrt{2}}-\frac{1}{2}=\frac{2}{2}=\frac{\sqrt{2}-1}{2}\)

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2011
The acceleration vs distance graph for a particle moving with initial velocity $5 \mathrm{~m} / \mathrm{s}$ is shown in the fiture. The velocity of the particle at $x=35 \mathrm{~m}$ will be

Options:
A $20.62 \mathrm{~m} / \mathrm{s}$
B $20 \mathrm{~m} / \mathrm{s}$
C $25 \mathrm{~m} / \mathrm{s}$
D $50 \mathrm{~m} / \mathrm{s}$

The correct answer is:
$25 \mathrm{~m} / \mathrm{s}$

Solution:
Area of graph
$$
\begin{array}{l}
=\int \mathrm{adx}=\int \frac{\mathrm{vdv}}{\mathrm{dx}} \cdot \mathrm{d} \mathrm{x}==\int_{\mathrm{V}_{\mathrm{i}}}^{\mathrm{v}_{\mathrm{f}}} \mathrm{v} \mathrm{d} \mathrm{v} \\
=\left[\frac{\mathrm{v}^{2}}{2}\right]_{\mathrm{v}_{\mathrm{i}}}^{\mathrm{v}_{\mathrm{f}}}, 150+150=\frac{1}{2}\left[\mathrm{~V}_{\mathrm{f}}^{2}-\mathrm{V}_{\mathrm{i}}^{2}\right]
\end{array}
$$
$$
\begin{array}{l}
300 \times 2=V_{f}^{2}-(5)^{2} \\
V_{f}^{2}=600+25, V_{f}=25
\end{array}
$$

Topic: Physics, Motion In One Dimension, WBJEE, WBJEE 2021
The acid in which $\mathrm{O}-\mathrm{O}$ bonding is present is

Options:
A $\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{3}$
B $\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{6}$
C $\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}$
D $\mathrm{H}_{2} \mathrm{S}_{4} \mathrm{O}_{6}$

The correct answer is:
$\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}$

Solution:
$\mathrm{H}_{2} \mathrm{S}_{2} \mathrm{O}_{8}$ shows $-\mathrm{O}-\mathrm{O}-$ (peroxy) bonding as follows:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), WBJEE, WBJEE 2016
The addition of $\mathrm{HBr}$ to 2-pentene gives

Options:
A 2-bromopentane only
B 3-bromopentane only
C 2-bromopentane and 3-bromopentane
D 1-bromopentane and 3-bromopentane

The correct answer is:
2-bromopentane and 3-bromopentane

Solution:
Hints:

Topic: Chemistry, Hydrocarbons, WBJEE, WBJEE 2009
The amount of electrolytes required to coagulate a given amount of Agl colloidal solution (-ve charge) will be in the order

Options:
A $NaNO_{3}>\mathrm{Al}_{2}\left(\mathrm{NO}_{3}\right)_{3}>\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$
B $\mathrm{Al}_{2}\left(\mathrm{NO}_{3}\right)_{3}>\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}>\mathrm{NaNO}_{3}$
C $\mathrm{Al}_{2}\left(\mathrm{NO}_{3}\right)_{3}>\mathrm{NaNO}_{3}>\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}$
D $\mathrm{NaNO}_{3}>\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}>\mathrm{Al}_{2}\left(\mathrm{NO}_{3}\right)_{3}$

The correct answer is:
$\mathrm{NaNO}_{3}>\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}>\mathrm{Al}_{2}\left(\mathrm{NO}_{3}\right)_{3}$

Solution:
Since, AgI is a negatively charged sol, so according to Hardy Schulze law, higher the positive charge on the metal atoms, more is the tendency of metal atom to cause coagulation of negatively charged colloid. The order of positive charge is
$$
\mathrm{Na}^{+} < \mathrm{Ba}^{2+} < \mathrm{AI}^{3 +}
$$
Thus, the order of amount of electrolytes required is in the order
$$
\mathrm{NaNO}_{3}>\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2}>\mathrm{Al}_{2}\left(\mathrm{NO}_{3}\right)_{3}
$$
Note In paper, the formula of aluminium nitrait is incorrect. Correct formula of aluminium nitrea is $\mathrm{Al}_{2}\left(\mathrm{NO}_{3}\right)_{3}$

Topic: Chemistry, Surface Chemistry, WBJEE, WBJEE 2014
The amount of the heat released when \(20 \mathrm{ml} 0.5 \mathrm{M} \mathrm{NaOH}\) is mixed with \(100 \mathrm{ml} 0.1 \mathrm{M} \mathrm{HCl}\) is \(\mathrm{x} \mathrm{kJ}\). The heat of neutralization is

Options:
A \(-100 \mathrm{~xkJ} / \mathrm{mol}\)
B \(-50 \mathrm{~kJ} / \mathrm{mol}\)
C \(+100 \mathrm{~xkJ} / \mathrm{mol}\)
D \(+50 \mathrm{~xkJ} / \mathrm{mol}\)

The correct answer is:
\(-100 \mathrm{~xkJ} / \mathrm{mol}\)

Solution:
Hints: \(\underset{20 \times 05}{\mathrm{NaOH}}+\underset{100 \times 01}{\mathrm{HCl}} \rightarrow \mathrm{NaCl}+\underset{10 \text { millimole prodiral }}{\mathrm{H}_2 \mathrm{O}}\)
During formation of 10 millimole of \(\mathrm{H}_2 \mathrm{O}\) the heat released is \(\mathrm{xKJ}\). Therefore heat of neutralisation is \(-100 \times \mathrm{KJ} / \mathrm{mol}\) (heat released hence negative)

Topic: Chemistry, Thermodynamics (C), WBJEE, WBJEE 2011
The angle between a normal to the plane \(2 x-y+2 z-1=0\) and the X-axis is

Options:
A \(\cos ^{-1} \frac{2}{3}\)
B \(\cos ^{-1} \frac{1}{5}\)
C \(\cos ^{-1} \frac{3}{4}\)
D \(\cos ^{-1} \frac{1}{3}\)

The correct answer is:
\(\cos ^{-1} \frac{2}{3}\)

Solution:
Hint: Normal : 2, -1, 2
\(\text {X-axis : } 1,0,0\)
\(\therefore \theta=\cos ^{-1} \frac{2}{3}\)

Topic: Mathematics, Three Dimensional Geometry, WBJEE, WBJEE 2023
The angle between a pair of tangents drawn from a point $P$ to the circle
$x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+$
$13 \cos ^{2} \alpha=0$ is $2 \alpha .$ The equation of the locus of the point $P$ is

Options:
A $x^{2}+y^{2}+4 x+6 y+9=0$
B $x^{2}+y^{2}-4 x+6 y+9=0$
C $x^{2}+y^{2}-4 x-6 y+9=0$
D $x^{2}+y^{2}+4 x-6 y+9=0$

The correct answer is:
$x^{2}+y^{2}+4 x-6 y+9=0$

Solution:
We have equation of circle $x^{2}+y^{2}+4 x-6 y+9 \sin ^{2} \alpha+13 \cos ^{2} \alpha=0$
Here, $\begin{aligned} C &=(-2,3) \\ \text { Radius } &=\sqrt{(-2)^{2}+(3)^{2}-\left(9 \sin ^{2} \alpha+13 \cos ^{2} \alpha\right)} \\ &=\sqrt{4+9-9 \sin ^{2} \alpha-13 \cos ^{2} \alpha} \\ &=\sqrt{13-13\left(1-\sin ^{2} \alpha\right)-9 \sin ^{2} \alpha} \\=& \sqrt{13 \sin ^{2} \alpha-9 \sin ^{2} \alpha} \\=& \sqrt{4 \sin ^{2} \alpha}=2 \sin \alpha \\ \text { Here, } & \sin \alpha=\frac{A C}{P C} \end{aligned}$
$\Rightarrow \quad P C \sin \alpha=A C$
$\Rightarrow \quad P C^{2} \sin ^{2} \alpha=A C^{2}=(2 \sin \alpha)^{2}$
$\Rightarrow \quad\left[\left(h+2)^{2}+(k-3)^{2}\right] \sin ^{2} \alpha=4 \sin ^{2} \alpha\right.$
$\Rightarrow \quad(h+2)^{2}+(k-3)^{2}=4$
$\Rightarrow \quad h^{2}+4+4 h+k^{2}+9-6 k=4$
$\Rightarrow \quad h^{2}+k^{2}+4 h-6 k+9=0$
Hence, locus of a point is $x^{2}+y^{2}+4 x-6 y+9=0$

Topic: Mathematics, Circle, WBJEE, WBJEE 2018
The angle between the lines joining the foci of an ellipse to one particular extremity of the minor axis is $90^{\circ}$. The eccentricity of the ellipse is

Options:
A $\frac{1}{8}$
B $\frac{1}{\sqrt{3}}$
C $\sqrt{\frac{2}{3}}$
D $\sqrt{\frac{1}{2}}$

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

Solution:
$$
\begin{gathered}
\text { } \frac{b}{a e}=\tan \frac{\pi}{4} \\
\qquad \begin{array}{c}
a e \Rightarrow \frac{b}{a}=e \\
e^2=1-\frac{b^2}{a^2} \\
e^2=1-e^2
\end{array}
\end{gathered}
$$
$$
e^2=\frac{1}{2} \Rightarrow e=\frac{1}{\sqrt{2}}
$$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2009
The angle of intersection between the curves $y=[|\sin x|+|\cos x|]$ and $x^{2}+y^{2}=10,$ where $[x]$ denotes the greatest integer $\leq x,$ is

Options:
A $\tan ^{-1} 3$
B $\tan ^{-1}(-3)$
C $\tan ^{-1} \sqrt{3}$
D $\tan ^{-1}(1 / \sqrt{3})$

The correct answers are:
$\tan ^{-1} 3$

Solution:
Given, $y=\left[|\sin x|+|\cos x|\right.$ land $x^{2}+y^{2}=10$
We know that $(|\sin x|+|\cos x|) \in[1, \sqrt{2}]$
$\therefore$
$$
y=1
$$
The point of intersection of given curve is
$$
\begin{aligned}
& & x^{2}+1^{2} &=10 \\
\Rightarrow & & x^{2} &=9 \\
\Rightarrow & & x &=\pm 3
\end{aligned}
$$
$\therefore$ Point of intersection is (±3,1) $\begin{array}{lr}\text { Now, } & x^{2}+y^{2}=10 \\ \Rightarrow \quad 2 x+2 y \frac{d y}{d x} & =0\end{array}$
$\Rightarrow \quad \frac{d y}{d x}=-\frac{x}{y}$
At point (-3,1) $\frac{d y}{d x}=\frac{3}{1}=3 \Rightarrow m_{1}=3$
Slope of line $y=1$ is $m_{2}=0$
$\therefore$ Angle between two curves is
$$
\begin{aligned}
\tan \theta &=\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}=3 \\
\Rightarrow \quad \theta &=\tan ^{-1}(3)
\end{aligned}
$$

Topic: Mathematics, Circle, WBJEE, WBJEE 2014
The angles of a triangle are in the ratio 2: 3: 7 and the radius of the circumscribed ircle is $10 \mathrm{cm}$. The length of the smallest ide is

Options:
A $2 \mathrm{cm}$
B $5 \mathrm{cm}$
C $7 \mathrm{cm}$
D $10 \mathrm{cm}$

The correct answer is:
$10 \mathrm{cm}$

Solution:
Given, that angles of a triangles are $2 x, 3 x$ and
$7 x$
Since, $2 x+3 x+7 x=180$
$\Rightarrow 12 x=180^{\circ} \Rightarrow x=15^{\circ}$
So, angles are $30^{\circ}, 45^{\circ}$ and $105^{\circ}$
Now, $\frac{a}{\sin 30^{\circ}}=2 R$
$\Rightarrow \quad \frac{a}{\frac{1}{2}}=2 \times 10 \quad[\because R=10 \mathrm{cm}]$
$\Rightarrow \quad 2 a=20 \Rightarrow a=10 \mathrm{cm}$

Topic: Mathematics, Properties of Triangles, WBJEE, WBJEE 2019
The angular points of a triangle are $A(-1,-7)$ $B(5,1)$ and $C(1,4)$. The equation of the bisector of the angle $\angle A B C$ is

Options:
A $x=7 y+2$
B $($ b) $7 y=x+2$
C $y=7 x+2$
D $7 x=y+2$

The correct answer is:
$($ b) $7 y=x+2$

Solution:
Here, $A B=\sqrt{(5+1)^{2}+(1+7)^{2}}=\sqrt{36+64}=10$
$B C=\sqrt{\left(1-5)^{2}+(4-1)^{2}\right.}=\sqrt{16+9}=5$
By angle bisector theorem,
$$
A P: C P=10: 5=2: 1
$$
$P\left(\frac{2 \times 1+1 \times(-1)}{2+1}, \frac{2 \times 4+1 \times(-7)}{2+1}\right)=P\left(\frac{1}{3}, \frac{1}{3}\right)$
Required equation of $B P$ is $y-1=\frac{\frac{1}{3}-1}{\frac{1}{3}-5}(x-5)$
$\Rightarrow \quad y-1=\frac{-2}{-14}(x-5)$
$\Rightarrow \quad 7 y-7=x-5$
$\Rightarrow \quad 7 y=x+2$

Topic: Mathematics, Straight Lines, WBJEE, WBJEE 2018
The approximate value of $\sin 31^{\circ}$ is

Options:
A $>0.5$
B $>0.6$
C $ < 0.5$
D $ < 0.4$

The correct answer is:
$>0.5$

Solution:
We know that, $\sin 30^{\circ}=\frac{1}{2}=0.5$ In $1^{\text {st }}$ quadrant $\sin x$ is increasing function. $\sin 31^{\circ}>\sin 30^{\circ}$
$\Rightarrow \quad \sin 31^{\circ}>0.5$

Topic: Mathematics, Trigonometric Ratios & Identities, WBJEE, WBJEE 2018
The approximate value of \(\sqrt[5]{33}\) correct to 4 decimal places is

Options:
A 2.0000
B 2.1001
C 2.0125
D 2.0500

The correct answer is:
2.0125

Solution:
Hints: \(y=x^{1 / 5} \Delta y=\left(\frac{d y}{d x}\right) \Delta x=1 / 80 \times 1\)
\(y=2+1 / 80\)

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2011
The area bounded by the parabolas $y=4 x^2, y=\frac{x^2}{9}$ and the line $y=2$ is

Options:
A $\frac{5 \sqrt{2}}{3}$ sq. units
B $\frac{10 \sqrt{2}}{3}$ sq. units
C $\frac{15 \sqrt{2}}{3}$ sq. units
D $\frac{20 \sqrt{2}}{3}$ sq. units

The correct answer is:
$\frac{20 \sqrt{2}}{3}$ sq. units

Solution:
Hints: $y=4 x^2$
$$
y=\frac{x^2}{4}
$$
$$
\begin{aligned}
& A=\int_{\mathrm{r}}^2\left[\frac{\sqrt{y}}{2}-3 \sqrt{y}\right] d y=\left(\frac{1}{2}-3\right) \int_0^2 \sqrt{y} d y \\
& =\left(\frac{-\sqrt{\mathrm{y}}}{2}\right) \frac{5}{3}\left(\mathrm{y}^{3 / 2}\right)_0^2=-\frac{5}{3}(2 \sqrt{2}-0) \\
& =\left|-\frac{\sqrt{2}}{3}\right|=\frac{10 \sqrt{2}}{3} \text {, Area of bounded figure }=2 \mathrm{~A}=\frac{20 \sqrt{2}}{3} \\
&
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2010
The area bounded by the parabolas $y=4 x^{2}, y=\frac{x^{2}}{9}$ and the straight line $y=2$ is

Options:
A $\frac{20 \sqrt{2}}{3}$ sq. unit
B $10 \sqrt{5}$ sq. unit
C $\frac{10 \sqrt{3}}{7}$ sq. unit
D $10 \sqrt{2}$ sq. unit

The correct answer is:
$\frac{20 \sqrt{2}}{3}$ sq. unit

Solution:
$y=4 x^{2}, y=\frac{x^{2}}{9}, y=2$
$A=2 \int_{0}^{2}\left(3 \sqrt{y}-\frac{\sqrt{y}}{2}\right) d y=2 \int_{0}^{2} \frac{5 \sqrt{y}}{2} d y=\left.5 \frac{y^{3 / 2}}{3 / 2}\right|_{0} ^{2}=\frac{10}{3}\left[2^{3 / 2}\right]=\frac{20 \sqrt{2}}{3}$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2021
The area bounded by $y=x+1$ and $y=\cos x$ and the $X$ -axis, is

Options:
A 1 sq unit
B $\frac{3}{2}$ sq unit
C $\frac{1}{4}$ sq unit
D $\frac{1}{8}$ sq unit

The correct answers are:
$\frac{3}{2}$ sq unit

Solution:
We have, $y=x+1 \text { and } y=\cos x$

$\begin{aligned} \text { Require area } &=\int_{-1}^{0}(x+1) d x+\int_{0}^{\pi / 2} \cos x d x \\ &=\left[\frac{x^{2}}{2}+x\right]_{-1}^{0}+[\sin x]_{0}^{\pi / 2} \\ &=\left[(0)-\left(\frac{1}{2}-1\right)\right]+[1-0] \\ &=\frac{1}{2}+1=\frac{3}{2} \text { sq unit. } \end{aligned}$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2019
The area bounded by \(y^2=4 x\) and \(x^2=4 y\) is

Options:
A \(\frac{20}{3}\) sq. unit
B \(\frac{16}{3}\) sq. unit
C \(\frac{14}{3}\) sq. unit
D \(\frac{10}{3}\) sq. unit

The correct answer is:
\(\frac{16}{3}\) sq. unit

Solution:
\(\begin{aligned} & \text { Hints: } A=\int_0^4\left(\sqrt{4 \mathrm{x}}-\frac{\mathrm{x}^2}{4}\right) \mathrm{dx} \\ & \left.=2 \cdot \frac{2}{3}\left(\mathrm{x}^{3 / 2}\right)_0^4-\frac{1}{4.3}\right)\left(\mathrm{x}^3\right)_0^4 \\ & =\frac{4}{3}\left[4^{3 / 2}-0\right]-\frac{1}{4.3}\left[4^3-0\right] \\ & =\frac{4^{5 / 2}}{3}-\frac{16}{3}=\frac{32-16}{3}=\frac{16}{3}\end{aligned}\)

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2011
The area enclosed between \(y^2=x\) and \(y=x\) is

Options:
A \(\frac{2}{3}\) sq. units
B \(\frac{1}{2}\) units
C \(\frac{1}{3}\) units
D \(\frac{1}{6}\) units

The correct answer is:
\(\frac{1}{6}\) units

Solution:
\(\begin{aligned} & \text {Hints: } A=\int_0^1(\sqrt{x}-x) d x \\ & =\frac{2}{3}\left(x^{3 / 2}\right)_0^1-\frac{1}{2}\left(x^2\right)_0^1=\frac{2}{3}[1-0]-\frac{1}{2}[1-0] \\ & =\frac{2}{3}-\frac{1}{2}=\frac{4-3}{6}=\frac{1}{6}\end{aligned}\)

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2011
The area enclosed by $y=3 x-5, y=0, x=3$ and $x=5$ is

Options:
A 12 sq. units
B 13 sq. units
C $13 \frac{1}{2}$ sq. units
D $14 \mathrm{sq}$. units

The correct answer is:
$14 \mathrm{sq}$. units

Solution:
Hints: $A=\int_3^5(3 x-5) d x$
$=\frac{3}{2}\left(x^2\right)_3^5-5(x)_3^5,=\frac{3}{2}[25-9]-5(5-3)$
$\frac{3}{2} \cdot 16-5(2)=24-10=14$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2010
The area enclosed by $y=\sqrt{5-x^{2}}$ and
$y=|x-1|$ is

Options:
A $\left(\frac{5 \pi}{4}-2\right)$ sq units
B $\left(\frac{5 \pi-2}{2}\right)$ sq units
C $\left(\frac{5 \pi}{4}-\frac{1}{2}\right)$ sq units
D $\left(\frac{\pi}{2}-5\right)$ sq units

The correct answer is:
$\left(\frac{5 \pi}{4}-\frac{1}{2}\right)$ sq units

Solution:
The graphs of $y=|x-1|$ and $y=\sqrt{5-x^{2}}$ are shown in the figure and the shaded region is the required region bounded by the two curves.

Let $A$ be the area bounded by the given curves. Then
$$
\begin{aligned}
A &=\int_{-1}^{1}\left(\sqrt{5-x^{2}}+x-1\right) d x+\int_{1}^{2}\left(\sqrt{5-x^{2}}-x+1\right) d x \\
&=\int_{-1}^{1} \sqrt{5-x^{2}} d x+\int_{1}^{2} \sqrt{5-x^{2}} d x +\int_{-1}^{1}(x-1) d x+\int_{1}^{2}(-x+1) d x \\
&=\int_{-1}^{2} \sqrt{5-x^{2}} d x+\left[\frac{x^{2}}{2}-x\right]^{1}_{-1}+\left[-\frac{x^{2}}{2}+x\right]_{1}^{2} \\
&=\left[\frac{1}{2} x \sqrt{5-x^{2}}+\frac{5}{2} \sin ^{-1} \frac{x}{\sqrt{5}}\right]_{-1}^{2}-\frac{5}{2} \\
&=1+\frac{5}{2} \sin ^{-1} \frac{2}{\sqrt{5}}+1+\frac{5}{2} \sin ^{-1} \frac{1}{\sqrt{5}}-\frac{5}{2} \\
&=-\frac{1}{2}+\frac{5}{2} \sin ^{-1}\left(\frac{2}{\sqrt{5}} \times \sqrt{1-\frac{1}{5}}+\frac{1}{\sqrt{5}} \sqrt{1-\frac{4}{5}}\right) \\
&=-\frac{1}{2}+\frac{5}{2} \sin ^{-1}(1)=\frac{5 \pi}{4}-\frac{1}{2} \mathrm{squnits}
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2016
The area of the figure bounded by the parabola $x=-2 y^{2}, x=1-3 y^{2}$ is

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

The correct answers are:
$\frac{4}{3}$ square unit

Solution:
Hint:
$\mathrm{A}=\int_{-1}^{1}\left(1-3 \mathrm{y}^{2}\right)-\left(-2 \mathrm{y}^{2}\right) \mathrm{d} \mathrm{y}$
$\Rightarrow \mathrm{A}=2 \int_{0}^{1}\left(1-\mathrm{y}^{2}\right) \mathrm{d} \mathrm{y}=\frac{4}{3}$

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2020
The area of the figure bounded by the parabolas $x=-2 y^{2}$ and $x=1-3 y^{2}$ is

Options:
A $\frac{4}{3} \mathrm{sq}$ units
B $\frac{2}{3} \mathrm{sq}$ units
C $\frac{3}{7}$ sq units
D $\frac{6}{7}$ sq units

The correct answer is:
$\frac{4}{3} \mathrm{sq}$ units

Solution:
We have, $x=-2 y^{2}$
$\Rightarrow \quad y^{2}=-\frac{1}{2} x$ ...(i)
and $\quad x=1-3 y^{2}$
$\Rightarrow y^{2}=-\frac{1}{3}(x-1)$ ...(ii)
From Eqs. (i) and (ii),we get.
$\begin{aligned} &-\frac{1}{2} x=-\frac{1}{3}(x-1) \\ \Rightarrow & 3 x=2(x-1) \\ \Rightarrow & 3 x=2 x-2 \\ \Rightarrow & x=-2 \\ \Rightarrow & y^{2}=-\frac{1}{2}(-2)=1 \\ \Rightarrow & y=\pm 1 \end{aligned}$
$\therefore$ Point of intersection of two curves is $(-2 \pm 1)$

$\therefore$ Required Area $=2 \int_{0}^{1} \sin x d x$
$=2\left[y-\frac{y^{3}}{3}\right]_{0}^{1}$
$=2\left[1-\frac{1}{3}\right]$
$=\frac{4}{3} \mathrm{sq}$ units

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2017
The area of the region bounded by the curve $y=x^3$ its tangent at (1,1) and X-axis is

Options:
A $\frac{1}{12} \mathrm{sq}$ unit
B $\frac{1}{6} \mathrm{sq}$ unit
C $\frac{2}{17} \mathrm{sq}$ unit
D $\frac{2}{15} \mathrm{sq}$ unit

The correct answer is:
$\frac{1}{12} \mathrm{sq}$ unit

Solution:
We have, $y=x^{3}$ and $A(1,1)$
$\therefore$ $\frac{d y}{d x}=3 x^{2}$
On putting $x=1$ in $\mathrm{Eq}$. $\frac{d y}{d x}=3(1)^{2}=3$
$\therefore$ Equation of tangent at $A(1,1)$ is
$y-1=3(x-1) \Rightarrow y=3 x-2$

$\therefore$ Required area $=\int_{0}^{1} x^{3} d x-\int_{2 / 3}^{1}(3 x-2) d x$
$=\left[\frac{x^{4}}{4}\right]_{0}^{1}-\left[\frac{3 x^{2}}{2}-2 x\right]_{2 / 3}^{1}$
$=\frac{1}{4}-\left[\left(\frac{3}{2}-2\right)-\left(\frac{2}{3}-\frac{4}{3}\right)\right]$
$=\frac{1}{12} \mathrm{sq}$ unit

Topic: Mathematics, Area Under Curves, WBJEE, WBJEE 2015