datavorous/entrance-exam-dataset huggingface dataset:
Paragraph:
A uniform thin cylindrical disk of mass $M$ and radius $R$ is attached to two identical massless springs of spring constant $k$ which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance $d$ from its centre. The axle is massless and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is $L$. The disk is initially at its equilibrium position with its centre of mass $(C M)$ at a distance Lfrom the wall. The disk rolls without slipping with velocity $\mathbf{v}_0=v_0 \hat{\mathbf{i}}$ The coefficient of friction is $\mu$.
Question:
The centre of mass of the disk undergoes simple harmonic motion with angular frequency $\omega$ equal to

Options:
A
$\sqrt{\frac{k}{M}}$

B
$\sqrt{\frac{2 k}{M}}$

C
$\sqrt{\frac{2 k}{3 M}}$

D
$\sqrt{\frac{4 k}{3 M}}$

The correct answer is:
$\sqrt{\frac{4 k}{3 M}}$

Solution:
$$
\begin{aligned}
& F_{\text {net }}=-\left(\frac{4 k}{3}\right) \cdot x \\
& \therefore a=\frac{F_{\text {net }}}{M}=-\left(\frac{4 k}{3 M}\right) x=-\omega^2 x \\
& \therefore \omega=\sqrt{\frac{4 k}{3 M}}
\end{aligned}
$$
$\therefore$ correct option is (d).

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
A uniform thin cylindrical disk of mass $M$ and radius $R$ is attached to two identical massless springs of spring constant $k$ which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance $d$ from its centre. The axle is massless and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is $L$. The disk is initially at its equilibrium position with its centre of mass $(C M)$ at a distance Lfrom the wall. The disk rolls without slipping with velocity $\mathbf{v}_0=v_0 \hat{\mathbf{i}}$ The coefficient of friction is $\mu$.
Question:
The net external force acting on the disk when its centre of mass is at displacement $x$ with respect to its equilibrium position is

Options:
A
$-k x$

B
$-2 k x$

C
$-\frac{2 k x}{3}$

D
$-\frac{4 k x}{3}$

The correct answer is:
$-\frac{4 k x}{3}$

Solution:
$\therefore \quad \frac{2 k x-f}{M}=R\left[\frac{f \cdot R}{\frac{1}{2} M R^2}\right]$
Solving this equation, we get $f=\frac{2 k x}{3}$
$$
\therefore\left|F_{\text {net }}\right|=2 k x-f=2 k x-\frac{2 k x}{3}=\frac{4 k x}{3}
$$
This is opposite to displacement.
$$
\therefore \quad F_{\text {net }}=-\frac{4 k x}{3}
$$
$\therefore$ correct option is (d).

Topic: Physics, Oscillations, JEE Main
Paragraph:
A uniform thin cylindrical disk of mass $M$ and radius $R$ is attached to two identical massless springs of spring constant $k$ which are fixed to the wall as shown in the figure. The springs are attached to the axle of the disk symmetrically on either side at a distance $d$ from its centre. The axle is massless and both the springs and the axle are in a horizontal plane. The unstretched length of each spring is $L$. The disk is initially at its equilibrium position with its centre of mass $(C M)$ at a distance Lfrom the wall. The disk rolls without slipping with velocity $\mathbf{v}_0=v_0 \hat{\mathbf{i}}$ The coefficient of friction is $\mu$.
Question:
The net external force acting on the disk when its centre of mass is at displacement $x$ with respect to its equilibrium position is

Options:
A
$-k x$

B
$-2 k x$

C
$-\frac{2 k x}{3}$

D
$-\frac{4 k x}{3}$

The correct answer is:
$-\frac{4 k x}{3}$

Solution:
$\therefore \quad \frac{2 k x-f}{M}=R\left[\frac{f \cdot R}{\frac{1}{2} M R^2}\right]$
Solving this equation, we get $f=\frac{2 k x}{3}$
$$
\therefore\left|F_{\text {net }}\right|=2 k x-f=2 k x-\frac{2 k x}{3}=\frac{4 k x}{3}
$$
This is opposite to displacement.
$$
\therefore \quad F_{\text {net }}=-\frac{4 k x}{3}
$$
$\therefore$ correct option is (d).

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2008 (Paper 2)
Paragraph:
A wooden cylinder of diameter $4 r$, height $H$ and density $\rho / 3$ is kept on a hole of diameter $2 r$ of a tank, filled with liquid of density $\rho$ as shown in the figure.
Question:
If height $h_2$ of water level is further decreased, then

Options:
A
cylinder will not move up and remains at its original position

B
for $h_2=h / 3$, cylinder again starts moving up

C
for $h_2=h / 4$, cylinder again starts moving up

D
for $h_2=h / 5$ cylinder again starts moving up

The correct answer is:
cylinder will not move up and remains at its original position

Solution:
For $h_2 < \frac{4 h}{9}$, buoyant force will further decrease. Hence, the cylinder remains at its original position.

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2006
Paragraph:
A wooden cylinder of diameter $4 r$, height $H$ and density $\rho / 3$ is kept on a hole of diameter $2 r$ of a tank, filled with liquid of density $\rho$ as shown in the figure.
Question:
The block in the above question is maintained at the position by external means and the level of liquid is lowered. The height $h_2$ when this external force reduces to zero is

Options:
A
$\frac{4 h}{9}$

B
$\frac{5 h}{9}$

C
remains same

D
$\frac{2 h}{3}$

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

Solution:
Again equating the forces, we get
$$
h_2=\frac{4 h}{9}
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2006
Paragraph:
A wooden cylinder of diameter $4 r$, height $H$ and density $\rho / 3$ is kept on a hole of diameter $2 r$ of a tank, filled with liquid of density $\rho$ as shown in the figure.
Question:
Now level of the liquid starts decreasing slowly. When the level of liquid is at a height $h_1$ above the cylinder, the block starts moving up. At what value of $h_1$, will the block rise?

Options:
A
$\frac{4 h}{9}$

B
$\frac{5 h}{9}$

C
$\frac{5 h}{3}$

D
remains same

The correct answer is:
$\frac{5 h}{3}$

Solution:
Let $A_1=$ Area of cross-section of cylinder $=4 \pi r^2$
$A_2=$ Area of base of cylinder in air $=\pi r^2$
and $A_3=$ Area of base of cylinder in water
$$
=A_1-A_2=3 \pi r^2
$$
Drawing free body diagram of cylinder.
Equation the net downward forces and net upward forces, we get
$$
h_1=\frac{5}{3} H
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Main
Paragraph:
A wooden cylinder of diameter $4 r$, height $H$ and density $\rho / 3$ is kept on a hole of diameter $2 r$ of a tank, filled with liquid of density $\rho$ as shown in the figure.
Question:
Now level of the liquid starts decreasing slowly. When the level of liquid is at a height $h_1$ above the cylinder, the block starts moving up. At what value of $h_1$, will the block rise?

Options:
A
$\frac{4 h}{9}$

B
$\frac{5 h}{9}$

C
$\frac{5 h}{3}$

D
remains same

The correct answer is:
$\frac{5 h}{3}$

Solution:
Let $A_1=$ Area of cross-section of cylinder $=4 \pi r^2$
$A_2=$ Area of base of cylinder in air $=\pi r^2$
and $A_3=$ Area of base of cylinder in water
$$
=A_1-A_2=3 \pi r^2
$$
Drawing free body diagram of cylinder.
Equation the net downward forces and net upward forces, we get
$$
h_1=\frac{5}{3} H
$$

Topic: Physics, Mechanical Properties of Fluids, JEE Advanced, JEE Advanced 2006
Paragraph:
An acylic hydrocarbon $P$, having molecular formula $\mathrm{C}_6 \mathrm{H}_{10}$, gave acetone as the only organic product through the following sequence of reactions, in which $Q$ is an intermediate organic compounds.
Question:
$$
\text { The structure of the compound } Q \text { is }
$$

Options:
A

B

C

D

The correct answer is:

Solution:
Explained in the beginning.

Topic: Chemistry, Aldehydes and Ketones, JEE Advanced, JEE Advanced 2011 (Paper 1)
Paragraph:
An acylic hydrocarbon $P$, having molecular formula $\mathrm{C}_6 \mathrm{H}_{10}$, gave acetone as the only organic product through the following sequence of reactions, in which $Q$ is an intermediate organic compounds.
Question:
$$
\text { The structure of compound } P \text { is }
$$

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
Paragraph:
An acylic hydrocarbon $P$, having molecular formula $\mathrm{C}_6 \mathrm{H}_{10}$, gave acetone as the only organic product through the following sequence of reactions, in which $Q$ is an intermediate organic compounds.
Question:
$$
\text { The structure of compound } P \text { is }
$$

Options:
A

B

C

D

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Advanced, JEE Advanced 2011 (Paper 1)
Paragraph:
Carbon-14 is used to determine the age of organic material. The procedure is based on the formation of ${ }^{14} \mathrm{C}$ by neutron capture in the upper atmosphere.
$$
{ }_7^{14} \mathrm{~N}+{ }_0^1 n \longrightarrow{ }_6^{14} \mathrm{C}+{ }_1 n^1
$$
${ }^{14} \mathrm{C}$ is absorbed by living organisms during photosynthesis. The ${ }^{14} \mathrm{C}$ content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of ${ }^{14} \mathrm{C}$ in the dead being, falls to the decay which ${ }^{14} \mathrm{C}$ undergoes.
$$
{ }_6^{14} \mathrm{C} \longrightarrow{ }_7^{14} \mathrm{~N}+\beta^{-}
$$
The half life period of ${ }^{14} \mathrm{C}$ is 5770 years. The decay constant $(\lambda)$ can be calculated by using the following formula $\lambda=\frac{0.693}{t_{1 / 2}}$.
The comparison of the $\beta^{-}$activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of ${ }^{14} \mathrm{C}$ to ${ }^{12} \mathrm{C}$ in living matter is $1: 10^{12}$.Question:
A nuclear explosion has taken place leading to increase in concentration of $C^{14}$ in nearby areas. $\mathrm{C}^{14}$ concentration is $C_1$ in nearby areas and $C_2$ in areas far away. If the age of the fossil is determined to be $T_1$ and $T_2$ at the places respectively then

Options:
A
The age of the fossil will increase at the place where explosion has taken and
$$
T_1-T_2=\frac{1}{\lambda} \ln \frac{C_1}{C_2}
$$

B
The age of the fossil will decrease at the place where explosion has taken place and $T_1-T_2=\frac{1}{\lambda} \ln \frac{C_1}{C_2}$

C
The age of fossil will be determined to be same

D
$\frac{T_1}{T_2}=\frac{C_1}{C_2}$

The correct answer is:
The age of the fossil will increase at the place where explosion has taken and
$$
T_1-T_2=\frac{1}{\lambda} \ln \frac{C_1}{C_2}
$$

Solution:
All radioactive decays are the examples of first order kinetics.
So, decay constant $\lambda=\frac{1}{T_1-T_2} \ln \frac{C_1}{C_2}$
$C_1$ is the concentration at $T_1$ time
$C_2$ is the concentration at $T_2$ time
So, $\quad T_1-T_2=\frac{1}{\lambda} \ln \frac{C_1}{C_2}$

Topic: Chemistry, Chemical Kinetics, JEE Advanced, JEE Advanced 2006
Paragraph:
Carbon-14 is used to determine the age of organic material. The procedure is based on the formation of ${ }^{14} \mathrm{C}$ by neutron capture in the upper atmosphere.
$$
{ }_7^{14} \mathrm{~N}+{ }_0^1 n \longrightarrow{ }_6^{14} \mathrm{C}+{ }_1 n^1
$$
${ }^{14} \mathrm{C}$ is absorbed by living organisms during photosynthesis. The ${ }^{14} \mathrm{C}$ content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of ${ }^{14} \mathrm{C}$ in the dead being, falls to the decay which ${ }^{14} \mathrm{C}$ undergoes.
$$
{ }_6^{14} \mathrm{C} \longrightarrow{ }_7^{14} \mathrm{~N}+\beta^{-}
$$
The half life period of ${ }^{14} \mathrm{C}$ is 5770 years. The decay constant $(\lambda)$ can be calculated by using the following formula $\lambda=\frac{0.693}{t_{1 / 2}}$.
The comparison of the $\beta^{-}$activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of ${ }^{14} \mathrm{C}$ to ${ }^{12} \mathrm{C}$ in living matter is $1: 10^{12}$.Question:
What should be the age of fossil for meaningful determination of its age?

Options:
A
6 years

B
6000 years

C
60,000 years

D
It can be used to calculate any age

The correct answer is:
6000 years

Solution:
Radio carbon dating method ceases to be accurate over periods longer than 30,000 years.

Topic: Chemistry, Chemical Kinetics, JEE Advanced, JEE Advanced 2006
Paragraph:
Carbon-14 is used to determine the age of organic material. The procedure is based on the formation of ${ }^{14} \mathrm{C}$ by neutron capture in the upper atmosphere.
$$
{ }_7^{14} \mathrm{~N}+{ }_0^1 n \longrightarrow{ }_6^{14} \mathrm{C}+{ }_1 n^1
$$
${ }^{14} \mathrm{C}$ is absorbed by living organisms during photosynthesis. The ${ }^{14} \mathrm{C}$ content is constant in living organism once the plant or animal dies, the uptake of carbon dioxide by it ceases and the level of ${ }^{14} \mathrm{C}$ in the dead being, falls to the decay which ${ }^{14} \mathrm{C}$ undergoes.
$$
{ }_6^{14} \mathrm{C} \longrightarrow{ }_7^{14} \mathrm{~N}+\beta^{-}
$$
The half life period of ${ }^{14} \mathrm{C}$ is 5770 years. The decay constant $(\lambda)$ can be calculated by using the following formula $\lambda=\frac{0.693}{t_{1 / 2}}$.
The comparison of the $\beta^{-}$activity of the dead matter with that of the carbon still in circulation enables measurement of the period of the isolation of the material from the living cycle. The method however, ceases to be accurate over periods longer than 30,000 years. The proportion of ${ }^{14} \mathrm{C}$ to ${ }^{12} \mathrm{C}$ in living matter is $1: 10^{12}$.Question:
Which of the following option is correct?

Options:
A
In living organisms, circulation of ${ }^{14} \mathrm{C}$ from atmosphere is high so the carbon content is constant in organism.

B
Carbon dating can be used to find out the age of earth crust and rocks.

C
Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content remains constant in living organisms.

D
Carbon dating can not be used to determine concentration of ${ }^{14} \mathrm{C}$ in dead beings.

The correct answer is:
Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content remains constant in living organisms.

Solution:
Radioactive absorption due to cosmic radiation is equal to the rate of radioactive decay, hence the carbon content as the ratio of $\mathrm{C}^{14}$ to $\mathrm{C}^{12}$ remains constant in living organism.

Topic: Chemistry, Chemical Kinetics, JEE Advanced, JEE Advanced 2006
Paragraph:
Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately $6.023 \times 10^{23}$ ) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept.
A $4.0$ molar aqueous solution of $\mathrm{NaCl}$ is prepared and $500 \mathrm{~mL}$ of this solution is electrolysed. This leads to the evolution of chlorine gas at one of electrodes (atomic mass: $\mathrm{Na}=23, \mathrm{Hg}=200 ; 1$ faraday $=96500$ coulombs).Question:
If the cathode is a Hg electrode, the maximum weight $(g)$ of amalgam formed from this solution is

Options:
A
200

B
225

C
400

D
446

The correct answer is:
446

Solution:
$\mathrm{Na}+\mathrm{Hg} \longrightarrow \mathrm{Na}-\mathrm{Hg}$
2 mole $\quad 2$ mole
2 moles of amalgam $=23 \times 2+2 \times 200=446 \mathrm{~g}$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately $6.023 \times 10^{23}$ ) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept.
A $4.0$ molar aqueous solution of $\mathrm{NaCl}$ is prepared and $500 \mathrm{~mL}$ of this solution is electrolysed. This leads to the evolution of chlorine gas at one of electrodes (atomic mass: $\mathrm{Na}=23, \mathrm{Hg}=200 ; 1$ faraday $=96500$ coulombs).Question:
The total charge (coulombs) required for complete electrolysis is

Options:
A
24125

B
48250

C
96500

D
193000

The correct answer is:
193000

Solution:
$2 \mathrm{Cl}^{-} \longrightarrow \mathrm{Cl}_2+2 e^{-}$
Total charge $=2 \times 96500$ Coulamb $=193000$ coulomb

Topic: Chemistry, Electrochemistry, JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
Chemical reactions involve interaction of atoms and molecules. A large number of atoms/molecules (approximately $6.023 \times 10^{23}$ ) are present in a few grams of any chemical compound varying with their atomic/molecular masses. To handle such large numbers conveniently, the mole concept was introduced. This concept has implications in diverse areas such as analytical chemistry, biochemistry, electrochemistry and radiochemistry. The following example illustrates a typical case, involving chemical/electrochemical reaction, which requires a clear understanding of the mole concept.
A $4.0$ molar aqueous solution of $\mathrm{NaCl}$ is prepared and $500 \mathrm{~mL}$ of this solution is electrolysed. This leads to the evolution of chlorine gas at one of electrodes (atomic mass: $\mathrm{Na}=23, \mathrm{Hg}=200 ; 1$ faraday $=96500$ coulombs).Question:
The total number of moles of chlorine gas evolved is

Options:
A
$0.5$

B
$1.0$

C
$2.0$

D
$3.0$

The correct answer is:
$1.0$

Solution:
$4.0 \mathrm{M} \mathrm{NaCl} .500 \mathrm{~mL}(0.5 \mathrm{~L})$
Moles of $\mathrm{NaCl}=4 \times 0.5=2$
2 moles $\mathrm{NaCl} \equiv 2$ moles $\mathrm{Cl}^{-}$
$$
\underset{2 \text { mole }}{2 \mathrm{Cl}^{-}} \longrightarrow \mathrm{Cl}_2
$$

Topic: Chemistry, Some Basic Concepts of Chemistry, JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:

Consider an evacuated cylindrical chamber of height $h$ having rigid conducting plates at the ends and an insulating curved surface as shown in the figure. A number of spherical balls made of a light weight and soft material and coated with a conducting material are placed on the bottom plate. The balls have a radius $r \ll h$. Now a high voltage source $(HV)$ is connected across the conducting plates such that the bottom plate is at $+V_{0}$ and the top plate at $-V_{0} .$ Due to their conducting surface, the balls will get charged, will become equipotential with the plate and are repelled by it. The balls will eventually collide with the top plate, where the coefficient of restitution can be taken to be zero due to the soft nature of the material of the balls. The electric field in the chamber can be considered to be that of a parallel plate capacitor. Assume that there are no collisions between the balls and the interaction between them is negligible. (Ignore gravity)

Question:

The average current in the steady state registered by the ammeter in the circuit will be

Options:
A Proportional to V012
B Proportional to V02
C Proportional to the potential V0
D Zero

The correct answer is:
Proportional to V02

Solution:
h=12 at2 [as u = 0]
2hmqE=time⇒ time= 2mq∆V
E=V0h
<current> =chargetime=qqV02mh2
q ∞ V0
<I>∞ V02

Topic: Physics, Current Electricity, JEE Advanced, JEE Advanced 2016 (Paper 2)
Paragraph:
Consider the circle $x^2+y^2=9$ and the parabola $y^2=8 x$. They intersect at $P$ and $Q$ in the first and the fourth quadrants, respectively. Tangents to the circle at $P$ and $Q$ intersect the $\mathrm{X}$-axis at $R$ and tangents to the parabola at $P$ and $Q$ intersect the $\mathrm{X}$-axis at $S$.Question:
The radius of the circumcircle of the $\triangle P R S$ is

Options:
A
5

B
$3 \sqrt{3}$

C
$3 \sqrt{2}$

D
$2 \sqrt{3}$

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

Solution:
Equation of circumcircle of $\triangle P R S$ is
$$
(x+1)(x-9)+y^2+\lambda y=0
$$
It will pass through $(1,2 \sqrt{2})$, then
$$
\begin{aligned}
-16+8+\lambda \cdot 2 \sqrt{2} & =0 \\
\lambda \quad \lambda & =\frac{8}{2 \sqrt{2}}=2 \sqrt{2}
\end{aligned}
$$
Equation of circumcircle is
$$
x^2+y^2-8 x+2 \sqrt{2} y-9=0
$$
Hence, its radius is $3 \sqrt{3}$.
ALITER
Let
$$
\begin{array}{ll}
\text { Let } & \angle P S R=\theta \\
\Rightarrow & \sin \theta=\frac{2 \sqrt{2}}{2 \sqrt{3}} \\
\Rightarrow & P R=6 \sqrt{2}=2 R \cdot \sin \theta
\end{array}
$$
$$
\Rightarrow \quad \sin \theta=\frac{2 \sqrt{2}}{2 \sqrt{3}}
$$
$$
\Rightarrow \quad R=3 \sqrt{3} \text {. }
$$

Topic: Mathematics, Parabola, JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
Consider the circle $x^2+y^2=9$ and the parabola $y^2=8 x$. They intersect at $P$ and $Q$ in the first and the fourth quadrants, respectively. Tangents to the circle at $P$ and $Q$ intersect the $\mathrm{X}$-axis at $R$ and tangents to the parabola at $P$ and $Q$ intersect the $\mathrm{X}$-axis at $S$.Question:
The radius of the incircle of the $\triangle P Q R$ is

Options:
A
4

B
3

C
$\frac{8}{3}$

D
2

The correct answer is:
2

Solution:
Radius of incircle is, $r=\frac{\Delta}{s}$
As $\Delta=16 \sqrt{2}$
$$
\begin{aligned}
& \therefore \quad s=\frac{6 \sqrt{2}+6 \sqrt{2}+4 \sqrt{2}}{2}=8 \sqrt{2} \\
& \therefore \quad r=\frac{16 \sqrt{2}}{8 \sqrt{2}}=2 \\
&
\end{aligned}
$$

Topic: Mathematics, Properties of Triangles, JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
Consider the circle $x^2+y^2=9$ and the parabola $y^2=8 x$. They intersect at $P$ and $Q$ in the first and the fourth quadrants, respectively. Tangents to the circle at $P$ and $Q$ intersect the $\mathrm{X}$-axis at $R$ and tangents to the parabola at $P$ and $Q$ intersect the $\mathrm{X}$-axis at $S$.Question:
The ratio of the areas of the $\triangle P Q S$ and $\triangle P Q R$ is

Options:
A
$1: \sqrt{2}$

B
$1: 2$

C
$1: 4$

D
$1: 8$

The correct answer is:
$1: 4$

Solution:
Coordinates of $P$ and $Q$ are $(1,2 \sqrt{2})$ and $(1,-2 \sqrt{2})$.
Area of $\triangle P Q R=\frac{1}{2} \cdot 4 \sqrt{2} \cdot 8=16 \sqrt{2}$
Area of $\triangle P Q S=\frac{1}{2} \cdot 4 \sqrt{2} \cdot 2=4 \sqrt{2}$
Ratio of area of $\triangle P Q S$ and $\triangle P Q R$ is $1: 4$.

Topic: Mathematics, Parabola, JEE Advanced, JEE Advanced 2007 (Paper 1)
Paragraph:
Consider the function $f:(-\infty, \infty) \rightarrow(-\infty, \infty)$ defined by $f(x)=\frac{x^2-a x+1}{x^2+a x+1} ; 0 < a < 2$Question:
Let $g(x)=\int_0^{e^x} \frac{f^{\prime}(t)}{1+t^2} d t$. Which of the following is true ?

Options:
A
$g^{\prime}(x)$ is positive on $(-\infty, 0)$ and negative on $(0, \infty)$

B
$g^{\prime}(x)$ is negative on $(-\infty, 0)$ and positive on $(0, \infty)$

C
$g^{\prime}(x)$ changes sign on both $(-\infty, 0)$ and $(0, \infty)$

D
$g^{\prime}(x)$ does not change sign on $(-\infty, \infty)$

The correct answer is:
$g^{\prime}(x)$ is negative on $(-\infty, 0)$ and positive on $(0, \infty)$

Solution:
$\because g^{\prime}(x)=\frac{f^{\prime}\left(e^x\right)}{1+\left(e^x\right)^2} \cdot e^x=2 a\left[\frac{e^{2 x}-1}{\left(e^{2 x}+a e^x+1\right)^2}\right]\left(\frac{e^x}{1+e^{2 x}}\right)$ $\begin{aligned} & g^{\prime}(x)=0 \text {, if } e^{2 x}-1 & =0, \\ \text { i.e., } & & x=0 \\ \text { If } & & x < 0, e^{2 x} < 1 \Rightarrow g^{\prime}(x) < 0 .\end{aligned}$

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2008 (Paper 2)