datavorous/entrance-exam-dataset huggingface dataset:
Paragraph:
p-amino-N, N-dimethylaniline is added to a strongly acidic solution of $X$. The resulting solution is treated with a few drops of aqueous solution of $Y$ to yield blue colouration due to the formation of methylene blue. Treatment of the aqueous solution of $Y$ with the reagent potassium hexacyanoferrate(II) leads to the formation of an intense blue precipitate. The precipitate dissolves an excess addition of the reagent. Similarly, treatment of the solution of $Y$ with the solution of potassium hexacyanoferrate(III) leads to a brown colouration due to the formation of $Z$.Question:
The compound $Y$, is

Options:
A
$\mathrm{MgCl}_2$

B
$\mathrm{FeCl}_2$

C
$\mathrm{FeCl}_3$

D
$\mathrm{ZnCl}_2$

The correct answer is:
$\mathrm{FeCl}_3$

Solution:
No Solution Available

Topic: Chemistry, Amines, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
p-amino-N, N-dimethylaniline is added to a strongly acidic solution of $X$. The resulting solution is treated with a few drops of aqueous solution of $Y$ to yield blue colouration due to the formation of methylene blue. Treatment of the aqueous solution of $Y$ with the reagent potassium hexacyanoferrate(II) leads to the formation of an intense blue precipitate. The precipitate dissolves an excess addition of the reagent. Similarly, treatment of the solution of $Y$ with the solution of potassium hexacyanoferrate(III) leads to a brown colouration due to the formation of $Z$.Question:
The compound $X$, is

Options:
A
$\mathrm{NaNO}_3$

B
$\mathrm{NaCl}$

C
$\mathrm{Na}_2 \mathrm{SO}_4$

D
$\mathrm{Na}_2 \mathrm{~S}$

The correct answer is:
$\mathrm{Na}_2 \mathrm{~S}$

Solution:
No Solution Available

Topic: Chemistry, Amines, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
p-amino-N, N-dimethylaniline is added to a strongly acidic solution of $X$. The resulting solution is treated with a few drops of aqueous solution of $Y$ to yield blue colouration due to the formation of methylene blue. Treatment of the aqueous solution of $Y$ with the reagent potassium hexacyanoferrate(II) leads to the formation of an intense blue precipitate. The precipitate dissolves an excess addition of the reagent. Similarly, treatment of the solution of $Y$ with the solution of potassium hexacyanoferrate(III) leads to a brown colouration due to the formation of $Z$.Question:
The compound $Z$, is

Options:
A
$\mathrm{Mg}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right]$

B
$\mathrm{Fe}\left[\mathrm{Fe}(\mathrm{CN})_6\right]$

C
$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$

D
$\mathrm{K}_2 \mathrm{Zn}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$

The correct answer is:
$\mathrm{Fe}\left[\mathrm{Fe}(\mathrm{CN})_6\right]$

Solution:
$p$-amino- $\mathrm{N}$, N-dimethylaniline reacts with sulphide ion $\left(\mathrm{S}^{2-}\right)$ in presence of $\mathrm{Fe}^{3+}$ ion to form blue coloured solution of methylene blue :

Topic: Chemistry, Amines, JEE Main
Paragraph:
p-amino-N, N-dimethylaniline is added to a strongly acidic solution of $X$. The resulting solution is treated with a few drops of aqueous solution of $Y$ to yield blue colouration due to the formation of methylene blue. Treatment of the aqueous solution of $Y$ with the reagent potassium hexacyanoferrate(II) leads to the formation of an intense blue precipitate. The precipitate dissolves an excess addition of the reagent. Similarly, treatment of the solution of $Y$ with the solution of potassium hexacyanoferrate(III) leads to a brown colouration due to the formation of $Z$.Question:
The compound $Z$, is

Options:
A
$\mathrm{Mg}_2\left[\mathrm{Fe}(\mathrm{CN})_6\right]$

B
$\mathrm{Fe}\left[\mathrm{Fe}(\mathrm{CN})_6\right]$

C
$\mathrm{Fe}_4\left[\mathrm{Fe}(\mathrm{CN})_6\right]_3$

D
$\mathrm{K}_2 \mathrm{Zn}_3\left[\mathrm{Fe}(\mathrm{CN})_6\right]_2$

The correct answer is:
$\mathrm{Fe}\left[\mathrm{Fe}(\mathrm{CN})_6\right]$

Solution:
$p$-amino- $\mathrm{N}$, N-dimethylaniline reacts with sulphide ion $\left(\mathrm{S}^{2-}\right)$ in presence of $\mathrm{Fe}^{3+}$ ion to form blue coloured solution of methylene blue :

Topic: Chemistry, Amines, JEE Advanced, JEE Advanced 2009 (Paper 1)
Paragraph:
Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is $x(t) v s p(t)$ curve in this plane. The arrow on the
curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative.
Question:
The phase space diagram for a ball thrown vertically up from ground is

Options:
A

B

C

D

The correct answer is:

Solution:
Momentum is first positive but decreasing. Displacement (or say position). is initially zero.
It will first increase. At highest point, momentum is zero and displacement is maximum. After that momentum is downwards (negative) and increasing but displacement is decreasing. Only (d) option satisfies these conditions.
Analysis of Question
(i) Question is simple.
(ii) Since all the graphs given in question are parabolic type, we need not to check the mathematical part of the question.

Topic: Physics, Motion In One Dimension, JEE Advanced, JEE Advanced 2011 (Paper 1)
Paragraph:
Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is $x(t) v s p(t)$ curve in this plane. The arrow on the
curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative.
Question:
Consider the spring-mass system, with the mass submerged in water, as shown in the figure. The phase space diagram for one cycle of this system is

Options:
A

B

C

D

The correct answer is:

Solution:
In all the given four figures, at mean position the position coordinate is zero. At the same time mass is starting from the extreme position in all fours cases. In figures (c) and (d), extreme position is more than the initial extreme position. But due to viscosity opposite should be the case. Hence, the answer should be either option (a) or option (b).
Correct answer is (b), because mass starts from positive extreme position (from uppermost position). Then, it will move downwards or momentum should be negative.
Analysis of Question
Question is simple but (a) and (b) options are confusing.

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2011 (Paper 1)
Paragraph:
Phase space diagrams are useful tools in analyzing all kinds of dynamical problems. They are especially useful in studying the changes in motion as initial position and momentum are changed. Here we consider some simple dynamical systems in one-dimension. For such systems, phase space is a plane in which position is plotted along horizontal axis and momentum is plotted along vertical axis. The phase space diagram is $x(t) v s p(t)$ curve in this plane. The arrow on the
curve indicates the time flow. For example, the phase space diagram for a particle moving with constant velocity is a straight line as shown in the figure. We use the sign convention in which position or momentum upwards (or to right) is positive and downwards (or to left) is negative.
Question:
The phase space diagram for simple harmonic motion is a circle centered at the origin. In the figure, the two circles represent the same oscillator but for different initial conditions, and $E_1$ and $E_2$ are the total mechanical energies respectively. Then,

Options:
A
$E_1=\sqrt{2} E_2$

B
$E_1=2 E_2$

C
$E_1=2 E_2$

D
$E_1=16 E_2$

The correct answer is:
$E_1=2 E_2$

Solution:
$E=\frac{1}{2} m \omega^2 A^2$
or
$$
\begin{aligned}
E & \propto A^2 \\
\frac{E_2}{E_1} & =\left(\frac{A_2}{A_1}\right)^2 \\
& =\left(\frac{a}{2 a}\right)^2
\end{aligned}
$$
or $\quad E_1=4 E_2$
$\therefore$ Correct option is (c).
Analysis of Question
(i) Question is simple.
(ii) Since, oscillator is same, $\omega_1=\omega_2$

Topic: Physics, Oscillations, JEE Advanced, JEE Advanced 2011 (Paper 1)
Paragraph:
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator automobiles.
A solution $M$ is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixtrue is $0.9$
Given Freezing point depression constant of water $\left(k^{\text {water }}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Freezing point depression constant of ethanol $\left(k_f^{\text {ethanol }}\right)=2.0 \mathrm{Kkg} \mathrm{mol}^{-1}$
Boiling point elevation constant of water $\left(k_b^{\text {water }}\right)=0.52 \mathrm{Kkg} \mathrm{mol}^{-1}$
Boiling point elevation constant of ethanol $\left(k_b^{\text {ethanol }}\right)=1.2 \mathrm{Kkg} \mathrm{mol}^{-1}$
Standard freezing point of water $=273 \mathrm{~K}$
Standard freezing point of ethanol $=155.7 \mathrm{~K}$
Standard boiling point of water $=373 \mathrm{~K}$
Standard boiling point of ethanol $=351.5 \mathrm{~K}$
Vapour pressure of pure water $=328 \mathrm{~mm}$ of $\mathrm{Hg}$
Vapour pressure of pure ethanol $=40 \mathrm{~mm}$ of $\mathrm{Hg}$
Molecular weight of water $=18 \mathrm{~g} \mathrm{~mol}^{-1}$
Molecular weight of ethanol $=46 \mathrm{~g} \mathrm{~mol}^{-1}$
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.Question:
The freezing point of the solution $M$ is

Options:
A
$268.7 \mathrm{~K}$

B
$268.5 \mathrm{~K}$

C
$234.2 \mathrm{~K}$

D
$150.9 \mathrm{~K}$

The correct answer is:
$150.9 \mathrm{~K}$

Solution:
Solution $M$ is mixture of ethanol and water.
Mole fraction of ethanol is $0.9 \Rightarrow$ Solvent is $\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}$
Mole fraction of water is $0.1 \Rightarrow \mathrm{H}_2 \mathrm{O}$ is solute
Molality of $\mathrm{H}_2 \mathrm{O}=\frac{n_2}{n_1 M_1}=\frac{0.1}{0.9 \times 46} \times 1000=2.415$
$$
\Delta T_f=k_f \cdot m=2 \times 2.415=4.83
$$
or Freezing point of solution $=155.7-4.83=150.87 \mathrm{~K}$

Topic: Chemistry, Solutions, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator automobiles.
A solution $M$ is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixtrue is $0.9$
Given Freezing point depression constant of water $\left(k^{\text {water }}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Freezing point depression constant of ethanol $\left(k_f^{\text {ethanol }}\right)=2.0 \mathrm{Kkg} \mathrm{mol}^{-1}$
Boiling point elevation constant of water $\left(k_b^{\text {water }}\right)=0.52 \mathrm{Kkg} \mathrm{mol}^{-1}$
Boiling point elevation constant of ethanol $\left(k_b^{\text {ethanol }}\right)=1.2 \mathrm{Kkg} \mathrm{mol}^{-1}$
Standard freezing point of water $=273 \mathrm{~K}$
Standard freezing point of ethanol $=155.7 \mathrm{~K}$
Standard boiling point of water $=373 \mathrm{~K}$
Standard boiling point of ethanol $=351.5 \mathrm{~K}$
Vapour pressure of pure water $=328 \mathrm{~mm}$ of $\mathrm{Hg}$
Vapour pressure of pure ethanol $=40 \mathrm{~mm}$ of $\mathrm{Hg}$
Molecular weight of water $=18 \mathrm{~g} \mathrm{~mol}^{-1}$
Molecular weight of ethanol $=46 \mathrm{~g} \mathrm{~mol}^{-1}$
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.Question:
Water is added to the solution $M$ such that the mole fraction of water in the solution becomes $0.9$ The boiling point of this solutions is

Options:
A
$380.4 \mathrm{~K}$

B
$376.2 \mathrm{~K}$

C
$375.5 \mathrm{~K}$

D
$354.7 \mathrm{~K}$

The correct answer is:
$376.2 \mathrm{~K}$

Solution:
$\chi_{\mathrm{H}_2 \mathrm{O}}=0.9$ (solvent) $\chi_{\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH}}=0.1$ (solute)
$$
\Delta T_b=k_b \times m=0.52 \times \frac{1.0 \times 1000}{0.9 \times 18}=3.2 \mathrm{~K}
$$
Boiling point, $T_b=373+3.2=376.2 \mathrm{~K}$

Topic: Chemistry, Solutions, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Applications of colligative properties are very useful in day-to-day life. One of its examples is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator automobiles.
A solution $M$ is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixtrue is $0.9$
Given Freezing point depression constant of water $\left(k^{\text {water }}\right)=1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$
Freezing point depression constant of ethanol $\left(k_f^{\text {ethanol }}\right)=2.0 \mathrm{Kkg} \mathrm{mol}^{-1}$
Boiling point elevation constant of water $\left(k_b^{\text {water }}\right)=0.52 \mathrm{Kkg} \mathrm{mol}^{-1}$
Boiling point elevation constant of ethanol $\left(k_b^{\text {ethanol }}\right)=1.2 \mathrm{Kkg} \mathrm{mol}^{-1}$
Standard freezing point of water $=273 \mathrm{~K}$
Standard freezing point of ethanol $=155.7 \mathrm{~K}$
Standard boiling point of water $=373 \mathrm{~K}$
Standard boiling point of ethanol $=351.5 \mathrm{~K}$
Vapour pressure of pure water $=328 \mathrm{~mm}$ of $\mathrm{Hg}$
Vapour pressure of pure ethanol $=40 \mathrm{~mm}$ of $\mathrm{Hg}$
Molecular weight of water $=18 \mathrm{~g} \mathrm{~mol}^{-1}$
Molecular weight of ethanol $=46 \mathrm{~g} \mathrm{~mol}^{-1}$
In answering the following questions, consider the solutions to be ideal dilute solutions and solutes to be non-volatile and non-dissociative.Question:
The vapour pressure of the solution $M$ is

Options:
A
$39.3 \mathrm{~mm} \mathrm{Hg}$

B
$36.0 \mathrm{~mm} \mathrm{Hg}$

C
$29.5 \mathrm{~mm} \mathrm{Hg}$

D
$28.8 \mathrm{~mm} \mathrm{Hg}$

The correct answer is:
$36.0 \mathrm{~mm} \mathrm{Hg}$

Solution:
Total vapour pressure, $p=p_A^{\circ} \chi_A$
$$
p=40 \times 0.9=36 \mathrm{~mm} \text { of } \mathrm{Hg}
$$
In the paragraph, it has been directed to take solute as non-volatile, thus $\mathrm{H}_2 \mathrm{O}$ do not contribute in the total vapour pressure.

Topic: Chemistry, Solutions, JEE Advanced, JEE Advanced 2008 (Paper 1)
Paragraph:
$$
\mathrm{RCONH}_2 \text { is converted into } \mathrm{RNH}_2 \text { by means of Hofmann bromamide degradation. }
$$
In this reaction, $R \mathrm{CONHBr}$ is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann degradation reaction is an intramolecular reaction.Question:
What are the constituent amines formed when the mixture of (i) and (ii) undergoes Hofmann bromamide degradation?

Options:
A

B

C

D

The correct answer is:

Solution:
$$
\text { Since the overall reaction is intermolecular, hence there will be no effect on product formation. }
$$

Topic: Chemistry, Amines, JEE Advanced, JEE Advanced 2006
Paragraph:
$$
\mathrm{RCONH}_2 \text { is converted into } \mathrm{RNH}_2 \text { by means of Hofmann bromamide degradation. }
$$
In this reaction, $R \mathrm{CONHBr}$ is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann degradation reaction is an intramolecular reaction.Question:
Which is the rate determining step in Hofmann bromamide degradation?

Options:
A
Formation of (i)

B
Formation of (ii)

C
Formation of (iii)

D
Formation of (iv)

The correct answer is:
Formation of (iv)

Solution:
The rate determining step is probably elimination of $\mathrm{Br}^{-}$to form isocyanate because it is slower step of given conversion.

Topic: Chemistry, Amines, JEE Advanced, JEE Advanced 2006
Paragraph:
$$
\mathrm{RCONH}_2 \text { is converted into } \mathrm{RNH}_2 \text { by means of Hofmann bromamide degradation. }
$$
In this reaction, $R \mathrm{CONHBr}$ is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann degradation reaction is an intramolecular reaction.Question:
What are the constituent amines formed when the mixture of (i) and (ii) undergoes Hofmann bromamide degradation?

Options:
A

B

C

D

The correct answer is:

Solution:
$$
\text { Since the overall reaction is intermolecular, hence there will be no effect on product formation. }
$$

Topic: Chemistry, Amines, JEE Main
Paragraph:
$$
\mathrm{RCONH}_2 \text { is converted into } \mathrm{RNH}_2 \text { by means of Hofmann bromamide degradation. }
$$
In this reaction, $R \mathrm{CONHBr}$ is formed from which this reaction has derived its name. Electron donating group at phenyl activates the reaction. Hofmann degradation reaction is an intramolecular reaction.Question:
How can the conversion of (i) to (ii) be brought about?

Options:
A
$\mathrm{KBr}$

B
$\mathrm{KBr}+\mathrm{CH}_3 \mathrm{ONa}$

C
$\mathrm{KBr}+\mathrm{KOH}$

D
$\mathrm{Br}_2+\mathrm{KOH}$

The correct answer is:
$\mathrm{Br}_2+\mathrm{KOH}$

Solution:

Topic: Chemistry, Amines, JEE Advanced, JEE Advanced 2006
Paragraph:
Read the following passage and answer the questions.
For every function $f(x)$ which is twice differentiable, these will be good approximation of $\int_a^b f(x) d x \cong\left(\frac{b-a}{2}\right)\{f(a)+f(b)\}$. Now, if we take $c=\frac{a+b}{2}$, then using above again, we get $\int_a^b f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x \cong \frac{b-a}{4}\{f(a)+f(b)+2 f(c)\}$ and so on.
We get approximation for value of $\int_a^b f(x) d x$.Question:
If $f^{\prime \prime}(x) < 0, \forall x \in(a, b), c(c, f(c))$ is point of maxima where $c \in(a, b)$, then $f^{\prime}(c)$ is

Options:
A
$\frac{f(b)-f(a)}{b-a}$

B
$3\left(\frac{f(b)-f(a)}{b-a}\right)$

C
$2\left(\frac{f(b)-f(a)}{b-a}\right)$

D
0

The correct answer is:
0

Solution:
$f^{\prime}(c)=0$.

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2006
Paragraph:
Read the following passage and answer the questions.
For every function $f(x)$ which is twice differentiable, these will be good approximation of $\int_a^b f(x) d x \cong\left(\frac{b-a}{2}\right)\{f(a)+f(b)\}$. Now, if we take $c=\frac{a+b}{2}$, then using above again, we get $\int_a^b f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x \cong \frac{b-a}{4}\{f(a)+f(b)+2 f(c)\}$ and so on.
We get approximation for value of $\int_a^b f(x) d x$.Question:
If $\lim _{t \rightarrow a} \frac{\int_a^t f(x) d x-\frac{(t-a)}{2}\{f(t)+f(a)\}}{(t-a)^3}=0$, then degree of polynomial function $f(x)$ at most is

Options:
A
0

B
1

C
3

D
2

The correct answer is:
1

Solution:
$$
\begin{aligned}
\lim _{t \rightarrow a} \frac{\int_a^t f(t) d t-\frac{(t-a)}{2}(f(t)+f(a))}{(t-a)^3} & =0 \\
\Rightarrow \quad \lim _{h \rightarrow 0} \frac{\int_a^{a+h} f(t) d t-\frac{h}{2}(f(a+h)+f(a))}{h^3} & =0 \\
\Rightarrow \quad \lim _{h \rightarrow 0} \frac{f(a+h)-\frac{1}{2}(f(a+h)+f(a))-\frac{h}{2}\left(f^{\prime}(a+h)\right)}{3 h^2} & =0
\end{aligned}
$$
$f^{\prime}(a+h)-\frac{1}{2} f^{\prime}(a+h)$
$\Rightarrow \quad \lim _{h \rightarrow 0} \frac{-\frac{1}{2} f^{\prime}(a+h)-\frac{h}{2} f^{\prime \prime}(a+h)}{6 h}=0$
$\Rightarrow \quad \lim _{h \rightarrow 0} \frac{-\frac{h}{2} f^{\prime}(a+h)}{6 h}=0$
$\Rightarrow \quad f^{\prime \prime}(a)=0, \forall a \in R$
$\Rightarrow(x)$ must have maximum degree 1.

Topic: Mathematics, Limits, JEE Advanced, JEE Advanced 2006
Paragraph:
Read the following passage and answer the questions.
For every function $f(x)$ which is twice differentiable, these will be good approximation of $\int_a^b f(x) d x \cong\left(\frac{b-a}{2}\right)\{f(a)+f(b)\}$. Now, if we take $c=\frac{a+b}{2}$, then using above again, we get $\int_a^b f(x) d x=\int_a^c f(x) d x+\int_c^b f(x) d x \cong \frac{b-a}{4}\{f(a)+f(b)+2 f(c)\}$ and so on.
We get approximation for value of $\int_a^b f(x) d x$.Question:
Good approximation of $\int_0^{\pi / 2} \sin x d x$, is

Options:
A
$\frac{\pi}{4}$

B
$\frac{\pi}{4}(\sqrt{2}+1)$

C
$\frac{\pi}{8}(\sqrt{2}+1)$

D
$\frac{\pi}{8}$

The correct answer is:
$\frac{\pi}{8}(\sqrt{2}+1)$

Solution:
$\int_0^{\pi / 2} \sin x d x=\frac{\frac{\pi}{2}-0}{4}\left(\sin 0+\sin \left(\frac{\pi}{2}\right)+2 \sin \left(\frac{0+\frac{\pi}{2}}{2}\right)\right)=\frac{\pi}{8}(1+\sqrt{2})$.

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2006
Paragraph:
Read the following passage and answer the questions.
Let $A B C D$ be a square of side length 2 units. $C_2$ is the circle through vertices $A, B, C, D$ and $C_1$ is the circle touching all the sides of square $A B C D$. $L$ is the line through $A$.Question:
A line $M$ through $A$ is drawn parallel to $B D$. Points $S$ moves such that its distances from the line $B D$ are the vertex $A$ are equal. If locus of $S$ cuts $M$ at $T_2$ and $T_3$ and $A C$ at $T_1$, then area of $\Delta T_1 T_2 T_3$ is

Options:
A
$\frac{1}{2}$ sq unit

B
$\frac{2}{3}$ sq unit

C
1 sq unit

D
2 sq unit

The correct answer is:
1 sq unit

Solution:
$\because \quad A G=\sqrt{2}$
$$
\therefore \quad A T_1=T_1 G=\frac{1}{\sqrt{2}}
$$
\{ as $A$ is the focus, $T_1$ is the vertex and $B D$ is the directrix of parabola.\}
Also, $T_2 T_3$ is latus rectum
$$
\begin{aligned}
& \therefore \quad T_2 T_3=4 \cdot \frac{1}{\sqrt{2}} \\
& \therefore \text { Area of } \Delta T_1 T_2 T_3=\frac{1}{2} \times \frac{1}{\sqrt{2}} \times \frac{4}{\sqrt{2}}=1 \text { sq unit }
\end{aligned}
$$

Topic: Mathematics, Parabola, JEE Advanced, JEE Advanced 2006
Paragraph:
Read the following passage and answer the questions.
Let $A B C D$ be a square of side length 2 units. $C_2$ is the circle through vertices $A, B, C, D$ and $C_1$ is the circle touching all the sides of square $A B C D$. $L$ is the line through $A$.Question:
A circle touches the line $L$ and the circle $C_1$ externally such that both the circles are on the same side of the line, then the locus of centre of the circle is

Options:
A
ellipse

B
hyperbola

C
parabola

D
parts of straight line

The correct answer is:
parabola

Solution:
Let $C$ be the centre of the required circle.
Now, draw a line parallel to $L$ at a distance of $r_1$ (radius of $C_1$ ) from it. Now, $C P_1=A C$
$\Rightarrow C$ lies on a parabola.

Topic: Mathematics, Parabola, JEE Advanced, JEE Advanced 2006