datavorous/entrance-exam-dataset huggingface dataset:
The maximum value of the expression $\frac{1}{\sin ^2 \theta+3 \sin \theta \cos \theta+5 \cos ^2 \theta}$ is

Options:

The correct answer is:
2

Solution:
Let
$$
\begin{aligned}
& f(\theta)=\frac{1}{\sin ^2 \theta+3 \sin \theta \cos \theta+5 \cos ^2 \theta} \\
& \text { Again let } \\
& g(\theta)=\sin ^2 \theta+3 \sin \theta \cos \theta+5 \cos ^2 \theta \\
& =\frac{1-\cos 2 \theta}{2}+5\left(\frac{1+\cos 2 \theta}{2}\right)+\frac{3}{2} \sin 2 \theta \\
& =3+2 \cos 2 \theta+\frac{3}{2} \sin 2 \theta \\
& \therefore \quad g(\theta)_{\min }=3-\sqrt{4+\frac{9}{4}}=3-\frac{5}{2}=\frac{1}{2} \\
& \therefore \quad f(\theta)=\frac{1}{g(\theta)_{\min }}=2
\end{aligned}
$$

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2010 (Paper 1)
The maximum value of the function $f(x)=2 x^3-15 x^2+36 x-48$ on the set $A=\left\{x \mid x^2+20 \leq 9 x\right\}$ is

Options:

The correct answer is:
5

Solution:
Given, $\quad A=\left\{x \mid x^2+20 \leq 9 x\right\}$ $=\{x \mid x \in[4,5]\}$
Now, $f^{\prime}(x)=6\left(x^2-5 x+6\right)$
Put $\quad f^{\prime}(x)=0 \Rightarrow x=2,3$
$f(2)=-20, f(3)=-21$,
$f(4)=-16, f(5)=7$
From graph, maximum of $f(x)$ on set $A$ is $f(5)=7$.

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2009 (Paper 2)
The minimum value of the sum of real numbers $a^{-5}, a^{-4}, 3 a^{-3}, 1, a^8$ and $a^{10}$ with $a>0$ is

Options:

The correct answer is:
8

Solution:
Using $A M \geq G M$,
$$
\begin{aligned}
& \frac{a^{-5}+a^{-4}+a^{-3}+a^{-3}+a^{-3}+1+a^8+a^{10}}{8} \\
& \geq\left(a^{-5} \cdot a^{-4} \cdot a^{-3} \cdot a^{-3} \cdot a^{-3} \cdot 1 \cdot a^8 \cdot a^{10}\right)^{\frac{1}{8}} \\
& \Rightarrow a^{-5}+a^{-4}+3 a^{-3}+1+a^8+a^{10} \geq 8 \cdot 1
\end{aligned}
$$
Hence, minimum value is 8 .

Topic: Mathematics, Sequences and Series, JEE Advanced, JEE Advanced 2011 (Paper 1)
The nitrogen oxide(s) that contain(s) N.N bond(s) is (are)

Options:
A
$\mathrm{N}_2 \mathrm{O}$

B
$\mathrm{N}_2 \mathrm{O}_3$

C
$\mathrm{N}_2 \mathrm{O}_4$

D
$\mathrm{N}_2 \mathrm{O}_5$

The correct answers are:
$\mathrm{N}_2 \mathrm{O}$
,
$\mathrm{N}_2 \mathrm{O}_3$
,
$\mathrm{N}_2 \mathrm{O}_4$

Solution:
(a)
(b)
(c)
(d)

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2009 (Paper 2)
The normal at a point $P$ on the ellipse $x^2+4 y^2=16$ meets the $x$-axis at $Q$. If $M$ is the mid-point of the line segment $P Q$, then the locus of $M$ intersects the latusrectum of the given ellipse at the points

Options:
A
$\left(\pm \frac{3 \sqrt{5}}{2}, \pm \frac{2}{7}\right)$

B
$\left(\pm \frac{3 \sqrt{5}}{2}, \pm \frac{\sqrt{19}}{4}\right)$

C
$\left(\pm 2 \sqrt{3}, \pm \frac{1}{7}\right)$

D
$\left(\pm 2 \sqrt{3}, \pm \frac{4 \sqrt{3}}{7}\right)$

The correct answer is:
$\left(\pm 2 \sqrt{3}, \pm \frac{1}{7}\right)$

Solution:
Given, $\quad \frac{x^2}{16}+\frac{y^2}{4}=1$
Here, $\quad a=4, b=2$
Equation of normal is
$$
\begin{aligned}
& 4 x \sec \theta-2 y \operatorname{cosec} \theta=12 \\
& M\left(\frac{7 \cos \theta}{2}, \sin \theta\right)=(h, k) \text { (say) } \\
& \therefore \quad h=\frac{7 \cos \theta}{2} \Rightarrow \cos \theta=\frac{2 h}{7} \\
& {\left[\because \cos ^2 \theta+\sin ^2 \theta=1\right]} \\
&
\end{aligned}
$$
Hence, locus is $\frac{4 x^2}{49}+y^2=1$
For given ellipse, $e^2=1-\frac{4}{16}=\frac{3}{4}$
$$
\begin{array}{rc}
\therefore \quad e=\frac{\sqrt{3}}{2} \\
\therefore \quad x=\pm 4 \times \frac{\sqrt{3}}{2}=\pm 2 \sqrt{3} \\
& {[\because x=\pm a e] \ldots \text { (ii) }}
\end{array}
$$
On solving Eqs. (i) and (ii), we get
$$
\begin{aligned}
& \frac{4}{49} \times 12+y^2=1 \Rightarrow y^2=1-\frac{48}{49}=\frac{1}{49} \\
& \Rightarrow \quad y=\pm \frac{1}{7}
\end{aligned}
$$
$\therefore$ Required points $\left(\pm 2 \sqrt{3}, \pm \frac{1}{7}\right)$.

Topic: Mathematics, Ellipse, JEE Main
The normal at a point $P$ on the ellipse $x^2+4 y^2=16$ meets the $x$-axis at $Q$. If $M$ is the mid-point of the line segment $P Q$, then the locus of $M$ intersects the latusrectum of the given ellipse at the points

Options:
A
$\left(\pm \frac{3 \sqrt{5}}{2}, \pm \frac{2}{7}\right)$

B
$\left(\pm \frac{3 \sqrt{5}}{2}, \pm \frac{\sqrt{19}}{4}\right)$

C
$\left(\pm 2 \sqrt{3}, \pm \frac{1}{7}\right)$

D
$\left(\pm 2 \sqrt{3}, \pm \frac{4 \sqrt{3}}{7}\right)$

The correct answer is:
$\left(\pm 2 \sqrt{3}, \pm \frac{1}{7}\right)$

Solution:
Given, $\quad \frac{x^2}{16}+\frac{y^2}{4}=1$
Here, $\quad a=4, b=2$
Equation of normal is
$$
\begin{aligned}
& 4 x \sec \theta-2 y \operatorname{cosec} \theta=12 \\
& M\left(\frac{7 \cos \theta}{2}, \sin \theta\right)=(h, k) \text { (say) } \\
& \therefore \quad h=\frac{7 \cos \theta}{2} \Rightarrow \cos \theta=\frac{2 h}{7} \\
& {\left[\because \cos ^2 \theta+\sin ^2 \theta=1\right]} \\
&
\end{aligned}
$$
Hence, locus is $\frac{4 x^2}{49}+y^2=1$
For given ellipse, $e^2=1-\frac{4}{16}=\frac{3}{4}$
$$
\begin{array}{rc}
\therefore \quad e=\frac{\sqrt{3}}{2} \\
\therefore \quad x=\pm 4 \times \frac{\sqrt{3}}{2}=\pm 2 \sqrt{3} \\
& {[\because x=\pm a e] \ldots \text { (ii) }}
\end{array}
$$
On solving Eqs. (i) and (ii), we get
$$
\begin{aligned}
& \frac{4}{49} \times 12+y^2=1 \Rightarrow y^2=1-\frac{48}{49}=\frac{1}{49} \\
& \Rightarrow \quad y=\pm \frac{1}{7}
\end{aligned}
$$
$\therefore$ Required points $\left(\pm 2 \sqrt{3}, \pm \frac{1}{7}\right)$.

Topic: Mathematics, Ellipse, JEE Advanced, JEE Advanced 2009 (Paper 2)
The number of $3 \times 3$ matrices $A$ whose entries are either 0 or 1 and for which the system $A\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$ has exactly two distinct solutions, is

Options:
A
0

B
$2^9-1$

C
168

D
2

The correct answer is:
0

Solution:
Since, $A\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}1 \\ 0 \\ 0\end{array}\right]$ is linear equation in three variables and that could have only unique, no solution or infinitely many solution.
$\therefore$ It is not possible to have two solutions.
Hence, number of matrices $A$ is zero.

Topic: Mathematics, Matrices, JEE Advanced, JEE Advanced 2010 (Paper 1)
The number of aldol reaction(s) that occurs in the given transformation is :

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Advanced, JEE Advanced 2012 (Paper 1)
The number of aldol reaction(s) that occurs in the given transformation is :

Options:
A 1
B 2
C 3
D 4

The correct answer is:
3

Solution:

Topic: Chemistry, Aldehydes and Ketones, JEE Main
The number of all possible values of $\theta$, where $0 < \theta < \pi$, for which the system of equations
$$
\begin{array}{r}
(y+z) \cos 3 \theta=(x y z) \sin 3 \theta \\
x \sin 3 \theta=\frac{2 \cos 3 \theta}{y}+\frac{2 \sin 3 \theta}{z}
\end{array}
$$
and $(x y z) \sin 3 \theta=(y+2 z) \cos 3 \theta$
$+y \sin 3 \theta$
have a solution $\left(x_0, y_0, z_0\right)$
with $\quad y_0 z_0 \neq 0$, is

Options:

The correct answer is:
3

Solution:
Given equations can be written as
$$
\begin{gathered}
x \sin 3 \theta-\frac{\cos 3 \theta}{y}-\frac{\cos 3 \theta}{z}=0 \\
x \sin 3 \theta-\frac{2 \cos 3 \theta}{y}-\frac{2 \sin 3 \theta}{z}=0 \\
\text { and } x \sin 3 \theta-\frac{2}{y} \cos 3 \theta \\
-\frac{1}{z}(\cos 3 \theta+\sin 3 \theta)=0
\end{gathered}
$$
Eqs. (ii) and (iii), implies
$$
\begin{aligned}
& 2 \sin 3 \theta=\cos 3 \theta+\sin 3 \theta \\
\Rightarrow & \sin 3 \theta=\cos 3 \theta \\
\therefore & \tan 3 \theta=1 \\
\Rightarrow & 3 \theta=\frac{\pi}{4}, \frac{5 \pi}{4}, \frac{9 \pi}{4} \text { or } \theta=\frac{\pi}{12}, \frac{5 \pi}{12}, \frac{9 \pi}{12} .
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Advanced, JEE Advanced 2010 (Paper 1)
The number of distinct real roots of $x^4-4 x^3+12 x^2+x-1=0$ is

Options:

The correct answer is:
2

Solution:
$f(x)=x^4-4 x^3+12 x^2+x-1$
$$
\begin{aligned}
f^{\prime}(x) & =4 x^3-12 x^2+24 x+1 \\
f^{\prime \prime}(x) & =12 x^2-24 x+24 \\
& =12\left(x^2-2 x+2\right) \\
& =12\left\{(x-1)^2+1\right\}>0, \text { for all } x
\end{aligned}
$$
$\Rightarrow f^{\prime}(x)$ is increasing.
Since, $f^{\prime}(x)$ is cubic and increasing. $\Rightarrow f^{\prime}(x)$ has only one real root and two imaginary roots.
$\therefore f(x)$ cannot have all distinct root.
$\Rightarrow$ Atmost 2 real roots.
Now, $f(-1)=15$,
$f(0)=-1$ and $f(1)=9$
$\therefore f(x)$ must have one root in $(-1,0)$ and other in $(0,1)$.
$\Rightarrow 2$ real roots.

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2011 (Paper 2)
The number of distinct real values of $\lambda$, for which the vectors $-\lambda^2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}$, $\hat{\mathbf{i}}-\lambda^2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+\hat{\mathbf{j}}-\lambda^2 \hat{\mathbf{k}}$ are coplanar, is

Options:
A
zero

B
one

C
two

D
three

The correct answer is:
two

Solution:
$$
\begin{aligned}
& \left|\begin{array}{ccc}
-\lambda^2 & 1 & 1 \\
1 & -\lambda^2 & 1 \\
1 & 1 & -\lambda^2
\end{array}\right|=0 \Rightarrow \lambda^6-3 \lambda^2-2=0 \\
& \Rightarrow \quad\left(1+\lambda^2\right)^2\left(\lambda^2-2\right)=0 \Rightarrow \lambda=\pm \sqrt{2} \\
&
\end{aligned}
$$

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2007 (Paper 1)
The number of hexagonal faces that are present in a truncated octahedron is

Options:

The correct answer is:
8

Solution:
The truncated octahedron is the 14 -faced Archimedean solid, with 14 total faces. 6 squares and 8 regular hexagons. The truncated octahedron is formed by removing the six right square pyramids one from each point of a regular octahedron as:

Topic: Chemistry, Solid State, JEE Advanced, JEE Advanced 2011 (Paper 2)
The number of neutrons emitted when ${ }_{92}^{235} \mathrm{U}$ undergoes controlled nuclear fission to ${ }_{54}^{142} \mathrm{Xe}$ and ${ }_{38}^{90} \mathrm{Sr}$ is

Options:

The correct answer is:
3

Solution:
${ }_{92} \mathrm{U}^{235} \longrightarrow{ }_{54} \mathrm{Xe}^{142}+{ }_{38} \mathrm{Sr}^{90}+3{ }_0 n^1$
Nuclear chemistry
Basic application
II

Topic: Chemistry, Structure of Atom, JEE Advanced, JEE Advanced 2010 (Paper 1)
The number of optically active products obtained from the complete ozonolysis of the given compound is :

Options:
A 0
B 1
C 2
D 4

The correct answer is:
0

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2012 (Paper 1)
The number of seven digit integers, with sum of the digits equal to 10 and formed by using the digits 1,2 and 3 only, is

Options:
A
55

B
66

C
77

D
88

The correct answer is:
77

Solution:
There are two possible cases arise:
Case I Five 1's, one 2's, one 3's
$\therefore$ Number of numbers $=\frac{7 !}{5 !}=42$
Case II Four 1's, three 2's
$\therefore$ Number of numbers $=\frac{7 !}{4 ! 3 !}=35$
$\therefore$ Total number of numbers $=42+35=77$

Topic: Mathematics, Permutation Combination, JEE Advanced, JEE Advanced 2009 (Paper 1)
The number of solutions of the pair of equations $2 \sin ^2 \theta-\cos 2 \theta=0$ and $2 \cos ^2 \theta-3 \sin \theta=0$ in the interval $[0,2 \pi]$ is

Options:
A
zero

B
one

C
two

D
four

The correct answer is:
two

Solution:
$2 \sin ^2 \theta-\cos 2 \theta=0 \Rightarrow \sin ^2 \theta=\frac{1}{4}$
Also, $2 \cos ^2 \theta=3 \sin \theta \Rightarrow \sin \theta=\frac{1}{2}$
$\Rightarrow$ Two solutions in $[0,2 \pi]$.

Topic: Mathematics, Trigonometric Equations, JEE Advanced, JEE Advanced 2007 (Paper 1)
The number of stereoisomers obtained by bromination of trans-2-butene is

Options:
A
1

B
2

C
3

D
4

The correct answer is:
1

Solution:

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2007 (Paper 2)
The number of structural isomers for $\mathrm{C}_6 \mathrm{H}_{14}$ is

Options:
A
3

B
4

C
5

D
6

The correct answer is:
5

Solution:
$$
\text { Molecular Formula } \mathrm{C}_6 \mathrm{H}_{14}
$$

Topic: Chemistry, General Organic Chemistry, JEE Advanced, JEE Advanced 2007 (Paper 1)
The number of values of $\theta$ in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\theta \neq \frac{n \pi}{5}$ for $n=0, \pm 1, \pm 2$ and $\tan \theta=\cot 5 \theta$ as well as $\sin 2 \theta=\cos 4 \theta$ is

Options:

The correct answer is:
3

Solution:
Given, $\tan \theta=\cot 5 \theta$
$$
\begin{aligned}
& \Rightarrow \quad \tan \theta=\tan \left(\frac{\pi}{2}-5 \theta\right) \\
& \Rightarrow \quad \frac{\pi}{2}-5 \theta=n \pi+\theta \\
& \Rightarrow \quad 6 \theta=\frac{\pi}{2}-n \pi \\
& \Rightarrow \quad \theta=\frac{\pi}{12}-\frac{n \pi}{6} \\
& \text { Also, } \cos 4 \theta=\sin 2 \theta=\cos \left(\frac{\pi}{2}-2 \theta\right) \\
& \Rightarrow \quad 4 \theta=2 n \pi \pm\left(\frac{\pi}{2}-2 \theta\right)
\end{aligned}
$$
Taking positive
$$
6 \theta=2 n \pi+\frac{\pi}{2} \Rightarrow \theta=\frac{n \pi}{3}+\frac{\pi}{12}
$$
Taking negative
$$
2 \theta=2 n \pi-\frac{\pi}{2} \Rightarrow \theta=n \pi-\frac{\pi}{4}
$$
Above values of $\theta$ suggests that there are only 3 common solutions.

Topic: Mathematics, Trigonometric Equations, JEE Advanced, JEE Advanced 2010 (Paper 1)
The number of water molecule(s) directly bonded to the metal centre in $\mathrm{CuSO}_4 \cdot 5 \mathrm{H}_2 \mathrm{O}$ is

Options:

The correct answer is:
4

Solution:
$\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_4\right] \mathrm{SO}_4 \cdot \mathrm{H}_2 \mathrm{O}$. Coordination number of $\mathrm{Cu}$ is 4 .

Topic: Chemistry, d and f Block Elements, JEE Advanced, JEE Advanced 2009 (Paper 2)
The oxidation number of $\mathrm{Mn}$ in the product of alkaline oxidative fusion of $\mathrm{MnO}_2$ is

Options:

The correct answer is:
6

Solution:
$2 \mathrm{MnO}_2+4 \mathrm{KOH}+\mathrm{O}_2 \stackrel{\text { Fusion }}{\longrightarrow} 2 \mathrm{~K}_2 \mathrm{MnO}_4+2 \mathrm{H}_2 \mathrm{O}$

Topic: Chemistry, d and f Block Elements, JEE Advanced, JEE Advanced 2009 (Paper 2)
The packing efficiency of the two-dimensional square unit cell shown below is

Options:
A
$39.27 \%$

B
$68.02 \%$

C
$74.05 \%$

D
$78.54 \%$

The correct answer is:
$78.54 \%$

Solution:
$a=(2 \sqrt{2} r)$
Packing fraction $=\frac{2 \times \pi r^2}{2(\sqrt{2} r)^2}=\frac{2 \pi r^2}{8 r^2}=\frac{\pi}{4}$
Solid state
Basic analysis (understanding)
II

Topic: Chemistry, Solid State, JEE Advanced, JEE Advanced 2010 (Paper 2)
The percentage of $p$-character in the orbitals forming $\mathrm{P}-\mathrm{P}$ bonds in $\mathrm{P}_4$ is

Options:
A
25

B
33

C
50

D
75

The correct answer is:
75

Solution:
Phosphorus atom is $s p^3$ hybridised in $\mathrm{P}_4$ usually.
Thus, $p$ character $75 \%$

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Advanced, JEE Advanced 2007 (Paper 1)
The periodic table consists of 18 groups. An isotope of copper, on bombardment with protons, undergoes a nuclear reaction yielding element $X$ as shown below. To which group, element $X$ belongs in the periodic table?

${ }_{29}^{63} \mathrm{Cu}+{ }_{1}^{1} \mathrm{H} \rightarrow 6_{0}^{1} n+{ }_{2}^{4} \alpha+2{ }_{1}^{1} \mathrm{H}+\mathrm{X}$

Options:

The correct answer is:
8

Solution:
${ }_{29}^{63} \mathrm{Cu}+{ }_{1}^{1} \mathrm{H} \rightarrow 6{ }_{0}^{1} n+{ }_{2}^{4} \mathrm{He}+2{ }_{1}^{1} \mathrm{H}+{ }_{Z}^{A} X$

Balancing the atomic mass and atomic number

$63+1=(6 \times 1)+4+2+A \Rightarrow A=52$

$29+1=(6 \times 0)+2+2+Z \Rightarrow Z=26$

Thus ${ }_{Z}^{A} X={ }_{26}^{52} X$ or ${ }_{26}^{52} \mathrm{Fe}$

Hence, $X$ belongs to group 8 in the periodic table.

Topic: Chemistry, Classification of Elements and Periodicity in Properties, JEE Advanced, JEE Advanced 2012 (Paper 1)
The point $P$ is the intersection of the straight line joining the points $Q(2,3,5)$ and $R(1,-1,4)$ with the plane $5 x-4 y-$ $z=1$. If $S$ is the foot of the perpendicular drawn from the point $T(2,1,4)$ to $Q R$, then the length of the line segment $P S$ is

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

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

Solution:
Equation of st. line joining $Q(2,3,5)$ and $R(1,-1,4)$ is

$\frac{x-2}{-1}=\frac{y-3}{-4}=\frac{z-5}{1}=\lambda$

Let $P(-\lambda+2,-4 \lambda+3,-\lambda+5)$

Since $P$ also lies on $5 x-4 y-z=1$

$\therefore-5 \lambda+10+16 \lambda-12+\lambda-5=1$

$\Rightarrow 12 \lambda=8 \Rightarrow \lambda=\frac{2}{3} \quad \therefore P=\left(\frac{4}{3}, \frac{1}{3}, \frac{13}{3}\right)$

Now let another point $S$ on $Q R$ be

$(-\mu+2,-4 \mu+3,-\mu+5)$

Since $S$ is the foot of perpendicular drawn from

$T(2,1,4)$ to $Q R$, where dr's of $S T$ are $\mu, 4 \mu-2, \mu-1$ and dr's of $Q R$ are $-1,-4,-1$

$\begin{array}{ll}

\therefore & -\mu-16 \mu+8-\mu+1=0 \Rightarrow 18 \mu=9 \Rightarrow \mu=\frac{1}{2} \\

\therefore & S=\left(\frac{3}{2}, 1, \frac{9}{2}\right)

\end{array}$

$\therefore \quad$ Distance between $P$ and $S$

$\begin{array}{l}

=\sqrt{\left(\frac{4}{3}-\frac{3}{2}\right)^{2}+\left(\frac{1}{3}-1\right)^{2}+\left(\frac{13}{3}-\frac{9}{2}\right)^{2}} \\

=\sqrt{\frac{1}{36}+\frac{4}{9}+\frac{1}{36}}=\frac{1}{\sqrt{2}}

\end{array}$

Topic: Mathematics, Three Dimensional Geometry, JEE Advanced, JEE Advanced 2012 (Paper 1)
The positive integer value of $n>3$ satisfying the equation
$$
\frac{1}{\sin \left(\frac{\pi}{n}\right)}=\frac{1}{\sin \left(\frac{2 \pi}{n}\right)}+\frac{1}{\sin \left(\frac{3 \pi}{n}\right)} \text { is }
$$

Options:

The correct answer is:
7

Solution:
Given, $n>3 \in$ Integer
$$
\begin{aligned}
& \text { and } \frac{1}{\sin \left(\frac{\pi}{n}\right)}=\frac{1}{\sin \left(\frac{2 \pi}{n}\right)}+\frac{1}{\sin \left(\frac{3 \pi}{n}\right)} \\
& \Rightarrow \frac{1}{\sin \frac{\pi}{n}}-\frac{1}{\sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}} \\
& \Rightarrow \quad \frac{\sin \frac{3 \pi}{n}-\sin \frac{\pi}{n}}{\sin \frac{\pi}{n} \cdot \sin \frac{3 \pi}{n}}=\frac{1}{\sin \frac{2 \pi}{n}}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad 2 \cos \left(\frac{2 \pi}{n}\right) \cdot \sin \frac{\pi}{n}=\frac{\sin \frac{\pi}{n} \cdot \sin \frac{3 \pi}{n}}{\sin \frac{2 \pi}{n}} \\
& \Rightarrow \quad 2 \sin \frac{2 \pi}{n} \cdot \cos \frac{2 \pi}{n}=\sin \frac{3 \pi}{n} \\
& \Rightarrow \quad \quad \sin \frac{4 \pi}{n}=\sin \frac{3 \pi}{n} \\
& \Rightarrow \frac{4 \pi}{n}=\pi-\frac{3 \pi}{n} \Rightarrow \frac{7 \pi}{n}=\pi \Rightarrow n=7
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Advanced, JEE Advanced 2011 (Paper 1)
The reaction of $P_2$ with $X$ leads selectively to $\mathrm{P}_4 \mathrm{O}_6$. The $\mathrm{X}$ is

Options:
A
dry $\mathrm{O}_2$

B
a mixture of $\mathrm{O}_2$ and $\mathrm{N}_2$

C
moist $\mathrm{O}_2$

D
$\mathrm{O}_2$ in the presence of aq. $\mathrm{NaOH}$

The correct answer is:
a mixture of $\mathrm{O}_2$ and $\mathrm{N}_2$

Solution:
White phosphorus on reaction with limited supply of oxygen gives lower oxide $P_4 \mathrm{O}_6$. Therefore, air $\left(\mathrm{O}_2+\mathrm{N}_2\right)$ is a good source for controlled supply of oxygen and the best choice for controlled oxidation of white phosphorus into lower oxide $\mathrm{P}_4 \mathrm{O}_6$.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2009 (Paper 1)
The reaction of white phosphorus with aqueous $\mathrm{NaOH}$ gives phosphine along with another phosphorus containing compound. The reaction type; the oxidation states of phosphorus in phosphine and the other product are respectively

Options:
A redox reaction; $-3$ and $-5$
B redox reaction; $+3$ and $+5$
C disproportionation reaction; $-3$ and $+5$
D disproportionation reaction; $-3$ and $+3$

The correct answer is:
disproportionation reaction; $-3$ and $+5$

Solution:

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2012 (Paper 2)
The reagent(s) for the following conversion,
is/are

Options:
A
alcoholic $\mathrm{KOH}$

B
alcoholic KOH followed by $\mathrm{NaNH}_2$

C
aqueous $\mathrm{KOH}$ followed by $\mathrm{NaNH}_2$

D
$\mathrm{Zn} / \mathrm{CH}_3 \mathrm{OH}$

The correct answer is:
alcoholic KOH followed by $\mathrm{NaNH}_2$

Solution:

Topic: Chemistry, Hydrocarbons, JEE Advanced, JEE Advanced 2007 (Paper 1)
The reagent(s) used for softening the temporary hardness of water is(are)

Options:
A
$\mathrm{Ca}_3\left(\mathrm{PO}_4\right)_2$

B
$\mathrm{Ca}(\mathrm{OH})_2$

C
$\mathrm{Na}_2 \mathrm{CO}_3$

D
$\mathrm{NaOCl}$

The correct answers are:
$\mathrm{Ca}(\mathrm{OH})_2$
,
$\mathrm{Na}_2 \mathrm{CO}_3$
,
$\mathrm{NaOCl}$

Solution:
Softening the temporary hardness of water means conversion of bicarbonate into calcium carbonate (precipitation)
$$
\begin{gathered}
\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2+\mathrm{Ca}(\mathrm{OH})_2 \longrightarrow \\
2 \mathrm{CaCO}_3 \downarrow+2 \mathrm{H}_2 \mathrm{O} \\
\mathrm{Ca}\left(\mathrm{HCO}_3\right)_2+\mathrm{Na}_2 \mathrm{CO}_3 \longrightarrow \\
\mathrm{CaCO}_3 \downarrow+2 \mathrm{NaHCO}_3
\end{gathered}
$$
Even $\mathrm{NaOCl}$ can be used, as in aqueous medium $\overline{\mathrm{O}} \mathrm{Cl}$ gets hydrolysed and makes solution alkaline enough to cause precipitation of $\mathrm{CaCO}_3$.
$$
\begin{aligned}
\mathrm{NaOCl}+\mathrm{H}_2 \mathrm{O} & \rightleftharpoons \mathrm{HOCl}+\mathrm{NaOH} \\
\mathrm{OH}^{-}+\mathrm{HCO}_3^{-} & \longrightarrow \mathrm{CO}_3^{2-}+\mathrm{H}_2 \mathrm{O}
\end{aligned}
$$
Representative elements (Hydrogen and its compounds)
Conceptual
II

Topic: Chemistry, Hydrogen, JEE Advanced, JEE Advanced 2010 (Paper 1)
The reversible expansion of an ideal gas under adiabatic and isothermal conditions is shown in the figure. Which of the following statement(s) is (are) correct?

Options:
A $T_{1}=T_{2}$
B $T_{3}>T_{1}$
C $w_{\text {isothermal }}>w_{\text {adiahatic }}$
D $\Delta U_{\text {isothermal }}>\Delta U_{\text {adiabatic }}$

The correct answers are:
$T_{1}=T_{2}$, $w_{\text {isothermal }}>w_{\text {adiahatic }}$, $\Delta U_{\text {isothermal }}>\Delta U_{\text {adiabatic }}$

Solution:
$T_{1}=T_{2}$ because process is isothermal.

Work done in adiabatic process is less than in isothermal process because area covered by isothermal curve is more than the area covered by the adiabatic curve.

In adiabatic process expansion occurs byusing internal energy, hence, it decreases while in isothermal process temperature remains constant, that's why no change in internal energy.

Topic: Chemistry, Thermodynamics (C), JEE Advanced, JEE Advanced 2012 (Paper 2)
The set of values of $\theta$ satisfying the inequation $2 \sin ^2 \theta-5 \sin \theta+2>0$, where $0 < \theta < 2 \pi$, is

Options:
A
$\left(0, \frac{\pi}{6}\right) \cup\left(\frac{5 \pi}{6}, 2 \pi\right)$

B
$\left[0, \frac{\pi}{6}\right] \cup\left[\frac{5 \pi}{6}, 2 \pi\right]$

C
$\left[0, \frac{\pi}{3}\right] \cup\left[\frac{2 \pi}{3}, 2 \pi\right]$

D
None of these

The correct answer is:
$\left(0, \frac{\pi}{6}\right) \cup\left(\frac{5 \pi}{6}, 2 \pi\right)$

Solution:
$$
\begin{aligned}
& \text { As, } 2 \sin ^2 \theta-5 \sin \theta+2>0 \\
& \Rightarrow(2 \sin \theta-1)(\sin \theta-2)>0 \\
& \quad[\text { where, }(\sin \theta-2) < 0 \text { for all } \theta \in R] \\
& \therefore \quad(2 \sin \theta-1) < 0
\end{aligned}
$$
$$
\begin{array}{ll}
\Rightarrow & \sin \theta < \frac{1}{2}, \text { shown as } \\
\therefore & \theta \in\left(0, \frac{\pi}{6}\right) \cup\left(\frac{5 \pi}{6}, 2 \pi\right)
\end{array}
$$

Topic: Mathematics, Trigonometric Equations, JEE Advanced, JEE Advanced 2006
The shape of $\mathrm{XeO}_{2} \mathrm{~F}_{2}$ molecule is

Options:
A trigonal bipyramidal
B square planar
C tetrahedral
D see-saw

The correct answer is:
see-saw

Solution:
$\mathrm{XeO}_{2} \mathrm{~F}_{2}$ has trigonal bipyramidal geometry, but due to presence of lone pair of electrons on equitorial position, its shape is see-saw.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Advanced, JEE Advanced 2012 (Paper 2)