datavorous/entrance-exam-dataset huggingface dataset:
Match the statements given in Column I with the intervals/union of intervals given in Column II.

Options:
A
(A) q, (B) r, (C) s, (D) t

B
(A) s, (B) t, (C) r, (D) q

C
(A) q, (B) r, (C) s, (D) r

D
(A) s, (B) t, (C) r, (D) r

The correct answer is:
(A) s, (B) t, (C) r, (D) r

Solution:
(A) Given, $|z|=1 \Rightarrow z \cdot \bar{z}=1$
$$
\therefore \quad \frac{2 i z}{1-z^2}=\frac{2 i z}{z \cdot \bar{z}-z^2}=\frac{2 i}{\bar{z}-z} \text {, }
$$
Let $\quad z=x+i y$
$$
\therefore z-\bar{z}=2 i y=\frac{2 i}{-2 i y}=-\frac{1}{y}
$$
where, $y=\sqrt{1-x^2}$
$\therefore \quad-1 \leq y \leq 1 \Rightarrow-1 \leq y$
and $y \leq 1 \Rightarrow-1 \geq \frac{1}{y}$ and $\frac{1}{y} \geq 1$
$\Rightarrow \operatorname{Re}\left(\frac{2 i z}{1-z^2}\right) \in(-\infty,-1] \cup[1, \infty)$
(B) $f(x)=\sin ^{-1}\left(\frac{8\left(3^{x-2}\right)}{1-3^{2(x-1)}}\right)$,
$$
\begin{aligned}
& \text { For domain, }-1 \leq \frac{8\left(3^{x-2}\right)}{1-3^{2(x-1)}} \leq 1 \\
& \Rightarrow \quad-1 \leq \frac{9 \cdot\left(3^{x-2}\right)-\left(3^{x-2}\right)}{1-3^{2(x-1)}} \leq 1 \\
& \therefore \quad-1 \leq \frac{3^x-3^{x-2}}{1-3^x \cdot 3^{(x-2)}} \leq 1 \\
& \frac{3^x-3^{x-2}}{1-3^x \cdot 3^{x-2}} \geq-1 \\
& \Rightarrow \frac{\left(3^x-1\right)\left(3^{x-2}-1\right)}{\left(3^{x-1}+1\right)\left(3^{x-1}-1\right)} \geq 0 \\
& \Rightarrow \quad x \in(-\infty, 0] \cup(1, \infty) \\
& \text { and } \frac{3^x-3^{x-2}}{1-3^x \cdot 3^{x-2}} \leq 1 \\
& \Rightarrow \frac{\left(3^{x-2}-1\right)\left(3^x+1\right)}{\left(3^{x-1}+1\right)\left(3^{x-1}-1\right)} \geq 0 \\
&
\end{aligned}
$$
and $x \in(-\infty, 1) \cup[2, \infty)$
$\therefore \quad x \in(-\infty, 0] \cup[2, \infty)$
(C) $f(\theta)=\left|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right|$
$$
\begin{aligned}
R_1 & \rightarrow R_1+R_3, \\
f(\theta) & =\left|\begin{array}{ccc}
0 & 0 & 2 \\
-\tan \theta & 1 & \tan \theta \\
-1 & -\tan \theta & 1
\end{array}\right| \\
& =2\left(\tan ^2 \theta+1\right)=2 \sec ^2 \theta \geq 2 \\
f(\theta) & \in[2, \infty)
\end{aligned}
$$
$$
\text { (D) } \begin{aligned}
f(x) & =x^{3 / 2}(3 x-10) ; x \geq 0 \\
f^{\prime}(x) & =x^{3 / 2} \cdot 3+\frac{3}{2} \cdot x^{1 / 2}(3 x-10) \\
& =3 x^{1 / 2}\left\{x+\frac{1}{2}(3 x-10)\right\}
\end{aligned}
$$
$$
\begin{aligned}
& =\frac{3}{2} x^{1 / 2}\{2 x+3 x-10\} \\
& =\frac{15}{2} x^{1 / 2}(x-2) \\
& \\
\therefore \quad & x \geq 2
\end{aligned}
$$

Topic: Mathematics, Functions, JEE Advanced, JEE Advanced 2011 (Paper 2)
$$
\text { Match the statements given in Column I with the values given in Column II. }
$$

Options:
A
(A) q, (B) p, (C) s, (D) s

B
(A) p, (B) p, (C) r, (D) q

C
(A) q, (B) r, (C) s, (D) q

D
(A) p, (B) s, (C) r, (D) s

The correct answer is:
(A) q, (B) p, (C) s, (D) s

Solution:
(A) $\therefore|\mathbf{a}|=\sqrt{1+3}=2$ $|\mathbf{b}|=\sqrt{1+3}=2$ $|\mathbf{c}|=\sqrt{12}=2 \sqrt{3}$
Using cosine law,
$$
\begin{aligned}
\cos C & =\frac{|\mathbf{a}|^2+|\mathbf{b}|^2-|\mathbf{c}|^2}{2|\mathbf{a}||\mathbf{b}|} \\
& =\frac{4+4-12}{2 \times 2 \times 2}=\frac{-4}{8}=\frac{-1}{2} \\
\Rightarrow \quad \angle C & =120^{\circ}=\frac{2 \pi}{3}
\end{aligned}
$$
(B)
$$
\begin{aligned}
& \int_a^b f(x) d x-3\left(\frac{x^2}{2}\right)_a^b=\left(a^2-b^2\right) \\
& \Rightarrow \int_a^b f(x) d x-\frac{3}{2}\left(b^2-a^2\right)=\left(a^2-b^2\right) \\
& \Rightarrow \int_a^b f(x) d x=\left(a^2-b^2\right)+\frac{3}{2}\left(b^2-a^2\right) \\
& =\frac{b^2-a^2}{2} \\
& \Rightarrow \int_a^b f(x) d x=\frac{b^2-a^2}{2} \\
&
\end{aligned}
$$
$$
f(x)=x \Rightarrow f\left(\frac{\pi}{6}\right)=\frac{\pi}{6}
$$
(C)
$$
\begin{aligned}
& \frac{\pi^2}{\log _e 3} \int_{7 / 6}^{5 / 6} \sec (\pi x) d x \\
& \Rightarrow \frac{\pi^2}{\log _e 3}\left\{\frac{\log |\sec \pi x+\tan \pi x|}{\pi}\right\}_{7 / 6}^{5 / 6} \\
& \Rightarrow \frac{\pi}{\log 3}\left\{\log \left|\sec \frac{5 \pi}{6}+\tan \frac{5 \pi}{6}\right|\right. \\
& \left.\quad-\log \left|\sec \frac{7 \pi}{6}+\tan \frac{7 \pi}{6}\right|\right\} \\
& \Rightarrow \frac{\pi}{\log 3}\left\{\log |\sqrt{3}|-\log \left|\frac{1}{\sqrt{3}}\right|\right\} \\
& \Rightarrow \frac{\pi}{\log 3}\{\log 3\}=\pi
\end{aligned}
$$
(D) $$
\begin{aligned}
& \left|\arg \frac{1}{(1-z)}\right|, \text { for }|z|=1 \\
& \Rightarrow\left|\arg (1-z)^{-1}\right| \\
& \Rightarrow|-\arg (1-z)| \Rightarrow|\arg (1-z)|
\end{aligned}
$$
From figure, $\arg (z-1)$ is maximum $=\pi$.

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2011 (Paper 2)
Match the statements in Column I with the properties Column II.

Options:
A
A-p, r; B-p, q, s; C-q, r; D-q, s

B
A-p; B-p, s; C-p; D-q, s

C
A-p, q; B-p, q; C-q, r; D-q, r

D
A-r, s; B-q; C-p; D-p, q, s

The correct answer is:
A-p, q; B-p, q; C-q, r; D-q, r

Solution:
(A) When two circles are intersecting they have a common normal and common tangent.
(B) Two mutually extemal circles have a common normal and common tangent.
(C) When one circle lies inside of other, then they have a common normal but no common tangent.
(D) Two branches of a hyperbola have a common normal but no common tangent.

Topic: Mathematics, Circle, JEE Advanced, JEE Advanced 2007 (Paper 2)
Match the statements of Column I with these in Column II.
[Note : Here $z$ takes values in the complex plane and $\operatorname{Im}(z)$ and $\operatorname{Re}(z)$ denote respectively, the imaginary part and the real part of $z$ ]

Options:
A
(A) q, (B) p, (C) p,s, (D) q,r,s

B
(A) q,r, (B) p, (C) p,s,t, (D) q,r,s,t

C
(A) q, (B) q, (C) p,s,t, (D) q,r,s,t

D
(A) q,r, (B) q, (C) p,s, (D) q,r,s

The correct answer is:
(A) q,r, (B) p, (C) p,s,t, (D) q,r,s,t

Solution:
(A) $z$ is equidistant from the points $i|z|$ and $-i|z|$, whose perpendicular bisector is $\operatorname{Im}(z)=0$.
(B) Sum of distance of $z$ from $(4,0)$ and $(-4,0)$ is a constant 10 , hence locus of $z$ is ellipse with semi-major axis 5 and focus at $(\pm 4,0), a e=4$.
$$
\therefore \quad e=\frac{4}{5}
$$
(C) $|z| \leq|w|+\left|\frac{1}{w}\right|=\frac{5}{2} < 3$
(D) $|z| \leq|w|+\left|\frac{1}{w}\right|=2$
$$
\therefore \quad \operatorname{Re}(z) \leq|z| \leq 2
$$

Topic: Mathematics, Complex Number, JEE Advanced, JEE Advanced 2010 (Paper 2)
Match the statements/expressions given in Column I with the values given in Column II.

Options:
A
(A) s, (B) q,s, (C) p,r,s,t, (D) q

B
(A) p, (B) q,s, (C) q,r,s,t, (D) r

C
(A) s, (B) s,t, (C) q,r,s, (D) r

D
(A) p, (B) s,t, (C) q,r,s,t, (D) q

The correct answer is:
(A) p, (B) q,s, (C) q,r,s,t, (D) r

Solution:
(a) Let $f(x)=x e^{\sin x}-\cos x$ $\overrightarrow{f^{\prime}}(x)=e^{\sin x}+x e^{\sin x} \cos x+\sin x \geq 0$ For interval $x \in\left(0, \frac{\pi}{2}\right)$, $f$ is strictly increasing.
$\therefore \quad f(0)=-1$ and $f\left(\frac{\pi}{2}\right)=\frac{\pi}{2} e \Rightarrow$ one solution
(b) Since, $\left|\begin{array}{lll}k & 4 & 1 \\ 4 & k & 2 \\ 2 & 2 & 1\end{array}\right|=0$ $\Rightarrow k(k-4)-4(0)+1(8-2 k)=0$ $\Rightarrow \quad k^2-6 k+8=0$ $\Rightarrow \quad k=2,4$
(c) Let $y=|x-1|+|x-2|+|x+1|$ $+|x+2|$
For solutions, $4 k \geq 6 \Rightarrow k \geq 3 / 2$
$\therefore$ Integer values of $k$ are $2,3,4,5$.
(d) Given, $\frac{d y}{d x}=y+1$
$\Rightarrow \quad \ln |(y+1)|=x+C$ $\Rightarrow \ln 2=C \Rightarrow \ln |y+1|=x+\ln 2$
Put $x=\ln 2$
$\therefore \quad \ln (y+1)=\ln 2+\ln 2=\ln 4$
$y+1=4 \Rightarrow y=3$

Topic: Mathematics, Functions, JEE Advanced, JEE Advanced 2009 (Paper 2)
Match the statements/expressions given in Column I with the values given in Column II.

Options:
A
(A) s, (B) q,s, (C) p,r,s,t, (D) q

B
(A) p, (B) q,s, (C) q,r,s,t, (D) r

C
(A) s, (B) s,t, (C) q,r,s, (D) r

D
(A) p, (B) s,t, (C) q,r,s,t, (D) q

The correct answer is:
(A) p, (B) q,s, (C) q,r,s,t, (D) r

Solution:
(a) Let $f(x)=x e^{\sin x}-\cos x$ $\overrightarrow{f^{\prime}}(x)=e^{\sin x}+x e^{\sin x} \cos x+\sin x \geq 0$ For interval $x \in\left(0, \frac{\pi}{2}\right)$, $f$ is strictly increasing.
$\therefore \quad f(0)=-1$ and $f\left(\frac{\pi}{2}\right)=\frac{\pi}{2} e \Rightarrow$ one solution
(b) Since, $\left|\begin{array}{lll}k & 4 & 1 \\ 4 & k & 2 \\ 2 & 2 & 1\end{array}\right|=0$ $\Rightarrow k(k-4)-4(0)+1(8-2 k)=0$ $\Rightarrow \quad k^2-6 k+8=0$ $\Rightarrow \quad k=2,4$
(c) Let $y=|x-1|+|x-2|+|x+1|$ $+|x+2|$
For solutions, $4 k \geq 6 \Rightarrow k \geq 3 / 2$
$\therefore$ Integer values of $k$ are $2,3,4,5$.
(d) Given, $\frac{d y}{d x}=y+1$
$\Rightarrow \quad \ln |(y+1)|=x+C$ $\Rightarrow \ln 2=C \Rightarrow \ln |y+1|=x+\ln 2$
Put $x=\ln 2$
$\therefore \quad \ln (y+1)=\ln 2+\ln 2=\ln 4$
$y+1=4 \Rightarrow y=3$

Topic: Mathematics, Functions, JEE Main
Match the statements/expressions given in Column I with the values given in Column II.

Options:
A
(A) q,s, (B) p,r,s,t, (C) t, (D) r

B
(A) p,s, (B) q,r,s, (C) s, (D) q

C
(A) q,s, (B) q,r,s, (C) t, (D) r

D
(A) p,s, (B) p,r,s,t, (C) t, (D) q

The correct answer is:
(A) q,s, (B) p,r,s,t, (C) t, (D) r

Solution:
(a) $2 \sin ^2 \theta+\sin ^2 2 \theta=2$
$\Rightarrow \quad \sin ^2 2 \theta=2 \cos ^2 \theta$
$\Rightarrow \quad 4 \sin ^2 \theta \cos ^2 \theta=2 \cos ^2 \theta$
$\Rightarrow \quad \cos ^2 \theta=0$ or $\sin ^2 \theta=\frac{1}{2}$
$\Rightarrow \quad \cos \theta=0$ or $\sin \theta=\pm \frac{1}{\sqrt{2}}$
$\Rightarrow \quad \theta=\pm \frac{\pi}{4}$ or $\frac{\pi}{2}$
(b) $f(x)=\left[\frac{6 x}{\pi}\right] \cos \left[\frac{3 x}{\pi}\right]$
Possible points of discontinuity of $\left[\frac{6 x}{\pi}\right]$ are
$$
\begin{gathered}
\frac{6 x}{\pi}=n, n \in I \\
\Rightarrow \quad x=\frac{n \pi}{6} \Rightarrow x=\frac{\pi}{6}, \frac{\pi}{3}, \frac{\pi}{2}, \pi \\
\lim _{x \rightarrow \pi^{-} / 6} f(x)=0 \cos 0=0 \\
\lim _{x \rightarrow \pi^{+} / 6} f(x)=1 \cos 0=1
\end{gathered}
$$
$\therefore$ Discontinuous at $x=\frac{\pi}{6}$.
Similarly, discontinuous at $x=\frac{\pi}{3}, \frac{\pi}{2}, \pi$.
(c) Here, $V=\left\|\begin{array}{lll}1 & 1 & 0 \\ 1 & 2 & 0 \\ 1 & 1 & \pi\end{array}\right\|=\pi$ cubic unit
(d) Given, $\mathbf{a}+\mathbf{b}+\sqrt{3} \mathbf{c}=\mathbf{0}$
$$
\begin{array}{ll}
\Rightarrow & \mathbf{a}+\mathbf{b}=-\sqrt{3} \mathbf{c} \\
\Rightarrow & |\mathbf{a}+\mathbf{b}|^2=\mid \sqrt{3} \mathbf{c}^2
\end{array}
$$
$$
\begin{array}{lrl}
\Rightarrow & a^2+b^2+2 \mathbf{a} \cdot \mathbf{b}=3 c^2 \\
\Rightarrow & 2+2 \cos \theta=3 \\
\Rightarrow & \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}
\end{array}
$$

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2009 (Paper 2)
$$
\text { Match the statements/expressions in Column I with the open intervals in Column II. }
$$

Options:
A
(A) p,q,t, (B) q,t, (C) r,s, (D) p,s

B
(A) p,q,s, (B) p,t, (C) p,q,r,t, (D) s

C
(A) p,q,s, (B) q,t, (C) r,s,t, (D) p,s

D
(A) p,q,t, (B) p,t, (C) p,q,r,t, (D) s

The correct answer is:
(A) p,q,s, (B) p,t, (C) p,q,r,t, (D) s

Solution:
(a) Given, $(x-3)^2 \cdot y^{\prime}+y=0$
$$
\begin{array}{rlrl}
\Rightarrow & \frac{d y}{d x} & =-\frac{y}{(x-3)^2} \\
\Rightarrow & & \int \frac{d y}{y} & =-\int \frac{d x}{(x-3)^3} \\
\Rightarrow & & \ln y & =\frac{1}{(x-3)}+\ln C \\
\Rightarrow & & y & =C e^{\frac{1}{x-3}}, C \neq 0
\end{array}
$$
$\therefore$ Domain of $y$ is $x \in R-\{3\}$
Aliter Given differential equation is homogeneous linear differential equation and has $x=3$ as a singular point, hence $x=3$ cannot be in domain of solution.
(b) Let $I=\int_1^5(x-1)(x-2)(x-3)(x-4)$ $(x-5) d x$
Let $x-3=t \Rightarrow d x=d t$
$\therefore I=\int_{-2}^2(t+2)(t+1) t(t-1)(t-2) d t$
$\because$ Integrand is an odd function.
$\therefore \quad I=0$
Aliter Let
$I=\int_1^5(x-1)(x-2)(x-3)(x-4)$
$(x-5) d x$
Using, $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$ $I=\int_1^5(5-x)(4-x)(3-x)(2-x)$ $(1-x) d x$
On adding Eqs. (i) and (ii) we get 2I $=0 \Rightarrow I=0$
(c) Let $f(x)=\cos ^2 x+\sin x$
$\Rightarrow f^{\prime}(x)=-2 \cos x \sin x+\cos x$ $=\cos x(1-2 \sin x)=0$ (say)
Sign scheme for first derivative
Points of local maxima are $\frac{\pi}{6}, \frac{5 \pi}{6}$.
Aliter $y=\cos ^2 x+\sin x$
$$
\Rightarrow \quad y=\frac{5}{4}-\left(\sin x-\frac{1}{2}\right)^2
$$
For $y$ to be maximum,
$$
\begin{aligned}
& \left(\sin x-\frac{1}{2}\right)^2=0 \Rightarrow \sin x=\frac{1}{2} \\
\Rightarrow \quad & x=n \pi+(-1)^n \frac{\pi}{6}, n \in I
\end{aligned}
$$
(d) Let $y=\tan ^{-1}(\sin x+\cos x)$
$$
\Rightarrow \quad \frac{d y}{d x}=\frac{\cos x-\sin x}{1+(\sin x+\cos x)^2}
$$
Clearly, by graph, $\cos x>\sin x$ is true for option(s).

Topic: Mathematics, Definite Integration, JEE Main
$$
\text { Match the statements/expressions in Column I with the open intervals in Column II. }
$$

Options:
A
(A) p,q,t, (B) q,t, (C) r,s, (D) p,s

B
(A) p,q,s, (B) p,t, (C) p,q,r,t, (D) s

C
(A) p,q,s, (B) q,t, (C) r,s,t, (D) p,s

D
(A) p,q,t, (B) p,t, (C) p,q,r,t, (D) s

The correct answer is:
(A) p,q,s, (B) p,t, (C) p,q,r,t, (D) s

Solution:
(a) Given, $(x-3)^2 \cdot y^{\prime}+y=0$
$$
\begin{array}{rlrl}
\Rightarrow & \frac{d y}{d x} & =-\frac{y}{(x-3)^2} \\
\Rightarrow & & \int \frac{d y}{y} & =-\int \frac{d x}{(x-3)^3} \\
\Rightarrow & & \ln y & =\frac{1}{(x-3)}+\ln C \\
\Rightarrow & & y & =C e^{\frac{1}{x-3}}, C \neq 0
\end{array}
$$
$\therefore$ Domain of $y$ is $x \in R-\{3\}$
Aliter Given differential equation is homogeneous linear differential equation and has $x=3$ as a singular point, hence $x=3$ cannot be in domain of solution.
(b) Let $I=\int_1^5(x-1)(x-2)(x-3)(x-4)$ $(x-5) d x$
Let $x-3=t \Rightarrow d x=d t$
$\therefore I=\int_{-2}^2(t+2)(t+1) t(t-1)(t-2) d t$
$\because$ Integrand is an odd function.
$\therefore \quad I=0$
Aliter Let
$I=\int_1^5(x-1)(x-2)(x-3)(x-4)$
$(x-5) d x$
Using, $\int_a^b f(x) d x=\int_a^b f(a+b-x) d x$ $I=\int_1^5(5-x)(4-x)(3-x)(2-x)$ $(1-x) d x$
On adding Eqs. (i) and (ii) we get 2I $=0 \Rightarrow I=0$
(c) Let $f(x)=\cos ^2 x+\sin x$
$\Rightarrow f^{\prime}(x)=-2 \cos x \sin x+\cos x$ $=\cos x(1-2 \sin x)=0$ (say)
Sign scheme for first derivative
Points of local maxima are $\frac{\pi}{6}, \frac{5 \pi}{6}$.
Aliter $y=\cos ^2 x+\sin x$
$$
\Rightarrow \quad y=\frac{5}{4}-\left(\sin x-\frac{1}{2}\right)^2
$$
For $y$ to be maximum,
$$
\begin{aligned}
& \left(\sin x-\frac{1}{2}\right)^2=0 \Rightarrow \sin x=\frac{1}{2} \\
\Rightarrow \quad & x=n \pi+(-1)^n \frac{\pi}{6}, n \in I
\end{aligned}
$$
(d) Let $y=\tan ^{-1}(\sin x+\cos x)$
$$
\Rightarrow \quad \frac{d y}{d x}=\frac{\cos x-\sin x}{1+(\sin x+\cos x)^2}
$$
Clearly, by graph, $\cos x>\sin x$ is true for option(s).

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2009 (Paper 1)
Match the Statements/Expressions in Column I with the Statements/Expressions in Column II.

Options:
A
(A) r, (B) q,r, (C) s, (D) p,r

B
(A) s, (B) q,s, (C) r, (D) p,q

C
(A) s, (B) q,r, (C) s, (D) p,q

D
(A) r, (B) q,s, (C) r, (D) p,r

The correct answer is:
(A) r, (B) q,s, (C) r, (D) p,r

Solution:
(A) Let $y=\frac{x^2+2 x+4}{x+2}$
$$
\begin{aligned}
& \Rightarrow x^2+(2-y) x+(4-2 y)=0 \\
& \Rightarrow \quad(2-y)^2-4(4-2 y) \geq 0 \\
& \Rightarrow \quad y^2+4 y-12 \geq 0 \\
& \Rightarrow \quad y \leq-6, y \geq 2 \\
&
\end{aligned}
$$
$\therefore$ Minimum value of $y$ is 2 .
(B) Since, $(A+B)(A-B)=(A-B)(A+B)$
$$
\begin{aligned}
& \Rightarrow \quad A^2-A B+B A-B^2=A^2+A B-B A-B^2 \\
& \Rightarrow \quad A B=B A \\
& \text { and } \quad(A B)^t=(-1)^k A B \\
& \Rightarrow \quad B^t A^t=(-1)^k A B \\
& \Rightarrow \quad-B A=(-1)^k A B \quad\left[\because B^t=-B, A^t=A\right] \\
& \Rightarrow \quad B A=(-1)^{k+1} A B \\
& \Rightarrow \quad(-1)^{k+1}=1 \\
&
\end{aligned}
$$
$\therefore k+1$ is even or $k$ is odd.
(C) $1 < 2^{\left(-k+3^{-a}\right)} < 2 \Rightarrow 0 < -k+3^{-a} < 1$
Given, $a=\log _3 \log _3 2 \Rightarrow 3^a=\log _3 2$
$$
\begin{array}{lc}
\Rightarrow & 3^{-a}=\log _2 3 \\
\therefore & k < \log _2 3 < 2 \\
\text { and } & 1+k>\log _2 3>1 \Rightarrow k>0
\end{array}
$$
From Eqs. (ii) and (iii), $0 < k < 2 \Rightarrow k=1$
$[\because k$ is an integer]
$$
\begin{array}{rlrl}
\text { (D) } & \sin \theta & =\cos \phi \\
\Rightarrow & \cos \left(\frac{\pi}{2}-\theta\right) & =\cos \phi \\
\Rightarrow & \frac{\pi}{2}-\theta & =2 n \pi \pm \phi, n \in Z \\
\Rightarrow & & \theta \pm \phi-\frac{\pi}{2} & =-2 n \pi, n \in Z \\
\Rightarrow & & \frac{1}{\pi}\left(\theta \pm \phi-\frac{\pi}{2}\right) & =-2 n, n \in Z
\end{array}
$$

Topic: Mathematics, Basic of Mathematics, JEE Advanced, JEE Advanced 2008 (Paper 2)
Match the transformation in Column I with appropriate options in Column II.

Options:
A
(A) p,q,r,s, (B) r,s, (C) t, (D) p,q

B
(A) p,s, (B) p,s, (C) s, (D) p,q,s

C
(A) p,s, (B) q,r, (C) s, (D) p,q

D
(A) p,r,s, (B) r,s, (C) t, (D) p,q,t

The correct answer is:
(A) p,r,s, (B) r,s, (C) t, (D) p,q,t

Solution:
(A) $\mathrm{CO}_2(\mathrm{~s}) \longrightarrow \mathrm{CO}_2(\mathrm{~g})$
It is just a phase transition (sublimation) as no chemical change has occurred. Sublimation is always endothermic. Product is gas, more disordered, hence $\Delta s$ is positive.
(B) $\mathrm{CaCO}_3(\mathrm{~s}) \longrightarrow \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})$
It is a chemical decomposition, not a phase change. Thermal decomposition occur at the expense of energy, hence endothermic. Product contain a gaseous species, hence, $\Delta S>0$.
(C) $2 \mathrm{H} \longrightarrow \mathrm{H}_2(\mathrm{~g})$
A new $\mathrm{H}-\mathrm{H}$ covalent bond is being formed, hence, $\Delta H < 0$.
Also, product is less disordered than reactant, $\Delta S < 0$.
(D) Allotropes are considered as different phase, hence $\mathrm{P}_{\text {(white, solid) }} \rightarrow \mathrm{P}_{\text {(red, solid) }}$ is a phase transition as well as allotropic change.
Also, red phosphorus is more ordered than white phosphorus, $\Delta S < 0$.

Topic: Chemistry, Thermodynamics (C), JEE Advanced, JEE Advanced 2011 (Paper 2)
Options:
A
$g(x)$ has local maxima at $x=1+\log _e 2$ and local minima at $x=e$

B
$f(x)$ has local maxima at $x=1$, and local minima at $x=2$

C
$g(x)$ has no local minima

D
$f(x)$ has no local maxima

The correct answers are:
$g(x)$ has local maxima at $x=1+\log _e 2$ and local minima at $x=e$
,
$f(x)$ has local maxima at $x=1$, and local minima at $x=2$

Solution:
Here, $f(x)=\left\{\begin{array}{cc}e^x, & 0 \leq x \leq 1 \\ 2-e^{x-1}, & 1 < x \leq 2 \\ x-e, & 2 < x \leq 3\end{array}\right.$ and $\quad g(x)=\int_0^x f(t) d t$
$$
\Rightarrow \quad g^{\prime}(x)=f(x)
$$
where, $g^{\prime}(x)=0$
$\Rightarrow \quad x=1+\log _e 2$
and $\quad x=e$.
Also, $\quad g(x)=\left\{\begin{array}{cc}-e^{x-1}, & 1 < x \leq 2 \\ 1, & 2 < x \leq 3\end{array}\right.$
$g\left(1+\log _e 2\right)=-e^{\log _e 2} < 0$, hence at $x=1+\log _e 2, g(x)$ has a local maximum.
Also, $g(e)=1>0$, hence at $x=e, g(x)$ has a local minima.
$\because f(x)$ is discontinuous at $x=1$, then we get
Local maxima at $x=1$ and local minima at $x=2$.

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2006
$\mathrm{MgSO}_4$ on reaction with $\mathrm{NH}_4 \mathrm{OH}$ and $\mathrm{Na}_2 \mathrm{HPO}_4$ forms a white crystalline precipitate. What is its formula?

Options:
A
$\mathrm{Mg}\left(\mathrm{NH}_4\right) \mathrm{PO}_4$

B
$\mathrm{Mg}_3\left(\mathrm{PO}_4\right)_2$

C
$\mathrm{MgCl}_2 \cdot \mathrm{MgSO}_4$

D
$\mathrm{MgSO}_4$

The correct answers are:
$\mathrm{Mg}\left(\mathrm{NH}_4\right) \mathrm{PO}_4$

Solution:
$\mathrm{MgSO}_4$ on reaction with $\mathrm{Na}_2 \mathrm{HPO}_4$ (disodium hydrogen phosphate) in presence of $\mathrm{NH}_4 \mathrm{OH}$ to give white precipitate of magnesium ammonium phosphate $\left(\mathrm{Mg}\left(\mathrm{NH}_4\right) \mathrm{PO}_4\right)$.

Topic: Chemistry, s Block Elements, JEE Advanced, JEE Advanced 2006
Most advanced invertebrates are

Options:
A Arthropoda
B Annelida
C Mollusca
D Cephalopoda

The correct answer is:
Cephalopoda

Solution:
Cephalopods such as octopus, cuttlefish and squid (coleoids) are the most advanced invertebrates among the given options. They belong to class-Cephalopoda of phylum - Mollusca. These organisms have a prominent head and a set arms or tentacles. They are widely regarded as the most intelligent of invertebrates and have well-developed senses and larger brains. Their nervous system is the most complex of all invertebrates.

Topic: Biology, Animal Kingdom, JIPMER, JIPMER 2019
Most materials have the refractive index, $n>1$. So, when a light ray from air enters a naturally occurring material, then by Snell's

law, $\frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{n_{2}}{n_{1}}$, it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, $\mathrm{n}=c / v=\pm \sqrt{\varepsilon_{r} \mu_{r}}$, where $c$ is the speed of electromagnetic waves in vacuum, $v$ its speed in the medium, $\varepsilon_{r}$ and $\mu_{r}$ are the relative permittivity and permeability of the medium respectively. In normal materials, both $\varepsilon_{r}$ and $\mu_{r}$, are positive, implying positive $n$ for the medium. When both $\varepsilon_{r}$ and $\mu_{r}$ are negative, one must choose the negative root of $n$. Such negative refractive index materials can now be artificially prepared and are called metamaterials. They exhibit significantly different optical behavior, without violating any physical laws. Since $n$ is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials.

Question: Choose the correct statement.

Options:
A The speed of light in the meta-material is $v=c|n|$
B The speed of light in the meta-material is $v=\frac{c}{|n|}$
C The speed of light in the meta-material is $v=c$.
D The wavelength of the light in the meta-material $\left(\lambda_{m}\right)$ is given by $\lambda_{m}=\lambda_{\text {air }}|n|$, where $\lambda_{\text {air }}$ is wavelength of the light in air.

The correct answer is:
The speed of light in the meta-material is $v=\frac{c}{|n|}$

Solution:
Speed of light in a medium, $V=\frac{C}{n}$ $n$ for meta-material $=|n|$

$\therefore \quad$ Speed of light in the meta-material $V=\frac{C}{|n|}$.

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2012 (Paper 2)
Most materials have the refractive index, $n>1$. So, when a light ray from air enters a naturally occurring material, then by Snell's

law, $\frac{\sin \theta_{1}}{\sin \theta_{2}}=\frac{n_{2}}{n_{1}}$, it is understood that the refracted ray bends towards the normal. But it never emerges on the same side of the normal as the incident ray. According to electromagnetism, the refractive index of the medium is given by the relation, $\mathrm{n}=c / v=\pm \sqrt{\varepsilon_{r} \mu_{r}}$, where $c$ is the speed of electromagnetic waves in vacuum, $v$ its speed in the medium, $\varepsilon_{r}$ and $\mu_{r}$ are the relative permittivity and permeability of the medium respectively. In normal materials, both $\varepsilon_{r}$ and $\mu_{r}$, are positive, implying positive $n$ for the medium. When both $\varepsilon_{r}$ and $\mu_{r}$ are negative, one must choose the negative root of $n$. Such negative refractive index materials can now be artificially prepared and are called metamaterials. They exhibit significantly different optical behavior, without violating any physical laws. Since $n$ is negative, it results in a change in the direction of propagation of the refracted light. However, similar to normal materials, the frequency of light remains unchanged upon refraction even in meta-materials.

Question: For light incident from air on a meta-material, the appropriate ray diagram is

Options:
A
B
C
D

The correct answer is:

Solution:
Let $n_{1}=$ refractive index of air and $n_{2}=$ refractive index of meta material which is negative

From Snell's law, $\frac{n_{2}}{n_{1}}=\frac{\sin \theta_{1}}{\sin \theta_{2}}$

$\because \quad n_{2}$ negative $\therefore \theta_{2}$ isalso negative hence graph (c) is correct.

Topic: Physics, Ray Optics, JEE Advanced, JEE Advanced 2012 (Paper 2)