datavorous/entrance-exam-dataset huggingface dataset:
Let $\mathrm{S}$ be the area of the region enclosed by $y=e^{-x^{2}}$, $y=0, x=0$ and $x=1$; then

Options:
A $S \geq \frac{1}{e}$
B $S \geq 1-\frac{1}{e}$
C $S \leq \frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right)$
D $S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$

The correct answers are:
$S \geq \frac{1}{e}$, $S \geq 1-\frac{1}{e}$, $S \leq \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right)$

Solution:
The given curve $y=e^{-x^{2}}$ Draw a rough sketch of curve at $x=0, y=1$ and at $x=1, y=1 / e$ $\because y=e^{-x^{2}}$ $\Rightarrow \frac{d y}{d x}=-2 x e^{-x^{2}} < 0 \quad \forall x \in(0,1)$

$\therefore y=e^{-x^{2}}$ is decreasing on $(0,1)$

Hence its graph is as shown in figure given below

Now, $\mathrm{S}=$ area exclosed by curve $=\mathrm{XYCO}$

and area of rectangle $\mathrm{OCYL}=\frac{1}{e}$

Clearly $S>\frac{1}{e} \quad \therefore$ a is true.

For $x \in[0,1] \Rightarrow x^{2} < x$

$\begin{array}{l}

\Rightarrow-x^{2}>-x \quad \Rightarrow e^{-x^{2}} \geq e^{-x} \forall x \in[0,1] \\

\Rightarrow \int_{0}^{1} e^{-x^{2}} d x>\int_{0}^{1} e^{-x} d x=1-\frac{1}{e} \\

\Rightarrow S>1-\frac{1}{e} \quad \therefore \text { (b) is true. }

\end{array}$

Now $S < $ area of rectangle $\mathrm{XADO}+$ area of rectangle $\mathrm{ZDCN}$ $\Rightarrow S < \frac{1}{\sqrt{2}} \times 1+\left(1-\frac{1}{\sqrt{2}}\right) \frac{1}{\sqrt{e}}$ $\therefore S < \frac{1}{\sqrt{2}}+\frac{1}{\sqrt{e}}\left(1-\frac{1}{\sqrt{2}}\right) \quad \because(\mathrm{d})$ is true

Also as $\frac{1}{4}\left(1+\frac{1}{\sqrt{e}}\right) < 1-\frac{1}{e} \quad \therefore$ (c) is incorrect.

Topic: Mathematics, Area Under Curves, JEE Advanced, JEE Advanced 2012 (Paper 1)
Let $S$ be the focus of the parabola $y^{2}=8 \mathrm{x}$ and let $P Q$ be the common chord of the circle $x^{2}+y^{2}-2 x-4 y=0$ and the given parabola. The area of the triangle $P Q S$ is

Options:

The correct answer is:
4

Solution:
Given parabola $y^{2}=8 x$

and circle $x^{2}+y^{2}-2 x-4 y=0$ pass through the origin

$\therefore$ One end of common chord PQ is origin. Say $\mathrm{P}(0,0)$

Let $\mathrm{Q}$ be the point $\left(2 t^{2}, 4 t\right)$, then it will satisfy the equation of circle.

$\therefore \quad 4 t^{4}+16 t^{2}-4 t^{2}-16 t=0$

$\Rightarrow t^{4}+3 t^{2}-4 t=0 \Rightarrow t\left(t^{3}+3 t-4\right)=0$

$\Rightarrow t(t-1)\left(t^{2}+t-4\right)=0 \Rightarrow t=0$ or 1

For $t=0$, we get point $P$, therefore $t=1$ gives point $Q$ as $(2,4)$.

We also observe here that $\mathrm{P}(0,0)$ and $\mathrm{Q}(2,4)$ are end points of diameter of the given circle and focus of the parabola is the point $S(2,0)$.

$\therefore \quad$ area $(\Delta \mathrm{PQS})=\frac{1}{2} \times P S \times Q S=\frac{1}{2} \times 2 \times 4=4$

Topic: Mathematics, Parabola, JEE Main
Let $S$ be the focus of the parabola $y^{2}=8 \mathrm{x}$ and let $P Q$ be the common chord of the circle $x^{2}+y^{2}-2 x-4 y=0$ and the given parabola. The area of the triangle $P Q S$ is

Options:

The correct answer is:
4

Solution:
Given parabola $y^{2}=8 x$

and circle $x^{2}+y^{2}-2 x-4 y=0$ pass through the origin

$\therefore$ One end of common chord PQ is origin. Say $\mathrm{P}(0,0)$

Let $\mathrm{Q}$ be the point $\left(2 t^{2}, 4 t\right)$, then it will satisfy the equation of circle.

$\therefore \quad 4 t^{4}+16 t^{2}-4 t^{2}-16 t=0$

$\Rightarrow t^{4}+3 t^{2}-4 t=0 \Rightarrow t\left(t^{3}+3 t-4\right)=0$

$\Rightarrow t(t-1)\left(t^{2}+t-4\right)=0 \Rightarrow t=0$ or 1

For $t=0$, we get point $P$, therefore $t=1$ gives point $Q$ as $(2,4)$.

We also observe here that $\mathrm{P}(0,0)$ and $\mathrm{Q}(2,4)$ are end points of diameter of the given circle and focus of the parabola is the point $S(2,0)$.

$\therefore \quad$ area $(\Delta \mathrm{PQS})=\frac{1}{2} \times P S \times Q S=\frac{1}{2} \times 2 \times 4=4$

Topic: Mathematics, Parabola, JEE Advanced, JEE Advanced 2012 (Paper 1)
Let $S_k, k=1,2, \ldots, 100$, denote the sum of the infinite geometric series whose first term is $\frac{k-1}{k !}$ and the common ratio is $\frac{1}{k}$. Then the value of $\frac{100^2}{100 !}+\sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right|$ is

Options:

The correct answer is:
4

Solution:
We have, $S_k=\frac{\frac{k-1}{k !}}{1-\frac{1}{k}}=\frac{1}{(k-1) !}$
Now,
$$
\begin{aligned}
& \left(k^2-3 k+1\right) S_k=\{(k-2)(k-1)-1\} \\
& =\frac{1}{(k-3) !}-\frac{1}{(k-1) !} \\
& \Rightarrow \sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right| \\
& =1+1+2-\left(\frac{1}{99 !}+\frac{1}{98 !}\right)=4-\frac{100^2}{100 !}
\end{aligned}
$$
$$
\Rightarrow \quad \frac{100^2}{100 !}+\sum_{k=1}^{100}\left|\left(k^2-3 k+1\right) S_k\right|=4
$$

Topic: Mathematics, Sequences and Series, JEE Advanced, JEE Advanced 2010 (Paper 1)
Let $S_n=\sum_{k=0}^n \frac{n}{n^2+k n+k^2}$ and $T_n=\sum_{k=0}^{n-1} \frac{n}{n^2+k n+k^2}$, for $n=1,2,3, \ldots$, then

Options:
A
$S_n < \frac{\pi}{3 \sqrt{3}}$
B
$S_n>\frac{\pi}{3 \sqrt{3}}$

C
$T_n < \frac{\pi}{3 \sqrt{3}}$
D
$T_n>\frac{\pi}{3 \sqrt{3}}$

The correct answers are:
$S_n < \frac{\pi}{3 \sqrt{3}}$,
$T_n>\frac{\pi}{3 \sqrt{3}}$

Solution:
Given,
$$
\begin{aligned}
S_n & =\sum_{k=0}^n \frac{n}{n^2+k n+k^2} \\
& =\sum_{k=0}^n \frac{1}{n} \cdot\left(\frac{1}{1+\frac{k}{n}+\frac{k^2}{n^2}}\right) < \lim _{n \rightarrow \infty} \sum_{k=0}^n \frac{1}{n}\left(\frac{1}{1+\frac{k}{n}+\left(\frac{k}{n}\right)^2}\right) \\
& =\int_0^1 \frac{1}{1+x+x^2} d x=\left[\frac{2}{\sqrt{3}} \tan ^{-1}\left(\frac{2}{\sqrt{3}}\left(x+\frac{1}{2}\right)\right)\right]_0^1 \\
& =\frac{2}{\sqrt{3}} \cdot\left(\frac{\pi}{3}-\frac{\pi}{6}\right)=\frac{\pi}{3 \sqrt{3}}
\end{aligned}
$$
i.e., $\quad S_n < \frac{\pi}{3 \sqrt{3}}$
Similarly, $T_n>\frac{\pi}{3 \sqrt{3}}$.

Topic: Mathematics, Definite Integration, JEE Advanced, JEE Advanced 2008 (Paper 1)
Let the eccentricity of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ be reciprocal to that of the ellipse $x^2+4 y^2=4$. If the hyperbola passes through a focus of the ellipse, then

Options:
A
the equation of the hyperbola is $\frac{x^2}{3^2}-\frac{y^2}{2^2}=1$

B
a focus of the hyperbola is $(2,0)$

C
the eccentricity of the hyperbola is $\sqrt{\frac{5}{3}}$

D
the equation of the hyperbola is $x^2-3 y^2=3$

The correct answers are:
a focus of the hyperbola is $(2,0)$
,
the equation of the hyperbola is $x^2-3 y^2=3$

Solution:
Here, equation of ellipse
$$
\begin{aligned}
& \frac{x^2}{4}+\frac{y^2}{1}=1 \\
& \Rightarrow \quad e^2=1-\frac{b^2}{a^2}=1-\frac{1}{4}=\frac{3}{4} \\
& \therefore \quad e=\frac{\sqrt{3}}{2} \text { and focus }(\pm a e, 0) \\
& =(\pm \sqrt{3}, 0) \\
&
\end{aligned}
$$
For hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, $e_1^2=1+\frac{b^2}{a^2}$ where, $e_1^2=\frac{1}{e^2}=\frac{4}{3}$
$$
\Rightarrow \quad 1+\frac{b^2}{a^2}=\frac{4}{3} \Rightarrow \frac{b^2}{a^2}=\frac{1}{3}
$$
and hyperbola passes through $(\pm \sqrt{3}, 0)$. Now, $\quad \frac{3}{a^2}=1 \Rightarrow a^2=3$
From Eqs. (i) and (ii), we get $b^2=1$
$\therefore$ Equation of hyperbola is
$$
\frac{x^2}{3}-\frac{y^2}{1}=1
$$
Focus is $(\pm a e, 0)$.
Now, $\quad\left(\pm \sqrt{3} \cdot \frac{2}{\sqrt{3}}, 0\right) \Rightarrow(\pm 2,0)$
Hence, both options (b) and (d) are correct.

Topic: Mathematics, Hyperbola, JEE Advanced, JEE Advanced 2011 (Paper 1)
Let the function $g:(-\infty, \infty) \rightarrow\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ be given by $g(u)=2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}$. Then, $g$ is

Options:
A
even and is strictly increasing in $(0, \infty)$

B
odd and is strictly decreasing in $(-\infty, \infty)$

C
odd and is strictly increasing in $(-\infty, \infty)$

D
Neither even nor odd, but is strictly increasing in $(-\infty, \infty)$

The correct answer is:
odd and is strictly increasing in $(-\infty, \infty)$

Solution:
$\therefore \quad g(u)=2 \tan ^{-1}\left(e^u\right)-\frac{\pi}{2}$
for $\quad u \in(-\infty, \infty)$
and
$$
\begin{aligned}
g(-u) & =2 \tan ^{-1}\left(e^{-u}\right)-\frac{\pi}{2} \\
& =2\left(\cot ^{-1}\left(e^u\right)\right)-\frac{\pi}{2} \\
& =2\left(\frac{\pi}{2}-\tan ^{-1}\left(e^u\right)\right)-\frac{\pi}{2} \\
& =\frac{\pi}{2}-2 \tan ^{-1}\left(e^u\right)=-g(u)
\end{aligned}
$$
$\therefore \quad g(-u)=-g(u)$
$\Rightarrow g(u)$ is an odd function.

Topic: Mathematics, Application of Derivatives, JEE Advanced, JEE Advanced 2008 (Paper 2)
Let the straight line $x=b$ divide the area enclosed by $y=(1-x)^2, y=0$ and $x=0$ into two parts $R_1(0 \leq x \leq b)$ and $R_2(b \leq x \leq 1)$ such that $R_1-R_2=\frac{1}{4}$. Then, $b$ equals to

Options:
A
$\frac{3}{4}$

B
$\frac{1}{2}$

C
$\frac{1}{3}$

D
$\frac{1}{4}$

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

Solution:
Here, area between 0 to $b$ is $R_1$ and $b$ to
$$
\begin{aligned}
& 1 \text { is } R_2 \\
& \therefore \int_0^b(1-x)^2 d x-\int_b^1(1-x)^2 d x=\frac{1}{4} \\
& \Rightarrow \quad\left(\frac{(1-x)^3}{-3}\right)_0^b-\left(\frac{(1-x)^3}{-3}\right)_b^1=\frac{1}{4} \\
& \Rightarrow-\frac{1}{3}\left\{(1-b)^3-1\right\}+\frac{1}{3}\left\{0-(1-b)^3\right\} \\
& =\frac{1}{4}
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad-\frac{2}{3}(1-b)^3=-\frac{1}{3}+\frac{1}{4}=-\frac{1}{12} \\
& \Rightarrow \quad(1-b)^3=\frac{1}{8} \\
& \Rightarrow \quad(1-b)=\frac{1}{2} \Rightarrow b=\frac{1}{2} \\
&
\end{aligned}
$$

Topic: Mathematics, Area Under Curves, JEE Advanced, JEE Advanced 2011 (Paper 1)
Let the vectors $\mathbf{P Q}, \mathbf{Q R}, \mathbf{R S}, \mathbf{S T}, \mathbf{T U}$ and UP represent the sides of a regular hexagon.
Statement I PQ $\times(\mathbf{R S}+\mathbf{S T}) \neq \mathbf{0}$
Statement II $\mathbf{P Q} \times \mathbf{R S}=\mathbf{0}$ and $\mathbf{P Q} \times \mathbf{S Q} \neq \mathbf{0}$

Options:
A
Statement I is true, Statement II is true; Statement II is a correct explanation for Statement I

B
Statement I is true, Statement II is true; Statement II is not a correct explanation for Statement I

C
Statement I is true, Statement II is false

D
Statement I is false, Statement II is true

The correct answer is:
Statement I is true, Statement II is false

Solution:
Since, $\mathbf{P Q}$ is not parallel to $\mathbf{T R}$
$\because$ TR is resultant of $\mathbf{S R}$ and $\mathbf{S T}$ vectors.
$\Rightarrow \quad \mathbf{P Q} \times(\mathbf{R S}+\mathbf{S T}) \neq \mathbf{0}$
But for Statement II, we have $\mathbf{P Q} \times \mathbf{R S}=\mathbf{0}$
which is not possible as $\mathbf{P Q}$ not parallel to $\mathbf{R S}$.
Hence, Statement I is true and Statement II is false.

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2007 (Paper 1)
Let two non-collinear unit vectors $\mathbf{a}$ and $\hat{\mathbf{b}}$ form an acute angle.
A point $P$ moves so that at any time $t$ the position vector $\mathbf{O P}$ (where, $O$ is the origin) is given by $\mathbf{a} \cos t+\hat{\mathbf{b}} \sin t$. When $P$ is farthest from origin $O$, let $M$ be the length of $\mathbf{O P}$ and $\hat{\mathbf{u}}$ be the unit vector along $\mathbf{O P}$. Then,

Options:
A
$\hat{\mathbf{u}}=\frac{\hat{\mathbf{a}}+\hat{\mathbf{b}}}{|\hat{\mathbf{a}}+\hat{\mathbf{b}}|}$ and $M=(1+\hat{\mathbf{a}} \cdot \hat{\mathbf{b}})^{1 / 2}$

B
$\hat{\mathbf{u}}=\frac{\hat{\mathbf{a}}-\hat{\mathbf{b}}}{|\hat{\mathbf{a}}-\hat{\mathbf{b}}|}$ and $M=(1+\hat{\mathbf{a}} \cdot \hat{\mathbf{b}})^{1 / 2}$

C
$\hat{\mathbf{u}}=\frac{\hat{\mathbf{a}}+\hat{\mathbf{b}}}{|\hat{\mathbf{a}}+\hat{\mathbf{b}}|}$ and $M=(1+2 \hat{\mathbf{a}} \cdot \hat{\mathbf{b}})^{1 / 2}$

D
$\hat{\mathbf{u}}=\frac{\hat{\mathbf{a}}-\hat{\mathbf{b}}}{|\mathbf{\mathbf { a }}-\hat{\mathbf{b}}|}$ and $M=(1+2 \hat{\mathbf{a}} \cdot \hat{\mathbf{b}})^{1 / 2}$

The correct answer is:
$\hat{\mathbf{u}}=\frac{\hat{\mathbf{a}}+\hat{\mathbf{b}}}{|\hat{\mathbf{a}}+\hat{\mathbf{b}}|}$ and $M=(1+\hat{\mathbf{a}} \cdot \hat{\mathbf{b}})^{1 / 2}$

Solution:
$\mathbf{O P}=\hat{\mathbf{a}} \cos t+\hat{\mathbf{b}} \sin t$
$$
\begin{aligned}
& \Rightarrow \quad|\mathbf{O P}|=\sqrt{\left(\hat{\mathbf{a}} \cdot \hat{\mathbf{a}} \cos ^2 t+\hat{\mathbf{b}} \cdot \hat{\mathbf{b}} \sin ^2 t+2 \hat{\mathbf{a}} \cdot \hat{\mathbf{b}} \sin t \cos t\right)} \\
& \Rightarrow \quad|\mathbf{O P}|=\sqrt{1+2 \hat{\mathbf{a}} \cdot \hat{\mathbf{b}} \cdot \sin t \cos t} \\
& \Rightarrow \quad|\mathbf{O P}|=\sqrt{1+\hat{\mathbf{a}} \cdot \hat{\mathbf{b}} \cdot \sin 2 t} \\
& \Rightarrow \quad|\mathbf{O P}|_{\max }=\sqrt{1+\hat{\mathbf{a}} \cdot \hat{\mathbf{b}}} \text { at } \sin 2 t=1 \Rightarrow t=\frac{\pi}{4} \\
& \Rightarrow \mathbf{O P}\left(\text { at } t=\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}(\hat{\mathbf{a}}+\hat{\mathbf{b}}) \\
&
\end{aligned}
$$
$\therefore$ Unit vector along $\mathbf{O P}$ at $\left(t=\frac{\pi}{4}\right)=\frac{\hat{\mathbf{a}}+\hat{\mathbf{b}}}{|\hat{\mathbf{a}}+\hat{\mathbf{b}}|}$

Topic: Mathematics, Vector Algebra, JEE Advanced, JEE Advanced 2008 (Paper 2)
Let $\left(x_0, y_0\right)$ be the solution of the following equations $(2 x)^{\log 2}=(3 y)^{\log 3}$, $3^{\log x}=2^{\log y}$, then $x_0$ is equal to

Options:
A
$\frac{1}{6}$

B
$\frac{1}{2}$

C
$\frac{1}{2}$

D
6

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

Solution:
Taking log on both sides,
$$
\log 2 \cdot \log (2 x)=\log 3(\log 3 y)
$$
$$
\begin{array}{ll}
\Rightarrow & \log 2\{\log 2+\log x\} \\
& =\log 3\{\log 3+\log y\} \\
\text { and } & \log x \cdot \log 3=\log y \log 2 \\
& \log y=\frac{\log x \cdot \log 3}{\log 2}
\end{array}
$$
From Eqs. (i) and (ii), we get
$$
\begin{aligned}
& \log 2\{\log 2+\log x\} \\
&=\log 3 \cdot\left\{\log 3+\frac{\log x \cdot \log 3}{\log 2}\right\} \\
& \Rightarrow \quad(\log 2)^2+\log 2 \cdot \log x \\
&=(\log 3)^2+\frac{(\log 3)^2}{(\log 2)} \cdot \log x \\
& \Rightarrow \quad \log x\left\{\frac{(\log 3)^2}{\log 2}-\log 2\right\} \\
&=(\log 2)^2-(\log 3)^2
\end{aligned}
$$
$$
\begin{aligned}
& \Rightarrow \quad \log x=-\log 2=\log 2^{-1} \\
& \therefore \quad x=\frac{1}{2}
\end{aligned}
$$

Topic: Mathematics, Basic of Mathematics, JEE Advanced, JEE Advanced 2011 (Paper 1)
Let $X$ and $Y$ be two events such that $P(X \mid Y)=\frac{1}{2}$, $P(Y / X)=\frac{1}{3}$ and $P(X \cap Y)=\frac{1}{6}$. Which of the following is (are) correct?

Options:
A $P(X \cup Y)=\frac{2}{3}$
B $X$ and $Y$ are independent
C $X$ and $Y$ are not independent
D $P\left(X^{c} \cap Y\right)=\frac{1}{3}$

The correct answers are:
$P(X \cup Y)=\frac{2}{3}$, $X$ and $Y$ are independent

Solution:
$\because P(X / Y)=\frac{P(X \cap Y)}{P(Y)} \Rightarrow \frac{1}{2}=\frac{1 / 6}{P(Y)} \Rightarrow P(Y)=\frac{1}{3}$

Similarly, $\mathrm{P}(Y / X)=\frac{P(X \cap Y)}{P(X)}$

$\Rightarrow \frac{1}{3}=\frac{1 / 6}{P(X)} \Rightarrow P(X)=\frac{1}{2}$

(a) $P(X \cup Y)=P(X)+P(Y)-P(X \cap Y)=\frac{1}{2}+\frac{1}{3}-\frac{1}{6}=\frac{2}{3}$

$\therefore$ (a) is true.

(b) $\because P(X \cap Y)=P(X) P(Y)$

$\Rightarrow X$ and $Y$ are independent events.

$\therefore$ (b) is true.

But (c) is not true.

(d) $\mathrm{P}\left(X^{C} \cap Y\right)=P\left(X^{C}\right) \times P(Y)=\frac{1}{2} \times \frac{1}{3}=\frac{1}{6}$

$\therefore$ (d) is not true.

Topic: Mathematics, Probability, JEE Advanced, JEE Advanced 2012 (Paper 2)
Let $\int(x)=\left\{\begin{array}{r}x^{2}\left|\cos \frac{\pi}{x}\right|, \quad x \neq 0 \\ 0, \quad x=0\end{array}, x \in R\right.$ then $\int$ is

Options:
A differentiable both at $x=0$ and at $x=2$
B differentiable at $x=0$ but not differentiable at $x=2$
C not differentiable at $x=0$ but differentiable at $x=2$
D differentiable neither at $x=0$ nor at $x=2$

The correct answer is:
differentiable at $x=0$ but not differentiable at $x=2$

Solution:
$f^{\prime}\left(0^{+}\right)=\lim _{h \rightarrow 0} \frac{f(0+h)-f(0)}{h}$

$\begin{aligned}

=\lim _{h \rightarrow 0} \frac{h^{2}\left|\cos \frac{\pi}{h}\right|}{h} &=\lim _{h \rightarrow 0} h\left|\cos \frac{\pi}{h}\right| \\

&=0 \times \text { some finite value }=0

\end{aligned}$

$\begin{array}{c}

=0 \times \text { some finite value }=0 \\

\text { and } f^{\prime}\left(0^{-}\right)=\lim _{h \rightarrow 0} \frac{f(0-h)-f(0)}{-h}=\lim _{h \rightarrow 0} \frac{h^{2}\left|\cos \frac{\pi}{-h}\right|}{-h} \\

=\lim _{h \rightarrow 0}-h\left|\cos \frac{\pi}{h}\right|=0 \times \text { some finite value }=0

\end{array}$

$\begin{array}{l}

\because f^{\prime}\left(0^{+}\right)=f^{\prime}\left(0^{-}\right) \therefore f \text { is differentiable at } x=0 \\

\text { Now } f^{\prime}\left(2^{+}\right)=\lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h} \\

\quad=\lim _{h \rightarrow 0} \frac{(2+h)^{2}\left|\cos \frac{\pi}{2+h}\right|-4\left|\cos \frac{\pi}{2}\right|}{h}

\end{array}$

$=\lim _{h \rightarrow 0} \frac{(2+h)^{2}\left(\cos \frac{\pi}{2+h}\right)}{h}$

$=\lim _{h \rightarrow 0} \frac{(2+h)^{2}}{h} \sin \left(\frac{\pi}{2}-\frac{\pi}{2+h}\right)$

$=\lim _{h \rightarrow 0} \frac{(2+h)^{2}}{h} \sin \left(\frac{\pi h}{2(2+h)}\right)$

$=\lim _{h \rightarrow 0} \frac{(2+h)^{2}}{h} \times \frac{\sin \left(\frac{\pi h}{2(2+h)}\right)}{\left(\frac{\pi h}{2(2+h)}\right)} \times \frac{\pi h}{2(2+h)}=\pi$

and $f^{\prime}\left(2^{-}\right)=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}$

$=\lim _{h \rightarrow 0} \frac{(2-h)^{2}\left|\cos \left(\frac{\pi}{2-h}\right)\right|-0}{-h}$

$=\lim _{h \rightarrow 0} \frac{-(2-h)^{2} \cos \left(\frac{\pi}{2-h}\right)}{-h}$

$=\lim _{h \rightarrow 0} \frac{(2-h)^{2} \sin \left(\frac{\pi}{2}-\frac{\pi}{2-h}\right)}{h}$

$=\lim _{h \rightarrow 0} \frac{(2-h)^{2}}{h} \times \frac{\sin \left(\frac{-\pi h}{2(2-h)}\right)}{\left(\frac{-\pi h}{2(2-h)}\right)} \times\left(\frac{-\pi h}{2(2-h)}\right)=-\pi$

$\because f^{\prime}\left(2^{+}\right) \neq f^{\prime}\left(2^{-}\right), \therefore f$ is not differentiable at $x=2 .$

Topic: Mathematics, Continuity and Differentiability, JEE Advanced, JEE Advanced 2012 (Paper 1)
Let $(x, y)$ be any point on the parabola $y^2=4 x$. Let $P$ be the point that divides the line segment from $(0,0)$ to $(x, y)$ in the ratio $1: 3$. Then, the locus of $P$ is

Options:
A
$x^2=y$

B
$y^2=2 x$

C
$y^2=x$

D
$x^2=2 y$

The correct answer is:
$y^2=x$

Solution:
$$
\text { By section formula, }
$$
$h=\frac{x+0}{4}, k=\frac{y+0}{4}$
$\therefore \quad x=4 h$ and $y=4 k$
Substituting in $y^2=4 x$,
$(4 k)^2=4(4 h) \Rightarrow k^2=h$
or $y^2=x$ is required locus.

Topic: Mathematics, Parabola, JEE Main
Let $(x, y)$ be any point on the parabola $y^2=4 x$. Let $P$ be the point that divides the line segment from $(0,0)$ to $(x, y)$ in the ratio $1: 3$. Then, the locus of $P$ is

Options:
A
$x^2=y$

B
$y^2=2 x$

C
$y^2=x$

D
$x^2=2 y$

The correct answer is:
$y^2=x$

Solution:
$$
\text { By section formula, }
$$
$h=\frac{x+0}{4}, k=\frac{y+0}{4}$
$\therefore \quad x=4 h$ and $y=4 k$
Substituting in $y^2=4 x$,
$(4 k)^2=4(4 h) \Rightarrow k^2=h$
or $y^2=x$ is required locus.

Topic: Mathematics, Parabola, JEE Advanced, JEE Advanced 2011 (Paper 2)
Let $(x, y)$ be such that $\sin ^{-1}(a x)+\cos ^{-1}(y)+\cos ^{-1}(b x y)=\frac{\pi}{2}$. Match the statements in Column I with the values in Column II.

Options:
A
A-p; B-q; C-q; D-p

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

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

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

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

Solution:
(A) If $a=1$ and $b=0$, then $\sin ^{-1} x+\cos ^{-1} y=0$
$$
\Rightarrow \quad \sin ^{-1} x=-\cos ^{-1} y \Rightarrow x^2+y^2=1
$$
(B) If $a=1$ and $b=1$, then
$$
\begin{array}{rlrl}
& & \sin ^{-1} x+\cos ^{-1} y+\cos ^{-1} x y & =\frac{\pi}{2} \\
\Rightarrow & & \cos ^{-1} x-\cos ^{-1} y & =\cos ^{-1} x y \\
\Rightarrow & x y+\sqrt{1-x^2} \sqrt{1-y^2} & =x y \quad \text { [taking sine on both the sides] }
\end{array}
$$
(C) If $a=1$ and $b=2$, then
$$
\begin{aligned}
& & \sin ^{-1} x+\cos ^{-1} y+\cos ^{-1}(2 x y) & =\frac{\pi}{2} \\
\Rightarrow & & \cos ^{-1} x-\cos ^{-1} y & =\cos ^{-1}(2 x y) \\
\Rightarrow & & x y+\sqrt{1-x^2} \sqrt{1-y^2} & =2 x y \\
\Rightarrow & & x^2+y^2 & =1
\end{aligned}
$$
[on squaring]
(D) If $a=2$ and $b=2$, then
$$
\begin{aligned}
& & \sin ^{-1}(2 x)+\cos ^{-1}(y)+\cos ^{-1}(2 x y) & =\frac{\pi}{2} \\
\Rightarrow & & 2 x y+\sqrt{1+4 x^2} \sqrt{1-y^2} & =2 x y \\
\Rightarrow & & \left(4 x^2-1\right)\left(y^2-1\right) & =0 .
\end{aligned}
$$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
Let $(x, y)$ be such that $\sin ^{-1}(a x)+\cos ^{-1}(y)+\cos ^{-1}(b x y)=\frac{\pi}{2}$. Match the statements in Column I with the values in Column II.

Options:
A
A-p; B-q; C-q; D-p

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

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

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

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

Solution:
(A) If $a=1$ and $b=0$, then $\sin ^{-1} x+\cos ^{-1} y=0$
$$
\Rightarrow \quad \sin ^{-1} x=-\cos ^{-1} y \Rightarrow x^2+y^2=1
$$
(B) If $a=1$ and $b=1$, then
$$
\begin{array}{rlrl}
& & \sin ^{-1} x+\cos ^{-1} y+\cos ^{-1} x y & =\frac{\pi}{2} \\
\Rightarrow & & \cos ^{-1} x-\cos ^{-1} y & =\cos ^{-1} x y \\
\Rightarrow & x y+\sqrt{1-x^2} \sqrt{1-y^2} & =x y \quad \text { [taking sine on both the sides] }
\end{array}
$$
(C) If $a=1$ and $b=2$, then
$$
\begin{aligned}
& & \sin ^{-1} x+\cos ^{-1} y+\cos ^{-1}(2 x y) & =\frac{\pi}{2} \\
\Rightarrow & & \cos ^{-1} x-\cos ^{-1} y & =\cos ^{-1}(2 x y) \\
\Rightarrow & & x y+\sqrt{1-x^2} \sqrt{1-y^2} & =2 x y \\
\Rightarrow & & x^2+y^2 & =1
\end{aligned}
$$
[on squaring]
(D) If $a=2$ and $b=2$, then
$$
\begin{aligned}
& & \sin ^{-1}(2 x)+\cos ^{-1}(y)+\cos ^{-1}(2 x y) & =\frac{\pi}{2} \\
\Rightarrow & & 2 x y+\sqrt{1+4 x^2} \sqrt{1-y^2} & =2 x y \\
\Rightarrow & & \left(4 x^2-1\right)\left(y^2-1\right) & =0 .
\end{aligned}
$$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Advanced, JEE Advanced 2007 (Paper 2)
Let $(x, y, z)$ be points with integer coordinates satisfying the system of homogeneous equations $3 x-y-z=0,-3 x+z=0,-3 x+2 y+z=0$. Then, the number of such points for which $x^2+y^2+z^2 \leq 100$ is

Options:

The correct answer is:
7

Solution:
Given, $\quad 3 x-y-z=0$
and
$$
\begin{array}{r}
-3 x+2 y+z=0 \\
-3 x+z=0
\end{array}
$$
On adding Eqs. (i) and (ii), we get $y=0$
So, $\quad 3 x=z$
Now, $\quad x^2+y^2+z^2 \leq 100$
$$
\begin{array}{lc}
\Rightarrow & x^2+(3 x)^2+0 \leq 100 \\
\Rightarrow & 10 x^2 \leq 100 \\
\Rightarrow & x^2 \leq 10 \\
\therefore & x=-3,-2,-1,0,1,2,3
\end{array}
$$
So, number of such 7 points are possible.

Topic: Mathematics, Determinants, JEE Advanced, JEE Advanced 2009 (Paper 2)
Let $y^{\prime}(x)+y(x) g^{\prime}(x)=g(x) g^{\prime}(x) y(0)=0, x \in R$, where $f^{\prime}(x)$ denotes $\frac{d f(x)}{d x}$ and $g(x)$ is a given non-constant differentiable function on $R$ with $g(0)=g(2)=0$. Then, the value of $y$ (2) is...

Options:

The correct answer is:
0

Solution:
$\frac{d y}{d x}+y \cdot g^{\prime}(x)=g(x) g^{\prime}(x)$
$$
\mathrm{IF}=e^{\int g^{\prime}(x) d x}=e^{g(x)}
$$
$\therefore$ Solution is
$$
y\left(e^{g(x)}\right)=\int g(x) \cdot g^{\prime}(x) \cdot e^{g(x)} d x+C
$$
Put $g(x)=t, g^{\prime}(x) d x=d t$
$$
\begin{aligned}
& y\left(e^{g(x)}\right)=\int t \cdot e^t d t+C \\
= & t \cdot e^t-\int 1 \cdot e^t d t+C=t \cdot e^t-e^t+C \\
& y e^{g(x)}=(g(x)-1) e^{g(x)}+C \quad \ldots(\mathrm{i})
\end{aligned}
$$
Given, $y(0)=0, g(0)=g(2)=0$
$\therefore$ Eq. (i) becomes,
$$
\begin{aligned}
& \quad y(0) \cdot e^{g(0)}=(g(0)-1) \cdot e^{g(0)}+C \\
& \Rightarrow \quad 0=(-1) \cdot 1+C \Rightarrow C=1 \\
& \therefore \quad y(x) \cdot e^{g(x)}=(g(x)-1) e^{g(x)}+1 \\
& \Rightarrow \quad y(2) \cdot e^{g(2)}=(g(2)-1) e^{g(2)}+1 \\
& \text { where, } \quad g(2)=0 \\
& \Rightarrow \quad y(2) \cdot 1=(-1) \cdot 1+1 \Rightarrow y(2)=0
\end{aligned}
$$

Topic: Mathematics, Differential Equations, JEE Advanced, JEE Advanced 2011 (Paper 2)