datavorous/entrance-exam-dataset huggingface dataset:
Let the equation of an ellipse be $\frac{x^{2}}{144}+\frac{y^{2}}{25}=1$ Then, the radius of the circle with centre $(0, \sqrt{2})$ and passing through the foci of the ellipse is

Options:
A 9
B 7
C 11
D 5

The correct answer is:
11

Solution:
Given equation of ellipse is
Here,
$$
\begin{array}{r}
\frac{x^{2}}{144}+\frac{y^{2}}{25}=1 \\
a^{2}=144 \text { and } b^{2}=25
\end{array}
$$
Now.
$$
e=\sqrt{1-\frac{b^{2}}{a^{2}}}=\sqrt{1-\frac{25}{144}}
$$
$$
=\sqrt{\frac{119}{144}}=\frac{\sqrt{119}}{12}
$$
Foci of an ellipse $=(\pm a e, 0)$
$$
\begin{array}{l}
=\left(\pm 12 \times \frac{\sqrt{119}}{12}, 0\right) \\
=(\pm \sqrt{119}, 0)
\end{array}
$$
Since, the circle with centre $(0, \sqrt{2})$ and passing through foci $(\pm \sqrt{119}, 0)$ of the ellipse.
$\therefore$ Radius of circle $=\sqrt{(\sqrt{119}-0)^{2}+(0-\sqrt{2})^{2}}$
$$
=\sqrt{119+2}=\sqrt{121}=11
$$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2014
Let the foci of the ellipse $\frac{x^{2}}{9}+y^{2}=1$ subtend a right angle at a point $P$. Then, the locus of $P$ is

Options:
A $x^{2}+y^{2}=1$
B $x^{2}+y^{2}=2$
C $x^{2}+y^{2}=4$
D $x^{2}+y^{2}=8$

The correct answer is:
$x^{2}+y^{2}=8$

Solution:
$\frac{x^{2}}{9}+\frac{y^{2}}{1}=1 \Rightarrow e=\sqrt{1-\frac{1}{9}}=\frac{2 \sqrt{2}}{3}$
0) ie, $(\pm 2 \sqrt{2}, 0)$ Two foci are $(\pm a e,$ Let $P(h, k)$ by any point on the ellipse
$\therefore$
$\frac{k-0}{h-2 \sqrt{2}} \times \frac{k-0}{h+2 \sqrt{2}}=-1$
[From given condition]
$\Rightarrow \quad h^{2}-8=-k^{2}$
$\Rightarrow \quad x^{2}+y^{2}=8$

Topic: Mathematics, Ellipse, WBJEE, WBJEE 2012
Let the number of elements of the sets $A$ and be $p$ and $q$, respectively. Then, the number relations from the set $A$ to the set $B$ is

Options:
A $2^{p+q}$
B $2^{\rho q}$
C $p+q$
D $p q$

The correct answer is:
$2^{\rho q}$

Solution:
Given number of elements of $A$ and $B$ are $p$ and q, respectively.
$\therefore$ The number of relations from the set $A$ to the $\operatorname{set} B$ is $2^{p q}$

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2014
Let the relation $R_{1}$ be defined on $R$ as $a R_{1}b$ if $1+a b > 0 .$ Then

Options:
A $R_{1}$ is reflexive only.
B $R_{1}$ is equivalence relation.
C $R_{1}$ is reflexive and transitive but not symmetric
D $R_{1}$ is reflexive and symmetric but not transitive.

The correct answer is:
$R_{1}$ is reflexive and symmetric but not transitive.

Solution:
We observe the following properties:
Reflexivity : Let a be an arbitrary element of $R$, Then, $a \in R$ $\Rightarrow 1+a \cdot a=1+a^{2}>0 \quad\left[\because a^{2}>0\right.$ for all $\left.a \in R\right]$
$\Rightarrow \quad(a, a) \in R_{1}$
$\left[\right.$ By def. of $\left.R_{1}\right]$
Thus, $(a, a) \in R_{1}$ for all $a \in R$. So, $R_{1}$ is reflexive on $R$.
Symmetry : Let $(a, b) \in R$ Then, $(a, b) \in R_{1}$
$\Rightarrow \quad 1+a b>0$
$[\because a b=b a$ for all $]$
$\Rightarrow \quad 1+b a>0 \Rightarrow(b, a) \in R_{1}$
Thus, $(a, b) \in R_{1} \neq(b, a) \in R,$ for all $a, b \in R$
So, $R_{1}$ is symmetric on $R$.
Transitivity : we observe that $\left(1, \frac{1}{2}\right) \in R_{1}$ and $\left(\frac{1}{2},-1\right) \in R_{1}$ but (1,-1)$\notin R_{1}$ because
$1+1 \times(-1)=0 \gg 0$
So, $R_{1}$ is not transitive on $R$.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2019
Let the tangent and normal at any point $P\left(at^2\right.$, 2at), $(a > 0)$, on the parabola $y^2=4 a x$ meet the axis of the parabola at $T$ and $\mathrm{G}$ respectively. Then the radius of the circle through $\mathrm{P}, \mathrm{T}$ and $\mathrm{G}$ is

Options:
A $a\left(1+t^2\right)$
B $\left(1+t^2\right)$
C $a\left(1-t^2\right)$
D $\left(1-\mathrm{t}^2\right)$

The correct answer is:
$a\left(1+t^2\right)$

Solution:
$P\left(a t^2, 2 a t\right), T\left(-a t^2, 0\right), G\left(2 a+a t^2, 0\right)$
Slope of $P T \times$ Slope of $P G=-1$
$\therefore$ TG is a diameter of the circle through points $\mathrm{P}, \mathrm{T}$ and $\mathrm{G}$.
$\therefore \text { radius }=\frac{1}{2} \mathrm{TG}=\mathrm{a}\left(1+\mathrm{t}^2\right)$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2022
Let $v_{n}$ and $E_{n}$ be the respective speed and energy of an electron in the $n$ th orbit of radius $r_{n}$, in a hydrogen atom, as predicted by Bohr's model. Then.

Options:
A plot of $\frac{E_{n} r_{n}}{E_{1} r_{1}}$ as a function of $n$ is a straight line of slope 0
B plot of $\frac{r_{n} v_{n}}{r_{1} v_{1}}$ as a function of $n$ is a straight line of slope 1
C plot of $\ln \left(\frac{r_{n}}{r_{1}}\right)$ as a function of $\ln (n)$ is a straight line of slope 2
D plot of In $\left(\frac{r E_{1}}{E_{n} r_{1}}\right)$ as a function of $\ln (n)$ is a straight line of slope 4

The correct answers are:
plot of $\frac{E_{n} r_{n}}{E_{1} r_{1}}$ as a function of $n$ is a straight line of slope 0, plot of $\frac{r_{n} v_{n}}{r_{1} v_{1}}$ as a function of $n$ is a straight line of slope 1, plot of $\ln \left(\frac{r_{n}}{r_{1}}\right)$ as a function of $\ln (n)$ is a straight line of slope 2, plot of In $\left(\frac{r E_{1}}{E_{n} r_{1}}\right)$ as a function of $\ln (n)$ is a straight line of slope 4

Solution:
We know that, $v \propto \frac{1}{n}$ (speed)
$E_{n} \propto \frac{1}{n^{2}} \quad(\text { energy })$
$r_{n} \propto n^{2}$
$\therefore E_{n T_{n}} \propto n^{0}$
$\therefore E_{n} r_{n} \propto E_{1} r_{1}$
$\frac{E_{1} r_{1}}{E_{1} r_{1}}$
$r_{n} v_{s}=n^{2} \times \frac{1}{n} \propto n = constant$
$\therefore \frac{r_{n} v_{n}}{r_{1} v_{1}}=n$
$r_{\mathrm{e}} \sim n^{2}$
$\therefore \frac{r_{\mathrm{n}}}{r_{1}}=n^{2}$
$\log \left(\frac{r_{\mathrm{n}}}{r_{1}}\right)=2 \log n \quad(\because$ slope $=2)$
$\frac{r_{n}}{E_{n}}=n^{4}$
$\therefore \frac{r_{n}}{E_{n}} \times \frac{E_{1}}{r_{1}}=n^{4}$
$\log \left(\frac{r_{1} E_{1}}{E_{\alpha} r_{1}}\right)=4 \log (n)(\because$ slope $=4)$

Topic: Physics, Atomic Physics, WBJEE, WBJEE 2017
Let $x_{1}, x_{2}$ be the roots of $x^{2}-3 x+a=0$ and $x_{3}, x_{4}$ be the roots of $x^{2}-12 x+b=0$ If $x_{1} < x_{2} < x_{3} < x_{4}$ and $x_{1}, x_{2}, x_{3}, x_{4}$ are in GP. then ab equals

Options:
A $\frac{24}{5}$
B 64
C 16
D 8

The correct answers are:
64

Solution:
We have,
$x_{1}, x_{2}$ be the roots of equation $x^{2}-3 x+a=0$
$\therefore \quad x_{1}+x_{2}=3$ and $x_{1} x_{2}=a$
Also, $x_{3}, x_{4}$ be the roots of equation $x^{2}-12 x+b=0$
$\therefore x_{3}+x_{4}=12$ and $x_{3} x_{4}=b$
Again, $x_{1}, x_{2}, x_{3}, x_{4}$ are in GP
$\therefore \quad x_{1}=A, x_{2}=A R, x_{3}=A R^{2}, x_{4}=A R^{3}$
Now, $x_{1}+x_{2}=3$
$\Rightarrow \quad A(1+R)=3$
and $\quad x_{3}+x_{4}=12$
$\Rightarrow \quad A R^{2}(1+R)=12$
From Eqs. (i) and (ii), we have
$R^{2}=4 \Rightarrow R=\pm 2$
When, $R=2, A=1$ and when, $R=-2,A=-3$
$\therefore$ Numbers are either 1,2,4,8 or -3,6,-12,24 But $x_{1} < x_{2} < x_{3} < x_{4}$
So, $x_{1}=1, x_{2}=2, x_{3}=4, x_{4}=8$
$\begin{aligned} \therefore a b &=x_{1} x_{2} x_{3} x_{4} \\ &=1 \times 2 \times 4 \times 8=64 \end{aligned}$

Topic: Mathematics, Quadratic Equation, WBJEE, WBJEE 2019
Let $x_{1}, x_{2}, \ldots, x_{13}$ be 15 distinct numbers chosen from $1,2,3, \ldots, 15 .$ Then, the value of $\left(x_{1}-1\right)\left(x_{2}-1\right)\left(x_{3}-1\right) \ldots\left(x_{15}-1\right)$ is

Options:
A always $\leq 0$
B 0
C akays even
D aways odd

The correct answer is:
0

Solution:
Since, $x_{1}, x_{2} \ldots, x_{15}$ are among $1,2,3, \ldots, 15$. So, $x_{1}$ can take any value from $1,2,3, \ldots, 15$. Among these values, one of the number must be 1, hence product will be 0

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2015
Let \(X\) be a nonvoid set. If \(\rho_1\) and \(\rho_2\) be the transitive relations on \(X\), then

Options:
A \(\rho_1, \rho_2\) is transitive relation
B \(\rho_1 \cdot \rho_2\) is not transitive relation
C \(\rho_1 \cdot \rho_2\) is equivalence relation
D \(\rho_1 \cdot \rho_2\) is not any relation on \(X\)

The correct answer is:
\(\rho_1 \cdot \rho_2\) is not transitive relation

Solution:
Hint: Fact

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2023
Let $[x]$ denote the greatest integer less than or equal to $x$ for any real number $x$. Then, $\lim _{n \rightarrow \infty} \frac{[n \sqrt{2}]}{n}$ is equal to

Options:
A 0
B 2
C $\sqrt{2}$
D 1

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

Solution:
We have. $n \sqrt{2}-1 < [n \sqrt{2]} \leq n \sqrt{2} \quad[\because x-1 \leq[x] \leq x]$
$\Rightarrow \quad \sqrt{2}-\frac{1}{n} < [n \sqrt{2}] \leq 1$
$\therefore$ By Sandwich theorem. $\lim _{n \rightarrow \infty}\left(\sqrt{2}-\frac{1}{n}\right)=\sqrt{2}$

Topic: Mathematics, Limits, WBJEE, WBJEE 2014
Let $[x]$ denote the greatest integer less than or equal to $x$, then the value of the integral $\int_{-1}^{1}(|x|-2[x]) d x$ is equal to

Options:
A 3
B 2
C -2
D -3

The correct answer is:
3

Solution:
Let $I=\int_{-1}^{1}(|x|-2[x]) d x$
$=\int_{-1}^{0}(|x|-2[x]) d x+\int_{0}^{1}(|x|-2[x]) d x$
$=\int_{-1}^{0}(-x-2(-1)) d x+\int_{0}^{1}(x-2(0)) d x$
$\quad=\int_{-1}^{0}(-x+2) d x+\int_{0}^{1} x d x$
$\quad=\left(-\frac{x^{2}}{2}+2 x\right)_{-1}^{0}+\left[\frac{x^{2}}{2}\right]_{0}$
$\quad=-\left(-\frac{1}{2}+2(-1)\right)+\frac{1}{2}$
$\quad=\frac{1}{2}+2+\frac{1}{2}=1+2=3$

Topic: Mathematics, Indefinite Integration, WBJEE, WBJEE 2012
Let $[x]$ denotes the greatest integer less than or equal to $x$. Then, the value of $\alpha$ for which the function $f(x)=\left\{\begin{array}{c}\frac{\sin \left[-x^{2}\right]}{\left[-x^{2}\right]}, x \neq 0 \\ \alpha, \quad x=0\end{array}\right.$ is continuous at $x=0,$ is

Options:
A $\alpha=0$
B $\alpha=\sin (-1)$
C $\alpha=\sin (1)$
D $\alpha=1$

The correct answer is:
$\alpha=\sin (1)$

Solution:
We have, $f(x)=\left\{\begin{array}{ll}\frac{\sin \left(-x^{2}\right)}{[-x^{2}]} & , x \neq 0 \\ \alpha, & x=0\end{array}\right.$
Now, $\lim _{x \rightarrow 0} \frac{\sin \left(-x^{2}\right)}{\left.\mid-x^{2}\right]}=\frac{\sin (-1)}{(-1)}=\sin (1)$
Since, $f(x)$ is continuous at $x=0$.
$\Rightarrow \lim _{x \rightarrow 0} f(x)=f(0) \Rightarrow \sin (1)=\alpha$

Topic: Mathematics, Continuity and Differentiability, WBJEE, WBJEE 2015
Let $\phi(x)=f(x)+f(1-x)$ and $f^{\prime \prime}(x) < 0$ in $[0,1]$, then

Options:
A $\phi$ is monotonic increasing in $\left[0, \frac{1}{2}\right]$ and monotonic decrasing in $\left[\frac{1}{2}, 1\right]$
B $\phi$ is monotonic increasing in $\left[\frac{1}{2}, 1\right]$ and monotonic decrasing in $\left[0, \frac{1}{2}\right]$
C $\phi$ is neither increasing nor decreasing in any sub interval of $[0,1]$
D $\phi$ is increasing $[0,1]$

The correct answer is:
$\phi$ is monotonic increasing in $\left[0, \frac{1}{2}\right]$ and monotonic decrasing in $\left[\frac{1}{2}, 1\right]$

Solution:
Hint:
$\phi^{\prime}(x)=f^{\prime}(x)-f^{\prime}(1-x)$
$f^{\prime}(x)-f^{\prime}(1-x) \geq 0$ (for monotonic increasing)
$f^{\prime}(x) \geq f^{\prime}(1-x), x \leq 1-x\left(\because f^{\prime}(x)\right.$ is decreasing)
$\mathrm{x} \leq \frac{1}{2} \Rightarrow \phi(\mathrm{x})$ is monotonic increasing in $\left[0, \frac{1}{2}\right]$ and monotonic decreasing in $\left[\frac{1}{2}, 1\right]$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2020
Let $x_{n}=\left(1-\frac{1}{3}\right)^{2}\left(1-\frac{1}{6}\right)^{2}\left(1-\frac{1}{10}\right)^{2} \ldots$ $\left(1-\frac{1}{\frac{n(n+1)}{2}}\right)^{2}, n \geq 2$ Then, the value of $\lim _{n \rightarrow \infty} x_{n}$ is

Options:
A $1 / 3$
B $1 / 9$
C $1 / 81$
D 0

The correct answer is:
$1 / 9$

Solution:
We have, $x_{n}=\left[\left(1-\frac{1}{3}\right)\left(1-\frac{1}{6}\right)\right.$ $$
\left.\left(1-\frac{1}{10}\right) \ldots\left(1-\frac{2}{n(n+1)}\right)\right]^{2}
$$
$\begin{aligned} \Rightarrow x_{n} &=\left[\prod_{n=2}^{n}\left(\frac{n^{2}+n-2}{n(n+1)}\right)\right]^{2} \\ &=\left[\prod_{n=2}^{n}\left(\frac{(n+2)(n-1)}{n(n+1)}\right)\right]^{2} \\ &=\left[\prod_{n=2}^{n}\left(\frac{n+2}{n+1}\right) \cdot \prod_{n=2}^{n}\left(\frac{n-1}{n}\right)\right]^{2} \\ \Rightarrow &\left[\prod_{n=2}^{n}\left(\frac{n+2}{n+1}\right)\right]^{2}\left[\prod_{n=2}^{n}\left(\frac{n-1}{n}\right)\right]^{2} \\ \Rightarrow \quad x_{n}=&\left(\frac{4}{3} \cdot \frac{5}{4} \cdot \frac{6}{5} \ldots \frac{n+2}{n+1}\right)^{2}\left(\frac{1}{2} \cdot \frac{2}{3} \cdot \frac{3}{4} \ldots \frac{n-1}{n}\right)^{2} \\ \Rightarrow \quad & x_{n}=\left(\frac{n+2}{3}\right)^{2}\left(\frac{1}{n}\right)^{2} \Rightarrow x_{n}=\frac{1}{9}\left(\frac{n+2}{n}\right)^{2} \\=& \quad \lim _{n \rightarrow \infty} x_{n}=\frac{1}{9}(1+0)^{2}=\frac{1}{9} \end{aligned}$

Topic: Mathematics, Limits, WBJEE, WBJEE 2015
Let $X_{n}=\left\{z=x+i y:|z|^{2} \leq \frac{1}{n}\right\}$ for all integers $n \geq 1 .$ Then, $\underset{n=1}{\cap} X_{n}$ is

Options:
A a singleton set
B not a finite set
C an empty set
D a finite set with more than one element

The correct answer is:
a singleton set

Solution:
Given, $X_{n}=\left\{z=x+i y:|z|^{2} \leq \frac{1}{n}\right\}$
$=\left\{x^{2}+y^{2} \leq \frac{1}{n}\right\}$
$\therefore$
$X_{1}=\left\{x^{2}+y^{2} \leq 1\right\}$
$\begin{array}{c}
X_{2}=\left\{x^{2}+y^{2} \leq \frac{1}{2}\right\} \\
X_{3}=\left\{x^{2}+y^{2} \leq \frac{1}{3}\right\} \\
\ldots \\
\qquad \begin{aligned}
X_{\infty} &=\left\{x^{2}+y^2 \leq 0\right\} \\
\bigcap_{n=1}^{\infty} x_n &=X_{1} \cap X_{2} \cap X_{3} \cap \ldots \cap X_{\infty} \\
&=\left\{x^{2}+y^{2}=0\right\}
\end{aligned}
\end{array}$
Hence, $\underset{n=1}{\cap} X_{n}$ is a singleton set.

Topic: Mathematics, Sets and Relations, WBJEE, WBJEE 2014
Let $\mathrm{y}=\frac{1}{1+\mathrm{x}+\ln \mathrm{x}}$, Then

Options:
A $x \frac{d y}{d x}+y=x$
B $x \frac{d y}{d x}=y(y \ln x-1)$
C $x^{2} \frac{d y}{d x}=y^{2}+1-x^{2}$
D $x\left(\frac{d y}{d x}\right)^{2}=y-x$

The correct answer is:
$x \frac{d y}{d x}=y(y \ln x-1)$

Solution:
Hint:
$x \frac{d y}{d x}=y(y \ln x-1)$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2020
Let $y=\left(\frac{3^{x}-1}{3^{x}+1}\right) \sin x+\log _{e}(1+x) \quad x>-1$ Then, at $x=0, \frac{d y}{d x}$ equals

Options:
A 1
B 0
C -1
D -2

The correct answer is:
1

Solution:
Given, $y=\left(\frac{3^{x}-1}{3^{x}+1}\right) \sin x+\log _{e}(1+x), x>-1$
Differentiating w.r.t. $x$, we get $\frac{d y}{d x}=\frac{d}{d x}\left[\left(\frac{3^{x}-1}{3^{x}+1}\right) \sin x\right]+\frac{d}{d x} \log _{e}(1+x)$
$=\left(\frac{3^{x}-1}{3^{x}+1}\right) \frac{d}{d x} \sin x+\sin x \frac{d}{d x}\left(\frac{3^{x}-1}{3^{x}+1}\right)$
$=\frac{1}{1+x} \frac{d}{d x}(1+x)$
$=\left(\frac{3^{x}-1}{3^{x}+1}\right) \cos x+\sin x$
$\frac{\left(3^{x}+1\right) \frac{d}{d x}\left(3^{x}-1\right)-\left(3^{x}-1\right) \frac{d}{d x}\left(3^{x}+1\right)}{\left(3^{x}+1\right)^{2}}$
$+\frac{1}{1+x}$
$=\left(\frac{3^{x}-1}{3^{x}+1}\right) \cos x+\sin x\left(3^{x}+1\right)\left(3^{x} \log _{e} 3\right)$
$\quad \frac{-\left(3^{x}-1\right)\left(3^{x} \log _{t} 3\right)}{\left(3^{x}+1\right)^{2}}+\frac{1}{1+x}$
$\therefore \quad\left(\frac{d y}{d x}\right)_{\text {at } x=0}=0+0+\frac{1}{1+0}=1$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2012
Let $y$ be the solution of the differential equation
$x \frac{d y}{d x}=\frac{y^{2}}{1-y \log x}$ satisfying $y(1)=1 .$ Then, $y$
satisfies

Options:
A $y=x^{y-1}$
B $y=x^{y}$
C $y=x^{y+1}$
D $y=x^{y+2}$

The correct answer is:
$y=x^{y}$

Solution:
Let $y=x^{y} \Rightarrow \log y=y \log x$
Differentiating both sides w.r.t. $x$, we get $\frac{1}{y} \frac{d y}{d x}=y \times \frac{1}{x}+\log x \frac{d y}{d x}$
$\Rightarrow \quad \frac{1}{y} \frac{d y}{d x}-\log x \frac{d y}{d x}=\frac{y}{x}$
$\Rightarrow \quad \frac{d y}{d x}\left(\frac{1}{y}-\log x\right)=\frac{y}{x}$
$\Rightarrow \quad \frac{d y}{d x}\left(\frac{1-y \log x}{y}\right)=\frac{y}{x}$
$\Rightarrow \quad \frac{x d y}{d x}=\frac{y^{2}}{(1-y \log x)}$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2012
Let $y=e^{x^{2}}$ and $y=e^{x^{2}} \sin x$ be two given curves. Then, angle between the tangents to the curves at any point of their intersection

Options:
A 0
B $\pi$
C $\frac{\pi}{2}$
D $\frac{\pi}{4}$

The correct answer is:
0

Solution:
For intersecting points, $e^{x^{2}}=e^{x^{2}} \sin x$ $\Rightarrow \quad e^{x^{2}}(\sin x-1)=0$
$\Rightarrow \quad e^{x^{2}}=0 \quad$ or $\sin x=1$
But $\quad e^{x^{2}} \neq 0 \Rightarrow \sin x=1$
$\Rightarrow$
$$
x=\frac{\pi}{2}
$$
Now.
$$
y=e^{x^{2}}
$$
$\therefore \quad \frac{d y}{d x}=e^{x^{2}} \cdot 2 x=2 x e^{x^{2}}$
Also, $y=e^{x^{2}} \sin x$
$\frac{d y}{d x}=e^{x^{2}} \cdot 2 x \sin x+e^{x^{2}} \cos x=2 x e^{x^{2}}$
$[\because \sin x=1, \cos x=0]$
Since, both the curves has equal slope. Hence, angle between the tangents at intersecting point is $0 .$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2015
Let $y=f(x)=2 x^{2}-3 x+2$. The differential of $y$ when $x$ changes from 2 to $1.99$ is

Options:
A $0.01$
B $0.18$
C $-0.05$
D $0.07$

The correct answer is:
$-0.05$

Solution:
Hint:
$\Delta x=-0.01 \Rightarrow \Delta y=f^{\prime}(x) \Delta x \Rightarrow \Delta y=f^{\prime}(2)(-0.01) \Rightarrow \Delta y=-5 \times 0.01 \Rightarrow \Delta y=-0.05$

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2020
Let $\mathrm{y}=\frac{\mathrm{x}^{2}}{(\mathrm{x}+1)^{2}(\mathrm{x}+2)}$. Then $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}$ is

Options:
A $2\left[\frac{3}{(x+1)^{4}}-\frac{3}{(x+1)^{3}}+\frac{4}{(x+2)^{3}}\right]$
B $3\left[\frac{2}{(x+1)^{3}}+\frac{4}{(x+1)^{2}}-\frac{5}{(x+2)^{3}}\right]$
C $\frac{6}{(x+1)^{3}}-\frac{4}{(x+1)^{2}}+\frac{3}{(x+1)^{3}}$
D $\frac{7}{(x+1)^{3}}-\frac{3}{(x+1)^{2}}+\frac{2}{(x+1)^{3}}$

The correct answers are:
$2\left[\frac{3}{(x+1)^{4}}-\frac{3}{(x+1)^{3}}+\frac{4}{(x+2)^{3}}\right]$

Solution:
Hint:
By partial fraction technique
$\begin{array}{l}
y=\frac{x^{2}}{(x+1)^{2}(x+2)}=\frac{4}{(x+2)}-\frac{3}{(x+1)}+\frac{1}{(x+1)^{2}} \\
\Rightarrow y^{\prime \prime}=\frac{6}{(x+1)^{4}}-\frac{6}{(x+1)^{3}}+\frac{8}{(x+2)^{3}}
\end{array}$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2020
Let $y(x)$ be a solution of $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y-4 x^{2}=0$ and $y(0)=-1 .$ Then y(1) s equal to

Options:
A $\frac{1}{2}$
B $\frac{1}{3}$
C $\frac{1}{6}$
D 1

The correct answer is:
$\frac{1}{6}$

Solution:
We have. $\left(1+x^{2}\right) \frac{d y}{d x}+2 x y-4 x^{2}=0$
$\Rightarrow \quad \frac{d y}{d x}+\left(\frac{2 x}{1+x^{2}}\right) y=\frac{4 x^{2}}{1+x^{2}}$
Here. $\mathrm{IF}=e^{\frac{1}{1+x^{2}}}=e^{\log \left(1+x^{2}\right)}=1+x^{2}$
$\therefore \quad y\left(1+x^{2}\right)=\int\left(1+x^{2}\right) \times \frac{4 x^{2}}{\left(1+x^{2}\right)} d x+C$
$\Rightarrow y\left(1+x^{2}\right)=\int 4 x^{2} d x+C$
$\Rightarrow \quad y\left(1+x^{2}\right)=\frac{4 x^{3}}{3}+C$
$\Rightarrow \quad y\left(1+x^{2}\right)=\frac{4 x^{3}}{3}-1$
$$
[y(0)=-1]
$$
$\Rightarrow \quad y=\frac{4 x^{3}}{3\left(1+x^{2}\right)}-\frac{1}{1+x^{2}}$
$\therefore \quad y(1)=\frac{4}{6}-\frac{1}{2}=\frac{1}{6}$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2018
Let $z_{1}$ and $z_{2}$ be complex numbers such that $z_{1} \neq z_{2}$ and $\left|z_{1}\right|=\left|z_{2}\right| .$ If $\operatorname{Re}\left(z_{1}\right)>0$ and
$\operatorname{Im}\left(z_{2}\right) < 0,$ then $\frac{z_{1}+z_{2}}{z_{1}-z_{2}}$ is

Options:
A one
B real and positive
C real and negative
D purely imaginary

The correct answer is:
purely imaginary

Solution:
Let $z_{1}=x_{1}+i y_{1}$ and $z_{2}=x_{2}+i y_{2}$
$\operatorname{Re}\left(z_{1}\right)>0 \Rightarrow x_{1}>0$
$\begin{array}{lr}\text { and } & \operatorname{Im}\left(z_{2}\right) < 0 \\ \Rightarrow & y_{2} < 0 \\ \text { Given, } & \left|z_{1}\right|=\left|z_{2}\right|\end{array}$
$\Rightarrow \quad\left|z_{1}\right|^{2}=\left|z_{2}^{2}\right|$
$\Rightarrow \quad z_{1} \bar{z}_{1}=z_{2} \bar{z}_{2}$
Now, $\left(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right)+\left(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right)$
$=\left(\frac{z_{1}+z_{2}}{z_{1}-z_{2}}\right)+\left(\frac{\bar{z}_{1}+\bar{z}_{2}}{\bar{z}_{1}-\bar{z}_{2}}\right)$
$=\frac{z_{1} \bar{z}_{1}+z_{2} \bar{z}_{1}-z_{1} \bar{z}_{2}-z_{2} \bar{z}_{2}+z_{1} \bar{z}_{1}+z_{1} \bar{z}_{2}-z_{2} \bar{z}_{1}+z_{2} \bar{z}_{2}}{\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)}$
$=\frac{2\left(\left|z_{1}\right|^{2}-\left|z_{2}\right|^{2}\right)}{\left(z_{1}-z_{2}\right)\left(\bar{z}_{1}-\bar{z}_{2}\right)}=0 \quad\left(\because\left|z_{1}\right|^{2}=\left|z_{2}\right|^{2}\right)$
$=\frac{z_{1}+z_{2}}{z_{1}-z_{2}}$ is purely imaginary.

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2018
Let $z_{1}$ and $z_{2}$ be two imaginary roots of $z^{2}+p z+q=0$, where $p$ and $q$ are real. The points $z_{1}, z_{2}$ and origin form an equilateral triangle if

Options:
A $\mathrm{p}^{2}>3 \mathrm{q}$
B $p^{2} < 3 q$
C $p^{2}=3 q$
D $p^{2}=q$

The correct answer is:
$p^{2}=3 q$

Solution:
Hint:
$\begin{aligned} & \mathrm{O}^{2}+\mathrm{z}_{1}^{2}+\mathrm{z}_{2}^{2}=\mathrm{z}_{1} \mathrm{z}_{2} \\ & \Rightarrow \mathrm{z}_{1}^{2}+\mathrm{z}_{2}^{2}=\mathrm{z}_{1} \mathrm{z}_{2} \\ & \Rightarrow\left(\mathrm{z}_{1}+\mathrm{z}_{2}\right)^{2}=3 \mathrm{z}_{1} \mathrm{z}_{2} \\ & \Rightarrow \mathrm{p}^{2}=3 \mathrm{q} \end{aligned}$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2020
Let $z_1$ and $z_2$ be two non-zero complex numbers. Then

Options:
A Principal value of $\arg \left(z_1 z_2\right)$ may not be equal to Principal value of arg $z_1+$ Principal value of arg $z_2$
B Principal value of $\arg \left(z_1 z_2\right)=$ Principal value of $\arg z_1+$ Principal value of $\arg z_2$
C Principal value of $\arg \left(z_1 / z_2\right)=$ Principal value of $\arg z_1-$ Principal value of $\operatorname{argz} z_2$
D Principal value of $\arg \left(z_1 / z_2\right)$ may not be $\arg z_1-\arg z_2$

The correct answers are:
Principal value of $\arg \left(z_1 z_2\right)$ may not be equal to Principal value of arg $z_1+$ Principal value of arg $z_2$, Principal value of $\arg \left(z_1 / z_2\right)$ may not be $\arg z_1-\arg z_2$

Solution:
$\arg \left(z_1 z_2\right)=\operatorname{argz_{1}}+\operatorname{argz_{2}}+2 k \pi$
$\arg \left(z_1 / z_2\right)=\operatorname{argz_{1}}-\operatorname{argz_{2}}+2 \mathrm{k} \pi$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2022
Let $z_{1}$ be a fixed point on the circle of radius 1 centred at the origin in the argand plane and $z_{1} \neq \pm 1 .$ Consider an equilateral triangle inscribed in the circle with $z_{1}, z_{2}, z_{3}$ as the
direction. Then, $z_{1} z_{2} z_{3}$ is equal to

Options:
A $z_{1}^{2}$
B $z_{1}^{3}$
C $z_{1}^{4}$
D $z_{1}$

The correct answer is:
$z_{1}^{3}$

Solution:
Given, $z_{1} \neq\pm 1$
Since, $z_{2}$ and $z_{3}$ can be obtained by rotating
vector representing through $\frac{2 \pi}{3}$ and $\frac{4 \pi}{3}$
respectively
$\therefore \quad z_{2}=z_{1} \omega$
and
$$
z_{3}=z_{1} \omega^{2}
$$
$$
\begin{aligned}
\therefore \quad z, z_{2} z_{3} &=z_{1} \times z_{1} \omega \times z_{1} \omega^{2} \\
&=z_{1}^{3} \omega^{3} \\
&=z_{1}^{3}
\end{aligned}
$$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2014
Let $z_{1}, z_{2}$ be two fixed complex numbers in the argand plane and $z$ be an arbitrary point satisfying $\left|z-z_{1}\right|+\left|z-z_{2}\right|=2\left|z_{1}-z_{2}\right|$. Then, the locus of $z$ will be

Options:
A an ellipse
B a straight line joining $z_{1}$ and $z_{2}$
C a parabola
D a bisector of the line segment joining $z_1$ and $z_2$

The correct answer is:
an ellipse

Solution:
We know that $\left|z-z_{1}\right|+\left|z-z_{2}\right|=k$ will represent an ellipse, if $\left|z_{1}-z_{2}\right| < k$
Hence. the equation $\left|z-z_{1}\right|+\left|z-z_{2}\right|$ $=2\left|z_{1}-z_{2}\right|$ represent an ellipse.

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2014
Let $z$ be a complex number such that the principal value of argument, arg $z>0$ Then, arg $z-\arg (-z)$ is

Options:
A $\frac{\pi}{2}$
B $\pm \pi$
C $\pi$
D $-\pi$

The correct answer is:
$\pi$

Solution:
$\begin{aligned} \therefore \arg (z)-\arg (-z) &=\theta-(-\pi+\theta) \\ &=\pi \end{aligned}$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2019