datavorous/entrance-exam-dataset huggingface dataset:
If $x=1+\frac{1}{2 \times 1 !}+\frac{1}{4 \times 2 !}+\frac{1}{8 \times 3 !}+\ldots$ and
$y=1+\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}+\frac{x^{6}}{3 !}+\ldots$
Then, the value of loge $y$ is

Options:
A $e$
B $\mathrm{e}^{2}$
C 1
D $\frac{1}{e}$

The correct answer is:
$e$

Solution:
Given,
$x=1+\frac{1}{2 \times 1 !}+\frac{1}{4 \times 2 !}+\frac{1}{8 \times 3 !}+\ldots$
$\Rightarrow \quad x=1+\frac{(1 / 2)^{\prime}}{1 !}+\frac{(1 / 2)^{2}}{2 !}+\frac{(1 / 2)^{3}}{3 !}+\ldots$
$\Rightarrow \quad x=e^{1 / 2} \Rightarrow x^{2}=e$
and $\quad y=1+\frac{x^{2}}{1 !}+\frac{x^{4}}{2 !}+\frac{x^{6}}{3 !}+\ldots$
$\Rightarrow \quad y=1+\frac{\left(x^{2}\right)^{1}}{1 !}+\frac{\left(x^{2}\right)^{2}}{2 !}+\frac{\left(x^{2}\right)^{3}}{3 !}+\ldots$
$\Rightarrow \quad y=e^{x^{2}}=e^{e} \quad$ [from Eq. (i)]
Taking log on both sides, we get
log$y=e$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2013
If $x_1$ be the size of the magnified image and $x_2$ the size of the diminished image in Lens Displacement Method, then the size of the object is :

Options:
A $\sqrt{x_1 x_2}$
B $x_1 x_2$
C $x_1^2 x_2$
D $x_1 x_2^2$

The correct answer is:
$\sqrt{x_1 x_2}$

Solution:
No Solution Available

Topic: Physics, Experimental Physics, WBJEE, WBJEE 2010
If $x^2+y^2=1$ then

Options:
A $y y^{\prime \prime}-\left(2 y^{\prime}\right)^2+1=0$
B $y y^{\prime \prime}+\left(y^{\prime}\right)^2+1=0$
C $y y^{\prime \prime}-\left(y^{\prime}\right)^2-1=0$
D $y y^{\prime \prime}+\left(2 y^{\prime}\right)^2+1=0$

The correct answer is:
$y y^{\prime \prime}+\left(y^{\prime}\right)^2+1=0$

Solution:
Hints: $2 x+2 y y^{\prime}=0$
$$
\begin{aligned}
& x+y y^{\prime}=0 \\
& 1+y y^{\prime \prime}+\left(y^{\prime}\right)^2=0
\end{aligned}
$$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2009
If $x^{2}+y^{2}=a^{2}$, then $\int_{0}^{a} \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x=$

Options:
A $2 \pi a$
B $\pi a$
C $\frac{1}{2} \pi \mathrm{a}$
D $\frac{1}{4} \pi \mathrm{a}$

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

Solution:
Hint:
$y_{1}=-\frac{x}{y}$
$\int_{0}^{a} \sqrt{1+\frac{x^{2}}{y^{2}}} d x=a \int_{0}^{a} \frac{1}{y} \cdot d x=a \int_{0}^{a} \frac{d x}{\sqrt{a^{2}-x^{2}}}$
$=a\left[\sin ^{-1}\left(\frac{x}{a}\right)\right]_{0}^{]}=a \pi / 2$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2020
If $x=a t+b t^{2},$ where $x$ is in metre $(\mathrm{m})$ and $t$ is in hour $(\mathrm{h})$, then unit of $b$ will be

Options:
A $\frac{\mathrm{m}^{2}}{\mathrm{h}}$
B $m$
C $\frac{m}{h}$
D $\frac{m}{h^{2}}$

The correct answer is:
$\frac{m}{h^{2}}$

Solution:
Given, $x=a t+b t^{2}$
$\begin{array}{l}
{[x]=\left[b t^{2}\right]} \\
\text { Unit of } b=\frac{x}{t^{2}}=\frac{\text { metre }}{(h)^{2}}=\frac{m}{h^{2}}
\end{array}$

Topic: Physics, Units and Dimensions, WBJEE, WBJEE 2016
If $x$ and $y$ are digits such that $17 !=3556xy428096000,$ then $x+y$ equals

Options:
A 15
B 6
C 12
D 13

The correct answer is:
15

Solution:
Given, $171=3556 \times 4428096000$
Since, 17! is divisible by $9,$ so sum of the digits $(48+x+y)$ must be divisible by 9
So, $x+y$ can be 15 or 6 . Also, 17! is divisible by $11,$ so $|10+x-y \mid$ must be multiple of 11 or $0 .$ The only possibility is
$$
\begin{array}{l}
|x-y|=1 \\
\therefore \quad x+y=15
\end{array}
$$

Topic: Mathematics, Permutation Combination, WBJEE, WBJEE 2015
If $x \frac{d y}{d x}+y=\frac{x f(x y)}{f^{\prime}(x y)}$, then $|f(x y)|$ is equal to

Options:
A $k e^{x^{2} / 2}$
B $\mathrm{ke}^{\mathrm{y}^{2} / 2}$
C $\mathrm{ke}^{\mathrm{x}^{2}}$
D $\mathrm{ke}^{\mathrm{y}^{2}}$

The correct answer is:
$k e^{x^{2} / 2}$

Solution:
$x \frac{d y}{d x}+y=\frac{x f(x y)}{f^{\prime}(x y)} \Rightarrow \frac{x d y+y d x}{d x}=\frac{x f(x y)}{f^{\prime}(x y)} \Rightarrow \frac{f^{\prime}(x y) d(x y)}{f(x y)}=x d x \Rightarrow \ln f(x y)=\frac{x^{2}}{2}+C \Rightarrow|f(x y)|=k e^{x^{2} / 2}$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2021
If $x \frac{d y}{d x}+y=x \frac{f(x y)}{f^{\prime}(x y)}$, then $|f(x y)|$ is equal to

Options:
A $\mathrm{Ce}^{\mathrm{x}^2 / 2}$
B $\mathrm{Ce}^{\mathrm{x}^2}$
C $\mathrm{Ce}^{2 x^2}$
D $\mathrm{Ce}^{\mathrm{x}^2 / 3}$

The correct answer is:
$\mathrm{Ce}^{\mathrm{x}^2 / 2}$

Solution:
$x \frac{d y}{d x}+y=x \frac{f(x y)}{f^{\prime}(x y)} \Rightarrow \frac{d(x y)}{d x}=x \frac{f(x y)}{f^{\prime}(x y)} \Rightarrow \frac{f^{\prime}(x y)}{f(x y)} d(x y)=x d x \Rightarrow \ln |f(x y)|=\frac{x^2}{2}+k \Rightarrow f(x y)=C e^{\frac{x^2}{2}}$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2022
If $[x]$ denotes the greatest integer less than or equal to $x$, then the value of the integral $\int_{0}^{2} x^{2}[x] d x$ equals

Options:
A $\frac{5}{3}$
B $\frac{7}{3}$
C $\frac{8}{3}$
D $\frac{4}{3}$

The correct answer is:
$\frac{7}{3}$

Solution:
Let.
$$
\begin{aligned}
I &=\int_{0}^{2} x^{2}[x] d x \\
&=\int_{0}^{1} x^{2} \cdot 0 d x+\int_{1}^{2} x^{2} \times 1 d x \\
&=\left[\frac{x^{3}}{3}\right]_{1}^{2}=\left[\frac{8}{3}-\frac{1}{3}\right]=\frac{7}{3}
\end{aligned}
$$

Topic: Mathematics, Definite Integration, WBJEE, WBJEE 2016
If $x$ is a positive real number different from 1 such that $\log _{a} x, \log _{b} x, \log _{c} x$ are in $\mathrm{AP}$, then

Options:
A $b=\frac{a+c}{2}$
B $b=\sqrt{a c}$
C $c^{2}=(a c)^{log_{a} b}$
D None of these

The correct answer is:
$c^{2}=(a c)^{log_{a} b}$

Solution:
Given, $\log _{a} x, \log _{b} x$ and $log_{c}$ $x$ are in $A P$.
$\therefore \quad 2 \log _{b} x=\log _{a} x+\log _{c} x$
$\begin{aligned} 2 \log _{b} x &=\frac{1}{\log _{x} a}+\frac{1}{\log _{x} c} \\ \frac{2}{\log _{x} b} &=\frac{\log _{x} a c}{\log _{x} a \log _{x} c} \\ 2 \log _{x} c &=\frac{\log _{x} b}{\log _{x} a}\left(\log _{x} a c\right) \\ 2 \log _{x} c &=\log _{a} b \cdot \log _{x} a c \\ \log _{x} c^{2} &=\log _{x}(a c)^{\log _{a} b} \\ c^{2} &=(a c)^{\log _{a} b} \end{aligned}$

Topic: Mathematics, Sequences and Series, WBJEE, WBJEE 2016
If $X$ is a random variable such that $\sigma(X)=2.6$ then $\sigma(1-4 x)$ is equal to

Options:
A 7.8
B $-10.4$
C 13
D 10.4

The correct answer is:
10.4

Solution:
Given that,
$$
\begin{aligned}
\sigma(X) &=2.6=\sqrt{\operatorname{var}(x)} \\
\sigma(1-4 x) &=\sqrt{\operatorname{var}(1-4 x)} \\
&=\sqrt{16 \operatorname{var}(x)} \\
&=4 \times \sqrt{\operatorname{var}(x)}=4 \times 2.6=10.4
\end{aligned}
$$

Topic: Mathematics, Statistics, WBJEE, WBJEE 2019
If $x+\log _{10}\left(1+2^{x}\right)=x \log _{10} 5+\log _{10} 6,$ then
the value of $x$ is

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

The correct answer is:
1

Solution:
We have,
$\begin{aligned} x+\log _{10}\left(1+2^{x}\right) &=x \log _{10} 5+\log _{10} 6 \\ \Rightarrow \log _{10}\left(1+2^{x}\right) &=x \log _{10} 5+\log _{10} 6-x \\ &=\log _{10} 5^{x}+\log _{10} 6-x \log _{10} 10 \\ &=\log _{10}\left(5^{x} \cdot 6\right)-\log _{10} 10^{x} \\ \Rightarrow \quad & \log _{10}\left(1+2^{x}\right)=\log _{10}\left(\frac{5^{x} \cdot 6}{10^{x}}\right) \end{aligned}$
$\Rightarrow \quad 1+2^{x}=\frac{5^{x} \cdot 6}{10^{x}}=\frac{6}{2^{x}}$
$\Rightarrow \quad 2^{x}\left(1+2^{x}\right)=6$
$\Rightarrow \quad t(1+t)=6$$\quad \quad \quad\left(\operatorname{let} 2^{x}=t\right)$
$\Rightarrow \quad t^{2}+t-6=0$
$\Rightarrow \quad(t+3)(t-2)=0$
$t+3=0$
or $t-2=0$
$\begin{array}{l}
\Rightarrow \quad t=2 \\
2^{x}=2 \Rightarrow x=1
\end{array}$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2018
If $x=\log _a b c, y=\log _b c a, z=\log _c a b$, then the value of $\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}$ will be

Options:
A $x+y+z$
B 1
C $a b+b c+c a$
D $\mathrm{abc}$

The correct answer is:
1

Solution:
Hints: $1+x=\log _a a+\log _a b c=\log _a a b c$
$\begin{aligned}
& \frac{1}{1+x}=\log _{a b c} a, \text { Similarly } \frac{1}{1+y}=\log _{a b c} b \\
& \frac{1}{1+z}=\log _{a b c} c \text {, Ans. }=\log _{(a b c)} a b c=1
\end{aligned}$

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2009
If $x \sin \left(\frac{y}{x}\right) d y=\left[y \sin \left(\frac{y}{x}\right)-x\right] d x, x>0$ and $y(1)=\frac{\pi}{2}$ then the value of $\cos \left(\frac{y}{x}\right)$ is

Options:
A 1
B $\log x$
C $\mathrm{e}$
D 0

The correct answer is:
$\log x$

Solution:
Hint:
$\operatorname{sinv}\left(v+x \cdot \frac{d v}{d x}\right)=v \sin v-1$
$\Rightarrow \mathrm{x} \sin \mathrm{v} \cdot \frac{\mathrm{dv}}{\mathrm{dx}}=-1$
$\Rightarrow \int \sin v \mathrm{~d} v=-\int \frac{\mathrm{dx}}{\mathrm{x}}$
$\Rightarrow-\cos v=-\log x+\mathrm{c}$
at $x=1 ; \mathrm{y}=\frac{\pi}{2} ; \mathrm{c}=0$
$\cos \left(\frac{\mathrm{y}}{\mathrm{x}}\right)=\log \mathrm{x}$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2020
If $\Delta(x)=\left|\begin{array}{ccc}x-2 & (x-1)^2 & x^3 \\ x-1 & x^2 & (x+1)^3 \\ x & (x+1)^2 & (x+2)^3\end{array}\right|$, then coefficient of $x$ in $\Delta(x)$ is

Options:
A 2
B -2
C 3
D -4

The correct answer is:
-2

Solution:
co-eff of $x$ in $\Delta(x)=\frac{\Delta^{\prime}(0)}{1 !}=-2$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2022
If $x, y$ and $z$ are greater than 1 , then the value of $\left|\begin{array}{ccc}1 & \log _{x} y & \log _{x} z \\ \log _{y} x & 1 & \log _{y} z \\ \log _{z} x & \log _{z} y & 1\end{array}\right|$ is

Options:
A $\log x \cdot \log y \cdot \log z$
B $\log x+\log y+\log z$
C 0
D $1-\{(\log x) \cdot(\log y) \cdot(\log z)\}$

The correct answer is:
0

Solution:
Let $\Delta=\left|\begin{array}{ccc}1 & \log _{x}y & \log _{x} z \\ \log _{y} x & 1 & \log _{y} z \\ \log _{z} x & \log _{z} y & 1\end{array}\right|$
$=\left|\begin{array}{lll}\frac{\log x}{\log x} & \frac{\log y}{\log x} & \frac{\log z}{\log x} \\ \frac{\log x}{\log y} & \frac{\log y}{\log y} & \frac{\log z}{\log y} \\ \frac{\log x}{\log z} & \frac{\log y}{\log z} & \frac{\log 2}{\log z}\end{array}\right|$
On taking $\frac{1}{\log x} \frac{1}{\log y} \frac{1}{\log z}$ common from $R_{1}, R_{2}$ and $R_{3}$
we got
$$
=\frac{1}{\log x \cdot \log y \cdot \log z}\left|\begin{array}{lll}
\log x & \log y & \log z \\
\log x & \log y & \log z \\
\log x & \log y & \log z
\end{array}\right|
$$
$\therefore \Delta=0$

Topic: Mathematics, Determinants, WBJEE, WBJEE 2016
If \(x^2+y^2=4\), then \(y \frac{d y}{d x}+x=\)

Options:
A 4
B 0
C 1
D -1

The correct answer is:
0

Solution:
Hints: \(x+y \frac{d y}{d x}=0\)

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2011
If \(x+\frac{1}{x}=2 \cos \theta\), then for any integer \(n, x^n+\frac{1}{x^n}=\)

Options:
A \(2 \cos n \theta\)
B \(2 \sin n \theta\)
C \(2 i \cos n \theta\)
D \(2 i \sin n \theta\)

The correct answer is:
\(2 \cos n \theta\)

Solution:
Hints: \(x+\frac{1}{x}=2 \cos \theta\)
Let \(x=\cos \theta+1 \sin \theta\)
\(\frac{1}{x}=\cos \theta-1 \sin \theta\)
Thus \(x^n+\frac{1}{x^n}=2 \cos n \theta\)

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2011
If \(x=\sin \theta\) and \(y=\sin k \theta\), then \(\left(1-x^2\right) y_2-x y_1-\alpha y=0\), for \(\alpha=\)

Options:
A \(\mathrm{k}\)
B \(-\mathrm{k}\)
C \(-\mathrm{k}^2\)
D \(\mathrm{k}^2\)

The correct answer is:
\(-\mathrm{k}^2\)

Solution:
Hint : \(x=\sin \theta\)
\(y=\sin k \theta\)
\(\frac{d y}{d x}=\frac{d y / d \theta}{d x / d \theta}=\frac{k \cos k \theta}{\cos \theta}=y_1 \ldots\ldots (I)\)
\(\begin{aligned}
& y_2=\frac{d^2 y}{d x^2}=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \times \frac{d \theta}{d x}=\left[\frac{-k^2 \sin (k \theta) \cdot \cos \theta+k \cos (k \theta) \sin \theta}{\cos ^2 \theta}\right] \times \frac{1}{\cos \theta} \\
& y_2=\frac{k \cos (k \theta) \sin \theta-k^2 \sin (k \theta) \cos \theta}{\cos ^3 \theta} \ldots \ldots \text { (II) }
\end{aligned}\)
also, \(1-x^2=1-\sin ^2 \theta=\cos ^2 \theta \ldots \ldots \text { (III) }\)
So, putting the values of \(1-x^2, y_1\) and \(y_2\) from (I), (II), (III) we get
\(\begin{aligned}
& \left(1-x^2\right) y_2-x y_1=\alpha y \\
& \Rightarrow\left(\cos ^2 \theta\right)\left[\frac{k \cos (k \theta) \sin \theta-k^2 \sin (k \theta) \cdot \cos \theta}{\cos ^3 \theta}\right]-\frac{k \sin \theta \cos (k \theta)}{\cos \theta}=\alpha(\sin k \theta) \\
& \Rightarrow \frac{k \cos (k \theta) \sin \theta-k^2 \sin (k \theta) \cos \theta-k \sin \theta \cos (k \theta)}{\cos \theta}=\alpha \sin (k \theta) \\
& \Rightarrow-k^2 \sin (k \theta)=\alpha \sin (k \theta)
\end{aligned}\)
So, \(\alpha=-k^2\)

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2023
If $y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right) \ldots\left(1+x^{2 n}\right),$ then the value of $\left(\frac{d y}{d x}\right)$ at $x=0$ is

Options:
A 0
B -1
C 1
D 2

The correct answer is:
1

Solution:
Given,
$y=(1+x)\left(1+x^{2}\right)\left(1+x^{4}\right)+\ldots+\left(1+x^{2 n}\right)$
$\Rightarrow \log y=\log (1+x)+\log \left(1+x^{2}\right)$ $+\log \left(1+x^{4}\right)+\ldots+\log \left(1+x^{2n}\right)$
$\Rightarrow \frac{1}{y} \frac{d y}{d x}=\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{3 x^{2}}{1+x^{3}}+\ldots+\frac{2 n x^{2 n-1}}{1+x^{2 n}}$
$\Rightarrow \frac{d y}{d x}=y\left[\frac{1}{1+x}+\frac{2 x}{1+x^{2}}+\frac{3 x^{2}}{1+x^{3}}+\ldots+\frac{2 n x^{2 n-1}}{1+x^{2 n}}\right]$
$\therefore \quad\left(\frac{d y}{d x}\right)_{x=0}=y\left(\frac{1}{1+0}\right)=1$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2016
If $y=(1+x)\left(1+x^2\right)\left(1+x^4\right) \ldots\left(1+x^{2 n}\right)$ then the value of $\left(\frac{d y}{d x}\right)_{x=0}$ is

Options:
A 0
B -1
C 1
D 2

The correct answer is:
1

Solution:
Hints: T-log \& Differentiate
$$
\begin{aligned}
& \frac{d y}{d x}=y\left[\frac{1}{1+x}+\frac{2 x}{1+x^2}+\ldots\right] \text { Put } x=0 \\
& \frac{d y}{d x}=1
\end{aligned}
$$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2010
If $y^{\prime \prime}-3 y^{\prime}+2 y=0$ where $y(0)=1, y^{\prime}(0)=0$, then the value of $y$ at $x=\log , 2$ is

Options:
A 1
B -1
C 2
D 0

The correct answer is:
0

Solution:
$$
\begin{aligned}
& \text { Hints: } \frac{d^2 y}{d x^2}-3 \frac{d y}{d x}+2 y=0 \\
& m^2-3 m+2=0, \quad y=A e^x+B e^{2 x} \\
& m=1, m=2, y^1=A e^x+2 B e^{2 x} \\
& y=0, A+B=1 \quad A+2 B=0, A=2, B=-1 \\
& y=2 e^x-e^{2 x} \\
& y=0 \quad \text { at } x=\ell n 2
\end{aligned}
$$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2010
If $y=4 x+3$ is parallel to a tangent to the parabola $y^{2}=12 x$, then its distance from the normal parallel to the given line is

Options:
A $\frac{213}{\sqrt{17}}$
B $\frac{219}{\sqrt{17}}$
C $\frac{211}{\sqrt{17}}$
D $\frac{210}{\sqrt{17}}$

The correct answer is:
$\frac{219}{\sqrt{17}}$

Solution:
Given equation of parabola is
$$
y^{2}=12 x
$$
On differentiating both sides w.r.t. $x,$ we get
$$
2 y \frac{d y}{d x}=12 \Rightarrow \frac{d y}{d x}=\frac{6}{y}
$$
Since, the normal to the curve is parallel to the line $y=4 x+3$
$\therefore$ Slope of normal curve $=$ Slope of line
$\Rightarrow \quad-\frac{Y}{6}=4$
$\Rightarrow \quad y=-24$
From Eq. (i), we get $(-24)^{2}=12 x$
$\Rightarrow \quad 24 \times 24=12 x$
$\Rightarrow \quad x=48$
$\therefore$ Normal point on a curve is (48,-24)
$\therefore$ Distance from (48,-24) to the line $4 x-y+3=0$ is
$\frac{4 \times 48+24+3}{\sqrt{4^{2}+1^{2}}}=\frac{219}{\sqrt{17}}$

Topic: Mathematics, Parabola, WBJEE, WBJEE 2014
If $\sqrt{y}=\cos ^{-1} x$, then it satisfies the differential equation $\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=c,$ where $c$ is equal to

Options:
A 0
B 3
C 1
D 2

The correct answer is:
2

Solution:
Given, $\sqrt{y}=\cos ^{-1} x \Rightarrow y=\left(\cos ^{-1} x\right)^{2}$
On differentiating both sides w.r.t. $x$, we get
$$
\frac{d y}{d x}=2\left(\cos ^{-1} x\right) \times \frac{-1}{\sqrt{1-x^{2}}}
$$
Again, differentiating both sides w.r.t. $x$, we get
$\frac{d^{2} y}{d x^{2}}=-2\left[\begin{array}{c}\sqrt{1-x^{2}} \times \frac{-1}{\sqrt{1-x^{2}}} \\ -\cos ^{-1} x \times\left(\frac{1}{2}\right) \frac{(-2 x)}{\left(1-x^{2}\right)^{1 / 2}} \\ \hline\left(\sqrt{1-x^{2}}\right)^{2}\end{array} \right ].$
$=-2\left[\frac{-1+\frac{x \cos ^{-1} x}{\left(1-x^{2}\right)^{1 / 2}}}{\left(1-x^{2}\right)}\right]$
$\frac{d^{2} y}{d x^{2}}=\left[\frac{2-\frac{2 x \cos ^{-1} x}{\left(1-x^{2}\right)^{1 / 2}}}{\left(1-x^{2}\right)}\right]$
$\Rightarrow \quad\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}=2+x \frac{d y}{d x}$
$\Rightarrow \quad\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=2$
But, it is given
$\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=c$
$\therefore c=2$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2014
If $y=e^{m \sin ^{-1} x}$ then $\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-k y=0,$ where $k$ is
equal to

Options:
A $m^{2}$
B 2
C -1
D $-m^{2}$

The correct answer is:
$m^{2}$

Solution:
$y=e^{msin^{-1}x}$ $\Rightarrow \quad \frac{d y}{d x}=e^{msin^{-1}x} \cdot \frac{m}{\sqrt{1-x^{2}}}$
$\Rightarrow \sqrt{1-x^{2}} \frac{d y}{d x}=m e^{msin^{-1}x}$
$\Rightarrow \sqrt{1-x^{2}} \frac{d y}{d x}=m y$
$\Rightarrow \sqrt{1-x^{2}} \frac{d^{2} y}{d x^{2}}+\frac{1}{2 \sqrt{1-x^{2}}}(-2 x) \frac{d y}{d x}=m \frac{d y}{d x}$
$\Rightarrow\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=m \sqrt{1-x^{2}} \frac{d y}{d x}$
$\Rightarrow\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}=m \cdot(m y)$
$\Rightarrow\left(1-x^{2}\right) \frac{d^{2} y}{d x^{2}}-x \frac{d y}{d x}-m^{2} y=0$
$\therefore$
$k=m^{2}$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2017
If $y=e^{\tan ^{-1} x}$ then

Options:
A $\left(1+x^2\right) y_2+(2 x-1) y_1=0$
B $\left(1+x^2\right) y_2+2 x y=0$
C $\left(1+x^2\right) y_2-y_1=0$
D $\left(1+x^2\right) y_2+3 x y_1+4 y=0$

The correct answer is:
$\left(1+x^2\right) y_2+(2 x-1) y_1=0$

Solution:
$\frac{d y}{d x}=e^{\tan ^{-1} x} \times \frac{1}{1+x^2}$
$\Rightarrow\left(1+x^2\right) y^{\prime}=y$
or $2 x y^{\prime}+\left(1+x^2\right) y^{\prime \prime}=y^{\prime}$
or $\left(1+x^2\right) y^{\prime \prime}+y^{\prime}(2 x-1)=0$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2022
If $y=e^{-x} \cos 2 x$, then which of the following differential equation is satisfied?

Options:
A $\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+5 y=0$
B $\frac{d^{2} y}{d x^{2}}+5 \frac{d y}{d x}+2 y=0$
C $\frac{d^{2} y}{d x^{2}}-5 \frac{d y}{d x}-2 y=0$
D $\frac{\sigma^{2} y}{d x^{2}}+2 \frac{d y}{d x}-5 y=0$

The correct answer is:
$\frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+5 y=0$

Solution:
Given, $y=e^{-x} \cos 2 x$
$\therefore \quad \frac{d y}{d x}=e^{-x}(-\sin 2 x) 2+\cos 2 x \cdot e^{-x}(-1)$
$\Rightarrow \quad \frac{d y}{d x}=-2 \sin 2 x \cdot e^{-x}-y$
$\Rightarrow \quad \frac{d y}{d x}+y=-2 \sin 2 x \cdot e^{-x}$
$\Rightarrow \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=-2\left[\sin 2 x \cdot e^{-x}(-1)\right. \left.e^{-x} 2 \cos 2 x\right]$
$=2 \sin 2 x \cdot e^{-x}-4 y \quad$ |from Eq. $\left.(0)\right]$
$\Rightarrow \frac{d^{2} y}{d x^{2}}+\frac{d y}{d x}=-\frac{d y}{d x}-y-4 y$
$\Rightarrow \quad \frac{d^{2} y}{d x^{2}}+2 \frac{d y}{d x}+5 y=0$

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2015
If $y=\tan ^{-1} \sqrt{\frac{1-\sin x}{1+\sin x}}$, then the value of $\frac{d y}{d x}$ at $x=\frac{\pi}{6}$ is

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

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

Solution:
$$
\text { Hints : } \begin{aligned}
& y=\tan ^{-1} \sqrt{\frac{1-\cos \left(\frac{\pi}{2}-x\right)}{1+\cos \left(\frac{\pi}{2}-x\right)}} \\
& =\tan ^{-1} \sqrt{\frac{2 \sin ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}{2 \cos ^2\left(\frac{\pi}{4}-\frac{x}{2}\right)}}=\tan ^{-1}\left|\tan \left(\frac{\pi}{4}-\frac{x}{2}\right)\right|=\left(\frac{\pi}{4}-\frac{x}{2}\right) \\
& \frac{d y}{d x}=-\frac{1}{2}
\end{aligned}
$$

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2009
If \(y=2 x^3-2 x^2+3 x-5\), then for \(x=2\) and \(\Delta x=0.1\) value of \(\Delta y\) is

Options:
A 2.002
B 1.9
C 0
D 09

The correct answer is:
1.9

Solution:
\(\text { Hints: } \frac{\mathrm{dy}}{\mathrm{dx}}=6 \mathrm{x}^2-4 \mathrm{x}+3 \quad \Delta \mathrm{y}=\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{x}-2} \Delta \mathrm{x}=1.9\)

Topic: Mathematics, Application of Derivatives, WBJEE, WBJEE 2011
If \(y=\frac{A}{x}+B x^2\), then \(x^2 \frac{d^2 y}{d x^2}=\)

Options:
A \(2 y\)
B \(y^2\)
C \(\mathrm{y}^3\)
D \(\mathrm{y}^4\)

The correct answer is:
\(2 y\)

Solution:
Hints: \(\frac{x^2 d^2 y}{d x^2}=2\left(A / x+B x^2\right)=2 y\)

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2011
If \(y=\frac{x}{\log _e|c x|}\) is the solution of the differential equation \(\frac{d y}{d x}=\frac{y}{x}+\phi\left(\frac{x}{y}\right)\), then \(\phi\left(\frac{x}{y}\right)\) is given by

Options:
A \(\frac{y^2}{x^2}\)
B \(-\frac{y^2}{x^2}\)
C \(\frac{x^2}{y^2}\)
D \(-\frac{x^2}{y^2}\)

The correct answer is:
\(-\frac{y^2}{x^2}\)

Solution:
Hint : \(y=\frac{x}{\ell n|c x|}\)
\(\begin{aligned}
& \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\ell \mathrm{n}|c x|-1}{(\ell \mathrm{n}|\mathrm{cx}|)^2}=\frac{1}{\ell \mathrm{n}|\mathrm{cx}|}-\frac{1}{(\ell \mathrm{n}|\mathrm{cx}|)^2} \\
& \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{y}}{\mathrm{x}}-\left(\frac{\mathrm{y}}{\mathrm{x}}\right)^2 \quad \therefore \phi\left(\frac{\mathrm{x}}{\mathrm{y}}\right)=-\frac{\mathrm{y}^2}{\mathrm{x}^2}
\end{aligned}\)

Topic: Mathematics, Differential Equations, WBJEE, WBJEE 2023
If \(y=\log ^n x\), where \(\log ^n\) means \(\log _e \log _e \log _e \ldots\) (repeated \(n\) times), then \(x \log x \log ^2 x \log ^3 x \ldots . . \log ^{n-1} x \log ^n x \frac{d y}{d x}\) is equal to

Options:
A \(\log x\)
B \(x\)
C 1
D \(\log ^{\mathrm{n}} x\)

The correct answer is:
\(\log ^{\mathrm{n}} x\)

Solution:
Hint : \(\because \frac{d y}{d x}=\frac{1}{x \log ^{n-1} x \quad \log ^{n-2} x \ldots . \log x}\)

Topic: Mathematics, Basic of Mathematics, WBJEE, WBJEE 2023
If Young's double slit experiment is done with white light, which of the following statements will be true?

Options:
A All the bright fringes will be coloured.
B All the bright fringes will be white.
C The central fringe will be white.
D No stable interference pattern will be visible.

The correct answer is:
The central fringe will be white.

Solution:
Young's double slit experiment with white light which consist wavelength from range $4000 Å$ to $7000 Å$
$\because$ At the centre of the screen, the path difference is zero for all wavelengths. The bright fringes of these wavelengths overlap at the centre. Thus, the white fringe at the centre is formed.

Topic: Physics, Wave Optics, WBJEE, WBJEE 2018
If \(y=\tan ^{-1} \frac{\sqrt{1+x^2}-1}{x}\), then \(y^{\prime}(1)=\)

Options:
A \(1 / 4\)
B \(1 / 2\)
C \(-1 / 4\)
D \(-1 / 2\)

The correct answer is:
\(1 / 4\)

Solution:
Hints: \(y=\tan ^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \quad\) Put \(x=\tan \theta\)
\(\begin{aligned} & =\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right)=\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \\ & =\tan ^{-1}\left(\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right)=\tan ^{-1} \tan \frac{\theta}{2} \\ & =\frac{\theta}{2}=\frac{1}{2} \cdot \tan ^{-1} x, y^{\prime}=\frac{1}{2\left(1+x^2\right)} \\ & y^{\prime}(1)=\frac{1}{2.2}=\frac{1}{4}\end{aligned}\)

Topic: Mathematics, Differentiation, WBJEE, WBJEE 2011
If $z_{1}=2+3 i$ and $z_{2}=3+4 i$ be two points on the complex plane. Then, the set of complex number $z$ satisfying $\left|z-z_{1}\right|^{2}+\left|z-z_{2}\right|^{2}$
$=\left|z_{1}-z_{2}\right|^{2}$ represents

Options:
A a straight line
B a point
C a circle
D a pair of straight line

The correct answer is:
a circle

Solution:
Given, $z_{1}=2+3 i$ and $z_{2}=3+4 i$
Now, we have $\left|z-z_{1}\right|^{2}+\left|z-z_{2}\right|^{2}=\left|z_{1}-z_{2}\right|^{2}$
(let $z=x+i y)$
$\Rightarrow|(x+i y)-(2+3 i)|^{2}+|(x+i y)-(3+4 i)|^{2}$
$\quad=|(2+3 i)-(3+4 i)|^{2}$
$\Rightarrow|(x-2)+i(y-3)|^{2}+|(x-3)+i(y-4)|^{2}$
$\quad=|-1-i|^{2}$
$\Rightarrow(x-2)^{2}+(y-3)^{2}+(x-3)^{2}+(y-4)^{2}=1+1$
$\Rightarrow \quad x^{2}+4-4 x+y^{2}+9-6 y$
$\quad+x^{2}+9-6 x+y^{2}+16-8 y=2$
$\Rightarrow \quad 2 x^{2}+2 y^{2}-10 x-14 y+36=0$
$\Rightarrow \quad x^{2}+y^{2}-5 x-7 y+18=0$
which represent a circle with centre $\left(\frac{5}{2}, \frac{7}{2}\right)$ and radius $\sqrt{\frac{25}{4}+\frac{49}{4}-18}=\frac{1}{\sqrt{2}}$

Topic: Mathematics, Complex Number, WBJEE, WBJEE 2013