datavorous/entrance-exam-dataset huggingface dataset:
If x, y, z are positive numbers in A.P. and tan-1x, tan-1y and tan-1z are also in A.P., then which of the following is correct.

Options:
A 6x=3y=2z
B 6x=4y=3z
C x=y=z
D 2x=3y=6z

The correct answer is:
x=y=z

Solution:
Given, tan-1x, tan-1y and tan-1z are in A.P., so

2tan-1y=tan-1x+tan-1z

⇒tan-1y-tan-1x=tan-1z-tan-1y

y−x1+xy=z−y1+yz ...i

Given x, y, z are in A.P.

i.e., 2y=x+z⇒y−x=z−y

Hence, from i

1+xy=1+yz

xy=yz either y=0 (rejected) or x=z

Hence, x = y = z.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2013 (07 Apr)
If x,y,z∈R, x+y+z=5, x2+y2+z2=9, then length of interval in which x lies is

Options:
A 8/3
B 4/3
C 2/3
D 1/3

The correct answer is:
4/3

Solution:
We have,

x+y+z=5

x2+y2+z2=9

⇒ x2+y2+(5-x-y)2=9

⇒ y2+x2-5x-5y+xy+8=0

⇒ y2+y(x-5)+x2-5x+8=0

⇒ Since y∈R

(x-5)2-4x2-5x+8≥0

⇒ x2-10x+25-4x2+20x-32≥0

⇒ 3x2-10x+7≤0

⇒ 3x2-7x-3x+7≤0

⇒ (3x-7)(x-1)≤0

⇒ x∈1,73
Length of interval =73-1=43

Topic: Mathematics, Quadratic Equation, JEE Main
If \( x=1 \) is a critical point of the function \( f(x)=\left(3 x^{2}+a x-2-a\right) e^{x} \), then

Options:
A \( x=1 \) and \( x=-\frac{2}{3} \) are local minima of \( f \)
B \( x=1 \) and \( x=-\frac{2}{3} \) is a local maxima of \( f \)
C \( x=1 \) is a local maxima and \( x=-\frac{2}{2} \) is a local minima of \( f \)
D \( x=1 \) is a local minima and \( x=-\frac{2}{3} \) are local maxima of \( f \)

The correct answer is:
\( x=1 \) is a local minima and \( x=-\frac{2}{3} \) are local maxima of \( f \)

Solution:
f(x)=3x2+ax-2-aex
f'(x)=3x2+ax-2-aex+ex(6x+a)=ex3x2+(a+6)x-2

∵ x=1 is a critical point ∴ f'(1)=0

∴3+a+6-2=0
a=-7

∴f'(x)=ex3x2-x-2=ex3x2-3x+2x-2=ex(3x+2)(x-1)

∴ maxima at x=-2 3∴ minima at x=1

Topic: Mathematics, Application of Derivatives, JEE Main
If \( x^{2}+\alpha y^{2}+2 \beta y=a^{2} \) represents a pair of perpendicular lines, then \( \beta \) equals to

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

The correct answer is:
a

Solution:
Let given line be
x2+αy2+2β-a2=0
Here, a=1, b=α, h=0, g=0, f=β, c= -a2
Condition for perpendicular line a+b=0
∴ 1+α=0 ⇒= -1
Condition of pair of lines
abc+2fgh-af2-bg2-ch2=0
∴ 1×α×-a2+0-1β2-0--a2 0=0
⇒ -a2α-β2=0
⇒ β2= -αa2
⇒ β2=--1a2 ∵ α=-1
⇒ β2=a2⇒ β=a

Topic: Mathematics, Straight Lines, JEE Main
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, JEE Main
If \( x^{3} d y+x y \cdot d x=x^{2} d y+2 y d x ; y(2)=e \) and \( x>1 \), then \( y(4) \) is equal to :

Options:
A \( \frac{\sqrt{e}}{2} \)
B \( \frac{1}{2}+\sqrt{e} \)
C \( \frac{3}{2} \sqrt{e} \)
D \( \frac{3}{2}+\sqrt{e} \)

The correct answer is:
\( \frac{3}{2} \sqrt{e} \)

Solution:
Given x3dy+xy·dx=x2dy+2ydx

⇒x3-x2dy=2-xydx

⇒dyy=2-xx3-x2dx

Integrating both sides with respect to x, we get

∫dyy=∫2-xx2x-1dx+k ............i

Let 2-xx2x-1=Ax+Bx2+Cx-1

⇒2-x=Axx-1+Bx-1+Cx2

Putting x=0⇒2=-B⇒B=-2

Putting x=1⇒2-1=C⇒C=1

Putting x=2⇒2-2=A21+B1+C22⇒2A+2=0⇒A=-1

From equation i, we get

∫dyy=∫-1x+-2x2+1x-1dx+k

⇒lny=-lnx+2x+lnx-1+k ..........ii

Given y2=e i.e. at x=2, y=e

⇒lne=-ln2+22+ln2-1+k

⇒1=-ln2+1+0+k

⇒k=ln2

Putting in equation ii, we get

lny=-lnx+2x+lnx-1+ln2

Now putting x=4, we get

lny=-ln4+24+ln4-1+ln2

⇒lny=-2ln2+12+ln3+ln2

⇒lny=-ln2+12+ln3

⇒lny=12+ln32

⇒ln2y3=12

⇒2y3=e12

⇒y=32e

⇒y4=32e

Topic: Mathematics, Differential Equations, JEE Main
If \( x=e^{t} \sin t, y=e^{t} \cos t, t \) is a parameter, then \( \frac{d^{2} y}{d x^{2}} \) at \( (1,1) \) is equal to

Options:
A \( -\frac{1}{2} \)
B \( -\frac{1}{4} \)
C \( 0 \)
D \( \frac{1}{2} \)

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

Solution:
Given that, x=etsint,y=etcost …1

At point 1,1,1=etsint,1=etcost

tant=1 ⇒t=π4

On differentiating Equation 1 w.r.t. x, we get

dydt=etcost-sint

and dxdt=etsint+cost

∴ dydx=dydtdxdt=cost-sintcost+sint

Again differentiating w.r.t. x, we get

d2ydx2=dxdtcost-sintcost-sintdtdx

=cost+sint−sint−cost−cost−sint−sint+costcost+sint2dtdx

=-2cost+sint2.1etsint+cost

= -2etcost+etsint.1cost+sint2

=-2x+y.1cost+sint2 from Equation1

At t=π4,x=1, y=1

∴ d2ydx2=-21+1.1cosπ4+sinπ42

=-112+12=-12

Topic: Mathematics, Differentiation, JEE Main
If \(x=e^{y+e^{y+e^{y+\ldots}}}\), then \(\frac{d y}{d x}=\)

Options:
A \(\frac{1-x}{x}\)
B \(\frac{1}{x}\)
C \(\frac{x}{1+x}\)
D \(\frac{1+x}{x}\)

The correct answer is:
\(\frac{1-x}{x}\)

Solution:
It is given that,
\(\begin{array}{lll}
& x =e^{y+e^{y+e^{y+\ldots}}} \\
\Rightarrow & x =e^{y+x} \\
\Rightarrow & \log _e x =x+y \Rightarrow y=\log _e x-x
\end{array}\)
On differentiating both sides with respect to ' \(x\) ', we get
\(\frac{d y}{d x}=\frac{1}{x}-1=\frac{1-x}{x}\)
Hence, option (a) is correct.

Topic: Mathematics, Differentiation, JEE Main
If \( x=\operatorname{asec}^{2} \theta, y=\operatorname{atan}^{2} \theta \) then \( \frac{d^{2} y}{d x^{2}}= \)

Options:
A \( 11 \)
B \( 04 \)
C \( 2 a \)
D \( 00 \)

The correct answer is:
\( 00 \)

Solution:
\( (D) \)
\( \frac{d x}{d \theta}=2 a \sec ^{2} \theta \tan \theta \)
\( \frac{d y}{d \theta}=2 a \tan \theta \sec ^{2} \theta \)
\( \therefore \frac{d y}{d x}=\frac{d y}{d x}=\frac{2 a \tan \theta \sec ^{2} \theta}{2 a \tan \theta \sec ^{2} \theta}=1 \)
\( \therefore \frac{d^{2} y}{d x^{2}}=0 \)

Topic: Mathematics, Differentiation, JEE Main
If \( x^{y}=e^{x-y} \) then \( \frac{d y}{d x} \) is equal to

Options:
A \( \frac{\log x}{\log (x-y)} \)
B \( \frac{e^{x}}{x^{x-y}} \)
C \( \frac{\log x}{(1+\log x)^{2}} \)
D \( \frac{1}{y}-\frac{1}{x-y} \)

The correct answer is:
\( \frac{\log x}{(1+\log x)^{2}} \)

Solution:
Given that, \( x^{y}=e^{x-y} \)
Taking log both sides, we get
\[
\begin{array}{l}
y \log x=x-y \\
y(1+\log x)=x \\
\Rightarrow y=\frac{x}{(1+\log x)}
\end{array}
\]
\[
\begin{array}{l}
\text { So, } \frac{d y}{d x}=\frac{(1+\log x)-x\left(\frac{1}{x}\right)}{(1+\log x)^{2}} \\
=\frac{\log x}{(1+\log x)^{2}}
\end{array}
\]

Topic: Mathematics, Differentiation, JEE Main
If y1/4+y-1/4=2x, and x2-1d2ydx2+αxdy dx+βy=0, then |α-β| is equal to _______.

Options:

The correct answer is:
17

Solution:
y14+1y14=2x

y142-2xy14+1=0

y14=x+x2-1& x-x2-1

So, 141y34 dydx=1+xx2-1

141y34dydx=y14x2-1

dydx=4yx2-1 ...1

Hence

d2ydx2=4x2-1y'-xyx2-1x2-1

x2-1y"=4x2-1y'-xyx2-1

x2-1y'=4x2-1y'-xyx2-1

x2-1y"=44y-xy'4 1

x2-1y"+xy'-16y=0

So α-β=17

Topic: Mathematics, Differentiation, JEE Main
If y1/4+y-1/4=2x, and x2-1d2ydx2+αxdy dx+βy=0, then |α-β| is equal to _______.

Options:

The correct answer is:
17

Solution:
y14+1y14=2x

y142-2xy14+1=0

y14=x+x2-1& x-x2-1

So, 141y34 dydx=1+xx2-1

141y34dydx=y14x2-1

dydx=4yx2-1 ...1

Hence

d2ydx2=4x2-1y'-xyx2-1x2-1

x2-1y"=4x2-1y'-xyx2-1

x2-1y'=4x2-1y'-xyx2-1

x2-1y"=44y-xy'4 1

x2-1y"+xy'-16y=0

So α-β=17

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2021 (27 Aug Shift 1)
If y=1x , then the value of dy1+y4+dx1+x4+3 is equal to

Options:
A 0
B 3
C 4
D -3

The correct answer is:
3

Solution:
y=1x⇒dydx=-1x2⇒x2dy + dx=0
⇒x2dy1+x4+dx1+x4=0
⇒dy1x4+1+dx1+x4=0
⇒dy1+y4+dx1+x4=0
⇒dy1+y4+dx1+x4+3=3

Topic: Mathematics, Differentiation, JEE Main
If y=1+xy+sin-1sin2x, then dydx at x=0 is

Options:
A 0
B ln⁡2
C 1
D 12

The correct answer is:
1

Solution:
dydx=1+xyy1+x+ln⁡1+x⋅dydx+sin2x1-sin4x
Putting x=0 & y=1, we get,

dydx=11+ln⁡1.dydx+0=1

Topic: Mathematics, Differentiation, JEE Main
If $y=2^{\log x}$, then $\frac{d y}{d x}$ is

Options:
A $2^{\log x} \cdot \log 2$
B $\frac{2^{\log x}}{\log 2}$
C $\frac{2^{\log x} \log 2}{x}$
D $\frac{2^{\log x}}{x}$

The correct answer is:
$\frac{2^{\log x} \log 2}{x}$

Solution:
We have,
$$
\begin{gathered}
y=2^{\log x} \\
\therefore \quad \frac{d y}{d x}=2^{\log x} \cdot \log 2 \cdot \frac{d}{d x}(\log x)=\frac{2^{\log x} \log 2}{x}
\end{gathered}
$$

Topic: Mathematics, Differentiation, JEE Main
If y2+logecos2x=y, x∈-π2,π2 then :

Options:
A y''(0)=0
B y'(0)+y"(0)=1
C y"(0)=2
D y'(0)+y"(0)=3

The correct answer is:
y"(0)=2

Solution:
Putting, x=0 in y2+logecos2x=y we get y=0,1

2y·y'+1cos2x·2cosx-sinx=y'

⇒2y·y'-2tanx=y'……1

y'0=0 for y=0&y=1.

Differentiating 1, 2y·y''+2y'2-2sec2x=y'',

y''0=-2 for y=0

y''0=2, for y=1

∴y''0=2

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2020 (03 Sep Shift 1)
If y=2+sin⁡x+2+sin⁡x+2+sin⁡x+…∞, then the value of dydx at x=0 is

Options:
A 0
B 2
C 12
D 13

The correct answer is:
13

Solution:
Given equation can be rewritten as
y=2+sin⁡x+y
⇒y-22=sin⁡x+y
⇒y2-4y+4=sin⁡x+y…i
Putting x=0 in the equation i, we get,
y2-4y+4=0+y⇒y2-5y+4=0
⇒y-1y-4=0
y=1 or y=4
∵y>2⇒y=4
Now differentiating equation i with respect to x, we get,
dydx=cos⁡x2y-5
Putting x=0, y=4

dydx=cos⁡024-5=13

Topic: Mathematics, Differentiation, JEE Main
If $y=2 \sin x+3 \cos x$ and $y+A \frac{d^2 y}{d x^2}=B$, then the values of $A$, $\mathrm{B}$ are respectively

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

The correct answer is:
1,0

Solution:
$$
\begin{aligned}
& y=2 \sin x+3 \cos x \\
& \therefore \frac{d y}{d x}=2 \cos x-3 \sin x \\
& \therefore \frac{d^2 y}{d x^2}=-2 \sin x-3 \cos x=-(2 \sin x+3 \cos x)=-y \\
& \therefore y+\frac{d^2 y}{d x^2}=0
\end{aligned}
$$
We have $y+A \frac{d^2 y}{d x^2}=B \Rightarrow A=1, B=0$

Topic: Mathematics, Differentiation, JEE Main
If yα=2tanα+cotα1+tan2α+1sin2α, α∈3π4,π, then dydα at α=5π6 is

Options:
A 4
B 43
C -4
D -14

The correct answer is:
4

Solution:
Simplifying using Trigonometric Identity, we get,

yα=1+cotα

or, yα=-1+cotα, as α∈3π4,π

Hence, dydα=cosec2α

∴ dydαα=5π6=4

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2020 (07 Jan Shift 1)
If y=2πx-1cosecx is the solution of the differential equation, dydx+pxy=-2πcosecx, 0<x<π2, then the function px is equal to :

Options:
A cotx
B cosecx
C secx
D tanx

The correct answer is:
cotx

Solution:
y=2πx-1cosec x

⇒dydx=2πcosec x-2xπ-1cosecx cotx cot⁡x

⇒dydx+2xπ-1cosecx cotx=2πcosecx

⇒dydx+ycotx=2πcosecx

⇒px=cotx

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (06 Sep Shift 2)
If $y=2^{x / \ln x}$ then $\frac{d y}{d x}$ at $x=e$ is

Options:
A $\mathrm{e}$
B $2^{\mathrm{e}} \log 2$
C $\log 2$
D 0

The correct answer is:
0

Solution:
$\log y=\frac{x}{\log x} \log 2$
$\left.\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\right]_{\mathrm{x}=\mathrm{e}}=0$

Topic: Mathematics, Differentiation, JEE Main
If y+3x=0 is the equation of a chord of the circle x2+y2-30x=0, then the equation of the circle with this chord as diameter is :

Options:
A x2+y2+3x-9y=0
B x2+y2-3x+9y=0
C x2+y2+3x+9y=0
D x2+y2-3x-9y=0

The correct answer is:
x2+y2-3x+9y=0

Solution:
To find point of intersection of given chord y=-3x with the given circle x2+y2-30x=0 eliminate y from two equations
Point of intersection
⇒ x2+9x2-30x=0
⇒ 10x x-3=0
⇒ x=0 or x=3
Therefore, y = 0 and y =-9 are the corresponding ordinates.
Point of intersection (0, 0), (3, -9)
Diametric form of equation of circle,
x x-3+yy+9=0
⇒x2+y2-3x+9y=0 .

Topic: Mathematics, Circle, JEE Main, JEE Main 2015 (10 Apr Online)
If $y=4 x-5$ is a tangent to the curve $y^2=\mathrm{p} x^3+\mathrm{q}$ at $(2,3)$, then $\mathrm{p}-\mathrm{q}$ is

Options:
A -5
B 5
C 9
D -9

The correct answer is:
9

Solution:
$y^2=\mathrm{p} x^3+\mathrm{q}... (i)$
Differentiating both sides w.r.t. $x$, we get
$\begin{gathered}
2 y \cdot \frac{\mathrm{d} y}{\mathrm{~d} x}=3 \mathrm{p} x^2 \\
\Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{3 \mathrm{p}}{2}\left(\frac{x^2}{y}\right) \\
\therefore \quad\left(\frac{\mathrm{d} y}{\mathrm{~d} x}\right)_{(2,3)}=\frac{3 \mathrm{p}}{2} \times \frac{4}{3}=2 \mathrm{p}
\end{gathered}$
Slope of the line $y=4 x-5$ is 4 .
Since the line touches the curve, their slopes are equal.
$\therefore \quad 2 \mathrm{p}=4 \Rightarrow \mathrm{p}=2$
Since $(2,3)$ lies on $y^2=\mathrm{p} x^3+\mathrm{q}$.
$\begin{array}{ll}
\therefore & 9=2 \times 8+q \Rightarrow q=-7 \\
\therefore & p-q=2+7=9
\end{array}$

Topic: Mathematics, Application of Derivatives, JEE Main
If y=4x-5 is tangent to the curve y2=px3+q at (2, 3) then (p,q) is

Options:
A (2, 7)
B (-2,Â 7)
C (-2,Â -7)
D (2, -7)

The correct answer is:
(2, -7)

Solution:
Curve is y2=px3+q
â´ 2ydydx=3px2
â dydx(2,3)=3p.42.3
â 4=2p
â p=2
Also, curve is passing through (2, 3)
â´ 9=8p+q
â q=-7
â´ p, q is 2, -7

Topic: Mathematics, Application of Derivatives, JEE Main
If y=acos⁡(log⁡x)+bsin⁡(log⁡x), where a & b are parameters, then x2y"+xy'is equal to

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

The correct answer is:
-y

Solution:
y=acosâ¡(logâ¡x)+bsinâ¡(logâ¡x)
On differentiating w.r.t. x, we get
y ′ = −asin(logx) x + bcos(logx) x
⇒x y ′ =−asin(logx)+bcos(logx)
Again, on differentiating w.r.t. x, we get
x y ″ + y ′ =−acos(logx) 1 x −bsin(logx) 1 x
⇒ x 2 y ″ + y ′ x=− acos(logx)+bsin(logx ]
⇒ x 2 y ″ +x y ′ =−y

Topic: Mathematics, Differentiation, JEE Main
If $y(\theta)=\frac{2 \cos \theta+\cos 2 \theta}{\cos 3 \theta+4 \cos 2 \theta+5 \cos \theta+2}$, then at $\theta=\frac{\pi}{2}, y^{\prime \prime}+y^{\prime}+y$ is equal to :

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

The correct answer is:
2

Solution:
$\begin{aligned} & y=\frac{2 \cos \theta+2 \cos ^2 \theta-1}{4 \cos ^3 \theta-3 \cos \theta+8 \cos ^2 \theta-4+5 \cos \theta+2} \\ & y=\frac{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)}{\left(2 \cos ^2 \theta+2 \cos \theta-1\right)(2 \cos \theta+2)} \\ & y=\frac{1}{2}\left(\frac{1}{1+\cos \theta}\right) \\ & \Rightarrow \theta=\frac{\pi}{2} \quad y=\frac{1}{2} \\ & y^{\prime}=\frac{1}{2}\left(\frac{-1}{(1+\cos \theta)^2} \times(-\sin \theta)\right) \\ & \Rightarrow \theta=\frac{\pi}{2} \quad y=\frac{1}{2}\end{aligned}$
$\begin{aligned} y^{\prime \prime} & =\frac{1}{2}\left[\frac{\cos \theta(1+\cos \theta)^2-\sin \theta(2)(1+\cos \theta)(-\sin \theta)}{(1+\cos \theta)^4}\right] \\ & \Rightarrow \theta=\frac{\pi}{2} \quad y=1\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2024 (05 Apr Shift 2)
If $y^{\cos x}=x^{\sin y}$, then $\frac{d y}{d x}=$

Options:
A $\frac{y(x \sin x \log y+\sin y)}{x(\cos x-y \log x \cos y)}$
B $\frac{y(x \sin x \log x-\sin y)}{x(\cos x+y \log x \cos y)}$
C $\frac{y(\sin y-x \log y)}{x(x-y \cos y(\log x))}$
D $\frac{y(\sin y+x \log y)}{x(x+y \cos y(\log x))}$

The correct answer is:
$\frac{y(x \sin x \log y+\sin y)}{x(\cos x-y \log x \cos y)}$

Solution:
It is given that
$y^{\cos x}=x^{\sin y}$
On taking logrithm both sides, we get $\cos x \log y=\sin y \log x$ On differentiating both side w.r.t. $x$, we get $\frac{1}{y}(\cos x) \frac{d y}{d x}-(\sin x) \log y$
$=\frac{1}{x} \sin y+(\log x)(\cos y) \frac{d}{d x}$
$\begin{aligned} & \Rightarrow \frac{d y}{d x}\left(\frac{1}{y} \cos x-(\log x) \cos y\right)=\frac{1}{x} \sin y+(\sin x) \log y \\ & \Rightarrow \frac{d y}{d x}=\frac{y(\sin y+x \sin x \log y)}{x(\cos x-y \log x \cos y)}\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
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, JEE Main
If $y=\cos ^{2} \frac{3 x}{2}-\sin ^{2} \frac{3 x}{2}$, then $\frac{d^{2} y}{d x^{2}}$ is

Options:
A $-3 \sqrt{1-y^{2}}$
B $9 y$
C $-9 y$
D $3 \sqrt{1-y^{2}}$

The correct answer is:
$-9 y$

Solution:
Given, $y=\cos ^{2} \frac{3 x}{2}-\sin ^{2} \frac{3 x}{2}$
$$
\begin{aligned}
&\Rightarrow \quad y=\cos ^{2} \frac{3 x}{2}-\left(1-\cos ^{2} \frac{3 x}{2}\right) \\
&\Rightarrow \quad y=2 \cos ^{2} \frac{3 x}{2}-1
\end{aligned}
$$
$$
\begin{aligned}
&\Rightarrow \frac{d y}{d x}=2-2 \cos \frac{3 x}{2}\left(-\sin \frac{3 x}{2}\right)\left(\frac{3}{2}\right) \\
&\Rightarrow \frac{d y}{d x}=-6 \cos \frac{3 x}{2} \sin \frac{3 x}{2} \\
&\Rightarrow \frac{d^{2} y}{d x^{2}}=-6\left[\cos \frac{3 x}{2}\left(\cos \frac{3 x}{2}\right) \cdot \frac{3}{2}-\sin \frac{3 x}{2}\right. \\
&\Rightarrow \frac{d^{2} y}{d x^{2}}=-9\left[\cos ^{2} \frac{3 x}{2}-\sin ^{2} \frac{3 x}{2}\right] \\
&\Rightarrow \frac{d^{2} y}{d x^{2}}=-9 y
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If $\mathrm{y}=\cos ^{2}\left(\frac{5 x}{2}\right)-\sin ^{2}\left(\frac{5 x}{2}\right)$, then $\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{d} x^{2}}\right)=$

Options:
A $-5 \sqrt{1-y^{2}}$
B $5 \sqrt{1-y^{2}}$
C $25 y$
D $-25 y$

The correct answer is:
$-25 y$

Solution:
Given
$$
\begin{aligned}
y &=\cos ^{2}\left(\frac{5 x}{2}\right)-\sin ^{2}\left(\frac{5 x}{2}\right) \\
y &=\cos \left(2 \times \frac{5 x}{2}\right) \Rightarrow y=\cos 5 x \\
\therefore \frac{d y}{d x} &=-5 \sin 5 x \Rightarrow \frac{d^{2} y}{d x^{2}} \\
&=-25 \cos 5 x=-25 y
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If y=cos⁡xcos⁡2xcos⁡4xcos⁡8x, then dydx at x=π2 is

Options:
A 1
B 0
C 4
D 16

The correct answer is:
1

Solution:
y=sin⁡16x16sin⁡x ∵ cos⁡Acos⁡2A.........cos⁡2n-1A=sin2nA2nsin⁡A
So, dydx=116sin⁡x⋅16cos⁡16x-sin⁡16xcos⁡xsin⁡x2
Putting x=π2, we get,
dydx=116⋅1.161-012=1

Topic: Mathematics, Differentiation, JEE Main
If $y \cos x+x \cos y=\pi$, then $y^{\prime \prime}(0)$ is

Options:
A $1$
B $\pi$
C $0$
D $-\pi$

The correct answer is:
$\pi$

Solution:
$y \cos x+x \cos y=\pi$
Differentiate both sides with respect to $x$, we get
$-y \sin x+\cos x \cdot y^{\prime}+x(-\sin y) y^{\prime}+\cos y$
Again differentiate with respect to $x$
$\begin{aligned} &-y^{\prime \prime} \sin x-y \cos x+\cos x \cdot y^{\prime \prime}+\sin x \cdot y^{\prime}-\sin y \cdot y^{\prime} \\ &-x\left[\cos y \cdot\left(y^{\prime}\right)^2+\sin y \cdot y^{\prime \prime}\right]-\sin y \cdot y^{\prime}\end{aligned}$
Putting $x=0$, we get $-y+y^{\prime \prime}-2 \sin y y^{\prime}=0$
$y^{\prime \prime}=y+2 y^{\prime} \sin y$
Since at $x=0, y=\pi ;\left(y^{\prime \prime}\right)_0=\pi$.

Topic: Mathematics, Differential Equations, JEE Main
If ydy dx=xy2x2+ϕy2x2ϕ'y2x2, x>0, ϕ>0, and y(1)=-1, then ϕy24 is equal to:

Options:
A 2ϕ1
B ϕ1
C 4ϕ2
D 4ϕ1

The correct answer is:
4ϕ1

Solution:
Given:yxdydx=y2x2+ϕy2x2ϕ'y2x2 ....1

Let yx=t

⇒y=xt

⇒dydx=t+x·dtdx

∴tt+xdtdx=t2+ϕt2ϕ't2

⇒xtdtdx=ϕt2ϕ't2

⇒t·ϕ't2ϕt2dt=1xdx
Integrating both sides

∫t·ϕ't2ϕt2dt=∫1xdx

Let ϕt2=p

⇒ϕ't2.2t=dp

⇒12∫1pdp=∫1xdx

⇒12lnp=lnx+C

⇒12lnϕt2=lnx+C

⇒12lnϕy2x2=lnx+C ...2

If x=1, y=-1 then C=12lnϕ1

Substituting value of C in 2

12lnϕy2x2=lnx+12lnϕ1

⇒lnϕy2x2=lnx2+lnϕ1

If x=2 then

lnϕy24=ln4+lnϕ1

SO, ϕy24=4ϕ1

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (31 Aug Shift 2)
If $y=e^{4 x} \cos 5 x$, then $\frac{d^{2} y}{d x^{2}}$ at $x=0$ is

Options:
A $-9$
B 9
C 8
D $-8$

The correct answer is:
$-9$

Solution:
Given $y=e^{4 x} \cos 5 x$
$\begin{aligned}
\therefore & \frac{\mathrm{dy}}{\mathrm{dx}}=\mathrm{e}^{4 \mathrm{x}}(-5 \sin 5 \mathrm{x})+\cos 5 \mathrm{x}\left(4 \mathrm{e}^{4 \mathrm{x}}\right) \quad=\mathrm{e}^{4 \mathrm{x}}(-5 \sin 5 \mathrm{x}+4 \cos 5 \mathrm{x}) \\
& \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}=\left[\mathrm{e}^{4 \mathrm{x}}(-25 \cos 5 \mathrm{x}-20 \sin 5 \mathrm{x})\right]+\left[(-5 \sin 5 \mathrm{x}+4 \cos 5 \mathrm{x}) \cdot\left(4 \mathrm{e}^{4} \mathrm{x}\right)\right] \\
&\left(\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}^{2}}\right)_{\mathrm{x}=0}=-25+(4 \times 4)=-25+16=-9
\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
If y=emsin-1â¡x and 1-x2dydx2=Ay2, then A= ________

Options:
A m
B –m
C m2
D -m2

The correct answer is:
m2

Solution:
Given, y=emsin-1â¡Â x
âÂ dydx=emsin-1â¡Â x⋅m⋅11-x2
âÂ dydx2=m2y21-x2
â´Â Â 1-x2dydx2=m2y2
â´Â Â A=m2

Topic: Mathematics, Differential Equations, JEE Main
If y=enx, then d2ydx2.d2xdy2 is equal to :

Options:
A ne-nx
B -ne-nx
C nenx
D 1

The correct answer is:
-ne-nx

Solution:
y=enx ⇒ dydx=nenx ∵ddxeax=aeax
⇒ d2ydx2=n2enx
dydx=nenx⇒dxdy=1nenx=1ne-nx
d2xdy2=ddydxdy
=ddxdxdy·dxdy
=ddx1ne-nx·e-nxn
=1n-ne-nx·e-nxn
=-1ne-2nx
⇒ d2ydx2.d2xdy2=n2enx×-e-2nxn
=-ne-nx

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2014 (09 Apr Online)
If $y=f(x)$ makes $+v e$ intercept of 2 and 0 unit on $x$ and $y$ axes and encloses an area of $3 / 4$ square unit with the axes then $\int_0^2 x f^{\prime}(x) d x ~~is$

Options:
A
3/2

B
1

C
5/4

D
-3/4

The correct answer is:
-3/4

Solution:
We have $\int_0^2 f(x) d x=\frac{3}{4} ;$ Now, $\int_0^2 x f^{\prime}(x) d x=x \int_0^2 f^{\prime}(x) d x-\int_0^2 f(x) d x$
$=[x f(x)]_0^2-\frac{3}{4}=2 f(2)-\frac{3}{4}=0-\frac{3}{4}(\because f(2)=0)=-\frac{3}{4}$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2002
If y=fx satisfies the conditions of Rolle's theorem in 2,6, then ∫ 2 6 f ′ x dx is equal to

Options:
A 2
B 0
C 4
D 6

The correct answer is:
0

Solution:
Since fx satisfies the conditions of Rolle's Theorem so f2=f6
Now, given integral =fx26=f6-f2=0

Topic: Mathematics, Application of Derivatives, JEE Main
If y=fx satisfies the differential equation dydx+2x1+x2y=3x21+x2 where f1=1, then f2 is equal to

Options:

The correct answer is:
1.8

Solution:
The given differential equation is a linear differential equation.

Integrating factor =e∫2x1+x2dx=1+x2
The solution of the differential equation is y1+x2=∫3x2dx=x3+C

∵y1=1⇒C=1
Hence, y=fx=x3+1x2+1

So, f2=95=1.8

Topic: Mathematics, Differential Equations, JEE Main
If y=fx satisfies the differential equationdydx+y=sinâ¡x+cosâ¡x and f0=2, then limxâââ¡fx-sinâ¡x is equal to

Options:
A e
B 2
C 0
D 1

The correct answer is:
0

Solution:
Given equation is a linear differential equation
Integrating factor =e∫dx=ex
So the equation is yex=∫exsin⁡x+cos⁡xdx=exsin⁡x+C
⇒fx=sin⁡x+Ce-x
Since f0=2⇒C=2
Hence, fx=sin⁡x+2e-x⇒fx-sin⁡x=2e-x
Therefore, limx→∞⁡fx-sin⁡x=0

Topic: Mathematics, Differential Equations, JEE Main
If y=lnxx, then d2ydx2=2x3lnx-λ, where λ is

Options:
A 32
B 23
C 12
D 13

The correct answer is:
32

Solution:
Using quotient rule, we get,
dydx=1-lnxx2
∴d2ydx2=x2-1x-1-lnx2xx4
=-1-2+2lnxx3
=2x3lnx-1+12

Topic: Mathematics, Differentiation, JEE Main
If $y=\log _{\cos x} \sin x$, then $\frac{d y}{d x}$ is equal to

Options:
A $\frac{(\cot x \log \cos x+\tan x \log \sin x)}{(\log \cos x)^{2}}$
B $\frac{(\tan x \log \cos x+\cot x \log \sin x)}{(\log \cos x)^{2}}$
C $\frac{(\cot x \log \cos x+\tan x \log \sin x)}{(\log \sin x)^{2}}$
D None of the above

The correct answer is:
$\frac{(\cot x \log \cos x+\tan x \log \sin x)}{(\log \cos x)^{2}}$

Solution:
Given, $y=\log _{\cos x} \sin x=\frac{\log \sin x}{\log \cos x}$
On differentiating w.r.t. $x$, we get
$$
\frac{d y}{d x}=\frac{\cot x \cdot \log \cos x+\tan x \cdot \log \sin x}{(\log \cos x)^{2}}
$$

Topic: Mathematics, Differentiation, JEE Main
If y=log10⁡x+logx⁡10+logx⁡x+log10⁡10, then dydx is equal to

Options:
A 1xloge⁡10-loge⁡10xloge⁡x2
B 1xloge⁡10-1xlog10⁡e
C 1xloge⁡10-loge⁡10xloge⁡x
D None of these

The correct answer is:
1xloge⁡10-loge⁡10xloge⁡x2

Solution:
y=log10⁡x+logx⁡10+logx⁡x+log10⁡10
=log10⁡x+loge⁡10logex+1+1
⇒ dydx=1xlog10⁡e-loge⁡10xloge⁡x2

Topic: Mathematics, Differentiation, JEE Main