datavorous/entrance-exam-dataset huggingface dataset:
Let \( X_{1}, X_{2}, X_{3} \ldots \) are in arithmetic progression with a common difference equal to \( d \) which is a two digit natural number.
\( y_{1}, y_{2}, y_{3} \ldots \) are in geometric progression with common ratio equal to \( 16 \). Arithmetic mean of \( X_{1}, X_{2} \ldots X_{n} \) is equal to
the arithmetic mean of \( y_{1}, y_{2} \ldots y_{n} \) which is equal to \( 5 \). If the arithmetic mean of \( X_{6}, X_{7} \ldots X_{n+5} \) is equal to the
arithmetic mean of \( y_{\mathrm{P}+1}, y_{\mathrm{P}+2} \ldots y_{\mathrm{P}+n} \), then \( d \) is equal to

Options:

The correct answer is:
15

Solution:
Mean X1,X2…Xn=n22X1+n-1dn=5

Mean of y1,y2....yn=y116n-115n=5

2X1+n-1d=10 ...1

y116n-1=75n ...2

Mean of X6,X7....Xn+5= Mean of yP+1,yP+2,yP+n

n22X6+n-1dn=yP+116n-115n=y116P16n-115n

X6+n-12d=16P75n15n=5×16P

X1+5d+n-12d=5×16P

5-n-12d+5d+n-12d=5×16P⇒d=16P-1

∵d is 2 digit natural number

⇒P=1, d=15

Topic: Mathematics, Sequences and Series, JEE Main
Let \( x^{2}+y^{2}-4 x-2 y-11=0 \) be a circle. A pair of tangents from the point \( (4,5) \) with a pair of radii form a quadrilateral of the area:

Options:
A 8 units
B 12 units
C 14 units
D 16 units

The correct answer is:
8 units

Solution:
Length of tangents from a point ( x 1 , y 1 ) to a circle x 2 + y 2 +2gx+2fy+c=0 is given by
x 1 2 + y 1 2 +2g x 1 +2f y 1 +c
The equation of the circle is, x 2 + y 2 −4x−2y−11=0
∴ its centre is ( 2,1 ) ,
and radius = 4+1+11 =4=BC
Length of the tangent from the point A(4, 5) is
= 16+25−16−10−11 = 4 =2=AB
∴ Area of quadrilateral ABCD is

=2(Area ΔABC)=2×12AB×BC

=2×12×2×4=8 square units

Topic: Mathematics, Circle, JEE Main
Let $y$ be an implicit function of $x$ defined by $x^{2 x}-2 x^x \cot y-1=0$. Then $y^{\prime}(1)$ equals

Options:
A
$-1$

B
1

C
$\log 2$

D
$-\log 2$

The correct answer is:
$-1$

Solution:
$x^{2 x}-2 x^x \cot y-1=0$
Now $x=1$,
$$
1-2 \cot y-1=0 \Rightarrow \cot y=0 \Rightarrow y=\frac{\pi}{2}
$$
Now differentiating eq. (1) w.r.t. ' $x$ '
$$
2 x^{2 x}(1+\log x)-2\left[x^x\left(-\operatorname{cosec}^2 y\right) \frac{d y}{d x}+\cot y x^x(1+\log x)\right]=0
$$
Now at $\left(1, \frac{\pi}{2}\right)$
$$
\begin{aligned}
& 2(1+\log 1)-2\left(1(-1)\left(\frac{d y}{d x}\right)_{\left(1, \frac{\pi}{2}\right)}+0\right)=0 \\
& \Rightarrow 2+2\left(\frac{d y}{d x}\right)_{\left(1, \frac{\pi}{2}\right)}=0 \Rightarrow\left(\frac{d y}{d x}\right)_{\left(1, \frac{\pi}{2}\right)}=-1
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2009
Let $y$ be the function which passes through $(1,2)$ having slope $(2 x+1)$. The area bounded between the curve and $x$-axis is

Options:
A $6$ sq. unit
B $5/6$ sq. unit
C $1 / 6$ sq. unit
D None of these

The correct answer is:
$1 / 6$ sq. unit

Solution:
$\begin{aligned} \frac{d y}{d x}=2 x+1 \Rightarrow & y=x^2+x+c \\ \Rightarrow y=x^2+x, & {[\boxtimes \quad c=0 \text { by putting } x=1, y=2) } \\ & \quad\left(x+\frac{1}{2}\right)^2=y+\frac{1}{4}, \text { which is a equation of parabola, whose vertices is, } V\left(\frac{-1}{2}, \frac{-1}{4}\right)\end{aligned}$
$\begin{aligned}& =\left|\int_{-1}^0\left(x^2+x\right) d x\right|=\left(\frac{x^3}{3}+\frac{x^2}{2}\right)_{-1}^0 \\
& =\left|\frac{-1}{3}+\frac{1}{2}\right|=\frac{1}{6}\end{aligned}$ sq. unit.

Topic: Mathematics, Area Under Curves, JEE Main
Let y=fx be a solution of the differential equation dydx=y2-x22xy ∀x,y>0. If f1=2, then f ′ 1 is equal to

Options:
A 2
B 52
C 54
D 34

The correct answer is:
34

Solution:
Given, dydx=y2-x22xy

Let y=vx
Then, v+xdvdx=v2-12v⇒xdvdx=v2-1-2v22v
⇒xdvdx=-12v2+1v
So, 2vv2+1dv+dxx=0

⇒xv2+1=C⇒x2+y2=Cx

Since, f1=2

⇒C=5
i.e. fx=5x−x2

⇒f'x=5−2x25x−x2⇒f'1=34

Topic: Mathematics, Differential Equations, JEE Main
Let y=fx be the solution of the differential equation dydx=-2xy-1 with f0=1, then limxââfx is equal to

Options:
A 12
B 0
C e
D 1

The correct answer is:
1

Solution:
dydx=-2xy-1
dydx+2xy=2x
Ι.F.=e∫2xdx=ex2
y⋅ex2=∫ex22xdx
y⋅ex2=ex2+C
∵y0=1⇒1=1+C⇒C=0
Hence, the solution is yex2=ex2
⇒y=1=fx
limx→∞fx=1

Topic: Mathematics, Differential Equations, JEE Main
Let y=f(x) be the solution of the differential equation y(x+1)dx-x2dy=0, y(1)=e. Then limx→0+f(x) is equal to

Options:
A 0
B

1e

C e2
D 1e2

The correct answer is:
0

Solution:
Given,

y(x+1)dx-x2dy=0

⇒x+1x2dx=dyy

⇒1x+1x2dx=dyy

Now integrating both side we get,

logex-1x=logey+c

Now on using y1=e we get, c=-2

So, the equation of curve becomes logex-1x=logey-2

⇒y=elnx-1x+2

Hence, limx→0+elnx-1-1x+2=e-∞=0.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (29 Jan Shift 1)
Let y=fx represent a parabola with focus -12,0 and directrix y=-12. Then S=x∈ℝ:tan-1fx+sin-1fx+1=π2:

Options:
A contains exactly two elements
B contains exactly one element
C is an infinite set
D is an empty set

The correct answer is:
contains exactly two elements

Solution:
Equation of parabola with focus (-12, 0) and directrix y=-12 is:

x+122=y+14

⇒y=x2+x

Put f(x)=y=x+x2 in S,

tan-1xx+1+sin-1x2+x+1=π2

⇒0≤x2+x+1≤1

⇒x2+x≤0 …1

Also,

x2+x≥0 …2

By 1 & 2, we have

x2+x=0⇒x=0,-1

S contains 2 element.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Let y=fx satisfies dydx=x+yx and fe=e, then the value of f1 is

Options:

The correct answer is:
0

Solution:
dydx=1+yx
dydx-yx=1
If =e-∫1xdx=1x
yx=∫1xdx⇒y=xlnx+cx=fx
fe=e+ce=e⇒c=0
fl=0

Topic: Mathematics, Differential Equations, JEE Main
Let y=fx=sin3π3cosπ32-4x3+5x2+132. Then, at x=1,

Options:
A 2y'+3π2y=0
B 2y'+3π2y=0
C 2y'-3π2y=0
D y'+3π2y=0

The correct answer is:
2y'+3π2y=0

Solution:
The given equation can be written as

y=sin3π3cosgx

where,

gx=π32-4x3+5x2+13/2

⇒g'x=π22-4x3+5x2+11/2-12x2+10x

⇒g'1=π222-2=-π

And,

g1=2π3=π-π3

Now,

y'=3sin2π3cosgx×cosπ3cosgx×π3-singxg'x

y'1=3sin2-π6·cosπ6·π3-sin2π3g'1

y'1=34·32·π3-32-π=3π216

y1=sin3π3cos2π3=-18

Therefore,

2y'1+3π2y1=0

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2023 (31 Jan Shift 1)
Let y=loge1-x21+x2,-1<x<1. Then at x=12, the value of 225y'-y" is equal to

Options:
A 732
B 746
C 742
D 736

The correct answer is:
736

Solution:
Let, fx=log1-x21+x2

⇒fx=log1-x2-log1+x2

⇒f'x=-2x1-x2-2x1+x2

⇒f'x=-2x1+x2+1-x21-x4

⇒f'x=4xx4-1

⇒f"x=x4-14-4x4x3x4-12

⇒f"x=4-3x4-1x4-12

⇒225f'x-f"x=2254xx4-1-4-3x4-1x4-12

Putting, x=12

⇒225f'12-f"12=2252-1516-4-316-1116-12

⇒225f'12-f"12=225-3215-4-1916-15162

⇒225f'12-f"12=225-3215+4×19×16225

⇒225f'12-f"12=-480+1216=736

⇒225f'12-f"12=736

⇒225y'12-y"12=736

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2024 (29 Jan Shift 2)
Let y=px be the parabola passing through the points –1, 0, 0, 1 and 1, 0. If the area of the region x, y:x+12+y-12≤1, y≤px is A, then 12π-4A is equal to ________ .

Options:

The correct answer is:
16

Solution:
Let, the parabola be y=ax2+bx+c,

Now given parabola is passing through -1,0, 0,1 & 1,0

Now putting the value in the above equation of parabola we get,

a-b+c=0 ......1, c=1 ....2 and a+b+c=0 ....3

Now solving above three equations we get,

Parabola as y=1-x2

Now area of region between y=1-x2 and x+12+y-12≤1 is given by,

Now from above diagram, required area will be,

A=∫-10-x2+1-x+12dx

⇒A=π4-13

Hence, 12(π-4A)=16

Topic: Mathematics, Area Under Curves, JEE Main, JEE Main 2023 (10 Apr Shift 1)
Let y = sin - 1 sin 8 - tan - 1 tan 1 0 + cos - 1 cos 1 2 - sec - 1 sec 9 + cot - 1 cot 6 - cosec - 1 cosec 7 . If y simplifies to a Ï + b then a-b is

Options:

The correct answer is:
53

Solution:
y = sin - 1 sin 8 - tan - 1 tan 1 0 + cos-1cos12
- sec - 1 sec 9 + cot - 1 cot 6 - cosec - 1 cosec 7
y = 3 Ï - 8 - 1 0 - 3 Ï + 4 Ï - 12 - 9 - 2 Ï + 6 - Ï - 7 - 2 Ï
y = 1 3 Ï - 4 0
a=13 and b=-40
a-b=53

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
Let yx=1+x1+x21+x41+x81+x16. Then y'-y" at x=-1 is equal to

Options:
A 976
B 464
C 496
D 944

The correct answer is:
496

Solution:
Given,

Function yx=1+x1+x21+x41+x81+x16

Now multiplying and dividing by 1-x both side we get,

1-xyx=1-x1+x1+x21+x41+x81+x16

⇒y1-x=1-x32

⇒y-xy=1-x32 ......(1)

Now differentiating equation 1 on both sides with respect to x we get,

y'-xy'-y=-32x31.......(2)

Now at x=-1

⇒y'(-1)=16

Now differentiating equation 2 with respect to x on both sides we get,

⇒y"-xy"-y'-y'=-3231x30

Now at x=-1

⇒y''(-1)=-480

So, at x=-1, y'-y"=16-(-480)=496

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Let y=x+2, 4y=3x+6 and 3y=4x+1 be three tangent lines to the circle (x-h)2+(y-k)2=r2. Then h+k is equal to :

Options:
A 5
B 5(1+2)
C 6
D 52

The correct answer is:
5

Solution:
Lines y=x+2, 4y=3x+6 and 3y=4x+1 are tangents to the circle (x-h)2+(y-k)2=r2.

Centre of the circle is h,k.

Equation of bisector of lines 4y=3x+6, 3y=4x+1 is:

4x-3y+15=±3x-4y+65

⇒4x-3y+1=±3x-4y+6

Taking positive sign, we get

4x-3y+1=3x-4y+6

⇒x+y=5

Since, centre h,k lies on the bisector, therefore

h+k=5

Topic: Mathematics, Circle, JEE Main, JEE Main 2023 (30 Jan Shift 1)
Let y=x3-6x2+9x+1 be an equation of a curve, then the x-intercept of the tangent to this curve whose slope is least, is

Options:

The correct answer is:
3

Solution:
dydx=3x2-4x+3=3x-22-1â¥-3
So minimum value is -3 when x=2
at x=2,y=3
Hence, equation of tangent is 3x+y=9
So, x -intercept =3

Topic: Mathematics, Application of Derivatives, JEE Main
Let y=x3-6x2+9x+1 be the equation of a curve, then the x-intercept of the tangent to this curve whose slope is least, is

Options:

The correct answer is:
3

Solution:
dydx=3x2-4x+3=3x-22-1â¥-3
So, minimum value is -3 when x=2
at x=2,y=3
Hence, equation of tangent is 3x+y=9
So, x -intercept =3

Topic: Mathematics, Application of Derivatives, JEE Main
Let $y(x)$ be a solution of $\frac{(2+\sin x}{(1+y)} \frac{d y)}{d x}=\cos x$. If $y(0)=2$, then $y\left(\frac{\pi}{2}\right)$ equals

Options:
A
$\frac{5}{2}$

B
2

C
$\frac{7}{2}$

D
3

The correct answer is:
$\frac{7}{2}$

Solution:
Given differential equation is
$$
\frac{(2+\sin x}{(1+y)} \cdot \frac{d y}{d x}=\cos x
$$
which can be rewritten as
$$
\frac{d y}{1+y}=\frac{\cos x}{2+\sin x} d x
$$
Integrate both the sides, we get
$$
\begin{aligned}
& \int \frac{d y}{1+y}=\int \frac{\cos x d x}{2+\sin x} \\
& \Rightarrow \log (1+y)=\log (2+\sin x)+\log C \\
& \Rightarrow 1+y=C(2+\sin x) \\
& \text { Given } y(0)=2 \\
& \Rightarrow 1+2=C[2+\sin 0] \Rightarrow C=\frac{3}{2}
\end{aligned}
$$
Now, $y\left(\frac{\pi}{2}\right)$ can be found as
$$
1+y=\frac{3}{2}\left(2+\sin \frac{\pi}{2}\right) \Rightarrow 1+y=\frac{9}{2}
$$
$$
\Rightarrow \quad y=\frac{7}{2}
$$
Hence, $y\left(\frac{\pi}{2}\right)=\frac{7}{2}$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2012 (07 May Online)
Let y(x) be the solution of the differential equation 2x2 dy+ey-2xdx=0, x>0. If y(e)=1, then y(1) is equal to:

Options:
A loge(2e)
B loge2
C 2
D 0

The correct answer is:
loge2

Solution:
dydx=-ey2x2+1x

e-ydydx=e-yx+-12x2

-e-ydydx+e-yx=12x2

Let e-y=t ...i

e-y-1dydx=dtdx

dtdx+tx=12x2

I.F=e∫1xdx=elnx=x

tx=∫12x2·xdx+C

Using equation 1
e-yx=12ℓnx+C

Given, ye=1

⇒e-1e=12+C⇒C=12

e-yx=12(1+ℓnx)

Put x=1 then y is ⇒y=ℓn2 or loge2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (26 Aug Shift 2)
Let y(x) be the solution of the differential equation (xlogx)dydx+y=2xlogx, (x≥1). Then y(e) is equal to

Options:
A 2e
B e
C 0
D 2

The correct answer is:
2

Solution:
Given, (xlogx)dydx+y=2xlogx

⇒dydx+1xlogxy=2
Integrating factor I.F=e∫Pdx=e∫1xlogxdx=e∫1/xlogxdx=eloglogx=logx

The solution of the equation becomes

y·e∫Pdx=∫Q·e∫Pdxdx+c, where, c is the constant of integration.

⇒ylog x=∫2log xdx+c

⇒ylog x=2x log x-x+c

Let, P1,y1 be any point on the curve.

⇒y1·0=20-1+c⇒c=2

⇒ylogx=2xlogx-x+2

Now, putting x=e, we get, y=2.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2015 (04 Apr)
Let yx is the solution of the differential equation x+2dydx-x+1y=2. If y0=-1, then the value of y2 is equal to

Options:
A e2+12
B e2-12
C 12-e2
D e2

The correct answer is:
e2-12

Solution:
Given differential equation is dydx-x+1x+2y=2x+2
I.F. =e-∫x+1x+2dx=e-xx+2
So the solution of the equation is yx+2e-x=-2e-x+C
Since y0=1, we have C=4

So y2=e2-12

Topic: Mathematics, Differential Equations, JEE Main
Let y=xlogex. If the value of dydx at x=e4 is k, then the value of 4e3k is
(use e=2.7)

Options:

The correct answer is:
13.5

Solution:
dydx=12xlogexddxxlogex
=12xlogexx×1x+1×logex
⇒dydxx=e4=12e4×41+4=54e2∵logee=1

Hence, 4e3k=4e354e2=5e=13.5

Topic: Mathematics, Differentiation, JEE Main
Let y=xxxâ¦ â , then dydx is equal to (given x>0 )

Options:
A yxy-1
B y2x(1-ylnx)
C yx(1+ylnx)
D None of these

The correct answer is:
y2x(1-ylnx)

Solution:
Given, y= x x x...∞ then y= x y
Taking ln on both the sides, we get
lny=ylnx
⇒ 1y.dydx=y.1x+dydx.lnx
⇒dydx1y-ln⁡x=yx
⇒ dydx=y2x(1-ylnx)

Topic: Mathematics, Differentiation, JEE Main
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 Main
Let y=y1x and y=y2x be the solution curves the differential equation dydx=y+7 with initial conditions y10=0 and y20=1 respectively. Then the curves y=y1x and y = y2x intersect at

Options:
A no point
B two points
C one point
D infinite number of points

The correct answer is:
no point

Solution:
We have been given thatdydx=y+7⇒dyy+7=dx

⇒ln|y+7|=x+c

⇒|y+7|=k.ex

⇒y=k·ex-7

⇒y10=0⇒k=7⇒y1x=7ex-1

⇒y20=1⇒1=k-7⇒8=k

⇒y2x=8ex-7

⇒y1x=y2x⇒8ex-7=7ex-7

Hence, No point of intersection.

This is the required option.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (13 Apr Shift 1)
Let y=y1x and y=y2x be two distinct solutions of the differential equation dydx=x+y, with y10=0 and y20=1 respectively. Then, the number of points of intersection of y=y1x and y=y2x is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
Given,

dydx=x+y⇒dydx-y=x

So, IF=e∫-1dx=e-x

Now solution is given by ye-x=∫xe-xdx

⇒ye-x=-xe-x-e-x+c

⇒y=-x-1+cex

Given, y10=0⇒c=1

∴ y1=-x-1+ex ⋯1

Also given y20=1⇒c=2

∴ y2=-x-1+2ex ⋯2

Now from equation 1 & 2 we get, y2-y1=ex>0 ∴ y2≠y1

∴ Number of points of intersection of y1 and y2 is zero.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (27 Jul Shift 1)
Let y=yt be a solution of the differential equation dydt+αy=γe-βt Where, α>0,β>0 and γ>0. Then Limt→∞ yt

Options:
A is 0
B does not exist
C is 1
D is -1

The correct answer is:
is 0

Solution:
Given,

dydt+αy=γ·e-βt

Which is a linear differential equation,

So, integrating factor will be I.F=e∫αdt=eαt

So, the solution of differential equation is given by

y×eαt=γ∫e-βt×eαt

⇒y×eαt=γ∫eα-βt

⇒y×eαt=γeα-βtα-β+c

⇒y=γe-βtα-β+ce-αt

Now finding limt→∞y we get,

limt→∞y=limt→∞γe-βtα-β+ce-αt

⇒limt→∞y=γ×0+c×0=0 as e-∞→0

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (25 Jan Shift 2)
Let y=yx be a function of x satisfying y1-x2=k-x1-y2 where k is a constant and y12=-14.Then dydx at x=12 , is equal to

Options:
A -54
B -52
C 25
D 52

The correct answer is:
-52

Solution:
x=12,y=-14⇒xy=-18

y.-2x21-x2+y'.1-x2=0-1.1-y2+x.-2y21-y2y'

-xy1-x2+y'1-x2=-1-y2+xy.y'1-y2

y'1-x2-xy1-y2=xy1-x2-1-y2

y'32+18.154=-18.34-154

y'45+1215=-1+4543

y'=-1523=-52

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Let y=y(x) be a solution curve of the differential equation y+1tan2xdx+tanxdy+ydx=0, x∈0,π2. If limx→0+xyx=1, then the value of yπ4 is:

Options:
A π4+1
B π4-1
C π4
D -π4

The correct answer is:
π4

Solution:
Given:

y+1tan2xdx+tanxdy+ydx=0

⇒y+1tan2x+ydx+tanxdy=0

⇒tanxdydx+y+1tan2x+y=0

⇒dydx+1+ytanx=-ycotx

⇒dydx+ytanx+cotx=-tanx

This is a linear differential equation of the form dydx+Pxy=Qx.

I.F =e∫(tanx+cotx)dx

=e∫tan2x+1tanxdx

=e∫sec2xtanxdx

=elogetanx

=tanx ∵x∈0,π2

Solution is

ytanx=∫-tan2xdx+c

⇒ytanx=∫1-sec2xdx+c

⇒ytanx=x-tanx+c

Now,

limx→0+xy=1

⇒limx→0+xtanxx-tanx+c=1

⇒1(0-0+c)=1⇒c=1

Then the function is

ytanx=x-tanx+1

Put x=π4,

yπ4tanπ4=π4-tanπ4+1

⇒yπ4=π4

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (26 Aug Shift 1)
Let y=yx be a solution of the differential equation, 1-x2dydx+1-y2=0,x<1. If y12=32, then y-12 is equal to

Options:
A 32
B -12
C 12
D -32

The correct answer is:
12

Solution:
dy1-y2+dx1-x2=0

⇒sin-1y+sin-1x=c
At,x=12, y=32⇒ c=π2⇒ sin-1y=cos-1x
Hence, y-12=sincos-1-12=12

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (08 Jan Shift 1)
Let y=yx be a solution of the differential equation (x cosx)dy+(xy sinx+y cosx-1)dx=0,0<x<π2. If π3yπ3=3, then π6y"π6+2y'π6 is equal to _______.

Options:

The correct answer is:
2

Solution:
Given,

x cosxdy+xy sinx+y cosx-1dx=0

⇒x cosxdydx+yx sinx+y cosx=1

⇒dydx+yx sinx+cosxx cosx=1x cosx

Which is a linear differential equation of form dydx+Pxy=Qx

Now we know that integrating factor will be e∫Pdx

⇒IF=e∫tanx+1xdx=elnsecx+lnx=x·secx

Hence, solution of linear differential equation is given by,

⇒y·x secx=∫x secxx cosxdx

⇒xy secx=tanx+C

Now using π3yπ3=3 we get,

⇒π3yπ3secπ3=tanπ3+C

⇒C=3

Hence, xy secx=tanx+3

⇒xyx=2sinx+π3

Now differentiating both side we get,

⇒xy'x+yx=2cosx+π3

Now again differentiating we get,

⇒xy"(x)+2y'(x)=-2sinx+π3

Thus π6y"π6+2y'π6=-2

Hence π6y"π6+2y'π6=2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (06 Apr Shift 1)
Let y=yx be solution of the differential equation logedydx=3x+4y, with y0=0. If y-23loge2=αloge2, then the value of α is equal to:

Options:
A -14
B 14
C 2
D -12

The correct answer is:
-14

Solution:
Given,

logedydx=3x+4y

⇒dydx=e3x·e4y

⇒∫e-4ydy=∫e3xdx

⇒e-4y-4=e3x3+C

Given,

y0=0

So,

-14-13=C⇒C=-712

So, the particular solution is

e-4y-4=e3x3-712

⇒e-4y=4e3x-7-3

⇒e4y=37-4e3x⇒4y=ln37-4e3x

4y=ℓn36 when x=-23ℓn2

⇒y=14ℓn12

⇒y=-14ℓn2

So, α=-14

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (27 Jul Shift 1)
Let y=yx be the solution curve of the differential equation dydx+1x2-1y=x-1x+112, x>1 passing through the point 2,13. Then 7y8 is equal to

Options:
A 11+6loge3
B 19
C 12-2loge3
D 19-6loge3

The correct answer is:
19-6loge3

Solution:
Given,

dydx+yx2-1=x-1x+112 is a linear differential equation.

Here I.F.=e∫dxx2-1=e12lnx-1x+1=x-1x+1

General solution will be yx-1x+1=∫x-1x+1dx+C

⇒yx-1x+1=x-2lnx+1+c

Given y2=13

⇒ c=2ln3-53

So, the equation of the curve is yx-1x+1=x-2lnx+1+2ln3-53

Now putting x=8, we get

7y83=8-4ln3+2ln3-53

⇒ 7y8=19-6ln3

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (28 Jul Shift 2)
Let y=yx be the solution curve of the differential equation dydx=yx1+x21+logex, x>0,y(1)=3. Then y2(x)9 is equal to :

Options:
A x25-2x32+logex3
B x22x32+logex3-3
C x23x31+logex2-2
D x27-3x32+logex2

The correct answer is:
x25-2x32+logex3

Solution:
Given differential equation can be rewritten as,

dydx-yx=y31+logex

1y3dydx-1xy2=1+logex

Let -1y2=t⇒2y3dydx=dtdx

∴dt2dx+tx=1+logex

⇒dtdx+2tx=21+logex ..........(1)

We know the solution of the differential equation dydx+Py=Q is given by,

ye∫Pdx=∫Qe∫Pdx+C (Where I.F.=e∫Pdx)

Therefore, on solving equation(1) we get,

I.F.=e∫2xdx=x2

-x2y2=231+logexx3-x33+C ......(2)

Given,y(1)=3

⇒19=-213-19+C

∴ C=59

Now putting value of C in equation(2) we get,

y29=x25-2x32+logex3

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (25 Jan Shift 1)
Let y=yx be the solution curve of the differential equation, y2-xdydx=1 , satisfying y0=1 . This curve intersects the X-axis at a point whose abscissa is

Options:
A 2-e
B -e
C 2
D 2+e

The correct answer is:
2-e

Solution:
dxdy+x=y2

This equation is a linear differential equation of the type dxdy+Px=Q, where P=1 and Q=y2.
Integrating Factor I.F. =e∫1dy=ey

Now solution of the differential equation is xI.F.=∫PI.F.dy+C
x·ey=∫y2·ey·dy=y2·ey-∫2y·ey·dy

⇒ x·ey=y2·ey-2y·ey+2ey+c

⇒ x=y2-2y+2+c·e-y

Put x=0, y=1

⇒ 0=1-2+2+ce

⇒ c=-e

Hence x=y2-2y+2+-e·e-y

On putting y=0 we get x=0-0+2+-ee-0

⇒ x=2-e

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (07 Jan Shift 2)
Let $y=y(x)$ be the solution of the differential equation $\left(1+x^2\right) \frac{d y}{d x}+y=e^{\tan ^{-1} x}$, $y(1)=0$. Then $y(0)$ is

Options:
A $\frac{1}{2}\left(e^{\pi / 2}-1\right)$
B $\frac{1}{2}\left(1-e^{\pi / 2}\right)$
C $\frac{1}{4}\left(1-e^{\pi / 2}\right)$
D $\frac{1}{4}\left(e^{\pi / 2}-1\right)$

The correct answer is:
$\frac{1}{2}\left(1-e^{\pi / 2}\right)$

Solution:
$\begin{aligned} & \frac{d y}{d x}+\frac{y}{1+x^2}=\frac{e^{\tan ^{-1} x}}{1+x^2} \\ & \text { I.F. }=e^{\int \frac{1}{1+x^2} d x}=e^{\tan ^{-1} x} \\ & y \cdot e^{\tan ^{-1} x}=\int\left(\frac{e^{\tan ^{-1} x}}{1+x^2}\right) e^{\tan ^{-1} x} \cdot d x \\ & \text { Let } \tan ^{-1} x=z \quad \therefore \frac{d x}{1+x^2}=d z \\ & \therefore y \cdot e^z=\int e^{2 z} d z=\frac{e^{2 z}}{2}+C \\ & y \cdot e^{\tan ^{-1} x}=\frac{e^{2 \tan ^{-1} x}}{2}+C \\ & \Rightarrow y=\frac{e^{\tan ^{-1} x}}{2}+\frac{C}{e^{\tan ^{-1} x}}\end{aligned}$
$\begin{aligned} & \because \mathrm{y}(1)=0 \Rightarrow 0=\frac{\mathrm{e}^{\pi / 4}}{2}+\frac{\mathrm{C}}{\mathrm{e}^{\pi / 4}} \Rightarrow \mathrm{C}=\frac{-\mathrm{e}^{\pi / 2}}{2} \\ & \therefore \mathrm{y}=\frac{\mathrm{e}^{\tan ^{-1} \mathrm{x}}}{2}-\frac{\mathrm{e}^{\pi / 2}}{2 \mathrm{e}^{\tan ^{-1} \mathrm{x}}} \\ & \therefore \mathrm{y}(0)=\frac{1-\mathrm{e}^{\pi / 2}}{2}\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (06 Apr Shift 1)
Let y=yx be the solution of the differential equation 1-x2dy=xy+x3+21-x2dx,-1<x<1
and y0=0. If ∫-12121-x2yxdx=k then k-1 is equal to

Options:

The correct answer is:
320

Solution:
Given,

1-x2dydx=xy+x3+21-x2

⇒dydx+-x1-x2y=x3+21-x2

IF=e∫-x1-x2dx=1-x2

yx·1-x2=x44+2x+c

y0=0⇒c=0

1-x2yx=x44+2x

So,required value =∫-1212x44+2xdx-14·2∫012x4dx

=110x5012=1320

k-1=320

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (27 Jun Shift 2)
Let y=y(x) be the solution of the differential equation 1-x2dy=xy+x3+231-x2dx, -1<x<1, y(0)=0. If y12=mn,m and n are coprime numbers, then m+n is equal to __________.

Options:

The correct answer is:
97

Solution:
Given,

1-x2dy=xy+x3+231-x2dx

⇒dydx-xy1-x2=x3+231-x21-x2

The above equation is a linear differential equation,

So, IF=e-∫x1-x2dx=e12ln1-x2=1-x2

Now, the solution of the differential equation is given by,

y1-x2=∫x3+231-x21-x2×1-x2dx

⇒y1-x2=3∫x3+2dx

⇒y1-x2=3x44+2x+c

Now, using y(0)=0 we get, c=0

Now, finding y12

y1-122=3124·4+2·12

⇒y=132+2

⇒y12=6532=mn

Hence, m+n=97

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (30 Jan Shift 1)