datavorous/entrance-exam-dataset huggingface dataset:
If the shortest distance between the straight lines 3(x-1)=6(y-2)=2(z-1) and 4(x-2)=2(y-λ)=(z-3), λ∈R is 138, then the integral value of λ is equal to:

Options:
A 3
B 2
C 5
D -1

The correct answer is:
3

Solution:
The given lines are 3(x-1)=6(y-2)=2(z-1) and 4(x-2)=2(y-λ)=(z-3)

And, they can be written as L1:(x-1)2=(y-2)1=(z-1)3 and

L2:(x-2)1=y-λ2=z-34.

The lines are in the form x-x1l=y-y1m=z-z1n, where x1, y1, z1 is a point on the line and l, m, n are the direction ratios of the vector parallel to the line.

Thus, the position vectors of the points on the line are respectively a1→=i^+2j^+k^ & a2→=2i^+λj^+3k^ and direction of the lines are along the vectors r→1=2i^+j^+3k^ and r2→=i^+2j^+4k^.

Thus, we have a→=a2→-a1→=i^+λ-2j^+2k^.

Shortest distance = Projection of a→ on r1→×r2→

=a→·r1→×r2→r1→×r2→

Now r1→×r2→=i^j^k^213124

⇒r1→×r2→=i^4-6-j^8-3+k^4-1

⇒r1→×r2→=-2i^-5j^+3k^

⇒r1→×r2→=-22+-52+32=4+25+9=38

And, a→·r1→×r→2=1λ-22213124

⇒a→·r1→×r→2=14-6-λ-28-3+24-1

⇒a→·r1→×r→2=-2-5λ+10+6

⇒a→·r1→×r→2=14-5λ

Given, the shortest distance is 138, hence, we have

138=14-5λ38

⇒14-5λ=1

⇒14-5λ=1 or 14-5λ=-1

⇒λ=135 or 3

∴ Integral value is λ=3.

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2021 (22 Jul Shift 1)
If the shortest distance of the parabola y2=4x from the centre of the circle x2+y2-4x-16y+64=0 is d, then d2 is equal to :

Options:
A 16
B 24
C 20
D 36

The correct answer is:
20

Solution:
Given: Equation of parabola is y2=4x and the equation of circle is x2+y2-4x-16y+64=0.

We know that, equation of normal to parabola is given by, y=mx-2m-m3.

Also, the centre of given circle is 2,8.

The normal of parabola is passing through center of circle.

⇒8=2m-2m-m3

⇒m=-2

So point P on parabola is given by,

am2,-2am=(4,4)

And C=(2,8)

⇒PC=4+16=20

d2=20.

Topic: Mathematics, Parabola, JEE Main, JEE Main 2024 (27 Jan Shift 1)
If the shortest wavelength in Lyman series of hydrogen atom is A, then the longest wavelength in Paschen series of He+ is

Options:
A 5A9
B 36A7
C 36A5
D 9A5

The correct answer is:
36A7

Solution:
1λ=RH×(Z)2 1n12−1n22
Shortest wavelength for Lyman series of H atom
n1=1 (Lyman series)
n2=∞
SO, 1λ1=1×RH11−1∞
λ1=1RH=A ∴λ=A
longest wavelength for Paschen series of He+ atom
n1=3
n2=4
1λ2=(2)2×RH 1(3)2−1(4)2
1λ2=4×RH 79×16=R×736
λ2=367RH=367A.

Topic: Chemistry, Structure of Atom, JEE Main, JEE Main 2017 (08 Apr Online)
If the sixth term in the expansion of 1x83+x2log10⁡x8 is 5600, then the value of x is

Options:
A 2
B 5
C 10
D 10

The correct answer is:
10

Solution:
Note that for log10â¡x to be defined, x>0.
We have, T6=T5+1=8C51x838-5x2log10â¡x5
â5600=8Ã7Ã63Ã21x8x10log10â¡x5
â100=x2log10â¡x5
â100x2=log10â¡x5
The curves y=100x2 and y=log10â¡x5 intersect in just one point 10,1
âx=10

Topic: Mathematics, Binomial Theorem, JEE Main
If the slab is brought inside the parallel plate capacitor with a speed v. Then the variation of dcdt with respect to v is dcdt=Rate change of capacitance -

Options:
A
B
C
D

The correct answer is:

Solution:
Ceq=C1+C2=Îµ0bxKd+Îµ0lâxbd
=Îµ0bdlb+xK-1
â´dcdt=Îµ0bdK-1 dxdt=Îµ0bdK-1V

Topic: Physics, Electrostatics, JEE Main
If the slit width is $2 \mathrm{~mm}$ and wavelength of light used is $4000 Å$, then Fresnel distance is nearly

Options:
A $2 \mathrm{~mm}$
B $10 \mathrm{~m}$
C $20 \mathrm{~km}$
D $2 \mu \mathrm{m}$

The correct answer is:
$10 \mathrm{~m}$

Solution:
(b) Slit width $\mathrm{a}=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}$. $\lambda=4000 \mathrm{~A}^{\circ}=4 \times 10^{-7} \mathrm{~m}$.
Fresnel distance: $\mathrm{Z}_{\mathrm{F}}=\frac{\mathrm{a}^2}{\lambda}=\frac{4 \times 10^{-6}}{4 \times 10^{-7}}=10$
$$
\mathrm{Z}_{\mathrm{F}}=10 \mathrm{~m}
$$

Topic: Physics, Wave Optics, JEE Main
If the slope of the tangent to the curve at any point $\mathrm{P}(x, y)$ is $\frac{y}{x}-\cos ^{2} \frac{y}{x}$, then the equation of a curve passing through $\left(1, \frac{\pi}{4}\right)$ is

Options:
A $\tan \left(\frac{y}{x}\right)+\log x=1$
B $\tan \left(\frac{y}{x}\right)+\log y=1$
C $\tan \left(\frac{x}{y}\right)+\log x=1$
D $\tan \left(\frac{x}{y}\right)+\log y=1$

The correct answer is:
$\tan \left(\frac{y}{x}\right)+\log x=1$

Solution:
According to the condition,

$\frac{\mathrm{dy}}{\mathrm{dx}}=\frac{y}{x}-\cos ^{2} \frac{y}{x} \ldots \text { (i) }$

This is a homogeneous differential equation Substituting $y=v x$, we get

$\begin{array}{l}

v+x \frac{\mathrm{dv}}{\mathrm{dx}}=v-\cos ^{2} v \\

\Rightarrow x \frac{\mathrm{dv}}{\mathrm{dx}}=-\cos ^{2} v \\

\Rightarrow \int \sec ^{2} \mathrm{vdv}=-\int \frac{\mathrm{dx}}{\mathrm{x}} \\

\Rightarrow \tan v=-\log x+\mathrm{C} \\

\Rightarrow \tan \frac{y}{x}+\log x=\mathrm{C}

\end{array}$

Substituting $\mathrm{x}=1, y=\frac{\pi}{4}$, we get $\mathrm{C}=1$

Thus, we get

$\tan \left(\frac{y}{x}\right)+\log x=1$

which is the required solution.

Topic: Mathematics, Differential Equations, JEE Main
If the solubility is represented as ' \( \mathrm{S} \) ' \( \mathrm{mol} \mathrm{L} \) and the solubility product as \( \mathrm{K}_{\mathrm{sp}} \), the relation between the two for \( \mathrm{Zr}_{3}\left(\mathrm{PO}_{4}\right)_{4} \)
is

Options:
A \( \mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 7} \)
B \( \mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{27}\right)^{1 / 3} \)
C \( \mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{m}}}{6912}\right)^{1 / 7} \)
D \( \mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{sp}}}{108}\right)^{1 / 5} \)

The correct answer is:
\( \mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{m}}}{6912}\right)^{1 / 7} \)

Solution:
Zr3PO44→3Zr+4+4PO4-3InitialSFinal03S4S

Ksp=3S34S4Ksp=27S3×64S4Ksp=6912S7S=Ksp69121/7

Topic: Chemistry, Ionic Equilibrium, JEE Main
If the solubility of a sparingly soluble salt of the type $B A_2$ (giving three ions on dissociation of a molecule) is $x$ moles per litre, then its solubility product is given by

Options:
A $x^2$
B $2x^3$
C $4x^2$
D $4x^3$

The correct answer is:
$4x^3$

Solution:
By formula $B A_2 \rightarrow B^{+}+2 A^{-}$
$K_{s p}=4 x^3$

Topic: Chemistry, Ionic Equilibrium, JEE Main
If the solubility of \( \mathrm{Ag}_{2} \mathrm{CrO}_{4} \) is \( \mathrm{S} \) moles/litre, its solubility product will be

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

The correct answer is:
\( 4 \mathrm{~S}^{3} \)

Solution:
Ag 2 CrO 4 S moles/ litre ⇌ 2 Ag 2 S + + CrO 4 - - S
K⁡ sp⁡ = Ag + 2 CrO 4 - - = 2 S 2 S = 4 S 3

Topic: Chemistry, Ionic Equilibrium, JEE Main
If the solubility of $\mathrm{MX}_2$ type electrolyte is $0.5 \times 10^{-4} \mathrm{~mol} \mathrm{~L}^{-1}$, then, find out $\mathrm{K}_{s p}$ of the electrolyte.

Options:
A $5 \times 10^{-12}$
B $25 \times 10^{-10}$
C $1 \times 10^{-13}$
D $5 \times 10^{-13}$

The correct answer is:
$5 \times 10^{-13}$

Solution:
An electrolyte \(\mathrm{MX}_2\) undergoes dissociation as follows :-
\(\mathrm{MX}_2 \rightleftharpoons \mathrm{M}^{+2}+2 \mathrm{X}^{-}\)
\(\begin{array}{|c|c|c|c|}
\hline \text {Concentration } & \mathrm{MX}_2 & \mathrm{M}^{+2} & \mathrm{X}^{-} \\
\hline \text {Initial concentration } & 1 & 0 & 0 \\
\hline \text {Concentration at Equilibrium } & 1-\mathrm{s} & \mathrm{s} & 2 \mathrm{~s} \\
\hline
\end{array}\)
Thus from the above condition we can say that,
\(\mathrm{K}_{\mathrm{sp}}=\mathrm{s} \times(2 \mathrm{~s})^2=4 \times(\mathrm{s})^3\)
Here, s (the solubility) is \(0.5 \times 10^{-4} \mathrm{~mole} / \mathrm{lit}\).
\(\begin{aligned}
& \therefore \mathrm{K}_{\mathrm{sp}}=4 \times\left(0.5 \times 10^{-4}\right)^3 \\
& \therefore \mathrm{K}_{\mathrm{sp}}=5 \times 10^{-13}
\end{aligned}\)

Topic: Chemistry, Solutions, JEE Main
If the solubility product of AB2 is 3.20×10-11 M3, then the solubility of AB2 in pure water is _______ ×10-4 mol L-1

[Assuming that neither kind of ion reacts with water]

Options:

The correct answer is:
2

Solution:
AB2⇌A2++2B−s2s

Ksp=4s3=3.20×10−11

So solubility =2×10−4 mol L−1

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2020 (06 Sep Shift 2)
If the solubility product of PbS is 8×10-28, then the solubility of PbS in pure water at 298 K is x×10-16 mol L-1. The value of x is____ (Nearest integer) [Given 2=1.41]

Options:

The correct answer is:
282

Solution:
PbSs⇌Pb2+aq+S2-aq

Let us consider the solubility of PbS is "S" M.

Thus,

Solubility product =Pb2+aqS2-aq

Ksp=Pb2+aqS2-aq ⋯ i

We know

Ksp=S2

S=Ksp=8×10-28=22×10-14

=2.82×10-14

=282×10-16

Topic: Chemistry, Ionic Equilibrium, JEE Main, JEE Main 2022 (29 Jul Shift 1)
If the solution curve of the differential equation 2x-10y3dy+ydx=0, passes through the points (0,1) and (2,β), then β is a root of the equation?

Options:
A y5-2y-2=0
B y5-y2-1=0
C 2y5-y2-2=0
D 2y5-2y-1=0

The correct answer is:
y5-y2-1=0

Solution:
Given that: 2x−10y3dy+ydx=0

⇒dxdy=10y2-2xy

⇒dxdy+2xy=10y2

⇒ I.F. =e2∫1ydx=y2

⇒x⋅y2=∫10y4dy+C

⇒x⋅y2=2y5+c

Put x=0 & y=1⇒c=−2

⇒x⋅y2=2y5−2

Passing Through (2,β)
⇒2⋅β2=2β5−2

⇒β5−β2−1=0
root of the equations
y5-y2-1=0

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (27 Aug Shift 2)
If the solution curve, of the differential equation dydx=x+y-2x-y passing through the point (2,1) is tan-1y-1x-1-1βlogeα+y-1x-12=logex-1, then 5β+α is equal to

Options:

The correct answer is:
11

Solution:
Given: dydx=x+y-2x-y

Now, finding the intersection point of x+y-2=0 and x-y=0.

⇒x+x-2=0, x=y

⇒x=1, y=1

⇒x,y≡1,1

Let, x=X+1, y=Y+1

⇒dYdX=X+YX-Y

Putting, Y=tX

⇒dYdX=t+XdtdX

⇒1+t1-t=t+XdtdX

⇒1+t-t+t21-t=XdtdX

⇒1XdX=1-t1+t2dt

⇒∫1XdX=∫1-t1+t2dt

⇒logX=tan-1t-12log1+t2+C

⇒logx-1=tan-1y-1x-1-12log1+y-1x-12+C

It is given that the solution curve passes through 2,1.

⇒log2-1=tan-11-12-1-12log1+1-12-12+C

⇒0=tan-10-12log1+C

⇒C=0

⇒α=1, β=2

⇒5β+α=11.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (27 Jan Shift 2)
If the solution curve of the differential equation tan-1y-xdy=1+y2dx passes through the point 1,0 then the abscissa of the point on the curve whose ordinate is tan1 is

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

The correct answer is:
2e

Solution:
tan-1y-xdy=1+y2dx

⇒dxdy+11+y2x=tan-1y1+y2

I.F. =e∫11+y2dy=etan-1y

General solution will be

xetan-1y=∫etan-1y×tan-1y1+y2dy

⇒xetan-1y=tan-1yetan-1y-etan-1y+C

Since the curve passes through 1,0 ⇒ C=2

So, the curve reduces to xetan-1y=tan-1yetan-1y-etan-1y+2

when y=tan1, xe=1e-e+2

⇒ x=2e

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (27 Jun Shift 2)
If the solution curve of the differential equation (y-2logex)dx+(xlogex2)dy=0, x>1 passes through the points e, 43 and (e4, α), then α is equal to _______

Options:

The correct answer is:
3

Solution:
Given,

(y-2logex)dx+(xlogex2)dy=0, x>1

⇒2xlnxdydx+y=2lnx where logex=lnx

⇒dydx+y2xlnx=1x

Which is linear differential equation,

So, I.F=e∫12xlnx=lnx

Now, solution of the equation is given by,

y·lnx=∫lnxxdx

⇒y·lnx=23lnx32+C ...(i)

Given, eq.(i) passes through point e, 43

So, C=23

Hence, solution of differential equation will be,

ylnx=23lnx32+23

Also given, above equation passes through point (e4, α)

αlne4=23lne432+23

⇒2α=23432+23

⇒α=13×8+13

⇒α=3

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2023 (08 Apr Shift 1)
If the solution curve y=y(x) of the differential equation 1+y21+logexdx+xdy=0,x>0 passes through the point (1,1) and ye=α-tan32β+tan32, then α+2β is

Options:

The correct answer is:
3

Solution:
Given: 1+y21+logexdx+xdy=0

⇒1+logexxdx=-11+y2dy

⇒∫1x+logxxdx+∫dy1+y2=0

⇒logx+(logx)22+tan-1y=C

This curve passes through the point 1,1.

⇒log1+(log1)22+tan-11=C

⇒C=π4

⇒logx+(logx)22+tan-1y=π4

Now, putting x=e

⇒loge+(loge)22+tan-1y=π4

⇒tan-1y=π4-32

⇒y=tanπ4-32

⇒y=tanπ4-tan321+tanπ4tan32

⇒y=1-tan321+tan32

Hence, on comparing we get,

⇒α=β=1

⇒α+2β=3

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (29 Jan Shift 1)
If the solution curve y=yx of the differential equation y2 dx+x2-xy+y2dy=0, which passes through the point 1,1 and intersects the line y=3x at the point α,3α, then value of loge3α is equal to

Options:
A π2
B π4
C π6
D π12

The correct answer is:
π12

Solution:
Given y2dx+x2-xy+y2dy=0

⇒ dxdy=-x2-xy+y2y2 ...i

Now let x=vy ⇒v+ydvdy=dxdy

Now putting in equation i we get

⇒v+ydvdy=-v2-v+1

⇒ ydvdy=-v2-1

⇒-dv1+v2=dyy

Now Integrating Both side

⇒-∫dy1+v2=∫dyy

⇒-tan-1v=logy+c ⇒-tan-1xy=logy+c

Also this curve passes through 1,1

⇒ tan-11=log1+c ⇒ c=-π4

⇒-tan-1xy=logy-π4

Now putting y=3x and x=α we get

=-tan-1x3x=log3α-π4 =-tan-113=log3α-π4

=-π6+π4=log3α

⇒log3α=π12

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (25 Jun Shift 1)
If the solution of $\frac{d y}{d x}=\frac{y^3 \cos \sqrt{x}}{\sqrt{x} e^{1 / y^2}}, y(0)=1$ is $\frac{1}{y^2}=\log _e(f(x))$, then $f(x)=$

Options:
A $4+4 \sin \sqrt{x}$
B $e \sin \sqrt{x}$
C $1-4 \sin \sqrt{x}$
D $e-4 \sin \sqrt{x}$

The correct answer is:
$e-4 \sin \sqrt{x}$

Solution:
$\frac{d y}{d x}=\frac{y^3 \cos \sqrt{x}}{\sqrt{x} e^{1 / y^2}}$
$\Rightarrow \int \frac{e^{1 / y^2}}{y^3} d y=\int \frac{\cos \sqrt{x}}{\sqrt{x}} d x$
On putting $\frac{1}{y^2}=t \Rightarrow-\frac{2}{y^3} d y=d t$
and on putting $u=\sqrt{x} \Rightarrow d u=\frac{d x}{2 \sqrt{x}}$
we get $\int \frac{e^t d t}{(-2)}=\int 2 \cos u d u$
$\begin{aligned} & \Rightarrow \quad-\frac{1}{2}\left(e^t\right)=2 \sin u+k \\ & \Rightarrow \quad-\frac{1}{2}\left(e^{1 / y^2}\right)=2 \sin \sqrt{x}+k\end{aligned}$
$\because y(0)=1$ means at $x=0, y=1$
$\Rightarrow \quad-\frac{1}{2}\left(e^{1 / y^1}\right)=2 \sin \sqrt{0}+k$
$\Rightarrow \quad k=-\frac{e}{2}$
$\therefore \quad-\frac{1}{2}\left(e^{1 / y^2}\right)=2 \sin \sqrt{x}-\frac{e}{2}$
Takin, $\log$ both sides with base $e$,
$\log \left(\frac{e^{1 / y^2}}{2}\right)=\log \left[(2 \sin \sqrt{x})-\frac{e}{2}\right]$
$\begin{aligned} & \Rightarrow \quad \log \left(-\frac{1}{2}\right)+\log \left(e^{1 / y^2}\right)=\log \left(2 \sin \sqrt{x}-\frac{e}{2}\right) \\ & \Rightarrow \quad \log 1-\log (-2)+\frac{1}{y^2}=\log \left(2 \sin \sqrt{x}-\frac{e}{2}\right) \\ & \Rightarrow \quad \frac{1}{y^2}=\log \left[\left(2 \sin \sqrt{x}-\frac{e}{2}\right) \times(-2)\right] \\ & \Rightarrow \quad \frac{1}{y^2}=\log (e-4 \sin \sqrt{x})\end{aligned}$
$\therefore \quad f(x)=e-4 \sin \sqrt{x}$

Topic: Mathematics, Differential Equations, JEE Main
If the solution of the differential equation 1+exydx+exy1-xydy=0 is x+kyexy=C (where, C is an arbitrary constant), then the value of k is equal to

Options:

The correct answer is:
1

Solution:
The given equation is
exydx-xydy+exydy+dx=0
or exyydxy+exydy+dx=0
⇒dexyy+dx=0
On integrating, we get,

exyy+x=C
⇒k=1

Topic: Mathematics, Differential Equations, JEE Main
If the solution of the differential equation (2x+3y-2)dx+(4x+6y-7)dy=0, y(0)=3, is αx+βy+3loge|2x+3y-γ|=6, then α+2β+3γ is equal to ______.

Options:

The correct answer is:
29

Solution:
Given: (2x+3y-2)dx+(4x+6y-7)dy=0

⇒dydx=-2x+3y-24x+6y-7 ...i

Let, 2x+3y-2=t

⇒2+3dydx=dtdx ...ii

Also, 4x+6y-4=2t

⇒4x+6y-7=2t-3 ...iii

Using i, ii and iii,

⇒13dtdx-2=-t2t-3

⇒dtdx-2=-3t2t-3

⇒dtdx=-3t2t-3+2

⇒dtdx=-3t+4t-62t-3

⇒dtdx=t-62t-3

⇒∫2t-3t-6dt=∫dx

⇒∫2t-12t-6+9t-6·dt=x+c

⇒2t+9log(t-6)=x+c

⇒2(2x+3y-2)+9log(2x+3y-8)=x+c ...iv

It is given that y(0)=3.

⇒2(0+9-2)+9log(0+9-8)=0+c

⇒14+9log(1)=c

⇒c=14

Using iv,

⇒4x+6y-4+9log(2x+3y-8)=x+14

⇒3x+6y+9log(2x+3y-8)=18

⇒x+2y+3log(2x+3y-8)=6

Comparing with αx+βy+3loge|2x+3y-γ|=6.

⇒α=1, β=2, γ=8

⇒α+2β+3γ=1+4+24

⇒α+2β+3γ=29

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (27 Jan Shift 1)
If the solution of the differential equation dydx=x3+xy2y3-yx2 is yk-xk=2x2y2+λ (where, λ is an arbitrary constant), then the value of k is

Options:
A 2
B 4
C 1
D 32

The correct answer is:
4

Solution:
ydyxdx=x2+y2y2-x2
Let, y2=Y;x2=Xâydyxdx=dYdX
Hence, the equation is dYdX=X+YY-X
âYdY-XdX=XdY+YdX
On integrating we get Y22-X22=â«dXY=XY+c
or Y2-X2=2XY+Î»
ây4-x4=2x2y2+Î»

Topic: Mathematics, Differential Equations, JEE Main
If the solution of the differential equation x2dy+2xydx=sin⁡xdx is xky+cos⁡x=C (where C is an arbitrary constant), then the value of k is equal to

Options:

The correct answer is:
2

Solution:
The given equation is
dx2y=sin⁡xdx
On integrating, we get,
x2y=-cos⁡x+C

or x2y+cos⁡x=C
⇒k=2

Topic: Mathematics, Differential Equations, JEE Main
If the solution of the differential equation y3x2cosâ¡x3dx+sinâ¡x3y2dy=x3dx is 2sinâ¡x3yk=x2+C ( where C is an arbitrary constant), then the value of k is equal to

Options:
A 3
B 2
C 1
D 4

The correct answer is:
3

Solution:
The given equation is
y3⋅3x2cos⁡x3dx+sin⁡x33y2dy=xdx
⇒dsinx3⋅y3=xdx
On integrating, we get,
sin⁡x3⋅y3=x22+C
⇒k=3.

Topic: Mathematics, Differential Equations, JEE Main
If the solution of the equation logcosxcotx+4logsinxtanx=1, x∈0,π2 is sin-1α+β2, where α,β are integers, then α+β is equal to:

Options:
A 3
B 5
C 6
D 4

The correct answer is:
4

Solution:
Given:

logcosxcotx+4logsinxtanx=1, x∈0,π2

⇒lncosx-lnsinxlncosx+4lnsinx-lncosxlnsinx=1

⇒1-lnsinxlncosx+41-lncosxlnsinx=1

⇒lnsinxlncosx+4lncosxlnsinx-4=0

⇒(lnsinx)2-4(lnsinx)(lncosx)+4(lncosx)2=0

⇒lnsinx-2lncosx2=0

⇒lnsinx=2lncosx

⇒lnsinx=lncos2x

⇒cos2x=sinx

⇒1-sin2x=sinx

⇒sin2x+sinx-1=0

⇒sinx=-1+52

So, α=-1, β=5

∴α+β=4

Topic: Mathematics, Basic of Mathematics, JEE Main, JEE Main 2023 (30 Jan Shift 1)
If the solution $y(x)$ of the given differential equation $\left(\mathrm{e}^y+1\right) \cos x \mathrm{~d} x+\mathrm{e}^y \sin x \mathrm{~d} y=0$ passes through the point $\left(\frac{\pi}{2}, 0\right)$, then the value of $\mathrm{e}^{y\left(\frac{\pi}{6}\right)}$ is equal to_________

Options:

The correct answer is:
3

Solution:
$\begin{aligned}
& \left(\mathrm{e}^{\mathrm{y}}+1\right) \cos \mathrm{x} \mathrm{dx}+\mathrm{e}^{\mathrm{y}} \sin \mathrm{x} d \mathrm{y}=0 \\
& \Rightarrow \mathrm{d}\left(\left(\mathrm{e}^{\mathrm{y}}+1\right) \sin \mathrm{x}\right)=0 \\
& \left(\mathrm{e}^{\mathrm{y}}+1\right) \sin \mathrm{x}=\mathrm{C}
\end{aligned}$
It passes through $\left(\frac{\pi}{2}, 0\right)$
$\Rightarrow \mathrm{c}=2$
Now, $x=\frac{\pi}{6}$
$\Rightarrow \mathrm{e}^{\mathrm{y}}=3$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2024 (06 Apr Shift 2)
If the source of light used in a Young's double slit experiment is changed from red to violet:

Options:
A consecutive fringe lines will come closer.
B the fringes will become brighter.
C the central bright fringe will become a dark fringe.
D the intensity of minima will increase.

The correct answer is:
consecutive fringe lines will come closer.

Solution:
β=λ·Dd

λR>λV

βR=λRDd and βV=λVDd

βR>βV

Fringe pattern will shrink.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2021 (24 Feb Shift 2)
If the speed of a wave doubles as it passes from shallow water into deeper water, its wavelength will be

Options:
A unchanged
B halved
C doubled
D quadrupled

The correct answer is:
doubled

Solution:
We have, speed $=$ frequency $\times$ wavelength
Also, when a wave passes from one medium to the other, its frequency remains constant.
$\therefore$ speed $\propto$ wavelength
Therefore, when the speed of a wave doubles, its wavelength also doubles.

Topic: Physics, Electromagnetic Waves, VITEEE, VITEEE 2008
If the standard deviation of $0,1,2,3$, wes 9 is $K$, then the standard deviation of $10,11,12,13 \ldots . .19$ is

Options:
A $K$
B $K+10$
C $K+\sqrt{10}$
D $10 K$

The correct answer is:
$K$

Solution:
Since each data increases by 10 ,
There will be no change in standard deviation because $\left(\mathrm{x}_{\mathrm{i}}-\mathrm{x}\right.$) remains same.
$\therefore$ The standard deviation of $10,11,12,13 \ldots\ldots$ 19 will be will be $K$.

Topic: Mathematics, Statistics, JEE Main
If the standard deviation of 0, 1, 2, ..., 9 is k, then the standard deviation of 10, 11, 12, ... , 19 is

Options:
A k
B k+10
C k + 1 0
D 10k

The correct answer is:
k

Solution:
Here, we see that 10 is added in each observation of the first data.
Since we know that SD does not depend on change of origin.
Hence, SD of second data is k.

Topic: Mathematics, Statistics, JEE Main
If the standard deviation of data is 12 and mean is 72 , then coefficient of variation is

Options:
A $15.67 \%$
B $14.67 \%$
C $13.67 \%$
D $16.67 \%$

The correct answer is:
$16.67 \%$

Solution:
$\begin{aligned} & \text { Coefficient of variation }=\frac{\text { Standard Deviation }}{\text { Mean }} \times 100 \\ & \frac{12}{72} \times 100 \%=16.67 \%\end{aligned}$

Topic: Mathematics, Statistics, JEE Main
If the standard deviation of first $n$ natural numbers is 2 , then the value of $n$ is

Options:
A 7
B 5
C 4
D 6

The correct answer is:
7

Solution:
$\begin{aligned} & \text { S.D. }=\sqrt{\frac{\sum x_i^2}{n}-(\bar{x})^2} \\ & \Rightarrow 2=\sqrt{\frac{\sum n^2}{n}-\left(\frac{\sum n}{n}\right)^2} \\ & \Rightarrow 4=\frac{n(n+1)(2 n+1)}{6 n}-\left(\frac{n(n+1)}{2 n}\right)^2 \\ & \Rightarrow 4=\frac{n+1}{2}\left\{\frac{2 n+1}{3}-\frac{n+1}{2}\right\} \\ & \Rightarrow 4=\frac{n+1}{2}\left(\frac{n-1}{6}\right) \\ & \Rightarrow n^2=49 \Rightarrow n=7\end{aligned}$

Topic: Mathematics, Statistics, JEE Main
If the standard deviation of n observations x1,x2,.....,xn is 5 and for another set of n observations y1,y2,.....,yn is 4, then the standard deviation of n observations x1-y1,x2-y2,.....,xn-yn is

Options:
A 1
B 52
C 5
D Data insufficient

The correct answer is:
Data insufficient

Solution:
S.D2=σx-y2
=1n∑i=1nxi-yi-x¯+y¯2
=1n∑i=1nxi-x¯2+1n∑i=1nyi-y¯2-2n∑i=1nxi-x¯yi-y¯
=σx2+σy2-2α where, 1n∑i=1nxi-x¯yi-y¯=α
As α is not known, therefore we cannot find σx-y2 or σx-y.

Hence, data is insufficient.

Topic: Mathematics, Statistics, JEE Main
If the standard deviation of the numbers -1, 0, 1, k is 5 where k>0, then k is equal to

Options:
A 6
B 453
C 2103
D 26

The correct answer is:
26

Solution:
S.D.=∑xi2n-∑xin2=5

⇒∑xi2n-∑xin2=5

⇒1+1+0+k24--1+0+1+k42=5

⇒2+k24-k216=5

⇒8+4k2-k216=5

⇒8+3k2=80

⇒3k2=72

⇒k2=24

⇒k=±26

Since, k>0

⇒k=26.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2019 (09 Apr Shift 1)
If the standard deviation of the numbers 2, 3, a and 11 is 3.5, then which of the following is true ?

Options:
A 3a2-34a+91=0.
B 3a2-23a+44=0.
C 3a2-26a+55=0.
D 3a2-32a+84=0.

The correct answer is:
3a2-32a+84=0.

Solution:
x	x2

2	4

3	9

a	a2

11	121

Sum→16+a	Sum→134+a2

σ=∑x24-∑xi42
134+a24-16+a42=3510
12134+a2-16+a24=72
536+4a2-256-a2-32a=14
⇒280+3a2-32a=196
⇒3a2-32a+84=0.

Topic: Mathematics, Statistics, JEE Main, JEE Main 2016 (03 Apr)
If the standard deviation of the numbers 2,4,a and 10 is 3.5, then which of the following is true?

Options:
A 3a2-23a+24=0
B 3a2-26a+46=0
C 3a2-32a+28=0
D 3a2-34a+45=0

The correct answer is:
3a2-32a+28=0

Solution:
S.D. =∑xi2n-∑xin2
∴494=4+16+a2+1004-16+a42
⇒3a2-32a+28=0

Topic: Mathematics, Statistics, JEE Main
If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant K for the reaction.
Zns+Cu2+aq⇌Zn2+aq+Cus
at 300 K is approximately:
R=8 JK-1mol-1, F=96000 C mol-1

Options:
A e-160
B e-80
C e160
D e320

The correct answer is:
e160

Solution:
∆G=∆Go+RTlnQ
At equilibrium ∆G=0, Q=k
0=∆Go+RT lnK
∆Go=-RT lnK ∴∆Go=-nFEo
So -nFEo=-RT lnk
′n′ factor for the above cell =2
So -2×96000×2=-8×300 lnk
lnk=160
k=e160

Topic: Chemistry, Electrochemistry, JEE Main, JEE Main 2019 (09 Jan Shift 2)
If the standard molar enthalpy change for combustion of graphite powder is -2.48×102 kJ mol-1, the amount of heat generated on combustion of 1 g of graphite powder in kJ (Nearest integer):

Options:

The correct answer is:
21

Solution:
1 mol graphite =12 gmC
Ans. =24812=20.67 kJ/gm heat evolved

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2021 (22 Jul Shift 1)
If the straight line 2x-3y+17=0 is perpendicular to the line passing through the points 7,17 and 15, β, then β equals :

Options:
A -5
B 353
C 5
D -353

The correct answer is:
5

Solution:
Slope of line 2x-3y+17=0 is 23.

Slope of a line passing through 7,17 and 15,β is 17-β7-15.

Since, both lines are perpendicular to each other, so product of slopes will be -1.

⇒17-β7-15×23=-1

⇒17-β-8=-32⇒17-β=12⇒β=5.

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2019 (12 Jan Shift 1)