datavorous/entrance-exam-dataset huggingface dataset:
If \( f(x) \) satisfies the differential equation \( \frac{d y}{d x}=(x-y)^{2} \) and given that \( y(1)=1 \), then

Options:
A \( -\ln \left|\frac{1-x+y}{1+x-y}\right|=2(x-1) \)
B \( \ln \left|\frac{2-y}{2-x}\right|=x+y-1 \)
C \( \ln \left|\frac{1-x+y}{1+x-y}\right|=2(x-1) \)
D \( \frac{1}{2} \ln \left|\frac{1-x+y}{1+x-y}\right|+\ln |x|=0 \)

The correct answer is:
\( -\ln \left|\frac{1-x+y}{1+x-y}\right|=2(x-1) \)

Solution:
x−y=t
1−dydx=dtdx
Hence 1−dtdx=t2
⇒1−t2=dtdx
⇒∫dt1−t2=∫dx
⇒12log1+tt−1=x+c
12logx−y+1x−y−1=x+c
At x=1 and y=1, we get c=−1
Hence 12logx−y+1x−y−1=x−1
Hence −ln1−x+y1+x−y=2x−1

Topic: Mathematics, Differential Equations, JEE Main
If \(f(x+2 y, x-2 y)=x y\), then \(f(x, y)\) is equal to

Options:
A \(\frac{1}{4} x y\)
B \(\frac{1}{4}\left(x^2-y^2\right)\)
C \(\frac{1}{8}\left(x^2-y^2\right)\)
D \(\frac{1}{2}\left(x^2+y^2\right)\)

The correct answer is:
\(\frac{1}{8}\left(x^2-y^2\right)\)

Solution:
\(\begin{aligned}\text{Hint: }
& x+2 y=a \\
& \frac{x-2 y=b}{2 x=a+b} \\
& 4 y=a-b
\end{aligned}\)
\(f(a, b)=\left(\frac{a+b}{2}\right)\left(\frac{a-b}{4}\right)=\frac{a^2-b^2}{8}\)

Topic: Mathematics, Functions, JEE Main
If \(f(x)=\frac{1-x}{1+x}\) the domain of \(f^{-1}(x)\) is

Options:
A \(\mathrm{R}\)
B \(\mathrm{R}-\{-1\}\)
C \((-\infty,-1)\)
D \((-1, \infty)\)

The correct answer is:
\(\mathrm{R}-\{-1\}\)

Solution:
Let \(f(x)=y\). Then \(\frac{1-x}{1+x}=y\)
\(\Rightarrow x=\frac{1-y}{1+y} \Rightarrow f^{-1}(y)=\frac{1-y}{1+y}\)
Thus, \(f^{-1}(x)=\frac{1-x}{1+x}\)
Clearly, \(\mathrm{f}^{-1}(\mathrm{x})\) is defined for \(1+\mathrm{x} \neq 0\).
Hence, domain of \(\mathrm{f}^{-1}(\mathrm{x})\) is \(\mathrm{R}-\{-1\}\)

Topic: Mathematics, Functions, JEE Main
If \( f(x)=\frac{x}{\sqrt{1+x^{2}}} \), then \( (f o f o f)(x) \) is

Options:
A \( \frac{3 x}{\sqrt{1+x^{2}}} \)
B \( \frac{x}{\sqrt{1+3 x^{2}}} \)
C \( \frac{3 x}{\sqrt{1-x^{2}}} \)
D none of these

The correct answer is:
\( \frac{x}{\sqrt{1+3 x^{2}}} \)

Solution:
Given:

fx=x1+x2

Then,

fofx=ffx=fxx2+1=x1+x21+x21+x2=x2x2+1

fofofx=fffx=fx2x2+1=x2x2+11+x22x2+1=x1+3x2

Topic: Mathematics, Functions, JEE Main
If \( f(x)=\left\{\begin{array}{cc}3^{x}, & -1 \leq x \leq 1 \\ 4-x, & 1 < x < 4\end{array}\right. \), then at \( x=1, f(x) \) will be

Options:
A Continuous but not differentiable
B Neither continuous nor differentiable
C Continuous and differentiable
D Differentiable but not continuous

The correct answer is:
Continuous but not differentiable

Solution:
Since f 1 - 0 = lim x → 1 3 x = 3
f 1 + 0 = lim x → 1 4 - x = 3
and f(1) = 31 = 3
∴ f 1 - 0 = f 1 + 0 = f 1
⇒ f x is continuous at x = 1
Again f ′ 1 + 0 = lim x → 1 + f x - f 1 x - 1 = lim x → 1 3 x - 3 x - 1 = lim h → 0 3 1 + h - 3 h = 3 lim h → 0 3 h - 1 h = 3 log 3
and f ′ 1 + 0 = lim x → 1 f x - f 1 x - 1 = lim x → 1 4 - x - 3 x - 1 = - 1
∴ f ′ 1 + 0 ≠ f ′ 1 - 0
⇒ f(x) is not differentiable at x = 1

Topic: Mathematics, Continuity and Differentiability, JEE Main
If \( f(x)=\left\{\begin{array}{cc}\mathrm{x}+2, & \mathrm{x}>0 \\ \mathrm{x}^{2}-2, & 0 \leq \mathrm{x} < 1, \text { then the number of points of discontinuity of }|f(x)| \text { is: } \\ \mathrm{x}, & \mathrm{x} \geq 1\end{array}\right. \)

Options:
A \( 1 \)
B \( 0 \)
C \( 2 \)
D none of these

The correct answer is:
\( 1 \)

Solution:
Given function f(x)=x+2, x>0x2-2, 0≤x<1x, x≥1.

∴f(x)=|x+2|, x<0|x2+2|, 0≤x<1|x|, x≥1.

Using properties of modulus function,

∴f(x)=-x-2, x>-2x+2, -2≤x<0x2+2, 0≤x<1x, x≥1.

For continuityof |f(x)| at x=1,

L.H.L.=f(1-)=12+2=3

R.H.L.=f(1+)=1

Hence, function is discontinuous at x=1.

Therefore, number of discontinuity is 1.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If \( f(x)=\left\{\begin{array}{cl}\frac{\sin (p+1) x+\sin x}{x} & , x < 0 \\ q & , x=0 \text { is continuous at } x=0, \text { then the ordered pair }(p, q) \text { is equal to } \\ \frac{\sqrt{x+x^{2}}-\sqrt{x}}{x^{3 / 2}} & , x>0\end{array}\right. \)

Options:
A \( \left(-\frac{3}{2},-\frac{1}{2}\right) \)
B \( \left(-\frac{1}{2}, \frac{3}{2}\right) \)
C \( \left(\frac{5}{2}, \frac{1}{2}\right) \)
D \( \left(-\frac{3}{2}, \frac{1}{2}\right) \)

The correct answer is:
\( \left(-\frac{3}{2}, \frac{1}{2}\right) \)

Solution:
∵ fx=sin⁡p+1x +sinxxqx2+x-xx3/2 :x<0:x=0:x>0

∵f is continuous at x=0

L.H.L=

limh→0 sinp+1-h+sin(-h)-h=limh→0sinp+1h+sinhh

=limh→0sinp+1hhp+1p+1+limh→0sinhh=p+1+1

=p+2

R.H.L= limh→0 h2+h- hh32× h2+h+hh2+h+h=limh→0h2+h-hh32 h12h+1+1

limh→011+h+1=12

⇒L.H.L.=R.H.L.=f0⇒p+2=12=q
=p=-32 & q=12

Topic: Mathematics, Continuity and Differentiability, JEE Main
If \( f(x)=[\tan x]^{2}, x \in\left(0, \frac{\pi}{3}\right) \), then \( f^{\prime}\left(\frac{\pi}{4}\right) \) (where, [.] is the greatest integer function)

Options:
A is equal to \( 1 \)
B is equal to \( 0 \)
C does not exist
D None of these

The correct answer is:
does not exist

Solution:
LHD: f'π4=limh→0+fπ4−h−fπ4−h
=limh→0+⁡tan⁡π4-h2⁡-tan⁡π42-h
=limh→0+⁡0-1-h→∞
∴ f'π4 does not exist.

Topic: Mathematics, Limits, JEE Main
If \( f(x+y)=f(x)+3 y^{2}+k x y, f(1)=1 \) and \( f(2)=6 \), then the value of \( \frac{1}{10} \int_{-1}^{2} f(x) d x \) is

Options:

The correct answer is:
1.3

Solution:
fx+y=fx+3y2+kxy..(i)

In (i), put x=1,f1+y=f1+3y2+ky

⇒fy+1=3y2+ky+1.......(ii)

In (ii), put y=1,f2=31+k+1=6

⇒k=2

Using (ii), fy+1=3y2+2y+1

Replace y→x-1

⇒fx=3x-12+2x-1+1=3x-12+2x-1

Now ∫-12fxdx=∫-123x-12+2x-1dx=x-13+x2-x-12=8+1+4-1-2+1=13

Topic: Mathematics, Definite Integration, JEE Main
If \( f(x+y)=f(x)+f(y)+|x| y+x^{2} y^{2}, \forall x, y \in R \) and \( f^{\prime}(0)=0 \), then

Options:
A not be differentiable at every non-zero \( x \).
B differentiable for all \( x \in R \).
C twice differentiable at \( x=0 \).
D none of the above.

The correct answer is:
differentiable for all \( x \in R \).

Solution:
Given:

fx+y=f(x)+f(y)+xy+x2y2

and f'(0)=0

Substitute, x=y=0

f(0+0)=f(0)+f(0)+0+0

⇒f(0)=0

f'(x)=limh→0fx+h-f(x)h

f'(x)=limh→0f(x)+f(h)+xh+x2h2-f(x)h

=limh→0f(h)h+x+x2h

=f'(0)+x

f'(x)=0+x=x

f'(x)=-xx<0xx≥0

Clearly f(x) is differentiable at all points.

Topic: Mathematics, Continuity and Differentiability, JEE Main
If $g:[-2,2] \rightarrow R$ where $g(x)=x^3+\tan x+\left[\frac{x^2+1}{P}\right]$ is a odd function then the value of parametric $P$ is

Options:
A $-5 \lt P \lt 5$
B $P \lt 5$
C $P\gt5$
D None of these

The correct answer is:
$P\gt5$

Solution:
$\begin{aligned} & g(x)=x^3+\tan x+\frac{x^2+1}{P} \\ & g(-x)=(-x)^3+\tan (-x)+\frac{(-x)^2+1}{P} \\ & g(-x)=-x^3-\tan x+\frac{x^2+1}{P} \\ & g(x)+g(-x)=0 \text { because } g(x) \text { is a odd function } \\ & \therefore\left[x^3+\tan x+\frac{x^2+1}{P}\right]+\left[-x^3-\tan x+\frac{x^2+1}{P}\right]=0 \\ & \Rightarrow \frac{2\left(x^2+1\right)}{P}=0 \Rightarrow 0 \leq \frac{x^2+1}{P} \lt 1 \text { because } x \in[-2,2] \\ & \Rightarrow 0 \leq \frac{5}{P} \lt 1 \Rightarrow P\gt5 .\end{aligned}$

Topic: Mathematics, Functions, JEE Main
If gcdm, n=1 and 12-22+32-42+....+20212-20222+20232=1012m2n then m2-n2 is equal to

Options:
A 240
B 200
C 220
D 180

The correct answer is:
240

Solution:
Let

S=12-22+32-42+....+20212-20222+20232

⇒S=1-21+2+3-43+4+....+2021-20222021+2022+20232

⇒S=-3+7+11+15+....+4043+20232

The number of terms in the bracket are 20222=1011⇒S=-101126+1010×4+20232

⇒S=-1011×2023+20232

⇒S=2023×1012

⇒S=172×7×1012

So,

m=17, n=7 and gcd17,7=1

Hence, m2-n2=172-72=240

Hence this is the correct option.

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2023 (06 Apr Shift 2)
If $\mathrm{g}_{\mathrm{E}}$ and $\mathrm{g}_{\mathrm{m}}$ are the accelerations due to gravity on the surfaces of the earth and the moon respectively and if Millikan's oil drop experiment could be performed on the two surfaces, one will find the ratio $\frac{\text { electronic charge on the moon }}{\text { electronic charge on the earth }}$ to be

Options:
A
$1$

B
$0$

C
$\mathrm{g}_{\mathrm{E}} / \mathrm{g}_{\mathrm{m}}$

D
$\mathrm{g_m}$ / $\mathrm{g_E}$

The correct answer is:
$1$

Solution:
No Solution Available

Topic: Physics, Electrostatics, JEE Main, JEE Main 2007
If ΔG°[HI(g)=-1.7 kJ] , the equilibrium constant for the reaction 2HI(g)⇌H2(g)+I2(g) at 25°C is

Options:
A 24
B 2
C 3.6
D 0.5

The correct answer is:
3.6

Solution:
ΔG°=-2.303RTlog⁡K
log⁡K=-ΔG°2.303RT=--1.7×1032.303×8.314×298=log K=+0.298
K1=K=0.298
H2+I2⇌2HI K1=(0.504)2
2HI⇌H2+I2 K2=?
K2=1K1=1(0.504)2=3.6

Topic: Chemistry, Chemical Equilibrium, JEE Main
If $\mathrm{g}$ is the acceleration due to gravity on the earth's surface, the gain in the potential energy of object of mass $m$ raised from the surface of the earth to a height equal to the radius $R$ of the earth is

Options:
A
$2 \mathrm{mgR}$

B
$\frac{1}{2} \mathrm{mgR}$

C
$\frac{1}{4} \mathrm{mgR}$

D
$\mathrm{mgR}$

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

Solution:
No Solution Available

Topic: Physics, Gravitation, JEE Main, JEE Main 2004
If g is the inverse of a function f and f'x=11+x5, then g'x is equal to

Options:
A 11+gx5
B 1+gx5
C 1+x5
D 5x4

The correct answer is:
1+gx5

Solution:
g is the inverse of f i.e. gx=f-1x
⇒fgx=x
Differentiating both the sides w.r.t x , we get
f'gx.g'x=1
⇒g'x=1f'gxas f'x=11+x5
⇒g'x=111+gx5
g'x=1+gx5

Topic: Mathematics, Functions, JEE Main, JEE Main 2014 (06 Apr)
If $g$ is the inverse of function $f$ and $f^{\prime}(x)=\sin x$, then $\mathrm{g}^{\prime}(\mathrm{x})$ is equal to

Options:
A $\operatorname{cosec}\{\mathrm{g}(\mathrm{x})\}$
B $\sin \{g(x)\}$
C $\frac{1}{\sin \{g(x)\}}$
D None of these

The correct answer is:
$\frac{1}{\sin \{g(x)\}}$

Solution:
Since, $g$ is the inverse of function $f$. Therefore, $\mathrm{g}(\mathrm{x})=\mathrm{f}^{-1}(\mathrm{x})$

$\Rightarrow \mathrm{f}[\mathrm{g}(\mathrm{x})]=\mathrm{x}$

$\Rightarrow$ fog $(\mathrm{x})=\mathrm{x},$ for all $\mathrm{x}$

Differentiate both side,w.r.tx

$\Rightarrow \frac{\mathrm{d}}{\mathrm{dx}}\{\operatorname{fog}(\mathrm{x})\}=\frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{x}),$ for all $\mathrm{x}$

$\Rightarrow \mathrm{f}^{\prime}[\mathrm{g}(\mathrm{x})] \mathrm{g}^{\prime}(\mathrm{x})=1,$ for all $\mathrm{x}$

$\Rightarrow \sin \{\mathrm{g}(\mathrm{x})\} \mathrm{g}^{\prime}(\mathrm{x})=1,$ for all $\mathrm{x}$

(By defn of $\left.f^{\prime}(x)\right)$

$\Rightarrow \mathrm{g}^{\prime}(\mathrm{x})=\frac{1}{\sin \{\mathrm{g}(\mathrm{x})\}}$

Topic: Mathematics, Differentiation, JEE Main
If $g(x)=\int_0^x \cos 4 t ~d t$, then $g(x+\pi)$ equals

Options:
A
$\frac{g(x)}{g(\pi)}$

B
$g(x)+g(\pi)$

C
$g(x)-g(\pi)$

D
None of these

The correct answer is:
$g(x)+g(\pi)$

Solution:
$g(x)=\int_0^x \cos 4 t ~d t$
$\Rightarrow g^{\prime}(x)=\cos 4 x \quad \Rightarrow g(x)=\frac{\sin 4 x}{4}+k \quad \Rightarrow g(x)=\frac{\sin 4 x}{4}[\because g(0)=0]$
$g(x+\pi)=g(x)+g(\pi)=g(x)-g(\pi)(\because g(\pi)=0)$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2012 (Offline)
If gx is a differentiable function such that â«1sinâ¡Î±x2gxdx=sinâ¡Î±-1, âÎ±â0,Ï2, then the value of g13 is equal to

Options:
A 4
B 43
C 32
D 9

The correct answer is:
9

Solution:
Differentially both sides w.r.t. Î± , we get
sin 2 α⋅g sinα ⋅cosα=cosα
âgsinâ¡Î±=1sin2â¡Î±
â´g13=9

Topic: Mathematics, Definite Integration, JEE Main
If gx=x2+x-1 and gofx=4x2-10x+5, then f54 is equal to

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

The correct answer is:
-12

Solution:
gfx=4542-10.54+5=-54
gf54=f254+f54-1
-54=f254+f54-1
f254+f54+14=0
f54+122=0
f54=-12

Topic: Mathematics, Functions, JEE Main, JEE Main 2020 (07 Jan Shift 1)
If gx=x2+x-1 and gofx=4x2-10x+5, then f54 is equal to

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

The correct answer is:
-12

Solution:
gfx=4542-10.54+5=-54
gf54=f254+f54-1
-54=f254+f54-1
f254+f54+14=0
f54+122=0
f54=-12

Topic: Mathematics, Functions, JEE Main
If G.M. $=18$ and A.M. $=27$, then H.M. is

Options:
A $\frac{1}{18}$
B $\frac{1}{12}$
C $12$
D $9 \sqrt{6}$

The correct answer is:
$12$

Solution:
$\mathbb{B} G^2=A H \Rightarrow(18)^2=27 H \Rightarrow H=12$.

Topic: Mathematics, Sequences and Series, JEE Main
If gravitational constant G, planks constant h and speed of light c considered as fundamental physical quantities then find dimensional formula of time

Options:
A G12 h12 c-52
B G1 h2 c-52
C G12 h12 c-2
D G2 h12 c-2

The correct answer is:
G12 h12 c-52

Solution:
Let
T=GÎ±LÎ²CÎ³
=M-1L3T-29⋅ML2T-1Î²LT-1Î³
âÂ -Î±+Î²=0
3â+2Î²+Î³=0
-2Î±-Î²-Î³=1
On solving we get
Î±=12,Â Î²=12,Â Î³=52

Topic: Physics, Units and Dimensions, JEE Main
If gravitational field due to uniform thin hemispherical shell at point P is I, then the magnitude of gravitational field at Q is (Mass of hemispherical shell is M, radius is R )

Options:
A GM2R2-I
B GM2R2+I
C GM4R-I
D 2I -GM2R2

The correct answer is:
GM2R2-I

Solution:
gravitational field due to II at P = gravitational field at I at Q

At P,

EI+EII=G2m2R2=GM2R2

EI=GM2R2-EII=GM2R2-I

Topic: Physics, Gravitation, JEE Main
If α=h4π2me2, then the correct expression to calculate of the circumference of the first orbit of hydrogen atom is

Options:
A 4πhα
B 2ÏhÎ±
C 4h2πα
D Both A and C

The correct answer is:
Both A and C

Solution:
circumference =2πr
2×π×n2h24π2mZe2
n=1
Z=1
and h4π2me2=α
Thus, 2×π×r
=4πhα
or 4h2πα

Topic: Chemistry, Structure of Atom, JEE Main
If $H$ is the harmonic mean between $p$ and $q$, then the value of $\frac{H}{p}+\frac{H}{q}$ is

Options:
A $2$
B $\frac{p q}{p+q}$
C $\frac{p+q}{p q}$
D None of these

The correct answer is:
$2$

Solution:
As given
$H=\frac{2 p q}{p+q}$
$\therefore \frac{H}{p}+\frac{H}{q}=\frac{2 q}{p+q}+\frac{2 p}{p+q}=\frac{2(p+q)}{p+q}=2$

Topic: Mathematics, Sequences and Series, JEE Main
If half-life of a substance is 5 yrs, then the total amount of substance left after 15 years, when initial amount is 64 grams is

Options:
A
16 grams

B
2 grams

C
32 grams

D
8 grams

The correct answer is:
8 grams

Solution:

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2002
If $\mathrm{HCl}$ is added over then what is formed

Options:
A

B

C

D None of these

The correct answer is:

Solution:

Topic: Chemistry, Hydrocarbons, JEE Main
If heat of dissociation of CHCl2COOH is 0.7 kcal/mole , the, ÎH for the reaction
CHCl2COOH+KOHâCHCl2COOH+H2O is

Options:
A –13 kcal
B +13 kcal
C –14.4 kcal
D –13.7 kcal

The correct answer is:
–13 kcal

Solution:
Enthalpy of neutralization of strong acid and strong base is â13.7âkcal/mol
ÎH= â13.7+0.7
=â13.0âkcal

Topic: Chemistry, Thermodynamics (C), JEE Main
If heat passes into a system, what happens to its internal energy?

Options:
A It decreases
B It remains unchanged
C It increases
D It becomes zero

The correct answer is:
It increases

Solution:
When heat passes into a system, it increases the internal energy of the system.

Topic: Chemistry, Thermodynamics (C), NEET
If Henle's loop were absent from mammalian nephron, which one of the following is to be expected?

Options:
A There will be no urine formation
B There will be hardly any change in the quality of urine formation
C The urine will be more concentrated
D The urine will be more dilute

The correct answer is:
The urine will be more dilute

Solution:
The main function of the Henle's loop is to absorb water from the tubular lumen, thus making the urine concentrated. If the loop of Henle becomes absent, then the urine becomes more dilute.

Topic: Biology, Excretory Products and their Elimination, NEET
If hydrogen electrode dipped in 2 solution of $\mathrm{pH}=3$ and $\mathrm{pH}=6$ and salt bridge is connected the e.m.f. of resulting cell is

Options:
A $0.177 \mathrm{~V}$
B $0.3 \mathrm{~V}$
C $0.052 \mathrm{~V}$
D $0.104 \mathrm{~V}$

The correct answer is:
$0.177 \mathrm{~V}$

Solution:
$E_{\text {cell }}=-0.059 \log \frac{10^{-6}}{10^{-3}}=-0.059 \log 10^{-3}$
$=-0.059 \times(-3)=0.177 \mathrm{~V}$.

Topic: Chemistry, Electrochemistry, JEE Main
If I=â«016x141+xdx=k+4tanâ1m, then 3k-m is equal to

Options:

The correct answer is:
6

Solution:
Let x=t4⇒dx=4t3dt
⇒I=∫02t.4t3dt1+t2
⇒I4=∫02t4-1+11+t2dt
⇒I4=∫02t2-1t2+1t2+1+1t2+1dt
=t33-t+tan-1t02
⇒I4=83-2+tan-12
∴I=83+4tan-12
⇒k=83, m=2
∴3k-m=6

Topic: Mathematics, Definite Integration, JEE Main
If $\mathrm{I}_0$ is the intensity of the principal maximum in the single slit diffraction pattern, then what will be its intensity when the slit width is doubled?

Options:
A
$2\mathrm{I}_0$

B
$4\mathrm{I}_0$

C
$\mathrm{I}_0$

D
$\mathrm{I}_0/2$

The correct answer is:
$\mathrm{I}_0$

Solution:
Maximum intensity is independent of slit width

Topic: Physics, Wave Optics, JEE Main, JEE Main 2005
If $\mathrm{I}_1=\int_0^1 2^{x^2} d x, I_2=\int_0^1 2^{x^3} d x, I_3=\int_1^2 2^{x^2} d x$ and $I_4=\int_1^2 2^{x^3} d x$ then

Options:
A
$I_2>I_1$

B
$I_1>I_2$

C
$I_3=I_4$

D
$I_3>I_4$

The correct answer is:
$I_1>I_2$

Solution:
$$
\begin{aligned}
& l_1=\int_0^1 2^{x^2} d x, l_2=\int_0^1 2^{x^3} d x, I_3=\int_0^1 2^{x^2} d x, I_4=\int_0^1 2^{x^3} d x \\
& \forall 0 < x < 1, x^2>x^3 \\
& \Rightarrow \int_0^1 2^{x^2} d x>\int_0^1 2^{x^3} d x \\
& \Rightarrow I_1>I_2 .
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2005
If $I_1=\int_0^1 e^{-x} \cos ^2 x d x ; I_2=\int_0^1 e^{-x^2} \cos ^2 x d x$ and $I_3=\int_0^1 e^{-x^3} d x$; then

Options:
A
$I_2>I_3>I_1$

B
$I_3>I_1>I_2$

C
$I_2>I_1>I_3$

D
$I_3>I_2>I_1$

The correct answer is:
$I_3>I_2>I_1$

Solution:
Given:
$$
\begin{aligned}
&I_1=\int_0^1 e^{-x} \cos ^2 x d x \\
&I_2=\int_0^1 e^{-x^2} \cos ^2 x d x \text { and } \\
&I_3=\int_0^1 e^{-x^3} d x \\
&\text { For } x \in(0,1) \\
&\Rightarrow x>x^2 \text { or }-x < -x^2 \\
&\text { and } x^2>x^3 \text { or }-x^2 < -x^3 \\
&\therefore e^{-x^2} < e^{-x^3} \text { and } e^{-x} < e^{-x^2} \\
&\Rightarrow e^{-x} < e^{-x^2} < e^{-x^3} \\
&\Rightarrow e^{-x^3}>e^{-x^2}>e^{-x} \\
&\Rightarrow I_3>I_2>I_1
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
If I1=∫0π2esin⁡x1+xcos⁡xdx and I2=∫0π2ecos⁡x1-xsin⁡xdx, then I1I2 is equal to (where x denotes the greatest integer less than or equal to x)

Options:

The correct answer is:
2

Solution:
I1=∫0π2esin⁡x+xesin⁡x.cos⁡xdx=xesin⁡xπ20=πe2
I2=∫0π2ecos⁡x+x-ecos⁡x.sin⁡xdx=xecos⁡xπ20=π2
Hence, I1I2=e⇒e=2

Topic: Mathematics, Definite Integration, JEE Main
If I1=∫02πsin3⁡xdx and I2=∫01ln⁡1x-1dx, then

Options:
A I1+I2>0
B I+I2<0
C I1<I2
D I1=I2

The correct answer is:
I1=I2

Solution:
I2=∫01ln⁡1-x-ln⁡xdx
=∫01ln⁡1-xdx-∫01ln⁡xdx
=∫01ln⁡xdx-∫01ln⁡xdx (using a+b-x property)
⇒I2=0
Also, I1=∫02πsin3⁡xdx
=∫02πsin3⁡2π-xdx
=-∫02πsin3⁡xdx
⇒I1=-I1
⇒I1=0
Hence, I1=I2

Topic: Mathematics, Definite Integration, JEE Main
If $\mathrm{I}_1=\int_0^{3 \pi} \mathrm{f}\left(\cos ^2 \mathrm{x}\right) \mathrm{dx}$ and $\mathrm{I}_2=\int_0^\pi \mathrm{f}\left(\cos ^2 \mathrm{x}\right) \mathrm{dx}$, then

Options:
A $\mathrm{I}_1=\mathrm{I}_2$
B $3 \mathrm{I}_1=\mathrm{I}_2$
C $\mathrm{I}_1=3 \mathrm{I}_2$
D $\mathrm{I}_1=5 \mathrm{I}_2$

The correct answer is:
$\mathrm{I}_1=3 \mathrm{I}_2$

Solution:
Hints : $I_1=3 \int_0^\pi \mathrm{f}\left(\cos ^2 x\right) d x=3 I \quad$ [period is $\pi$ ]

Topic: Mathematics, Definite Integration, JEE Main
If I1=â«0Ïxsinx1+cos2xdx, I2=â«0Ïxsin4xdx then, I1: I2 is equal to

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

The correct answer is:
4:3

Solution:
Given, I 1 = â« 0 Ï Ïâx sin Ïâx 1+ cos Ïâx 2 dx âââ â« 0 a f x dx= â« 0 a f aâx dx
=Ï â« 0 Ï sinx 1+ cos 2 x dx â â« 0 Ï xsinx 1+ cos 2 x dx
2 I 1 =Ï â« 0 Ï sinx 1+ cos 2 x dx =2Ï â« 0 Ï 2 sinx 1+ cos 2 x dx
ââ I 1 =Ï â« 0 Ï 2 sinx 1+ cos 2 x dx =Ï â« 0 1 dt 1+ t 2 â t=cosx
= Ï tan -1 t 0 1 = Ï 2 4
I2=â«0ÏÏ-xsin4xdx
=Ïâ«0Ïsin4xdx-I2
â 2I2=2Ïâ«0Ï/2sin4xdx=2ÏÂ·34Â·12Â·Ï2
âI2=316Ï2
Therefore, Iâ¡ 1:Iâ¡ 2 = 1 4 : 3 16 = 4:3

Topic: Mathematics, Definite Integration, JEE Main
If ∑i=15xi-6=5 and ∑i=15xi-62=25, then the standard deviation of observations 3x1+2, 3x2+2, 3x3+2, 3x4+2 and 3x5+2 is equal to

Options:

The correct answer is:
6

Solution:
Variance of 3xi+2=9var(xi)
=9∑xi25-∑xi52
=95-1
=36
⇒S.D=6

Topic: Mathematics, Statistics, JEE Main
If ∑i=19xi−5=9 and ∑i=19xi−52=45, then the standard deviation of the 9 items x1, x2,….,x9 is

Options:
A 3
B 9
C 4
D 2

The correct answer is:
2

Solution:
Given ∑ i=1 9 x i −45=9

⇒∑i=19xi=54

Also ∑i=19xi2−10xi+25=45

⇒∑i=19xi2−10×54+25×9=45

⇒∑xi2=45+540−225

=360
∴ σ2=∑xi2N−∑xiN2

= 360 9 − ( 54 9 ) 2

So, variance =40−36=4

∴ SD=2

Topic: Mathematics, Statistics, JEE Main, JEE Main 2018 (08 Apr)
If ∑i=120 20Ci-1 20Ci+20Ci-13=k21, then k equals

Options:
A 200
B 100
C 50
D 400

The correct answer is:
100

Solution:
Given ∑i=120 20Ci-1 20Ci+ 20Ci-13=k21
We have 20Ci-1 20Ci+ 20Ci-1

Using Cr-1n+Crn=Crn+1, we get

20Ci-1 20Ci+ 20Ci-1= 20Ci-1 21Ci

Now, using Crn=nr×Cr-1n-1, we get

20Ci-1 21Ci=20Ci-121i×20Ci-1=i21
Hence, the given sum is
∑i=120i3213=1213∑i=120i3

Now, using ∑r3r=1n=nn+122, we get

1213∑i=120i3=121320·2122=10021
Using the given sum, we get 10021=k21

⇒k=100.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2019 (10 Jan Shift 1)
If $I_1=\int_e^{e^2} \frac{d x}{\log x}$ and $I_2=\int_1^2 \frac{e^x}{x} d x_1$ then

Options:
A $l_1-l_2$
B $l_1 \gt I_2$
C $I_1 \lt I_2$
D None of these

The correct answer is:
$l_1-l_2$

Solution:
Put $\log x=u$ in $l_1$, so that $d x=x d u=e^u d u$ Also as $x=e$ to $e^2, u=1$ to 2
Thus, $I_1=\int_1^2 \frac{e^u}{u} d u=\int_1^2 \frac{e^x}{x} d x$. Hence, $I_1=I_2$.

Topic: Mathematics, Definite Integration, JEE Main
If Σi=1nxi-a=n and Σi=1nxi-a2=na, n, a>1, then the standard deviation of n observations x1, x2, …,xn is

Options:
A a-1

B n(a-1)

C n(a-1)

D (a-1)

The correct answer is:
(a-1)

Solution:
SD, σ=∑xi2n-∑xin2

SD is independent of change of origin, so, for given data
σ=Σxi-a2n-Σxi-an2
=nan-nn2=a-1

Topic: Mathematics, Statistics, JEE Main, JEE Main 2020 (06 Sep Shift 1)
If $i=\sqrt{-1}$, then $1+i^2+i^3-i^6+i^8$ is equal to

Options:
A $2-i$
B $1$
C $3$
D $-1$

The correct answer is:
$2-i$

Solution:
$1+i^2+i^3-i^6+i^8=1-1-i+1+1=2-i$.

Topic: Mathematics, Complex Number, JEE Main
If I=∫1+x41-x432dx=1fx+C (where, C is the constant of integration) and f2=-154, then the value of 2f12 is

Options:

The correct answer is:
3

Solution:
Given integral is
I=∫1x3+x1x2-x232dx
Let 1x2-x2=t
⇒-2x3-2xdx=dt
∴I=-12∫dtt32
=t-12+C
=11x2-x2+C
∴fx=1x2-x2
⇒f12=2-12=32
⇒2f12=3

Topic: Mathematics, Indefinite Integration, JEE Main
If i2=-1 and 1+i2n=1-i2m=1, ∀n,m∈N, then the minimum value of n+m is equal to

Options:

The correct answer is:
16

Solution:
Here, 1+i2=eiπ4 and 1-i2=e-iπ4
Now, eiπ4n=e-iπ4m=1

⇒nπ4=-mπ4=2kπ, ∀k∈Z
Hence, n=8k and m=-8k
So nmin=mmin=8

i.e. n+m=16

Topic: Mathematics, Complex Number, JEE Main
If i2=-1 and ∑r=1nir, ∀n∈N, is a non-zero real number, then n can be

Options:
A 100
B 201
C 302
D 403

The correct answer is:
403

Solution:
ip+ip+1+ip+2+ip+3=0, ∀p∈N
So,
∑r=1nir=0for n=4ki for n=4k+1i-1 for n=4k+2-1 for n=4k+3 ∀k∈N
Hence, n can be 403 for the sum to be non-zero real.

Topic: Mathematics, Complex Number, JEE Main
If i2=-1, then for a complex number Z the minimum value of Z+Z-3+Z+i+Z-3-2i occurs at

Options:
A Z=2
B Z=2+i
C Z=1
D Z=1+i

The correct answer is:
Z=1

Solution:
The minimum value of Z+Z-3 occurs at points on the line segment joining O and A
Similarly, the minimum value of Z+i+Z-3-2i occurs at a point on the line segment joining B and C.
â Minimum value of Z+Z-3+Z+i+Z-3-2i occurs at a point on the intersection of OA and BC

Hence, the minimum value occurs at Z=1

Topic: Mathematics, Complex Number, JEE Main
If $i^2=-1$, then sum $i+i^2+i^3+\ldots$ to 1000 terms is equal to

Options:
A $1$
B $-1$
C $i$
D $0$

The correct answer is:
$0$

Solution:
$\begin{aligned} & i+i^2+i^3+\ldots \ldots . \text { up to } 1000 \text { terms } \\ & =\frac{i\left(1-i^{1000}\right)}{1-i}=\frac{i\left(1-\left(i^4\right)^{250}\right)}{1-i}=\frac{i(1-1)}{1-i}=0\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
If $i^2=-1$, then the value of $\sum_{n=1}^{200} i^n$ is

Options:
A $50$
B $-50$
C $0$
D $100$

The correct answer is:
$0$

Solution:
$\sum_{n=1}^{200} i^n=i+i^2+i^3+\ldots .+i^{200}=\frac{i\left(1-i^{200}\right)}{1-i}$(since G.P.)
$=\frac{i(1-1)}{1-i}=0$

Topic: Mathematics, Complex Number, JEE Main