datavorous/entrance-exam-dataset huggingface dataset:
The initial mass of a rocket is 1000 kg. Calculate at what rate the fuel should be burnt so that the rocket is given an acceleration of, 20 m s-2. The gases come out at a relative speed of 500 m s-1, with respect to the rocket:

Use g=10 m s-2

Options:
A 10 kg s-1
B 60 kg s-1
C 500 kg s-1
D 6.0×102 kg s-1

The correct answer is:
60 kg s-1

Solution:
Here we are applying the equation, Fr−mg=ma

From Newton's second law, Fr=dmdtvr⇒dmdtvr=mg+ma

Put the given values in the above equations,

⇒dmdt500=1000(10+20)⇒dmdt=60 kg s-1.

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The initial pressure and volume of a given mass of an ideal gas $\left(\right.$ with $\frac{C_{p}}{C_{V}}=\gamma),$ taken in a cylinder fitted with a piston, are $p_{0}$ and $V_{0}$ respectively. At this stage the gas has the same temperature as that of the surrounding medium which is $T_{0}$. It is adiabatically compressed to a volume equal to $\frac{V_{0}}{2}$ Subsequently the gas is allowed to come to thermal equilibrium with the surroundings. What is the heat released to the surrounding?

Options:
A 0
B $(2^{\gamma-1}-1) \frac{p_{0} V_{0}}{\gamma-1}$
C $\gamma {p_{0}} {V_{0}} ln2$
D $\frac{{p_{0}} V_{0}}{2(\gamma-1)}$

The correct answers are:
$(2^{\gamma-1}-1) \frac{p_{0} V_{0}}{\gamma-1}$

Solution:
The equation of state for an ideal gas undergoing adiabatic process is given as,
$$
T V^{\gamma-1}=\text { constant }
$$
Let the temperature after adiabatic compression as given in the question be $T$ then,
$$
T_{0} V^{\gamma-1}=T\left(\frac{V_{0}}{2}\right)^{\gamma-1}
$$
$\Rightarrow \quad T=T_{0} 2^{\gamma-1}$
Now, heat released at volume $\frac{V_{0}}{2}$ to achieve
temperature $T_{0}$ The net heat released can be determined by the
equation.
$$
\begin{aligned}
\Delta Q &=\mu C_{V} \Delta T \\
&=\mu \times \frac{R}{\gamma-1} \quad\left(T_{0} 2^{\gamma-1}-T_{0}\right) \\
&=\frac{\mu R T_{0}}{\gamma-1}\left(2^{\gamma-1}-1\right)\left(\because \mu T_{0} R=p_{0} V_{0}\right)
\end{aligned}
$$
$\therefore$ Heat released $=\frac{p_{0} V_{0}}{\gamma-1}\left(2^{\gamma-1}-1\right)$

Topic: Physics, Thermodynamics, JEE Main
The initial pressure and volume of an ideal gas are P0 and V0. The final pressure of the gas when the gas is suddenly compressed to volume V04 will be:

(Given γ= ratio of specific heats at constant pressure and at constant volume.)

Options:
A

P0(4)γ

B

4P0

C

P0

D

P0(4)1γ

The correct answer is:

P0(4)γ

Solution:
As the gas is suddenly compressed the process is adiabatic.

So, PVγ=constant

⇒P0V0γ=P·V04γ

⇒P=4γ·P0

Topic: Physics, Thermodynamics, JEE Main, JEE Main 2023 (13 Apr Shift 2)
The initial speed of a bullet fired from a rifle is $630 \mathrm{~m} / \mathrm{s}$. The rifle is fired at the centre of a target $700 \mathrm{~m}$ away at the same level as the target. How far above the centre of the target ?

Options:
A
$1.0 \mathrm{~m}$

B
$4.2 \mathrm{~m}$

C
$6.1 \mathrm{~m}$

D
$9.8 \mathrm{~m}$

The correct answer is:
$6.1 \mathrm{~m}$

Solution:
Let ' $t$ ' be the time taken by the bullet to hit the target.
$$
\begin{aligned}
&\therefore \quad 700 \mathrm{~m}=630 \mathrm{~ms}^{-1} t \\
&\Rightarrow t=\frac{700 \mathrm{~m}}{630 \mathrm{~ms}^{-1}}=\frac{10}{9} \mathrm{sec}
\end{aligned}
$$
For vertical motion,
Here, $u=0$
$$
\begin{aligned}
&\therefore h=\frac{1}{2} g t^2 \\
&=\frac{1}{2} \times 10 \times\left(\frac{10}{9}\right)^2
\end{aligned}
$$
$$
=\frac{500}{81} \mathrm{~m}=6.1 \mathrm{~m}
$$
Therefore, the rifle must be aimed $6.1 \mathrm{~m}$ above the centre of the target to hit the target.

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2014 (11 Apr Online)
The initial speed of a projectile fired from ground is u. At the highest point during its motion, the speed of projectile is 32u. The time of flight of the projectile is:

Options:
A u2g
B ug
C 2ug
D 3ug

The correct answer is:
ug

Solution:
At maximum height, only horizontal velocity of the particle remains. Let the angle of project be θ. The horizontal velocity of the particle will remain same, therefore

ucosθ=3u2⇒θ=30°

Now, time of flight of projectile is

T=2usinθg=ug

Topic: Physics, Motion In Two Dimensions, JEE Main, JEE Main 2023 (31 Jan Shift 1)
The initial state of a certain gas is $\left(p_i, V_i, T_i\right)$. It undergoes expansion till its volume becomes $V_f$ Consider the following two cases
(a) the expansion takes place at constant temperature.
(b) the expansion takes place at constant pressure. Plot the $p-V$ diagram for each case. In which of the two cases, is the work done by the gas more?

Options:

None

Solution:
Consider the given diagram $p-V$, where variation is shown for each process.
(a) The expansion from $V_i$ to $V_f$ temperature $T_2$ remain constant so isothermal expansion i.e., $\rho_i V_i=\rho_f V_f$ constant $T$.
(b) The expansion is at costant pressure $P_i$, so process is isobaric process. Thus graph $p-V$ will be parallel to Vaxis till its volume becomes $V_f$ So process 2 is isothermal.
Since, work done $=$ area under the $p-V$ curve. Here, area under the $p-V$ curve 1 is more. So, work done is more when the gas expands in isobaric process.

Topic: Physics, Thermodynamics
The initial velocity vi required to project a body vertically upward from the surface of the earth to reach a height of 10R, where R is the radius of the earth, may be described in terms of escape velocity ve such that vi=xy×ve. The value of x will be

Options:

The correct answer is:
10

Solution:
-GMm11R=-GMmR+12mv2

⇒v=20GM11R

Topic: Physics, Gravitation, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The initial volume of a gas cylinder is $750.0 \mathrm{~mL}$. If the pressure of gas inside the cylinder changes from $840.0 \mathrm{~mm} \mathrm{Hg}$ to $360.0 \mathrm{~mm} \mathrm{Hg}$, the final volume the gas will be:

Options:
A
$1.750 \mathrm{~L}$

B
$3.60 \mathrm{~L}$

C
$4.032 \mathrm{~L}$

D
$7.50 \mathrm{~L}$

The correct answer is:
$1.750 \mathrm{~L}$

Solution:
According to Boyle's law
$$
\begin{aligned}
&\frac{V_1}{V_2}=\frac{P_2}{P_1} ; \frac{750}{V_2}=\frac{360}{840} \\
&V_2=1750 \mathrm{ml}=1.750 \mathrm{~L}
\end{aligned}
$$

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2014 (11 Apr Online)
The input resistance of a silicon transistor is $100 \Omega$. Base current is changed by $40 \mu \mathrm{A}$ which results in a change in collector current by $2 \mathrm{~mA}$. This transistor is used as a common-emitter amplifier with a load resistance of $4 \mathrm{k} \Omega$. The voltage gain of the amplifier is

Options:
A 2000
B 3000
C 4000
D 1000

The correct answer is:
2000

Solution:
We know
$$
V_g=\beta \frac{R_L}{R_i}
$$
$$
=\frac{\Delta I_C}{\Delta I_B} \frac{R_L}{R_i} \quad\left(\because \beta=\frac{\Delta I_C}{\Delta I_B}\right)
$$
On putting given values
$$
\begin{aligned}
& V_g=\frac{2 \times 10^{-3}}{40 \times 10^{-6}} \times \frac{4 \times 10^3}{100} \\
& V_g=2000
\end{aligned}
$$

Topic: Physics, Semiconductors, NEET, NEET 2012 (Mains)
The inradius of the triangle having sides 26, 28, 30 units is

Options:

The correct answer is:
8

Solution:
s=26+28+302=42
Δ=ss-as-bs-c
=4242-2642-2842-30
=42×16×14×12
=7×6×4×4×7×2×6×2
=7×6×4×2
r=Δs=7×6×4×242=8

Topic: Mathematics, Properties of Triangles, JEE Main
The instantaneous displacement of a simple harmonic oscillator is given by y=acos⁡ωt+π4. Its speed will be maximum at the time

Options:
A 2πω
B ω2π
C ωπ
D π4ω

The correct answer is:
π4ω

Solution:
Speed is maximum when y = 0
â´ 0=acosâ¡Ït+Ï4
Ït+ Ï 4 =( 2n+1 ) Ï 2
Ït=nÏ+ Ï 4
t= nÏ Ï + Ï 4Ï
for n=0,âât= Ï 4Ï

Topic: Physics, Oscillations, JEE Main
The instantaneous rate of disappearance of $\mathrm{MnO}_4^{-}$ion in the following reaction is $4.56 \times 10^{-3} \mathrm{Ms}^{-1}$
$2 \mathrm{MnO}_4^{-}+10 \mathrm{I}^{-}+16 \mathrm{H}^{+} \rightarrow 2 \mathrm{Mn}^{2+}+5 \mathrm{I}_2+8 \mathrm{H}_2 \mathrm{O}$
The rate of appearance $\mathrm{I}_2$ is :

Options:
A
$4.56 \times 10^{-4} \mathrm{Ms}^{-1}$

B
$1.14 \times 10^{-2} \mathrm{Ms}^{-1}$

C
$1.14 \times 10^{-3} \mathrm{Ms}^{-1}$

D
$5.7 \times 10^{-3} \mathrm{Ms}^{-1}$

The correct answer is:
$1.14 \times 10^{-2} \mathrm{Ms}^{-1}$

Solution:
Given $-\frac{d \mathrm{MnO}_4^{-}}{d t}=4.56 \times 10^{-3} \mathrm{Ms}^{-1}$
From the reaction given,
$\begin{aligned}
& -\frac{1}{2} \frac{\mathrm{dMnO}_4^{-}}{\mathrm{dt}}=\frac{4.56 \times 10^{-3}}{2} \mathrm{Ms}^{-1} \\
& -\frac{1}{2} \frac{d \mathrm{MnO}_4^{-}}{d t}=\frac{1}{5} \frac{d \mathrm{I}_2}{d t} \\
& \therefore-\frac{5}{2} \frac{d \mathrm{MnO}_4^{-}}{d t}=\frac{d \mathrm{I}_2}{d t}
\end{aligned}$
On substituting the given value
$\therefore \frac{d \mathrm{I}_2}{d t}=\frac{4.56 \times 10^{-3} \times 5}{2}=1.14 \times 10^{-2} \mathrm{M} / \mathrm{s}$

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2013 (09 Apr Online)
The instantaneous values of alternating current and voltages in a circuit are given as
$$
\begin{aligned}
& i=\frac{1}{\sqrt{2}} \sin (100 \pi t) \text { ampere } \\
& e=\frac{1}{\sqrt{2}} \sin (100 \pi t+\pi / 3) \text { volt }
\end{aligned}
$$
The average power in Watts consumed in the circuit is

Options:
A $\frac{1}{4}$
B $\frac{\sqrt{3}}{4}$
C $\frac{1}{2}$
D $\frac{1}{8}$

The correct answer is:
$\frac{1}{8}$

Solution:
Given equations
$$
i=\frac{1}{\sqrt{2}} \sin (100 \pi t)
$$
and $e=\frac{1}{\sqrt{2}} \sin (100 \pi t+\pi / 3)$
$\therefore \quad i_o=\frac{1}{\sqrt{2}}$ and $V_o=\frac{1}{\sqrt{2}}$
We know that average power
$$
\begin{aligned}
& P_{\mathrm{av}}=V_{\mathrm{rms}} \times i_{\mathrm{rms}} \cos \phi \\
&=\frac{1}{2} \times \frac{1}{2} \times \cos 60^{\circ} \\
& \qquad\left[\because i_{\mathrm{rms}}=\frac{i_o}{\sqrt{2}} \text { and } V_{\mathrm{rms}}=\frac{V_o}{\sqrt{2}}\right] \\
&=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \\
&=\frac{1}{8} \mathrm{~W}
\end{aligned}
$$

Topic: Physics, Alternating Current, JEE Main
The instantaneous values of alternating current and voltages in a circuit are given as
$$
\begin{aligned}
& i=\frac{1}{\sqrt{2}} \sin (100 \pi t) \text { ampere } \\
& e=\frac{1}{\sqrt{2}} \sin (100 \pi t+\pi / 3) \text { volt }
\end{aligned}
$$
The average power in Watts consumed in the circuit is

Options:
A $\frac{1}{4}$
B $\frac{\sqrt{3}}{4}$
C $\frac{1}{2}$
D $\frac{1}{8}$

The correct answer is:
$\frac{1}{8}$

Solution:
Given equations
$$
i=\frac{1}{\sqrt{2}} \sin (100 \pi t)
$$
and $e=\frac{1}{\sqrt{2}} \sin (100 \pi t+\pi / 3)$
$\therefore \quad i_o=\frac{1}{\sqrt{2}}$ and $V_o=\frac{1}{\sqrt{2}}$
We know that average power
$$
\begin{aligned}
& P_{\mathrm{av}}=V_{\mathrm{rms}} \times i_{\mathrm{rms}} \cos \phi \\
&=\frac{1}{2} \times \frac{1}{2} \times \cos 60^{\circ} \\
& \qquad\left[\because i_{\mathrm{rms}}=\frac{i_o}{\sqrt{2}} \text { and } V_{\mathrm{rms}}=\frac{V_o}{\sqrt{2}}\right] \\
&=\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2} \\
&=\frac{1}{8} \mathrm{~W}
\end{aligned}
$$

Topic: Physics, Alternating Current, NEET, NEET 2012 (Mains)
The instantaneous velocity of a particle moving in a straight line is given as v=αt+βt2, where α and β are constants. The distance travelled by the particle between 1 s and 2 s is:

Options:
A 3α+7β
B 32α+73β
C α2+β3
D 32α+72β

The correct answer is:
32α+73β

Solution:
V=αt+βt2

dsdt=αt+βt2

∫S1S2 ds=∫12αt+βt2dt

S2-S1=αt22+βt3312

As the particle is not changing direction So distance = displacement. Distance =α[4-1]2+β[8-1]3

=3α2+7β3

Topic: Physics, Motion In One Dimension, JEE Main, JEE Main 2021 (25 Jul Shift 2)
The instantaneous velocity of point B of the given rod of length 0.5 m is 3 m s-1 in the represented direction. What is the angular velocity (in rad s-1) of the rod for the minimum possible magnitude of the velocity of the end A?

Options:

The correct answer is:
5.2

Solution:
If rod is rotated about end A, then vertical component of velocity, vâ¥ of end a will be zero.
â´Ï=vsinâ¡60ol=3 v2l
=3Ã32Ã0.5=5.2 rad s-1

Topic: Physics, Rotational Motion, JEE Main
The integer k, for which the inequality x2-23k-1x+8k2-7>0 is valid for every x in R is:

Options:
A 4
B 2
C 3
D 0

The correct answer is:
3

Solution:
x2-23k-1x+8k2-7>0

Now, D<0

⇒ 43k-12-4×1×8k2-7<0

⇒ 9k2-6k+1-8k2+7<0

⇒ k2-6k+8<0

⇒ k-4k-2<0

⇒ k∈2,4

⇒k=3

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The integral ∫012ln1+2x1+4x2dx equals

Options:
A π4ln2
B π16ln2
C π8ln2
D π32ln2

The correct answer is:
π16ln2

Solution:
We have, I=∫012ln1+2x1+4x2dx

Let, 2x=tanθ⇒dx=12sec2θdθ

⇒I=∫0π4ln1+tanθsec2θ·12·sec2θdθ ∵ 1+tan2θ=sec2θ
=12∫0π4ln1+tanθdθ
=12∫0π4ln1+tanπ4-θdθ

∵∫0afxdx=∫0afa-xdx=12∫0π4ln1+1-tanθ1+tanθdθ

=12∫0π4ln21+tanθdθ=12ln2∫0π4dθ-∫0π4ln1+tanθdθ
⇒2I=ln2π4-2I

⇒I=π16ln2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2014 (09 Apr Online)
The integral ∫0π1+4 sin2x2-4 sinx2dx equals

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

The correct answer is:
4 3 - 4 - π 3

Solution:
∫0π1+4 sin2x2-4 sinx2dx
=∫0π2 sinx2-12dx
=∫0π2 sinx2-1dx
=2∫0πsinx2-12dx
=2 ∫0π/3sinx2-12dx+∫π/3πsinx2-12dx
=2 ∫0π/3-sinx2-12dx+∫π/3πsinx2-12dx
=2 2cosx2+x20π/3+-2cosx2-x2π/3π
= 2 2 3 2 + π 6 - 2 + - π 2 + 2 3 2 + π 6
= 2 3 + π 6 - 2 - π 2 + 3 + π 6
= 2 2 3 - π 6 - 2
= 4 3 - π 3 - 4 .

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2014 (06 Apr)
The integral ∫0π213+2sinx+cosxdx is equal to:

Options:
A tan-12
B tan-12-π4
C 12tan-12-π8
D 12

The correct answer is:
tan-12-π4

Solution:
Given,

I=∫0π2dx3+2sinx+cosx

⇒I=∫0π2dx3+22tanx21+tan2x2+1-tan2x21+tan2x2

⇒I=∫0π2sec2x2·dx2tan2x2+4tanx2+4

Put tanx2=t ⇒12sec2x2dx=dt, so

⇒I=∫01dtt+12+1

⇒I=tan-1t+101

⇒I=tan-12-π4

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (29 Jul Shift 1)
The integral ∫02x-1-xdx is equal to

Options:

The correct answer is:
1.5

Solution:
∫02x-1-xdx=∫011-x-xdx+∫12x-1-xdx

=∫0121-2xdx+∫1212x-1dx+∫12dx

=x-x2012+x2-x121+x12=12-14-0-0+1-1-14-12+2-1

=14+14+1=32=1.5

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The integral $\int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x$ is equal to :

Options:
A $3 \pi-50 \log _e 2+20 \log _e 5$
B $3 \pi-25 \log _e 2+10 \log _e 5$
C $3 \pi-10 \log _e(2 \sqrt{2})+10 \log _e 5$
D $3 \pi-30 \log _e 2+20 \log _e 5$

The correct answer is:
$3 \pi-50 \log _e 2+20 \log _e 5$

Solution:
$\begin{aligned} & I=\int_0^{\pi / 4} \frac{136 \sin x}{3 \sin x+5 \cos x} d x \\ & 136 \sin x=A(3 \sin x+5 \cos x)+B(3 \cos x-5 \sin x) \\ & 136=3 A-5 B...(i) \\ & 0=5 A+3 B...(ii)\end{aligned}$
$\begin{aligned} & 3 \mathrm{~B}=-5 \mathrm{~A} \Rightarrow \mathrm{B}=-\frac{5}{3} \mathrm{~A} \\ & 136=3 \mathrm{~A}-5\left(-\frac{5}{3} \mathrm{~A}\right) \\ & 136=3 \mathrm{~A}+\frac{25}{3} \mathrm{~A} \\ & 136=\frac{34 \mathrm{~A}}{3} \\ & \Rightarrow \mathrm{A}=\frac{136 \times 3}{34}=12 \\ & \mathrm{~B}=\frac{-5}{3}(12)=-20\end{aligned}$
$\begin{aligned} & I=\int_0^{\pi / 4} \frac{A(3 \sin x+5 \cos x)}{3 \sin x+5 \cos x}+\int_0^{\pi / 4} \frac{B(3 \cos x-5 \sin x)}{3 \sin x+5 \cos x} \\ & =A(x)_0^{\pi / 4}+B[\ln (3 \sin x+5 \cos x)]_0^{\pi / 4} \\ & =12\left(\frac{\pi}{4}\right)-20 \ln \left(\frac{3}{\sqrt{2}}+\frac{5}{\sqrt{2}}\right)-\ln (0+5) \\ & =3 \pi-20 \ln 4 \sqrt{2}+20 \ln 5 \\ & =3 \pi-20 \times \frac{5}{2} \ln 2+20 \ln 5 \\ & =3 \pi-50 \ln 2+20 \ln 5\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (05 Apr Shift 1)
The integral ∫1-13cosx-sinx1+23sin2xdx is equal to

Options:
A 12logetanx2+π12x2+π6+C
B logetanx2+π6x2+π3+C
C 12logetanx2+π6x2+π3+C
D 12logetanx2-π12tanx2-π6+C

The correct answer is:
12logetanx2+π12x2+π6+C

Solution:
Let I=∫1-13cosx-sinx1+23sin2xdx

Multiplying by 32 in numerator and denominator, we get

I=∫32-12cosx-sinx32+sin2xdx

=∫32-12cosx-sinxsinπ3+sin2xdx

=∫32cosx-12cosx-32sinx+12sinx2sinx+π6cosx-π6dx

=∫cosx-π6-sinx+π62sinx+π6cosx-π6dx

=12∫dxsinx+π6-∫dxcosx-π6

=12∫cosecx+π6dx-∫secx-π6dx

=12lntanx+π62-12lntanx-π62+π4+C

=12lntanx2+π12tanx2+π6+C

Topic: Mathematics, Indefinite Integration, JEE Main
The integral ∫1-13cosx-sinx1+23sin2xdx is equal to

Options:
A 12logetanx2+π12x2+π6+C
B logetanx2+π6x2+π3+C
C 12logetanx2+π6x2+π3+C
D 12logetanx2-π12tanx2-π6+C

The correct answer is:
12logetanx2+π12x2+π6+C

Solution:
Let I=∫1-13cosx-sinx1+23sin2xdx

Multiplying by 32 in numerator and denominator, we get

I=∫32-12cosx-sinx32+sin2xdx

=∫32-12cosx-sinxsinπ3+sin2xdx

=∫32cosx-12cosx-32sinx+12sinx2sinx+π6cosx-π6dx

=∫cosx-π6-sinx+π62sinx+π6cosx-π6dx

=12∫dxsinx+π6-∫dxcosx-π6

=12∫cosecx+π6dx-∫secx-π6dx

=12lntanx+π62-12lntanx-π62+π4+C

=12lntanx2+π12tanx2+π6+C

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2022 (26 Jul Shift 2)
The integral â« â1/2 1/2 ( [ x ]+ân( 1+x 1âx ) )dx is equal to (x is the greatest integer â¤x)

Options:
A − 1 2
B 1
C 2ℓn( 1 2 )
D 0

The correct answer is:
− 1 2

Solution:
Let, I= â« â1/2 1/2 ( [ x ]+ân( 1+x 1âx ) )dx
= â« â1/2 1/2 [ x ]dx+ â« â1/2 1/2 ân( 1+x 1âx )dx
= â« â1/2 0 â1dx+ â« 0 1/2 0dx+0 [ âµââlog( 1+x 1âx )isââanââoddââfunction ]
=[âx]-1/20=0â(12)=â12

Topic: Mathematics, Definite Integration, JEE Main
The integral ∫1+2cotxcosecx+cotxdx, 0<x<π2 is equal to

Options:
A 2logsinx2+c
B 4logsinx2+c
C 4logcosx2+c
D 2logcosx2+c

The correct answer is:
2logsinx2+c

Solution:
Let I=∫1+2cotxcosec x+cotxdx

⇒I=∫2cotxcosecx+cosec2x+cot2xdx

⇒I=∫cosec x+cot x2dx
⇒I=∫cosec x+cot xdx
⇒I=∫cosec x+cotxdx
⇒I=logtanx2+logsinx+c, where c is the constant of integration.
⇒I=logsinx2cosx2·2sinx2cosx2+c
⇒I=2logsinx2+c

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2017 (08 Apr Online)
The integral ∫12ex. xx 2+loge x dx equals :

Options:
A e4e+1
B 4e2-1
C e4e-1
D e2e-1

The correct answer is:
e4e-1

Solution:
Let y=exx

lny=x1+lnx

1ydydx=2+lnx

⇒ dy=exx2+lnx dx

∫12ex.xx2+logexdx

=y12

=exx12

=4e2−e

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The integral ∫1exe2x-exxloge x  dx is equal to

Options:
A 32-e-12e2
B 12-e-1e2
C -12+1e-12e2
D 32-1e-12e2

The correct answer is:
32-e-12e2

Solution:
Let I1=∫1exe2xloge x ⁡dx

Let xex=t

Taking natural logarithm on both sides, we get

x ln x e =ln t

⇒xln x−1=ln t .........i

x	t

1	1ln1-1=lnt⇒t=1e

e	elne-1=lnt⇒t=1

Differentiating equation i on both the sides, we get

ln x−1+x1xdx=1tdt

⇒ln x dx=1tdt

∴I1=∫1e1t21tdt=t221e1=12-12e2

and I2=∫1eexxloge x⁡dx

Let exx=t

Taking natural logarithm on both sides, we get

xlnex=lnt

⇒x1-lnx=lnt .........ii

x	t

1	11-ln1=lnt⇒t=e

e	e1-lne=lnt⇒t=1

Differentiating equation ii both sides, we get

⇒x0-1x+1-lnx.1dx=dtt (using product rule of differentiation)

⇒lnx=-dtt

∴I2=∫e1t.-dtt=-te1=-1--e=e-1

Hence, required integral is I1-I2

=12-12e2-e-1

=32-e-12e2.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The integral ∫1(x-1)3(x+2)54 dx is equal to : (where C is a constant of integration)

Options:
A 34x+2x-154+C
B 43x-1x+214+C
C 43x-1x+254+C
D 34x+2x-114+C

The correct answer is:
43x-1x+214+C

Solution:
I=∫dx(x-1)3(x+2)54

I=∫dx(x-1)34(x+2)54

I=∫dx(x-1)2x+2x-154

Let x+2x-1 =t

Differentiate on both sides

x-1-x+2x-12.dx = dt

⇒dxx-12 = -dt3

I=-13∫dtt54

= -13t-54+11-54+C

=43t-14+C

=431t14+C

= 43x-1x+214+C

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2021 (31 Aug Shift 1)
The integral ∫1+x-1xex+1xdx, is equal to

Options:
A x+1ex+1x+c
B -xex+1x+c
C x-1ex+1x+c
D xex+1x+c

The correct answer is:
xex+1x+c

Solution:
∫1+x-1xex+1xdx=∫ex+1xdx+∫x1-1x2ex+1xdx

Integrating ∫x⏟I.(1−1x2)ex+1x⏟IIdx, by parts

I=xex+1x-∫ex+1xdx+c

=∫1+x-1xex+1xdx=∫ex+1xdx+xex+1x-∫ex+1xdx+c

=xex+1x+c, where c is the constant of integration.

Topic: Mathematics, Indefinite Integration, JEE Main
The integral ∫1+x-1xex+1xdx, is equal to

Options:
A x+1ex+1x+c
B -xex+1x+c
C x-1ex+1x+c
D xex+1x+c

The correct answer is:
xex+1x+c

Solution:
∫1+x-1xex+1xdx=∫ex+1xdx+∫x1-1x2ex+1xdx

Integrating ∫x⏟I.(1−1x2)ex+1x⏟IIdx, by parts

I=xex+1x-∫ex+1xdx+c

=∫1+x-1xex+1xdx=∫ex+1xdx+xex+1x-∫ex+1xdx+c

=xex+1x+c, where c is the constant of integration.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2014 (06 Apr)
The integral ∫24logx2logx2+log6-x2dx is equal to

Options:
A 6
B 2
C 4
D 1

The correct answer is:
1

Solution:
I=∫24logx2logx2+log6-x2dx

I=∫24log6-x2log6-x2+logx2dx

∵∫abfxdx=∫abfa+b-xdx

2I=∫24logx2+log6-x2log6-x2+logx2dx

2I=∫241dx

2I=x24

∴ I=1

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2015 (04 Apr)
The integral ∫(2x-1)cos(2x-1)2+54x2-4x+6dx is equal to (where c is a constant of integration)

Options:
A 12sin(2x-1)2+5+c
B 12cos(2x+1)2+5+c
C 12cos(2x-1)2+5+c
D 12sin(2x+1)2+5+c

The correct answer is:
12sin(2x-1)2+5+c

Solution:
∫(2x-1)cos(2x-1)2+5(2x-1)2+5dx

(2x-1)2+5=t2

22x-1 2dx=2 tdt

2t2-5dx=tdt

So ∫t2-5cost2t2-5dt=12sint+c

=12sin(2x-1)2+5+c

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The integral $\int \frac{x d x}{2-x^2+\sqrt{2-x^2}}$ equals :

Options:
A
$\log \left|1+\sqrt{2+x^2}\right|+c$

B
$-\log \left|1+\sqrt{2-x^2}\right|+c$

C
$-x \log \left|1-\sqrt{2-x^2}\right|+c$

D
$x \log \left|1-\sqrt{2+x^2}\right|+c$

The correct answer is:
$-\log \left|1+\sqrt{2-x^2}\right|+c$

Solution:
$\quad \mathrm{I}=\int \frac{x d x}{2-x^2+\sqrt{2-x^2}}$
$$
\begin{aligned}
& \text { Put } t=\sqrt{2-x^2}, \frac{d t}{d x}=\frac{1}{2 \sqrt{2-x^2}} \cdot(-2 x) \\
& \begin{aligned}
& \Rightarrow-t d t=x d x \\
& \therefore \\
& \mathrm{I}=\int \frac{(-t) d t}{t^2+t}=-\int \frac{1}{t+1} d t=-\log |t+1| \\
&=-\log \left|\sqrt{2-x^2}+1\right|+c
\end{aligned}
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2013 (23 Apr Online)
The integral ∫2x3-1x4+xdx, is equal to

Options:
A 12loge(x3+1)2|x3|+C
B loge|x3+1|x2+C
C logex3+1x+C
D 12loge|x3+1|x2+C

The correct answer is:
logex3+1x+C

Solution:
I=∫2x3-1x4+xdx

⇒I=∫2x-1x2x2+1xdx

Put, x2+1x=t⇒2x-1x2dx=dt

∴I=∫dtt=loget+C

=logex2+1x+C

=logex3+1x+C

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2019 (12 Apr Shift 1)
The integral ∫2x12+5x9x5+x3+13dx, is equal to

Options:
A x52x5+x3+12+c
B -x102x5+x3+12+c
C -x5x5+x3+12+c
D x102x5+x3+12+c

The correct answer is:
x102x5+x3+12+c

Solution:
Given, I=∫2x12+5x9x5+x3+13dx
=∫2x12+5x9dxx151+1x2+1x53
=∫2x3+5x6dx1+1x2+1x53
Let, 1+1x2+1x5=t ⇒-2x3-5x6dx=dt
⇒2x3+5x6dx=-dt
I=-∫dtt3=12 . t2
I=1211+1x2+1x52+c=12x10x5+x3+12+c
Where c is the constant of integration.
=x102x5+x3+12+c

Topic: Mathematics, Indefinite Integration, JEE Main
The integral ∫2x12+5x9x5+x3+13dx, is equal to

Options:
A x52x5+x3+12+c
B -x102x5+x3+12+c
C -x5x5+x3+12+c
D x102x5+x3+12+c

The correct answer is:
x102x5+x3+12+c

Solution:
Given, I=∫2x12+5x9x5+x3+13dx
=∫2x12+5x9dxx151+1x2+1x53
=∫2x3+5x6dx1+1x2+1x53
Let, 1+1x2+1x5=t ⇒-2x3-5x6dx=dt
⇒2x3+5x6dx=-dt
I=-∫dtt3=12 . t2
I=1211+1x2+1x52+c=12x10x5+x3+12+c
Where c is the constant of integration.
=x102x5+x3+12+c

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2016 (03 Apr)
The integral ∫π43π4dx1+cosx is equal to

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

The correct answer is:
2

Solution:
Let, I=∫π43π4dx1+cosx

⇒I=∫π43π4dx1+2cos2x2-1

⇒I=12∫π43π4sec2x2 dx
⇒I=tanx2π43π4
⇒I=tan3π8-tanπ8

⇒I=2+1-2-1

⇒I=2.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2017 (02 Apr)
The integral ∫π6π3sec23x·cosec43xdx is equal to

Options:
A 376-356
B 343-313
C 356 -323
D 353-313

The correct answer is:
376-356

Solution:
Let I=∫π6π3sec23x·cosec43xdx

⇒I=∫π6π31cos23x·sin43xdx

⇒I=∫π6π3cos43xcos23+43x·sin43xdx

⇒I=∫π6π3sec2xtan43xdx

Let tanx=t, sec2xdx=dt and at x=π6, t=tanπ6=13 and at x=π3, t=tanπ3=3

⇒I=∫133dtt43

Using ∫xndx=xn+1n+1

⇒I=t-43+1-43+1133

⇒I=-3t-13133

⇒I=-33-13-13-13

⇒I=-3312-13-1312-13

=-31316-316

=-3316+3×316

=376-356.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (10 Apr Shift 2)