datavorous/entrance-exam-dataset huggingface dataset:
The uncertainty in momentum of an electron is $1 \times 10^{-5} \mathrm{kgm} / \mathrm{s}$. The uncertainty in its position will be $\left(h=6.63 \times 10^{-34} \mathrm{Js}\right)$

Options:
A $5.28 \times 10^{-30} \mathrm{~m}$
B $5.25 \times 10^{-28} \mathrm{~m}$
C $1.05 \times 10^{-26} \mathrm{~m}$
D $2.715 \times 10^{-30} \mathrm{~m}$

The correct answer is:
$5.28 \times 10^{-30} \mathrm{~m}$

Solution:
According to $\Delta x \times \Delta p=\frac{h}{4 \pi}$
$\Delta x=\frac{h}{\Delta p \times 4 \pi}=\frac{6.62 \times 10^{-34}}{1 \times 10^{-5} \times 4 \times 3.14}=5.27 \times 10^{-30} \mathrm{~m}$.

Topic: Chemistry, Structure of Atom, JEE Main
The uncertainty in the position of a moving bullet of mass $10 \mathrm{gm}$ is $10^{-5} \mathrm{~m}$. Calculate the uncertainty in its velocity

Options:
A $5.2 \times 10^{-28} \mathrm{~m} / \mathrm{sec}$
B $3.0 \times 10^{-28} \mathrm{~m} / \mathrm{sec}$
C $5.2 \times 10^{-22} \mathrm{~m} / \mathrm{sec}$
D $3 \times 10^{-22} \mathrm{~m} / \mathrm{sec}$

The correct answer is:
$5.2 \times 10^{-28} \mathrm{~m} / \mathrm{sec}$

Solution:
Uncertainty of moving bullet velocity $\Delta v=\frac{h}{4 \pi \times m \times \Delta v}=\frac{6.625 \times 10^{-34}}{4 \times 3.14 \times .01 \times 10^{-5}}$ $=5.2 \times 10^{-28} \mathrm{~m} / \mathrm{sec}$

Topic: Chemistry, Structure of Atom, JEE Main
The unequivocal proof of DNA as the genetic material came from the studies on a

Options:
A bacterium
B fungus
C viroid
D bacterial virus

The correct answer is:
bacterial virus

Solution:
The unequivocal (leaving no doubt; unambiguous) proof that DNA is the genetic material came from the experiments of Alfred Hershey and Martha Chase. They worked with viruses that infect bacteria called bacteriophages.

Topic: Biology, Molecular Basis of Inheritance, NEET, NEET 2011 (Mains)
The unit cell of copper corresponds to a face centered cube of edge length 3.596 Å with one copper atom at each lattice point. The calculated density of copper in kg/m3 is ___ . [Molar mass of Cu:63.54 g; Avogadro Number =6.022×1023 J]

Options:

The correct answer is:
9076

Solution:
FCC,

d=Z×MNA×a3=4×63.541000×6.022×1023×3.596×10-103

=9076 kg/m3

Topic: Chemistry, Solid State, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The unit of electric field is not equivalent to

Options:
A $N / C$
B $J / C$
C $V / m$
D $J / C-m$

The correct answer is:
$J / C$

Solution:
By unit and dimension - electric feild.
$\frac{N}{C}=\frac{\text { volt }}{\text { meter }}=\frac{\mathrm{J}}{\mathrm{c}-\mathrm{mt}}$
Therefore, dimensions of $\mathrm{E}=\mathrm{MLT}^{-3} \mathrm{~A}^{-1}$ which is a vector quantity. So, the correct answer is 2 : J/C
Because, it gives the answer work done per unit charge which is called Potential.

Topic: Physics, Electrostatics, JEE Main
The unit of inductance is

Options:
A Volt/ampere
B Joule/ampere
C Volt-sec/ampere
D Volt-ampere/sec

The correct answer is:
Volt-sec/ampere

Solution:
$e=L \frac{d i}{d t} \Rightarrow L=\frac{\text { Volt }- \text { sec }}{\text { Ampere }}$

Topic: Physics, Electromagnetic Induction, JEE Main
The unit of ionic product of water $K_w$ are

Options:
A $\mathrm{Mol}^{-1} L^{-1}$
B $\mathrm{Mol}^{-2} \mathrm{~L}^{-2}$
C $\mathrm{Mol}^{-2} \mathrm{~L}^{-1}$
D $\mathrm{Mol}^2 \mathrm{~L}^{-2}$

The correct answer is:
$\mathrm{Mol}^2 \mathrm{~L}^{-2}$

Solution:
Ionic Product of water:
\(\mathrm{H}_2 \mathrm{O}(\mathrm{l})+\mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+} \text {(aq. ) }+\mathrm{OH}^{-} \text {(aq. ) or } \mathrm{H}_2 \mathrm{O}(\mathrm{l}) \rightleftharpoons \mathrm{H}_3 \mathrm{O}^{+}(\mathrm{aq})+\mathrm{OH}^{-}(\text {aq) } \text {. }\)
As per laws of mass action: \(\mathrm{K}_{\mathrm{c}}=\frac{\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{H}_2 \mathrm{O}\right]}\)
\(\mathrm{K}_{\mathrm{c}} \times\left[\mathrm{H}_2 \mathrm{O}\right]=\mathrm{K}_{\mathrm{w}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right]\)
We can see that units of \(\mathrm{K}_{\mathrm{w}}\) are equal to units of \(\left[\mathrm{H}^{+}\right] \times\left[\mathrm{OH}^{-}\right]\)
Units of \(\left[\mathrm{H}^{+}\right]=\left[\mathrm{OH}^{-}\right]=\mathrm{mol} \mathrm{L}^{-1}\)
\(\therefore \mathrm{K}_{\mathrm{w}}=\left(\mathrm{mol} \mathrm{L}^{-1}\right)^2=\mathrm{mol}^2 \mathrm{~L}^{-2}\)

Topic: Chemistry, Ionic Equilibrium, JEE Main
The unit of magnetic flux is

Options:
A Weber $/ m^2$
B Weber
C Henry
D $Ampere/m$

The correct answer is:
Weber

Solution:
$\phi=\vec{B} \cdot \vec{A} \quad(\vec{A}=$ Area vector $)$
$(\vec{B}=$ Magnetic field)
S. I. Unit of Magnetic field is Tesla(T)
$\therefore$ Dimensional Formula of Magnetic flux will be $\mathrm{TL}^2$
And the SI unit of $\mathrm{TL}^2$ is Weber.

Topic: Physics, Electromagnetic Induction, JEE Main
The unit of pole strength is:

Options:
A $A m^2$
B Am
C $\frac{A^2}{m}$
D $\frac{A^2}{m^2}$

The correct answer is:
Am

Solution:
Pole strength is given by $\mathrm{m}=\mathrm{M} / 2 \mathrm{l}$.
Unit of magnetic dipole moment is $\mathrm{Am}^2$.
Therefore, unit of pole strength will be Am

Topic: Physics, Magnetic Properties of Matter, JEE Main
The unit of rate constant for a zero order reaction is

Options:
A $\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$
B $\mathrm{L} \mathrm{mol}^{-1} \mathrm{~s}^{-1}$
C $\mathrm{L}^2 \mathrm{~mol}^{-2} \mathrm{~s}^{-1}$
D $\mathrm{s}^{-1}$

The correct answer is:
$\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$

Solution:
For zero order reaction,
Rate $=k \cdot[\text { Reactants }]^0$
$\therefore$ Rate $=k$
and unit of $k=\mathrm{mol} \mathrm{L}^{-1} \mathrm{~s}^{-1}$.

Topic: Chemistry, Chemical Kinetics, NEET, NEET 2011 (Mains)
The unit of the van der Waals gas equation parameter 'a' in P+an2V2(V-nb)=nRT is:

Options:
A dm3 mol-1
B kg ms-1
C atm dm6 mol-2
D kg ms-2

The correct answer is:
atm dm6 mol-2

Solution:
Pressure correction ΔP=an2V2

a=ΔP×V2n2atm(Lit)2( mole )2=atmdm32( mole )2

a=atm·dm6·mol-2

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2021 (27 Aug Shift 1)
The unit of thermal conductivity is

Options:
A J m K-1
B J m-1 K-1
C W m K-1
D W m-1 K-1

The correct answer is:
W m-1 K-1

Solution:
∆Q∆t=KA∆T∆l
K=∆Q∆t ∆l∆T1A
∴ Unit of K is W m-1 K-1

Topic: Physics, Units and Dimensions, JEE Main
The unit vector perpendicular to the vectors $6 \hat{i}+2 \hat{j}+3 \hat{k}$ and $3 \hat{i}-6 \hat{j}-2 \hat{k}$ is $-$

Options:
A $\frac{2 \hat{i}-3 \hat{j}+6 \hat{k}}{7}$
B $\frac{2 \hat{\mathrm{i}}-3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}$
C $\frac{2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}$
D $\frac{2 \hat{i}+3 \hat{j}+6 \hat{k}}{7}$

The correct answer is:
$\frac{2 \hat{\mathrm{i}}+3 \hat{\mathrm{j}}-6 \hat{\mathrm{k}}}{7}$

Solution:
Unit vector perpendicular to both the given vectors is, $\frac{(6 \hat{i}+2 \hat{j}+3 \hat{k}) \times(3 \hat{i}-6 \hat{j}-2 \hat{k})}{|(6 \hat{i}+2 \hat{j}+3 \hat{k}) \times(3 \hat{i}-6 \hat{j}-2 \hat{k})|}=\frac{2 \hat{i}+3 \hat{j}-6 \hat{k}}{7}$

Topic: Mathematics, Vector Algebra, JEE Main
The unit vectors orthogonal to \(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}\) and coplanar with \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}}\) are

Options:
A \(\pm \frac{1}{\sqrt{5}}(2 \hat{\mathbf{i}}-\hat{\mathbf{k}})\)
B \(\pm \frac{1}{\sqrt{10}}(3 \hat{\mathbf{j}}-\hat{\mathbf{k}})\)
C \(\pm \frac{1}{\sqrt{13}}(2 \hat{\mathbf{i}}-3 \hat{\mathbf{j}})\)
D \(\pm \frac{1}{\sqrt{17}}(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-2 \hat{\mathbf{k}})\)

The correct answer is:
\(\pm \frac{1}{\sqrt{10}}(3 \hat{\mathbf{j}}-\hat{\mathbf{k}})\)

Solution:
A vector which is coplanar to the vectors \(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) and \(\hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}\) is
\(\mathbf{a}(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}})+\mathbf{b}(\hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})\)
\((2 a+b) \hat{\mathbf{i}}+(a-b) \hat{\mathbf{j}}+(a+b) \hat{\mathbf{k}}\) ...(i)
Eq. (i) is orthogonal to \(3 \hat{\mathbf{i}}+2 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}\)
\(\begin{aligned}
& 3(2 a+b)+2(a-b)+6(a+b) =0 \\
& 14 a+7 b =0 \\
& 2 a+b =0
\end{aligned}\)
Required unit vector,
\(\begin{aligned}
& = \pm \frac{(2 a+b) \hat{\mathbf{i}}+(a-b) \hat{\mathbf{j}}+(a+b) \hat{\mathbf{k}}}{\sqrt{(2 a+b)^2+(a-b)^2+(a+b)^2}} \\
& = \pm \frac{3 a \hat{\mathbf{j}}-a \hat{\mathbf{k}}}{\sqrt{(3 a)^2+a^2}}= \pm \frac{3 \hat{\mathbf{j}}-\hat{\mathbf{k}}}{\sqrt{10}}
\end{aligned}\)
Hence, option (b) is correct.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The unstable intermediate dichlorocarbene $\left[: \mathrm{CCl}_{2}\right]$ is formed during

Options:
A Kolbe's reaction
B Friedel-Crafts reaction
C Williamson synthesis
D Reimer-Tiemann reaction

The correct answer is:
Reimer-Tiemann reaction

Solution:
The unstable intermediate $\left[: \mathrm{CCl}_{2}\right]$ is formed in Reimer-Tiemann reaction which is an electrophilic substitution reaction and occurs through the following steps.
(i) Generation of electrophile
(ii) Electrophilic substitution

Topic: Chemistry, Alcohols Phenols and Ethers, JEE Main
The upper $\frac{3}{4}$ th portion of a vertical pole subtends an angle $\tan ^{-1} \frac{3}{5}$ at a point in the horizontal plane through its foot and at a distance $40 \mathrm{~m}$ from the foot.

Options:
A
$80 \mathrm{~m}$

B
$20 \mathrm{~m}$

C
$40 \mathrm{~m}$

D
$80 \mathrm{~m}$

The correct answer is:
$80 \mathrm{~m}$

Solution:
$\tan \beta=\frac{\tan \theta-\tan \alpha}{1+\tan \theta \cdot \tan \alpha}$ or $\frac{3}{5}=\frac{\frac{\mathrm{h}}{40}-\frac{\mathrm{h}}{160}}{1+\frac{\mathrm{h}}{40} \cdot \frac{\mathrm{h}}{160}}$ $\mathrm{h}^2-200 \mathrm{~h}+6400=0, \mathrm{~h}=40$ or 160 metre
Therefore possible height $=40$ metre

Topic: Mathematics, Heights and Distances, JEE Main, JEE Main 2003
The upper half of an inclined plane with an angle of inclination $\phi$, is smooth while the lower half is rough. A body starting from rest at the top of the inclined plane comes to rest at the bottom of the inclined plane. Then the coefficient of friction for the lower half is

Options:
A $2 \tan \phi$
B $\tan \phi$
C $2 \sin \phi$
D $2 \cos \phi$

The correct answer is:
$2 \tan \phi$

Solution:
For upper half
From equation,
$$
v^2=u^2+2 a s
$$
we have, $u=0$ (from rest), $s=l / 2$
$$
v^2=0+2(g \sin \phi) \cdot \frac{l}{2}
$$
For lower half,
$$
\begin{aligned}
& v=0 \text { and } a=g(\sin \phi-\mu \cos \phi), \\
& \Rightarrow \quad 0=u^2+2 g(\sin \phi-\mu \cos \phi) \cdot \frac{l}{2} \\
& \Rightarrow-g l \sin \phi=g l(\sin \phi-\mu \cos \phi) \\
& \Rightarrow \quad \mu \cos \phi=2 \sin \phi \quad \Rightarrow \mu=2 \tan \phi
\end{aligned}
$$

Topic: Physics, Laws of Motion, JEE Main
The upper half of an inclined plane with inclination $\phi$ is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is given by

Options:
A
$2 \sin \phi$

B
$2 \cos \phi$

C
$2 \tan \phi$

D
$\tan \phi$

The correct answer is:
$2 \tan \phi$

Solution:
$m g s \sin \phi=\mu m g \cos \phi \cdot \frac{s}{2}$

Topic: Physics, Laws of Motion, JEE Main, JEE Main 2005
The upper half of an inclined plane with inclination Ï is perfectly smooth while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom if the coefficient of friction for the lower half is

Options:
A tanÏ
B 2tanÏ
C 2cosϕ
D 2sinϕ

The correct answer is:
2tanϕ

Solution:
For the smooth upper half, using v2−u2=2as, we get
v2−02=2g(sinϕ)×l
v=2g (sinϕ)×l
For the lower half, using v2−u2=2as, we get
02−2g sin ϕ×l=−2(μk cosϕ−sinϕ)g×l
or sinϕ=μkcosϕ−sinϕ
μkcosϕ=2sinϕ
or μk=2tanϕ

Topic: Physics, Laws of Motion, JEE Main
The upper stratosphere consisting of the ozone layer, protects us from the sun's radiation that falls in the wavelength region of

Options:
A 200-315 nm
B 600-750 nm
C 400-550 nm
D 0.8-1.5 nm

The correct answer is:
200-315 nm

Solution:
The upper stratosphere consisting of the ozone layer, protects us from the sun's radiation that falls in the wavelength region of 200-315 nm.

Topic: Chemistry, Environmental Chemistry, JEE Main, JEE Main 2019 (12 Jan Shift 2)
The use of artificial sweetener saccharin has been discontinued, as it is:

Options:
A Carcinogenic
B Thermally unstable
C 550 times sweeter than sugar
D All of these

The correct answer is:
Carcinogenic

Solution:
The correct option is $\mathbf{1}$ Carcinogenic
Benzoic sulfimide which is also known as sodium saccharin discovered in 1879 and is used as an artificial sweetener with zero food energy. It is about 550 times sweeter than sucrose so used as sugar substitute. It is thermally very stable. It has been discontinued because it is considered a carcinogenic agent by FDA.

Topic: Chemistry, Chemistry in Everyday Life, JEE Main
The V-I characteristic of a diode is shown in the figure. The ratio of forward to reverse bias resistance is:

Options:
A 100
B 106
C 10-6
D 10

The correct answer is:
10-6

Solution:
Forward bias resistance R1=0.8-0.720-10×10-3=0.110×10-3=10 Ω.
Reverse bias resistanceR2=1010-6=107 Ω.
R1R2=10-6.

Topic: Physics, Semiconductors, JEE Main, JEE Main 2017 (08 Apr Online)
The V-I graphs for a conductor at temperature T1 and T2 are shown in the figure. T2-T1 is proportional to

Options:
A tanθ
B sinθ
C cot2Î¸
D cos2θ

The correct answer is:
cot2θ

Solution:
The slope of V-I graph gives resistance,
R1=tanθ, R2=tan(90-θ)=cotθ
R1=R0(1+αT1) and R2=R0(1+αT2)
∴R2-R1=R0α(T2-T1)
⇒cotθ-tanθ=R0α(T2-T1)
∴T2-T1∝cosθsinθ-sinθcosθ
∝2(cos2θ-sin2θ)2sinθ⋅cosθ
∝2cos2θsin2θ
∴T2-T1∝cot2θ

Topic: Physics, Current Electricity, JEE Main
The vacant space in the bcc unit cell is

Options:
A $32 \%$
B $23 \%$
C $26 \%$
D None of these

The correct answer is:
$32 \%$

Solution:
In $b c c$ structure $68 \%$ of the available volume is occupied by spheres. Thus vacant space is $32 \%$.

Topic: Chemistry, Solid State, JEE Main
The valency of inert gases is

Options:
A 5
B 1
C 0
D None of these

The correct answer is:
0

Solution:
Valency
- The combining power (or capacity) of an element is known as its valency.
Valency of inert gas
- The general electronic configuration of inert gas is $n s^2 n p^6$ except for helium.
- The electronic configuration of inert gases is stable.
- They do not lose, gain, or share electrons.
- Thus the valency of inert gas is zero.

Topic: Chemistry, p Block Elements (Group 15, 16, 17 & 18), JEE Main
The value closest to the thermal velocity of a Helium atom at room temperature $(300 \mathrm{~K})$ in $\mathrm{ms}^{-1}$ is:
$\left[\mathrm{k}_{\mathrm{B}}=1.4 \times 10^{-23} \mathrm{~J} / \mathrm{K} ; \mathrm{m}_{\mathrm{He}}=7 \times 10^{-27} \mathrm{~kg}\right]$

Options:
A
$1.3 \times 10^4$

B
$1.3 \times 10^5$

C
$1.3 \times 10^2$

D
$1.3 \times 10^3$

The correct answer is:
$1.3 \times 10^3$

Solution:
$$
\begin{aligned}
v &=\sqrt{3 k_{\mathrm{B}} T / m_{\mathrm{He}}}=\sqrt{\frac{3 \times 1.4 \times 10^{-23} \times 300}{7 \times 10^{-27}}} \\
&=1.34 \times 10^3 \mathrm{~m} / \mathrm{s}
\end{aligned}
$$

Topic: Physics, Kinetic Theory of Gases, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The value $\cos 105^{\circ}+\sin 105^{\circ}$ is

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

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

Solution:
$\begin{aligned}
\cos 105^{\circ}+\sin 105^{\circ} & =\cos \left(90^{\circ}+15^{\circ}\right)+\sin \left(90^{\circ}+15^{\circ}\right) \\
= & \cos 15^{\circ}-\sin 15^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}-\frac{\sqrt{3}-1}{2 \sqrt{2}}=\frac{2}{2 \sqrt{2}}=\frac{1}{\sqrt{2}}
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of $\int_0^1 \frac{8 \log (1+x)}{1+x^2} d x$ is

Options:
A
$\frac{\pi}{8} \log 2$

B
$\frac{\pi}{2} \log 2$

C
$\log 2$

D
$\pi \log 2$

The correct answer is:
$\pi \log 2$

Solution:
$\mathrm{I}=8 \int_0^1 \frac{\log (1+\mathrm{x})}{1+\mathrm{x}^2} \mathrm{dx}$
$=8 \int_0^{\frac{\pi}{4}} \frac{\log (1+\tan \theta)}{1+\tan ^2 \theta} \sec ^2 \theta d \theta(\operatorname{let} x=\tan \theta)$
$=8 \int_0^{\frac{\pi}{4}} \log \left(1+\tan \left(\frac{\pi}{4}-\theta\right)\right) d \theta=8 \int_0^{\frac{\pi}{4}} \log \left(1+\frac{1-\tan \theta}{1+\tan \theta}\right) d \theta=8 \int_0^{\frac{\pi}{4}} \log 2 d \theta-8 \int_0^{\frac{\pi}{4}} \log (1+\tan \theta) d \theta$
$=8 \log 2 \frac{\pi}{4}-I$
$2 \mathrm{I}=2 \pi \log 2$
$\mathrm{I}=\pi \log 2$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2011
The value of ∫0π2cos⁡2xcos22⁡xcos23⁡xcos⁡24xdx is equal to

Options:
A 0
B 12
C π2
D π4

The correct answer is:
0

Solution:
Let the value of integral is I, then,
Î=â«0Ï2cosâ¡2xcosâ¡22xcosâ¡23xcosâ¡24xdx
Î=â«0Ï2-cosâ¡2xcosâ¡22xcosâ¡23xcosâ¡24xdx (applying f(x)=f(a+b-x))
Adding, we get,

2Î=0âÎ=0

Topic: Mathematics, Definite Integration, JEE Main
The value of∫0π/2sgnsin2⁡x-sin⁡x+12dx is equal to, (where, sgnx denotes the signum function of x )

Options:
A 0
B 1
C π
D π2

The correct answer is:
π2

Solution:
⇒sin2⁡x-sin⁡x+12=sin⁡x-122+14>0 ∀x∈0,π2
∴ sgnsin2⁡x-sin⁡x+12=1
Thus, Ι=∫0π21dx=x0π2=π2

Topic: Mathematics, Definite Integration, JEE Main
The value of ∫0π/2sin3xsinx+cosxdx is:

Options:
A π-12
B Ï-28
C Ï-14
D Ï-24

The correct answer is:
π-14

Solution:
I=∫0π2sin3xsinx + cosxdx ...i

Using property of definite integration ∫abfxdx=∫abfa+b-xdx, we get

I=∫0π2sin3π2-xsinπ2-x+cosπ2-x dx

Using sinπ2-x=cosx & cosπ2-x=sinx,

⇒I=∫0π2cos3xcosx+sinx dx ...ii

On adding the equation i & ii

⇒2I=∫0π2sin3x+ cos3xsinx+cosxdx

⇒2I=∫0π2sinx+cosxsin2x+cos2x-sin xcosxsinx+cosxdx

⇒2I=∫0π2sin2x+cos2x-sin xcosxdx

Now, using sin2x+cos2x=1 & 2sinxcosx=sin2x, we get

⇒2I=∫0π21-12sin2xdx

⇒2I= x+cos2x40π2

⇒2I=π2+cos2π24-0+cos204

⇒2I=π2-14-14

⇒2I=π2-12

⇒I=π4-14=π-14.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (09 Apr Shift 1)
The value of ∫02πsin⁡2x1+cos⁡3xdx , where [t] denotes the greatest integer function is

Options:
A π
B 2π
C -π
D -2π

The correct answer is:
-π

Solution:
I=∫02πsin2x1+cos3xdx …1
I=∫02πsin2π-2x1+cos2π-3xdx

Applying ∫0afx=∫0afa-xdx

I=∫02π-sin2x1+cos3xdx …2

Adding 1 and 2

2I=∫02πsin⁡2x1+cos⁡3x+-sin⁡2x1+cos⁡3xdx
⇒ 2I=∫02π-1dx {∵ x+-x=0 :x∈I-1 :x∉I

⇒ 2I=-x02π

⇒ 2I=-2π

⇒ I= -π

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (10 Apr Shift 1)
The value of ∫02x2-2x+4sinx-12x2-4x+5dx is equal to

Options:

The correct answer is:
0

Solution:
Let, Ι=∫02x2-2x+4sinx-1dx2x2-2x+5
∫02x-12+3sinx-12x-12+3dx
Put x-1=t, dx=dt
I=∫-11t2+32t2+3sin⁡tdt=0
Since, this is an odd function

Topic: Mathematics, Definite Integration, JEE Main
The value of ∫02x44-x2dx∫02x24-x2dx is equal to

Options:

The correct answer is:
2

Solution:
∫02x44-x2dx=-12∫02x3⏟Ι-2x4-x2⏟ΙΙdx
Using integration by parts, we get
=-12x34-x2323202-∫023x24-x23232dx
=-120-0-∫022x24-x232dx=∫02x24-x232dx
=∫02x24-x24-x2dx
⇒∫02x44-x2dx=4∫02x24-x2dx-∫02x44-x2dx
⇒2∫02x44-x2dx=4∫02x24-x2dx
⇒∫02x44-x2dx∫02x24-x2dx=2

Topic: Mathematics, Definite Integration, JEE Main
The value of ∫02πxsin8⁡xsin8⁡x+cos8⁡xdx is equal to:

Options:
A 2π
B 2π2
C π2
D 4π

The correct answer is:
π2

Solution:
Given, I=∫02πxsin8⁡xsin8⁡x+cos8⁡xdx ...i

Use the property ∫abfxdx=∫abfa+b-xdx we get,

I=∫02π2π-xsin8⁡2π-xsin8⁡2π-x+cos8⁡2π-xdx=∫02π2π-xsin8⁡xsin8⁡x+cos8⁡xdx ...ii

Adding equation i & ii we get,

2I=∫02π2πsin8⁡xsin8⁡x+cos8⁡xdx

Use the property ∫02afxdx=2∫0af2a-xdx we get,

I=2∫0ππsin8⁡xsin8⁡x+cos8⁡xdx

Use the property ∫02afxdx=2∫0af2a-xdx we get,

I=4∫0π/2πsin8⁡xsin8⁡x+cos8⁡xdx ...iii

Use the property ∫abfxdx=∫abfa+b-xdx we get,

I=4∫0π/2πcos8⁡xsin8⁡x+cos8⁡xdx ...iv ∵sinπ2-x=cosx

Adding the equation iii & iv we get,

2I=4π∫0π/21dx

⇒I=2πx0π/2=π2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2020 (09 Jan Shift 1)
The value of ∫0π3log1+3tan⁡xdx is equal to

Options:
A πlog⁡2
B π2log⁡2
C π3log⁡2
D π4log⁡2

The correct answer is:
π3log⁡2

Solution:
Let, Î=â«0Ï3log1+3tanâ¡xdx
Î=â«0Ï3log1+3tanÏ3-xdx
=â«0Ï3log1+3tan3-tanâ¡x1+3tanâ¡xdx
=â«0Ï3log1+3tanâ¡x+3-3tanâ¡x1+3tanâ¡xdx
Î=â«0Ï3logâ¡4-log1+3tanâ¡xdx
Î=logâ¡4Ï3-Î
Î=Ï6log4=Ï3log2

Topic: Mathematics, Definite Integration, JEE Main
The value of â«012Ïsinâ¡t+-sinâ¡tdt is equal to (where, . denotes the greatest integer function)

Options:
A 12π
B -12π
C -10π
D -6π

The correct answer is:
-12π

Solution:
We know that, x+-x=-1,x∉Z0,x∈Z
So, ∫012πsin⁡t+-sin⁡tdt
=6∫02πsin⁡t+-sin⁡tdt
=6∫0π2-1dt+∫π2π-1dt+∫π3π2-1dt+∫3π22π-1dt
=6-1π2+π2+π2+π2
=-12π

Topic: Mathematics, Definite Integration, JEE Main
The value of 0.16log2.513+132+133+….∞ is __________

Options:

The correct answer is:
4

Solution:
0.16log2.513+132+133+….∞

=0.16log2.5131-13=425log5212

=52-2log5212=12-2=4

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The value of ∫0πcosx3dx is

Options:
A 23
B 0
C 43
D -43

The correct answer is:
43

Solution:
I=∫0πcosx3dx=2∫0π2cos3xdx

=2∫0π21-sin2xcosxdx

=2∫0π2cosxdx-∫0π2sin2xcosxdx

=2sinx-sin3x30π2

=21-13-0+0=43

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The value of ∫0∞dx1+x4 is equal to

Options:
A π22
B π2
C π2
D 2π2

The correct answer is:
π22

Solution:
Let, I=â«0âx2+1-x21+x4dx
I=â«0âx21+x4dx+â«0â1-x21+x4dx
I=I1+I2
I2=â«0â1x2-1x2+1x2dx
Let, x+1x=t
I2=â«ââdtt2-2=0
I=â«0âx21+x4dx also I=â«0â11+x4dx
So, 2I=â«0â1+x21+x4dx
I=12â«0â1+1x2x2+1x2dx
Let, x-1x=t
I=12â«-ââdtt2+2
I=122tan-1â¡t2-ââ
=122Ï2+Ï2=Ï22

Topic: Mathematics, Definite Integration, JEE Main
The value of ∫ 0 π ( ∑ r=0 3 a r cos 3−r x sin r x )dx depends upon

Options:
A a1 and a2
B a0 and a3
C a2 and a3
D a1 and a3

The correct answer is:
a1 and a3

Solution:
Let, I=â«0Ïâr=03arcos3-râ¡xsinrâ¡x dx
=â«0Ïa0cos3â¡x dx+â«0Ïa1cos2â¡xsinâ¡xdx
+â«0Ïa2cosâ¡xsin2â¡x dx+â«0Ïa3sin3â¡x dx
Since, â«02afxdx
=2â«0afxdx ,if f2a-x=f(x)0 ,if f2a-x=-f(x)
â´ Integral Ist and IIIrd become zero.
â´The given integral depends upon a1and a3.

Topic: Mathematics, Definite Integration, JEE Main
The value of

11!50!+13!48!+15!46!+….+149!2!+151!1! is

Options:
A 25050!
B 25051!
C 25151!
D 25150!

The correct answer is:
25051!

Solution:
Let

S=11!·50!+13!·48!+15!·46!+....+151!·1!

⇒S=151!51!1!·50!+51!3!·48!+51!5!·46!+....+51!51!·0!

⇒S=151!C151+C351+C551+....+C5151

⇒S=251-151!

⇒S=25051!

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2023 (01 Feb Shift 1)
The value of ∫-11cot-1x+x3+x5x4+x2+1dx is equal to

Options:
A π2
B π4
C 3π4
D π

The correct answer is:
π

Solution:
x+x3+x5x4+x2+1 is an odd function, let it be equal to λ,
So,
Ιsay=∫01cot-1(λ)+cot-1-λdx=π∫01dx=π

Topic: Mathematics, Definite Integration, JEE Main
The value of ∫-11sin-1⁡x+x5+x3-1cos2⁡xdx is equal to

Options:
A tan⁡1
B 0
C 2tan⁡1
D -2tan⁡1

The correct answer is:
-2tan⁡1

Solution:
Let, I=∫-11sin-1⁡x+x5+x3-1cos2⁡xdx
I=∫-11-1cos2⁡x⁡dx
I=-∫-11sec2⁡xdx
I=-tan⁡x-11
I=-2tan⁡1

Topic: Mathematics, Definite Integration, JEE Main
The value of ∫-11x2ex3dx, where t denotes the greatest integer ≤t, is :

Options:
A e+13
B e-13e
C 13e
D e+13e

The correct answer is:
e+13e

Solution:
I=∫-11x2ex3dx

=∫-10x2ex3dx+∫01x2ex3dx

=∫-10x2e-1dx+∫01x2e0dx

=1e×x330-1+x3301

=1e×0--13+13

=13e+13=1+e3e

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (25 Feb Shift 1)
The value of ∫-11x-xdx (where [.] denotes greatest integer function) is

Options:
A 0
B 1
C 2
D None of these

The correct answer is:
1

Solution:
â«â11(xâ[x])dx=â«â11xdxââ«â11[x]dx
=[x22]â11â[â«â10[x]dx+â«01[x]dx]
=12[1â1]â[â«â10(â1)dx+â«010.dx]
[Ifâ1â¤x<0, [x]=â1If0â¤x<1, [x]=0]
=0â[âx]â10â0=0â[â0â(1)]=1

Topic: Mathematics, Definite Integration, JEE Main
The value of ∫-1212x+1x-12+x-1x+12-212 dx is:

Options:
A loge4
B 2loge16
C loge16
D 4loge(3+22)

The correct answer is:
loge16

Solution:
We have,

∫-1212x+1x-12+x-1x+12-212 dx

=∫-1212x-1x+1-x+1x-12dx

=∫-1212-4xx2-12dx

=∫-12124x1-x22dx

=∫-121216x21-x22dx

=∫-12124x1-x2dx

=2∫0124x1-x2dx

=4∫0122x1-x2dx

=-4∫012-2x1-x2dx

=-4loge1-x2012

=-4loge1-12

=4loge2

=loge16

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (26 Aug Shift 1)