datavorous/entrance-exam-dataset huggingface dataset:
The integral ∫xxsinx+cosx2dx is equal to, (where C is a constant of integration):

Options:
A tanx-xsecxxsinx+cosx+C
B secx+xtanxxsinx+cosx+C
C secx-xtanxxsinx+cosx+C
D tanx+xsecxxsinx+cosx+C

The correct answer is:
tanx-xsecxxsinx+cosx+C

Solution:
∫x2    dxxsinx+cosx2=∫xcosx⋅xcosxxsinx+cosx2 dx

=xcosx∫xcosxxsinx+cosx2dx−∫ddxxsecx∫xcosxxsinx+cosx2 dx dx

=xcosx−1xsinx+cosx+∫sec2x  dx, ∵    ∫xcosxxsinx+cosx2dx=−1xsinx+cosx

& ddxxsecx=secx+xsecxtanx=secx1+xsinxcosx=sec2xxsinx+cosx

=−xcosxxsinx+cosx+sinxcosx+C

=tanx−xsecxxsinx+cosx+C

Topic: Mathematics, Indefinite Integration, JEE Main
The integral ∫xxsinx+cosx2dx is equal to, (where C is a constant of integration):

Options:
A tanx-xsecxxsinx+cosx+C
B secx+xtanxxsinx+cosx+C
C secx-xtanxxsinx+cosx+C
D tanx+xsecxxsinx+cosx+C

The correct answer is:
tanx-xsecxxsinx+cosx+C

Solution:
∫x2    dxxsinx+cosx2=∫xcosx⋅xcosxxsinx+cosx2 dx

=xcosx∫xcosxxsinx+cosx2dx−∫ddxxsecx∫xcosxxsinx+cosx2 dx dx

=xcosx−1xsinx+cosx+∫sec2x  dx, ∵    ∫xcosxxsinx+cosx2dx=−1xsinx+cosx

& ddxxsecx=secx+xsecxtanx=secx1+xsinxcosx=sec2xxsinx+cosx

=−xcosxxsinx+cosx+sinxcosx+C

=tanx−xsecxxsinx+cosx+C

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2020 (04 Sep Shift 1)
The integrated rate equation is $R t=\log C_0-\log C_t$. The straight line graph is obtained by plotting

Options:
A
time vs $\log C_t$

B
$\frac{1}{\text { time }}$ vs $C_t$

C
time vs $C_t$

D
$\frac{1}{\text { time }}$ vs $\frac{1}{C_t}$

The correct answer is:
time vs $\log C_t$

Solution:
No Solution Available

Topic: Chemistry, Chemical Kinetics, JEE Main, JEE Main 2002
The integrating factor of $\frac{x d y}{d x}-y=x^{4}-3 x$ is

Options:
A $\mathrm{x}$
B $\log x$
C $\frac{1}{x}$
D $-x$

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

Solution:
Since $x \frac{d y}{d x}-y=x^{4}-3 x$
$$
\therefore \quad \frac{d y}{d x}-\frac{y}{x}=x^{3}-3
$$
Hence
$$
I F=e^{\int P d x}=e^{-\int \frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}
$$

Topic: Mathematics, Differential Equations, JEE Main
The integrating factor of $\frac{x d y}{d x}-y=x^4-3 x$ is

Options:
A
x
B $\log x$
C $\frac{1}{x}$
D $-\mathbf{X}$

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

Solution:
$\begin{aligned}
& \text { (c) Since } x \frac{d y}{d x}-y=x^4-3 x \\
& \therefore \frac{d y}{d x}-\frac{y}{x}=x^3-3
\end{aligned}$
Hence $I F=e^{\int P d x}=e^{-\int \frac{1}{x} d x}=e^{-\log x}=\frac{1}{x}$

Topic: Mathematics, Differential Equations, JEE Main
The integrating factor of \( \frac{d y}{d x}+y=\frac{1+y}{x} \) is

Options:
A \( x e^{x} \)
B \( x e^{\frac{1}{x}} \)
C \( \frac{e^{x}}{x} \)
D \( \frac{x}{e^{x}} \)

The correct answer is:
\( \frac{e^{x}}{x} \)

Solution:
Given differential equation,
\[
\begin{array}{l}
\frac{d y}{d x}+y=\frac{1+y}{x} \\
\Rightarrow \frac{d y}{d x}+y\left(1-\frac{1}{x}\right)=\frac{1}{x}
\end{array}
\]
So, integrating factor is given by
I.F. \( =e^{\int\left(1-\frac{1}{x}\right) d x}=e^{x} e^{-\ln x}=\frac{e^{x}}{x} \)

Topic: Mathematics, Differential Equations, JEE Main
The integrating factor of the differential equation $\frac{d y}{d x}+\frac{1}{x} y=x^{3}-3$ is

Options:
A $-y$
B $y$
C $x$
D $-x$

The correct answer is:
$x$

Solution:
$\frac{d y}{d x}+\frac{1}{x} y=x^{3}-3 \quad$ is linear differentiatial equation
$\therefore \text { I.F. }=\mathrm{e}^{\int \frac{1}{x} \mathrm{dx}}=\mathrm{e}^{\log x}=\mathrm{x}$

Topic: Mathematics, Differential Equations, JEE Main
The integrating factor of the differential equation $\frac{d y}{d x}(x \log x)+y=2 \log x$ is given by

Options:
A $e^{x}$
B $\log x$
C $\log (\log x)$
D $x$

The correct answer is:
$\log x$

Solution:
$\begin{aligned} & \frac{d y}{d x}(x \log x)+y=2 \log x \\ & \Rightarrow \quad \frac{d y}{d x}+\frac{y}{x \log x}=\frac{2}{x} \end{aligned}$
Here, $\quad P=\frac{1}{x \log x}, Q=\frac{2}{x}$
IF
$$
=e^{\int P d x}=e^{\int \frac{d x}{x \log x}}
$$
$=\log x$

Topic: Mathematics, Differential Equations, JEE Main
The integrating factor of the differential equation $\frac{d y}{d x}=y \tan x-y^2 \sec x$,

Options:
A $\tan x$
B $\sec x$
C $\cos x$
D $\cot x$

The correct answer is:
$\sec x$

Solution:
The differential equation is $\frac{d y}{d x}-y \tan x=-y^2 \sec x$
I.F. $=e^{-\int \tan x d x}$
This is Bernoulli's equation i.e. reducible to linear equation. Dividing the equation by $y^2$, we get
$\frac{1}{y^2} \frac{d y}{d x}-\frac{1}{y} \tan x=-\sec x$ .....(i)
Put $\frac{1}{y}=Y \Rightarrow-\frac{1}{y^2} \frac{d y}{d x}=\frac{d Y}{d x}$
Equation (i) reduces to $-\frac{d Y}{d x}-Y \tan x=-\sec x$
$\Rightarrow \frac{d Y}{d x}+Y \tan x=\sec x$, which is a linear equation
Hence I.F. $=e^{\int \tan x d x}=\sec x$.

Topic: Mathematics, Differential Equations, JEE Main
The integrating factor of the differential equation \( \frac{\mathrm{d} y}{\mathrm{~d} x}+\frac{y}{(1-x) \sqrt{x}}=1-\sqrt{x} \) may be

Options:
A \( \frac{1-\sqrt{x}}{1+\sqrt{x}} \)
B \( \frac{1+\sqrt{x}}{1-\sqrt{x}} \)
C \( \frac{1-x}{1+x} \)
D \( \frac{\sqrt{x}}{1-\sqrt{x}} \)

The correct answer is:
\( \frac{1+\sqrt{x}}{1-\sqrt{x}} \)

Solution:
Given, dydx+y1-xx=1-x

∴ If =e∫11-xx dx

Put x=t

⇒ 12x dx=dt

∴ If =e∫21-t2 dt

=e22log1+t1-t=1+t1-t=1+x1-x

Topic: Mathematics, Differential Equations, JEE Main
The integrating factor of the differential equation \( \left(2 x+3 y^{2}\right) d y=y d x(y>0) \) is

Options:
A \( -\frac{1}{y^{2}} \)
B \( \frac{1}{y^{2}} \)
C \( e^{\frac{1}{y}} \)
D \( \frac{1}{x} \)

The correct answer is:
\( \frac{1}{y^{2}} \)

Solution:
\( (\mathrm{B}) \)
\( \Rightarrow \frac{\left.2 x+3 y^{2}\right) \mathrm{dy}}{y}=\mathrm{y} \mathrm{dx} \)
\( \Rightarrow \frac{2 x}{y}+3 y=\frac{d x}{d y} \)
\( \Rightarrow \frac{d x}{d y}-\frac{2 x}{y}=3 y \)
If \( =e^{\int-\frac{2}{y} d y}=e^{-2 \log y}=e^{\log y^{-2}}=\frac{1}{y^{2}} \)

Topic: Mathematics, Differential Equations, JEE Main
The integrating factor of the differential equation $\left(x^2-1 \frac{d y}{d x}+2\right) x y=x$ is

Options:
A
$\frac{1}{x^2-1}$

B
$x^2-1$

C
$\frac{x^2-1}{x}$

D
$\frac{x}{x^2-1}$

The correct answer is:
$x^2-1$

Solution:
Given differential equation is
$$
\begin{aligned}
& \left(x^2-1\right) \frac{d y}{d x}+2 x y=x \\
& \Rightarrow \quad \frac{d y}{d x}+\frac{2 x}{x^2-1} \cdot y=\frac{x}{x^2-1}
\end{aligned}
$$
This is in linear form.
$$
\begin{aligned}
& \text { Integrating factor }=\int \frac{2 x}{x^2-1} d x=\int_e \frac{d t}{t} \\
& \text { where } \begin{aligned}
& =x^2-1 \\
& =e^{\log t}=x^2-1
\end{aligned}
\end{aligned}
$$
Hence, required integrating factor $=x^2-1$.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2012 (26 May Online)
The integrating factor of the differential equation $(x \log x) \frac{d y}{d x}+y=2 \log x$ is

Options:
A $\log x$
B $\log (\log x)$
C $e^x$
D $x$

The correct answer is:
$\log x$

Solution:
I.F. $=e^{\int \frac{1}{x \log x} d x}=e^{\log (\log x)}=\log x$.

Topic: Mathematics, Differential Equations, JEE Main
The intensity at the maximum in a Young's double slit experiment is I0. Distance between two slits is d=5λ, where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance D=10d ?

Options:
A I0
B I04
C 34I0
D I02

The correct answer is:
I02

Solution:
In YDSE Imax=I0
Path difference at a point in front of one of shifts is
∆x=dyD=dd2D=d22D ( Here  y= d 2 )
∆x=d2210d=d20=5λ20=λ4
Path difference is
ϕ=2πλΔx=2πλλ4
ϕ=π2
So intensity at that point is
I=Imaxcos2θ2
I=I0cos2π4=I02

Topic: Physics, Wave Optics, JEE Main
The intensity of a light pulse travelling along a communication channel decrease exponentially with distance $\mathrm{x}$ according to the relation $\mathrm{I}=\mathrm{I}_0 \mathrm{e}^{-\alpha x}$, where $\mathrm{I}_0$ is the intensity at $x=0$ and $\alpha$ is the attenuation constant.
(a) Show that the intensity reduces by $75 \%$ after a distance of $\left(\frac{\ln 4}{\alpha}\right)$.
(b) Attenuation of a signal can be expressed in decibel $(\mathrm{dB})$ according to the relation $\mathrm{dB}=10 \log _{10}\left(\frac{\mathrm{I}}{\mathrm{I}_0}\right)$. What is the attenuation in $\mathrm{dB} / \mathrm{km}$ for an optical fibre in which the intensity falls by $50 \%$ over a distance of $50 \mathrm{~km}$ ?

Options:

None

Solution:
(a) As given that intensity of a light pulse $\mathrm{I}=\mathrm{I}_0 \mathrm{e}^{-\mathrm{ax}}$ where, $\mathrm{I}_0$ is the intensity at $x=0$ and $\alpha$ is constant. According to the question remaining Intensity reduced by $75 \%$ so, remaining
$$
\mathrm{I}=25 \% \text { of } \mathrm{I}_0=\frac{25}{100} \mathrm{I}_0=\frac{\mathrm{I}_0}{4} \text { So, }\left(\mathrm{I}=\frac{\mathrm{I}_0}{4}\right)
$$
Using the formula mentioned in the question,
$$
\begin{aligned}
&\mathrm{I}=\mathrm{I}_0 \mathrm{e}^{-\alpha \mathrm{x}} \\
&\frac{\mathrm{I}_0}{4}=\mathrm{I}_0 \mathrm{e}^{-\alpha \mathrm{x}}
\end{aligned}
$$
So, $\frac{1}{4}=\mathrm{e}^{-\alpha \mathrm{x}}$
Taking $\log$ on both sides, we get
$$
\begin{array}{ll}
(\ln 1-\ln 4)=\log _e e^{-a x} & \\
\ln 1-\ln 4=-\alpha x \ln e & \left(\because \ln { }_e=1\right) \\
0-\ln 4=-\alpha x & (\because \ln 1=0) \\
x=\frac{\ln 4}{\alpha} &
\end{array}
$$
So, at distance $\mathrm{x}=\frac{\ln 4}{\alpha}$, the intensity is reduced by $75 \%$ of initial intensity.
(b) Let $\alpha$ be the attenuation in $\mathrm{dB} / \mathrm{km}$. If $\mathrm{x}$ is distance travelled by signal,
Then, $10 \log _{10}\left(\frac{\mathrm{I}}{\mathrm{I}_0}\right)=-\alpha \mathrm{x}$
where, $\mathrm{I}_0$ is the intensity initially.
According to the question Intensity reduced by $50 \%$ so remain
$\mathrm{I}=\left(50 \%\right.$ of $\left.\mathrm{I}_0\right)=\frac{\mathrm{I}_0}{2}$ and $\mathrm{x}=50 \mathrm{~km}$
Putting the value of $x$ in Eq. (i), we get
$$
\begin{aligned}
&10 \log _{10}\left(\frac{\mathrm{I}_0}{2 \mathrm{I}_0}\right)=-\alpha \times 50 \\
&10[\log 1-\log 2]=-50 \alpha
\end{aligned}
$$
$$
\frac{10 \times 0.3010}{50}=\alpha, \alpha=0.0602 \mathrm{~dB} \text { per km }
$$
So, The attenuation of signal for an optical fibre channel. $\alpha=0.0602 \mathrm{~dB} / \mathrm{km}$

Topic: Physics, Communication System
The intensity of gamma radiation from a given source is I. On passing through $36 \mathrm{~mm}$ of lead, it is reduced to $\frac{I}{8}$. The thickness of lead which will reduce the intensity to $\frac{I}{2}$ will be

Options:
A
$6 \mathrm{~mm}$

B
$9 \mathrm{~mm}$

C
$18 \mathrm{~mm}$

D
$12 \mathrm{~mm}$

The correct answer is:
$12 \mathrm{~mm}$

Solution:
Use I $=I_0 e^{-\mu x}$

Topic: Physics, Nuclear Physics, JEE Main, JEE Main 2005
The intensity of light emerging from one of the slits in a Young's double slit experiment is found to be $1.5$ limes the intensity of light emerging from the other slit. What will be the approximate ratio of intensity of an interference maximum to that of an interference minimum

Options:
A $2.25$
B 98
C 5
D $9.9$

The correct answer is:
98

Solution:
Hint:
$I_{1}=1.51_{2}$
$\frac{\left.\right|_{1}}{I_{2}}=\frac{3}{2} \cdot \quad \frac{l_{\max }}{I_{\min }}=\left(\frac{\sqrt{L_{1}}+\sqrt{L_{2}}}{\sqrt{I_{1}}-\sqrt{L_{2}}}\right)^{2}=\left(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}\right)^{2}=98$

Topic: Physics, Waves and Sound, JEE Main
The intensity of magnetization of a bar magnet is $5.0 \times 10^{4} \mathrm{Am}^{-1}$. The magnetic length and the area of cross-section of the magnet are $12 \mathrm{cm}$ and $1 \mathrm{cm}^{2}$ respectively. The magnitude of magnetic moment of this bar magnet is (in SI unit)

Options:
A 0.6
B 1.3
C 1.24
D 2.4

The correct answer is:
0.6

Solution:
We know that Intensity of magnetisation
$I=\frac{M}{V}$
[where $M=$ magnetic moment, $V=$ volume]
So
$\begin{aligned}
M=I V &=5.0 \times 10^{4} \times \frac{12}{100} \times \frac{1}{(100)^{2}} \\
&=60 \times 10^{4} \times 10^{-6}=0.6 \mathrm{Am}^{2}
\end{aligned}$

Topic: Physics, Magnetic Properties of Matter, JEE Main
The intensity of Î³ - radiation from a given source is I. On passing through 36 mm of lead, it is reduced to I8. Assuming that the intensity in transmitted radiation varies exponentially with the thickness of the material, then the thickness of lead, which will reduce the intensity to I2, will be

Options:
A 6 mm
B 9 mm
C 18 mm
D 12 mm

The correct answer is:
12 mm

Solution:
On passing through lead, the intensity of γ - radiation decreases exponentially with the thickness traversed and is given by

I=I0e-μx , where μ is the absorption coefficient of the lead.

In the first case, when x=36 mm, I=I08

I08=I0 e-μ×36

⇒loge8=36μ

In the second case,

I=I02

Let x be the thickness of the lead sheet, then

I02=I0 e-μx

loge⁡2=μx

μx36μ=loge2loge8

x=36×loge⁡2loge⁡23=36×loge⁡23loge⁡2

x=12 mm

Topic: Physics, Electromagnetic Waves, JEE Main
The intensity of sound increases at night due to

Options:
A Increase in density of air
B Decreases in density of air
C Low temperature
D None of these

The correct answer is:
Increase in density of air

Solution:
We know that at night amount of carbon dioxide in atmosphere increases which raises the density of atmosphere
Since intensity is directly proportional to density, intensity of sound is more at night.

Topic: Physics, Waves and Sound, JEE Main
The intensity of the light from a bulb incident on a surface is 0.22 W m-2. The amplitude of the magnetic field in this light-wave is _____ ×10-9 T

(Given : Permittivity of vacuum ϵ0=8.85×10-12 C2 N-1 m-2, speed of light in vacuum c=3×108 m s-1)

Options:

The correct answer is:
43

Solution:
The intensity is defined as energy per unit time per unit area. Therefore, I=dEAdt.

Energy density is energy per unit volume. Therefore, Ud=dEAdx.

Dividing IUd=dxdt=c

⇒Ud=Ic=0.223×108=223×10-10 J m-3

Magnetic energy density will be half of total energy density, Therefore, Ub=113×10-10 J m-3.

Now energy density in magnetic field is given by, Ub=B22μ0. Therefore,

Brms22μ0=113×10-10 J m-3 where Brms2=B02=B022

⇒B024μ0=113×10-10⇒B0=4×4π×10-7×113×10-10⇒B0=4.29×10-8 T≃43×10-9 T

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2022 (29 Jun Shift 1)
The intensity ratio of two coherent sources of light is P. They are interfering in the same region and produce an interference pattern. Then the fringe visibility is

Options:
A 1 + P2P
B 2P1 + P
C P1 + P
D 2P1 + P

The correct answer is:
2P1 + P

Solution:
Visibility, V= Imax - Imin Imax+ Imin= 2I1I2I1 + I2
= 2I1I2I1I2 + 1= 2PP +1

Topic: Physics, Wave Optics, JEE Main
The interaction between π bond and lone pair of electrons present on an adjacent atom is responsible for

Options:
A Hyperconjugation
B Inductive effect
C Electromeric effect
D Resonance effect

The correct answer is:
Resonance effect

Solution:
Resonance Effect – The polarity induced in a molecule by the interaction of a lone pair of electrons with a pi bond or the interaction of two pi bonds in nearby atoms is known as the resonance effect.

The inductive and resonance effect can be effectively employed to explain the stability of carbocations, free radicals and carbanions and the acidic and basic strengths of various organic compounds.

Topic: Chemistry, General Organic Chemistry, JEE Main, JEE Main 2024 (29 Jan Shift 1)
The interaction between two species wherein one species is benefited and the other remains unaffected is called ___.

Options:
A Commensalism
B Mutualism
C Predation
D Camouflage

The correct answer is:
Commensalism

Solution:
The interaction between two species wherein one species is benefited and the other remains unaffected is called Commensalism

Topic: Biology, Organisms and Populations, NEET
The interaction energy of London forces between two particles is proportional to rx, where r is the distance between the particles. The value of x is :

Options:
A 6
B -6
C 3
D -3

The correct answer is:
-6

Solution:
In London force interaction energy is inversely proportional the sixth power of the distance between two interacting particles 1r6

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The intercept cut off from $y$-axis is twice that from $x$-axis by the line and line is passes through $(1,2)$ then its equation is

Options:
A $2 x+y=4$
B $2 x+y+4=0$
C $2 x-y=4$
D $2 x-y+4=0$

The correct answer is:
$2 x+y=4$

Solution:
Let the intercept be $a$ and $2 a$, then the equation of line is $\frac{x}{a}+\frac{y}{2 a}=1$, but it also passes through $(1,2)$,
therefore $a=2$
Hence the required equation is $2 x+y=4$

Topic: Mathematics, Straight Lines, JEE Main
The intercept of the latusrectum to the parabola $y^{2}=4 a x$ are $b$ and $k$, then $k$ is equal to

Options:
A $\frac{a b}{a-b}$
B $\frac{a}{b-a}$
C $\frac{b}{b-a}$
D $\frac{a b}{b-a}$

The correct answer is:
$\frac{a b}{b-a}$

Solution:
For latusrectum $P Q$.
$$
y\left(t_{1}+t_{2}\right)-2 x-2 a t_{1} t_{2}=0
$$
and $\quad t_{1} t_{2}=-1$
For any point $P(x, y)$, the focal distance is $a+x$. $\therefore \quad b=a+x=a+a t_{1}^{2}=a\left(1+t_{1}^{2}\right) \quad \ldots(\mathrm{i})$
$c=a+x=a+a t_{2}^{2}=a+\frac{a}{t_{1}^{2}} \quad\left(\because t_{1} t_{2}=-1\right)$
$=\frac{a\left(1+t_{1}^{2}\right)}{t_{1}^{2}}$...(ii)
$\frac{b}{c}=t_{1}^{2}$
$\therefore$ From Eq. (i), $b=a\left(1+\frac{b}{c}\right)$
$\Rightarrow \quad b=a+\frac{a b}{c}$
$c=\frac{a b}{b-a}$

Topic: Mathematics, Parabola, JEE Main
The intercept on the line $y=x$ by the circle $x^2+y^2-2 x=0$ is $A B$. Equation of the circle on $\mathrm{AB}$ as a diameter is

Options:
A
$x^2+y^2-x-y=0$

B
$x^2+y^2-x+y=0$

C
$\mathrm{x}^2+\mathrm{y}^2+\mathrm{x}+\mathrm{y}=0$

D
$x^2+y^2+x-y=0$

The correct answer is:
$x^2+y^2-x-y=0$

Solution:
Points of intersection of line $y=x$ with $x^2+y^2-2 x=0$ are $(0,0)$ and $(1,1)$ hence equation of circle having end points of diameter $(0,0)$ and $(1,1)$ is $x^2+y^2-x-y=0$

Topic: Mathematics, Circle, JEE Main, JEE Main 2004
The intercepts made on the x, y and z axes, by the plane which bisects the line joining the points 1,2,3 and -3,4,5 at right angles, are a, b and c respectively, then the ordered triplet a,b,c is

Options:
A -92,9,9
B 92,9,9
C 9,-92,9
D 9,92,9

The correct answer is:
-92,9,9

Solution:
Let, equation of the plane is xa+yb+zc=1.

Mid-point M of line joining P1,2,3 and Q-3,4,5 is -1,3,4.

It lies on the plane i.e.
-1a+3b+4c=1
Also, PQ is parallel to the normal of the plane, so,
1a-4=1b2=1c2=λ
⇒1a=-4λ, 1b=2λ, 1c=2λ
⇒4λ+6λ+8λ=1
⇒λ=118
i.e. a=-92, b=9, c=9
Hence, ordered pair is -92,9,9

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The intercepts on the x-axis made by tangents to the curve, y=∫0xt dt, x∈R, which are parallel to the line y=2x, are equal to

Options:
A ± 3
B ± 4
C Â± 1
D ± 2

The correct answer is:
± 1

Solution:
Given y=∫0xtdt
If x>0, then
y=∫0xtdt
⇒y=x22
Ff x<0, then y=− x 2 2
∴ y={x2/2,x>0-x2/2,x<0
Now, dydx={x,x>0-x,x<0
Slope of tangent =2
Hence x=2, y=2 or x=-2, y=-2
Eq. of tangent y=2x-2 or y=2x+2
Hence, x-intercept =1 or -1.

Topic: Mathematics, Application of Derivatives, JEE Main, JEE Main 2013 (07 Apr)
The interference pattern is obtained with two coherent light sources of intensity ratio 4:1. And the ratio Imax+IminImax-Imin is 5x. Then, the value of x will be equal to :

Options:
A 3
B 4
C 2
D 1

The correct answer is:
4

Solution:
The ratio of the intensities is I1I2=41

After taking square root of above equation on both sides, we get, I1I2=21

Apply componendo and dividendo,
⇒I1+I2I1-I2=2+12-1=31
⇒I1+I2I1-I22=ImaxImin=91⇒Imax+IminImax-Imin=108=54

Therefore, the answer is 4.

Topic: Physics, Wave Optics, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The interior angles of a polygon are in A.P. If the smallest angle be $120^{\circ}$ and the common difference be $5^{\circ}$, then the number of sides is

Options:
A $8$
B $10$
C $9$
D $6$

The correct answer is:
$9$

Solution:
Let the number of sides of the polygon be $n$
Then the sum of interior angles of the polygon
$=(2 n-4) \frac{\pi}{2}=(n-2) \pi$
Since the angles are in A.P. and $a=120^{\circ}, d=5$ therefore $\frac{n}{2}[2 \times 120+(n-1) 5]=(n-2) 180$
$\Rightarrow n^2-25 n+144=0 \Rightarrow(n-9)(n-16)=0 \Rightarrow n=9,16$
But $n=16$ gives $T_{16}=a+15 d=120^{\circ}+15.5^{\circ}=195^{\circ}$, which is impossible as interior angle cannot be greater
than $180^{\circ}$. Hence $n=9$.

Topic: Mathematics, Sequences and Series, JEE Main
The intermediate compound formed in the conversion of starch to glucose is

Options:
A Lactose
B Sucrose
C Maltose
D Fructose

The correct answer is:
Maltose

Solution:

Topic: Chemistry, Biomolecules, JEE Main
The intermediate formed during the addition of $\mathrm{HCl}$ to propene in the presence of peroxide is

Options:
A $\mathrm{CH}_{3} \mathrm{CHCH}_{2} \mathrm{CI}$
B $\mathrm{CH}_{2} \mathrm{CHCH}_{3}$
C $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2}$
D $\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{C} \mathrm{H}_{2}$

The correct answer is:
$\mathrm{CH}_{2} \mathrm{CHCH}_{3}$

Solution:
The addition of HCl to propene proceeds by ionic mechanism and not by free radical mechanism. Hence it forms intermediate carbonium ion.

Topic: Chemistry, General Organic Chemistry, JEE Main
The intermediate never formed during chain growth polymerization is

Options:
A
B
C
D

The correct answer is:

Solution:
Chain growth polymerisation can be free radical, cationic, anionic polymerisation.
Carbenes are never produced during chain growth polymerisation.

Topic: Chemistry, Polymers, JEE Main
The intermediate product (X) formed in the following reaction is
B2H6+6NH3→3X→heat2B3N3H6+12H2

Options:
A BHNH33+BH4-
B BH2NH34+BH4-
C BHNH34+BH4-
D BH2NH32+BH4-

The correct answer is:
BH2NH32+BH4-

Solution:
3B2H6Diborane+6NH3→low temp.3BH2NH32+BH4-(X)→473KΔ2B3N3H6Borazine+12H2

Topic: Chemistry, p Block Elements (Group 13 & 14), JEE Main
The intermediate solutions of constraints must be checked by substituting them back into

Options:
A Objective function
B Constraint equations
C Not required
D None of these

The correct answer is:
Constraint equations

Solution:
Constraint equations
This is because the intermediate solutions must satisfy the constraint equations to ensure they are valid solutions within the feasible region.

Topic: Mathematics, Linear Programming, JEE Main
The intermolecular interaction that is dependent on the inverse cube of distance between the molecules is:

Options:
A Hydrogen bond
B Ion-ion interaction
C Ion-dipole interaction
D London force

The correct answer is:
Hydrogen bond

Solution:
For Hydrogen bond which is a case of dipole-dipole interaction, intermolecular force
F∝1r3
r is distance between two dipoles

Topic: Chemistry, States of Matter, JEE Main, JEE Main 2015 (04 Apr)
The intermolecular potential energy for the moleculesA ,B , C and D given below suggests that:

Options:
A A-D has the shortest bond length
B A-A has the largest bond enthalpy
C D is more electronegative than other atoms.
D A-B has the stiffest bond.

The correct answer is:
A-B has the stiffest bond.

Solution:
Bond enthalpy of AB bond is highest so A-B bond is more strong and B is highest electronegative atom. Order of bond length ⇒A-A<A-B<A-C<A-D

Topic: Chemistry, Chemical Bonding and Molecular Structure, JEE Main, JEE Main 2020 (04 Sep Shift 1)
The internal energy change in a system that has absorbed $2 \mathrm{k}$ cal of heat and done $500 \mathrm{~J}$ of work is :

Options:
A $7900 \mathrm{~J}$
B $8900 \mathrm{~J}$
C $6400 \mathrm{~J}$
D $5400 \mathrm{~J}$

The correct answer is:
$7900 \mathrm{~J}$

Solution:
$\mathrm{dU}=\mathrm{Q}-\mathrm{W}=8400-500=7900 \mathrm{~J}$

Topic: Physics, Thermodynamics, NEET, NEET 2009 (Mains)
The internal energy change (in J) when 90 g of water undergoes complete evaporation at 100°C is ..............

(Given : ΔHvap for water at 373 K=41 kJ/mol, R=8.314 JK-1 mol-1)

Options:

The correct answer is:
189494

Solution:
Number of moles of water =9018=5 moles

ΔH=ΔU+ΔngRT

41000X5=ΔU+5×8.314×373

205000=ΔU+15505.61

ΔU=189494.39 J=189494 J

Topic: Chemistry, Thermodynamics (C), JEE Main, JEE Main 2020 (02 Sep Shift 1)