datavorous/entrance-exam-dataset huggingface dataset:
Light of wavelength $550 \mathrm{~nm}$ falls normally on a slit of width $22.0 \times 10^{-5} \mathrm{~cm}$. The angular position of the second minima from the central maximum will be (in radians)

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

B
$\frac{\pi}{12}$

C
$\frac{\pi}{4}$

D
$\frac{\pi}{6}$

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

Solution:
If angular position of $2^{\text {nd }}$ maxima from central maxima is $\theta$ then
$$
\begin{aligned}
&\sin \theta=\frac{(2 \mathrm{n}-1) \lambda}{2 \mathrm{a}}=\frac{3 \lambda}{2 \mathrm{a}}=\frac{3 \times 550 \times 10^{-9}}{2 \times 22 \times 10^{-7}} \\
&\therefore \quad \theta=\frac{\pi}{8} \mathrm{rad}
\end{aligned}
$$

Topic: Physics, Wave Optics, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
Light of wavelength 550 nm falls normally on a slit of width 22.0×10-5 cm. The angular position of the second minima from the central maximum will be (in radians):

Options:
A π4
B π8
C π12
D π6

The correct answer is:
π6

Solution:
asinθ=2λ, 2nd dark
⇒sinθ=2λa=2×550×10−922×10−7⇒θ≃π6 rad

Topic: Physics, Wave Optics, JEE Main, JEE Main 2018 (15 Apr)
Light of wavelength 600 nm is incident normally on a slit of width 0.2 mm. The angular width of central maxima in the diffraction pattern is (measured from minimum to minimum)

Options:
A 6Ã10-3 rad
B 4Ã10-3 rad
C 2.4Ã10-3 rad
D 4.5Ã10-3 rad

The correct answer is:
6Ã10-3 rad

Solution:
Î»=600 nm=600Ã10-9m
d=0.2 mm=0.2Ã10-3m
âµ Linear width of central maximum,
x=2Î¸D
Where, 2Î¸: Angular width of central maximum.
We know that, x=2Î»Dd
Angular width 2Î¸.D=2Î»Dd
â 2Î¸=2Î»d=2Ã600Ã10-90.2Ã10-3
=6Ã10-7Ã104
=6Ã10-3 rad

Topic: Physics, Wave Optics, JEE Main
Light of wavelength 600 nm is incident normally on a slit of width 0.2 mm. The angular width of the central maxima in the diffraction pattern is nÃ10-3 rad, then what is the value of n? (measured from minimum to minimum)

Options:

The correct answer is:
6

Solution:
Î»=600 nm=600Ã10-9m
d=0.2 mm=0.2Ã10-3m
âµ Linear width of central maximum,
x=2Î¸D
Where, 2Î¸: Angular width of central maximum.
We know that, x=2Î»Dd
Angular width 2Î¸.D=2Î»Dd
â 2Î¸=2Î»d=2Ã600Ã10-90.2Ã10-3
=6Ã10-7Ã104
=6Ã10-3 rad

Topic: Physics, Wave Optics, JEE Main
Light of wavelength 4000 â« is allowed to fall on a metal surface having work function 2 eV. The maximum velocity of the emitted electrons is (h=6.6Ã10-34 Js)

Options:
A 1.35Ã105 m s-1
B 2.7Ã105 m s-1
C 6.2Ã105 m s-1
D 8.1Ã105 m s-1

The correct answer is:
6.2Ã105 m s-1

Solution:
12mv2=hcÎ»-Ï (in eV)
=6.6Ã10-34Ã3Ã1084000Ã10-10Ã1.6Ã10-19-2
=3.1-2=1.1 eV=1.1Ã1.6Ã10-19 J
=1.76Ã10-19 J
v=1.76Ã10-19Ã29Ã10-31
=6.2Ã105 m s-1

Topic: Physics, Dual Nature of Matter, JEE Main
Light of wavelength $\lambda=5000 A$ falls normally on a narrow slit. A screen placed at a distance of $1 \mathrm{~m}$ from the slit and perpendicular to the direction of light. The first minima of the diffraction pattern is situated at 5 $\mathrm{mm}$ from the centre of central maximum. The width of the slit is

Options:
A $0.1 \mathrm{~mm}$
B $1.0 \mathrm{~mm}$
C $0.5 \mathrm{~mm}$
D $0.2 \mathrm{~mm}$

The correct answer is:
$0.1 \mathrm{~mm}$

Solution:
Position of $\mathrm{n}^{\text {th }}$ minima $x_n=\frac{n \lambda D}{d}$
$\begin{array}{ll}\Rightarrow & \quad 5 \times 10^{-3}=\frac{1 \times 5000 \times 10^{-10} \times 1}{d} \\ \Rightarrow & d=10^{-4} \mathrm{~m}=0.1 \mathrm{~mm}\end{array}$

Topic: Physics, Wave Optics, JEE Main
Light propagates 2 cm distance in a glass of refractive index 1.5 in time t0. In the same time t0, light propagates a distance of 2.25 cm in a medium. The refractive index of the medium is

Options:
A 43
B 32
C 83
D None of these

The correct answer is:
43

Solution:
For a given time, optical path remains constant
∴ μ 1 x 1 = μ 2 x 2
or 1 . 5 × 2 = μ 2 × 2 . 2 5
∴ μ 2 = 1 . 5 × 2 2 . 2 5 = 2 1 . 5 = 2 0 1 5 = 4 3

Topic: Physics, Wave Optics, NEET
Light travels in two media M1 and M2 with speeds 1.5×108 m s-1 and 2.0×108 m s-1 respectively. The critical angle between them is

Options:
A tan-137
B tan-123
C cos-134
D sin-123

The correct answer is:
tan-137

Solution:
Relative refractive index can be found as,

μ2μ1=v1v2=1.52=34.

When the angle of incident is greater than the critical angle total internal reflection occurs. If the critical angle is ic then,

sinic=μ2μ1=34

Now tanic=342-32=37

Therefore, ic=tan-137

Topic: Physics, Ray Optics, JEE Main, JEE Main 2022 (26 Jul Shift 2)
Light wave enters from medium 1 to medium 2. Its velocity in $2^{\text {nd }}$ medium is double from $1^{\text {st }}$. For total internal reflection the angle of incidence must be greater than

Options:
A $30^{\circ}$
B $60^{\circ}$
C $45^{\circ}$
D $90^{\circ}$

The correct answer is:
$30^{\circ}$

Solution:
$\frac{\mu_2}{\mu_1}=\frac{v_1}{v_2}=\frac{1}{2} \Rightarrow \frac{\mu_1}{\mu_2}=2\left(\mu_1\gt\mu_2\right)$
For total internal reflection ${ }_2 \mu_1=\frac{1}{\sin C} \Rightarrow \frac{\mu_1}{\mu_2}$
$=\frac{1}{\sin C} \Rightarrow 2=\frac{1}{\sin C} \Rightarrow C=30^{\circ}$

Topic: Physics, Ray Optics, JEE Main
Light wave travelling in air along x-direction is given by Ey=540sinπ×104x-ct V m-1. Then, the peak value of magnetic field of wave will be (Given c=3×108 m s-1)

Options:
A 18×10-7 T
B 54×10-7 T
C 54×10-8 T
D 18×10-8 T

The correct answer is:
18×10-7 T

Solution:
Given here, Ey=540sinπ×104x-ct V m-1

Here, peak value of electric field, E0=540 V m-1

Using relation between magnetic and electric field,

Peak value of magnetic field is B0=E0c=5403×108=18×10-7 T

Topic: Physics, Electromagnetic Waves, JEE Main, JEE Main 2022 (25 Jul Shift 2)
Light year is a unit of

Options:
A Time
B Mass
C Distance
D Energy

The correct answer is:
Distance

Solution:
Light year is a distance which light travels in one year.

Topic: Physics, Units and Dimensions, JEE Main
Lights of wavelengths Î»1=340 nm and Î»2=540 nm are incident on a metallic surface. If the ratio of the maximum speeds of the electrons ejected is 2, the work function of the metal is

Options:
A 2 eV
B 1.85 eV
C 1 eV
D 1.5 eV

The correct answer is:
1.85 eV

Solution:
λ1=350 nm, λ2=540 nm, V1V2=2⇒V1=2V and V2=V
hcλ1=ϕ+12m(2V)2
hcλ2=ϕ+12mV2
hcλ1-hcλ2×4=ϕ-4ϕ
hc(4540 nm-1340 nm)=3ϕ
ϕ=1.8 eV

Topic: Physics, Dual Nature of Matter, JEE Main
$\lim _{\theta \rightarrow 0} \frac{1-\cos \theta}{\theta^2}=$

Options:
A $1$
B $2$
C $1/2$
D $1/4$

The correct answer is:
$1/2$

Solution:
$\lim _{\theta \rightarrow 0} \frac{2 \sin ^2(\theta / 2)}{\theta^2}=\frac{1}{2}$.

Topic: Mathematics, Limits, JEE Main
$\lim _{\theta \rightarrow 0} \frac{\sin 3 \theta-\sin \theta}{\sin \theta}=$

Options:
A $1$
B $2$
C $1/3$
D $3/2$

The correct answer is:
$2$

Solution:
$\begin{aligned}
& \lim _{\theta \rightarrow 0}\left[3-4 \sin ^2 \theta\right]-1=2 . \\
& \text { Aliter : } \lim _{\theta \rightarrow 0} \frac{\sin 3 \theta-\sin \theta}{\sin \theta}=\lim _{\theta \rightarrow 0} \frac{\sin 3 \theta}{\sin \theta}-\lim _{\theta \rightarrow 0} \frac{\sin \theta}{\sin \theta} \\
&=\frac{3}{1}-1=2 .
\end{aligned}$
You may also apply $L$-Hospital rule.

Topic: Mathematics, Limits, JEE Main
$\lim _{\alpha \rightarrow \beta}\left[\frac{\sin ^2 \alpha-\sin ^2 \beta}{\alpha^2-\beta^2}\right]=$

Options:
A $0$
B $1$
C $\frac{\sin \beta}{\beta}$
D $\frac{\sin 2 \beta}{2 \beta}$

The correct answer is:
$\frac{\sin 2 \beta}{2 \beta}$

Solution:
$\lim _{\alpha \rightarrow \beta} \frac{\sin ^2 \alpha-\sin ^2 \beta}{\alpha^2-\beta^2}$
Applying L-Hospital's rule,
$\lim _{\alpha \rightarrow \beta} \frac{2 \sin \alpha \cos \alpha}{2 \alpha}=\lim _{\alpha \rightarrow \beta} \frac{\sin 2 \alpha}{2 \alpha}=\frac{\sin 2 \beta}{2 \beta}$

Topic: Mathematics, Limits, JEE Main
$\lim _{\alpha \rightarrow \pi / 4} \frac{\sin \alpha-\cos \alpha}{\alpha-\frac{\pi}{4}}=$

Options:
A $\sqrt{2}$
B $1 / \sqrt{2}$
C $1$
D None of these

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

Solution:
$\begin{aligned}
& \lim _{\alpha \rightarrow \pi / 4} \frac{\sin \alpha-\cos \alpha}{\alpha-\pi / 4} \\
& =\lim _{\alpha \rightarrow \pi / 4}\left\{\frac{\sqrt{2}\left(\sin \alpha \cdot \frac{1}{\sqrt{2}}-\cos \alpha \cdot \frac{1}{\sqrt{2}}\right)}{\left(\alpha-\frac{\pi}{4}\right)}\right\} \\
& =\sqrt{2} \lim _{\alpha \rightarrow \pi / 4} \frac{\sin \left(\alpha-\frac{\pi}{4}\right)}{\left(\alpha-\frac{\pi}{4}\right)}=\sqrt{2} \times 1=\sqrt{2} .
\end{aligned}$
Aliter : Apply L-Hospital's rule,
$\lim _{\alpha \rightarrow \pi / 4} \frac{\sin \alpha-\cos \alpha}{\alpha-(\pi / 4)}=\lim _{\alpha \rightarrow \pi / 4} \frac{\cos \alpha+\sin \alpha}{1}=\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}=\sqrt{2} .$

Topic: Mathematics, Limits, JEE Main
$\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}=$

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

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

Solution:
$\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}=\lim _{h \rightarrow 0} \frac{(\sqrt{x+h})^2-(\sqrt{x})^2}{h(\sqrt{x+h}+\sqrt{x})}=\frac{1}{2 \sqrt{x}} .$
Aliter : Apply L-Hospital rule,
$\lim _{h \rightarrow 0} \frac{\sqrt{x+h}-\sqrt{x}}{h}=\lim _{h \rightarrow 0} \frac{1}{2 \sqrt{x+h}}=\frac{1}{2 \sqrt{x}} .$

Topic: Mathematics, Limits, JEE Main
limn→∞1+1+12+……+1nn2n is equal to

Options:
A 1e
B 0
C 12
D 1

The correct answer is:
1

Solution:
It is 1∞ form

L=elimn→∞1+12+13+…..+1nn

L=elimn→∞1n+12n+13n+…..+1n2

∴ L=e0+0+0+…..+0

⇒L=e0=1

Topic: Mathematics, Limits, JEE Main, JEE Main 2021 (25 Feb Shift 1)
limn→∞11+n+12+n+13+n+…+12n is equal to :-

Options:
A 0
B loge2
C loge32
D loge23

The correct answer is:
loge2

Solution:
We have,

limn→∞11+n+12+n+13+n+…+12n

=limn→∞11+n+21+n+31+n+….+1n+n

=limn→∞∑ r=1n1n+r

=limn→∞ ∑r=1n1n11+rn

=∫0111+xdx

=loge1+x01

=loge2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (01 Feb Shift 1)
limn→∞11+n5+2425+n5+3435+n5+…+n4n5+n5=

Options:
A 15log3
B 13log5
C 12log5
D log25

The correct answer is:
log25

Solution:
limn→∞11+n5+2425+n5+3435+n5+…+n4n5+n5

=limn→∞∑r=1nr4r5+n5

=limn→∞1n∑r=1nrn4rn5+1=∫01x41+x5dx

=15ln1+x501=15ln2=ln25

Topic: Mathematics, Limits, JEE Main
limn→∞12n11-12n+11-22n+11-32n+….+11-2n-12n is equal to

Options:
A 12
B 1
C 2
D -2

The correct answer is:
2

Solution:
Let I=limn→∞12n11-12n+11-22n+11-32n+….+11-2n-12n

Let 2n=t and if n→∞ then t→∞

I=limn→∞1t∑r=1t-111-rt

I=∫01dx1-x (Using definite integral as limit of sum)

Applying ∫0afxdx=∫0afa-xdx, we get

I=∫01dxx

=2x01=2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2022 (25 Jul Shift 2)
limn→∞1n+nn+12+nn+22+…+n2n-12 is equal to

Options:
A 12
B 14
C 13
D 1

The correct answer is:
12

Solution:
limn→∞1n+nn+12+nn+22+…+n2n-12

=limn→∞∑r=0n-1nn+r2=limn→∞∑r=0n-1nn2+2nr+r2

=limn→∞1n∑r=0n-11r/n2+2r/n+1

=∫01dxx+12=-1x+101=12

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (25 Feb Shift 2)
limn→∞3n4+2+1n2+2+2n2+…+3-1n2 is equal to

Options:
A 12
B 193
C 0
D 19

The correct answer is:
19

Solution:
Let

L=limn→∞3n4+2+1n2+2+2n2+...+3-1n2

⇒L=limn→∞3n2+0n2+2+1n2+2+2n2+...+2+1-1n2

⇒L=3limn→∞1n∑r=0n-12+rn2

⇒L=3∫012+x2dx

⇒L=2+x301

⇒L=27-8

⇒L=19

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (30 Jan Shift 2)
$\lim _{n \rightarrow \infty}\left(4^n+5^n\right)^{1 / n}$ is equal to

Options:
A $4$
B $5$
C $e$
D None of these

The correct answer is:
$5$

Solution:
$\begin{aligned} & \text { Given limit }=\lim _{n \rightarrow \infty}\left(4^n+5^n\right)^{1 / n} \\ & =\lim _{n \rightarrow \infty} 5\left[\left\{1+\left(\frac{4}{5}\right)^n\right\}^{(5 / 4)^n}\right]^{(1 / n) \cdot(4 / 5)^n}=5 \cdot e^0=5 . \\ & \left(\because\left(\frac{4}{5}\right)^n \rightarrow 0 \text { as } n \rightarrow \infty\right)\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{n \rightarrow \infty} \frac{a^{n}+b^{n}}{a^{n}-b^{n}},$ where $a>b>1,$ is equal to

Options:
A -1
B 1
C 0
D None

The correct answer is:
1

Solution:
limit $=\lim _{n \rightarrow \infty} \frac{1+\left(\frac{b}{a}\right)^{n}}{1-\left(\frac{b}{a}\right)^{n}}=1,$

because $0<\frac{b}{a}<1$ implies

$\left(\frac{b}{a}\right)^{n} \rightarrow 0$ as $n \rightarrow \infty$

Topic: Mathematics, Limits, JEE Main
limn→∞n+1 n+2….3nn2n1n is equal to

Options:
A 9e2
B 3log3-2
C 18e4
D 27e2

The correct answer is:
27e2

Solution:
limn→∞n+1n+2….3nn2n1n
Let y=n+1n+2…. n+2nn2n1n
y=n+1n·n+2n……n+2nn1n
logy=1nlog1+1n+log1+2n+…..log1+2nn

=1n∑r=12nln1+r/n
Replace ∑→∫ & rn→x & 1n→dx

As n→∞

logy= ∫02log1+xdx

Integrating by parts, we get,

logy=∫021·log1+xdx

=x·log1+x02-∫0211+x xdx

=xlog1+x02- ∫021+x1+xdx+ ∫0211+xdx
=xlog1+x02-x02+log1+x02
=2log3-0-2-0+log3-log1
=3log3-2
Since logy =3log3-2=log27-log e2
=y=elog27e2=27e2

=27e2.

Topic: Mathematics, Limits, JEE Main, JEE Main 2016 (03 Apr)
$\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^{2 n} \frac{r}{\sqrt{n^2+r^2}}$ equals

Options:
A ${ }^{1+\sqrt{5}}$
B $-1+\sqrt{5}$
C $-1+\sqrt{2}$
D $1+\sqrt{2}$

The correct answer is:
$-1+\sqrt{5}$

Solution:
$L=\lim _{n \rightarrow \infty} \sum_{r=1}^{2 n} \frac{1}{n} \cdot \frac{r / n}{\sqrt{1+(r / n)^2}}=\int_0^2 \frac{x}{\sqrt{1+x^2}} d x=\sqrt{5}-1$

Topic: Mathematics, Definite Integration, JEE Main
$\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\ldots .+\frac{1}{n^2} \sec ^2 1\right]$ equals

Options:
A
$\frac{1}{2} \sec 1$

B
$\frac{1}{2} \operatorname{cosec} 1$

C
$\tan 1$

D
$\frac{1}{2} \tan 1$

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

Solution:
$\lim _{n \rightarrow \infty}\left[\frac{1}{n^2} \sec ^2 \frac{1}{n^2}+\frac{2}{n^2} \sec ^2 \frac{4}{n^2}+\frac{3}{n^2} \sec ^2 \frac{9}{n^2}+\ldots .+\frac{1}{n} \sec ^2 1\right]$ is equal to $\lim _{n \rightarrow \infty} \frac{r}{n^2} \sec ^2 \frac{r^2}{n^2}=\lim _{n \rightarrow \infty} \frac{1}{n} \cdot \frac{r}{n} \sec ^2 \frac{r^2}{n^2}$
$\Rightarrow$ Given limit is equal to value of integral $\int_0^1 x \sec ^2 x^2 d x$
or $\frac{1}{2} \int_0^1 2 x \sec x^2 d x=\frac{1}{2} \int_0^1 \sec ^2 t d t$
[put $\left.x^2=t\right]$ $=\frac{1}{2}(\tan t)_0^1=\frac{1}{2} \tan 1$.

Topic: Mathematics, Limits, JEE Main, JEE Main 2005
$\lim _{n \rightarrow \infty} \frac{n(2 n+1)^2}{(n+2)\left(n^2+3 n-1\right)}=$

Options:
A $0$
B $2$
C $4$
D $\infty$

The correct answer is:
$4$

Solution:
$\begin{gathered}\lim _{n \rightarrow \infty} \frac{n(2 n+1)^2}{(n+2)\left(n^2+3 n-1\right)}=\lim _{n \rightarrow \infty} \frac{4 n^3+4 n^2+n}{n^3+5 n^2+5 n-2} \\ =\lim _{n \rightarrow \infty} \frac{n^3\left(4+\frac{4}{n}+\frac{1}{n^2}\right)}{n^3\left(1+\frac{5}{n}+\frac{5}{n^2}-\frac{2}{n^3}\right)}=4\end{gathered}$

Topic: Mathematics, Limits, JEE Main
$\lim _{n \rightarrow \infty}\left(\frac{1^2}{n^3+1^3}+\frac{2^2}{n^3+2^3}+\ldots+\frac{1}{2 n}\right)=$

Options:
A $\log 2$
B $2 \log 2$
C $\frac{1}{2} \log 2$
D $\log \sqrt[3]{2}$

The correct answer is:
$\log \sqrt[3]{2}$

Solution:
$L=\lim _{n \rightarrow \infty}\left[\frac{1^2}{n^3+1^3}+\frac{2^2}{n^3+2^3}+\ldots+\frac{n^2}{n^3+n^3}\right]$
$=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^2}{n^3+r^3}=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{r^2}{n^3\left(1+\frac{r^3}{n^3}\right)}$
$=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{\left(\frac{r}{n}\right)^2}{1+\left(\frac{r}{n}\right)^3}=\int_0^1 \frac{x^2 d x}{1+x^3}$
$=\frac{1}{3} \log 2=\log \sqrt[3]{2}$

Topic: Mathematics, Limits, JEE Main
$\lim _{n \rightarrow \infty} \frac{1+2^4+3^4+\ldots n^4}{n^5}-\lim _{n \rightarrow \infty} \frac{1+2^3+3^3+\ldots n^3}{n^5}$

Options:
A
$\frac{1}{5}$

B
$\frac{1}{30}$

C
Zero

D
$\frac{1}{4}$

The correct answer is:
$\frac{1}{5}$

Solution:
$\operatorname{Lim}_{n \rightarrow \infty}\left\{\left(\frac{1}{n}\right)^4+\left(\frac{2}{n}\right)^4+\left(\frac{3}{n}\right)^4+\ldots \ldots \ldots .\left(\frac{n}{n}\right)^4\right\}-\operatorname{Lim}_{n \rightarrow \infty} \frac{1}{n}\left\{\frac{1}{n^4}+\frac{2^3}{n^4}+\ldots \ldots . \cdot \frac{n^3}{n^4}\right\}$
$\int_0^1(x)^4 d x-0=\left[\frac{x^5}{5}\right]_0^1=\frac{1}{5}$

Topic: Mathematics, Limits, JEE Main, JEE Main 2003
$\lim _{n \rightarrow \infty} \frac{1^{99}+2^{99}+3^{99}+\ldots \ldots . n^{99}}{n^{100}}=$

Options:
A $\frac{9}{100}$
B $\frac{1}{100}$
C $\frac{1}{99}$
D $\frac{1}{101}$

The correct answer is:
$\frac{1}{100}$

Solution:
$\begin{aligned} & \lim _{n \rightarrow \infty} \frac{1^{99}+2^{99}+\ldots .+n^{99}}{n^{100}}=\lim _{n \rightarrow \infty} \sum_{r=1}^n\left(\frac{r^{99}}{n^{100}}\right) \\ = & \lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n\left(\frac{r}{n}\right)^{99}=\int_0^1 x^{99} d x=\left[\frac{x^{100}}{100}\right]_0^1=\frac{1}{100} .\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main
limn→∞nn2+12+nn2+22+nn2+32+.. ….+15n2 is equal to

Options:
A π4
B tan-1⁡2
C π2
D tan-1⁡(3)

The correct answer is:
tan-1⁡2

Solution:
To find,

limn→∞nn2+12+nn2+22+…+nn2+2n2

=∑r=12nnn2+r2=1n∑r=12n11+rn2

=∫0211+x2dx=tan-1⁡x02

=tan-1⁡2.

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2019 (12 Jan Shift 2)
$$
\operatorname{Lim}_{n \rightarrow \infty} \frac{1^p+2^p+3^p+\ldots+n^p}{n^{p+1}}
$$
is

Options:
A
$\frac{1}{p+1}$

B
$\frac{1}{1-p}$

C
$\frac{1}{p}-\frac{1}{p-1}$

D
$\frac{1}{p+2}$

The correct answer is:
$\frac{1}{p+1}$

Solution:

Topic: Mathematics, Limits, JEE Main, JEE Main 2002
$\lim _{n \rightarrow \infty}\left(\frac{n}{n+y}\right)^n$ equals

Options:
A $0$
B $1$
C $1/v$
D $e^{-y}$

The correct answer is:
$e^{-y}$

Solution:
$\begin{aligned} \lim _{n \rightarrow \infty}\left(\frac{n}{n+y}\right)^n=\lim _{n \rightarrow \infty}\left(\frac{1}{1+\frac{y}{n}}\right)^n =\lim _{n \rightarrow \infty}\left(1+\frac{y}{n}\right)^{-n}= & \lim _{n \rightarrow \infty}\left[\left(1+\frac{y}{n}\right)^n\right]^{-1}=e^{-y}\end{aligned}$

Topic: Mathematics, Limits, JEE Main
$\lim _{n \rightarrow \infty} \sum_{r=1}^n \cot ^{-1}\left(r^2+\frac{3}{4}\right)=$

Options:
A $\cot ^{-1} 2$
B $\cot ^{-1} \frac{1}{3}$
C $\tan ^{-1} 2$
D $\tan ^{-1} \frac{1}{3}$

The correct answer is:
$\tan ^{-1} 2$

Solution:
$\begin{aligned} & \text { } \lim _{n \rightarrow \infty} \sum_{r=1}^n \cot ^{-1}\left(r^2+\frac{3}{4}\right) \\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \tan ^{-1}\left[\frac{1}{r^2+1-\frac{1}{4}}\right] \\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n \tan ^{-1}\left[\frac{\left(r+\frac{1}{2}\right)-\left(r-\frac{1}{2}\right)}{1+\left(r+\frac{1}{2}\right)\left(r-\frac{1}{2}\right)}\right] \\ & =\lim _{n \rightarrow \infty} \sum_{r=1}^n\left[\tan ^{-1}\left(r+\frac{1}{2}\right)-\tan ^{-1}\left(r-\frac{1}{2}\right)\right], \\ & \quad\left[\because \tan ^{-1}(a-b)=\tan ^{-1}\left[\frac{a-b}{1+a \cdot b}\right]\right]\end{aligned}$
$\begin{aligned} & =\lim _{n \rightarrow \infty}\left[\left(\tan ^{-1} \frac{3}{2}-\tan ^{-1}\left(\frac{1}{2}\right)\right)\right]+\left[\tan ^{-1} \frac{5}{2}-\tan ^{-1} \frac{3}{2}\right] \\ & \quad+\left[\tan ^{-1} \frac{7}{2}-\tan ^{-1} \frac{5}{2}\right] \ldots \\ & +\left[\tan ^{-1}\left(n+\frac{1}{2}\right)-\tan ^{-1}\left(n-\frac{1}{2}\right)\right] \\ & =\lim _{n \rightarrow \infty}\left[\tan ^{-1}\left(n+\frac{1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right)\right] \\ & =\lim _{n \rightarrow \infty}\left[\tan ^{-1}\left(n+\frac{1}{2}\right)-\tan ^{-1}\left(\frac{1}{2}\right)\right] \\ & =\tan ^{-1} \infty-\tan ^{-1} \frac{1}{2}=\frac{\pi}{2}-\tan ^{-1} \frac{1}{2} \\ & =\cot ^{-1} \frac{1}{2}=\tan ^{-1} 2\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n} e^{\frac{-}{n}}$ is

Options:
A
e

B
$e-1$

C
$1-e$

D
$e+1$

The correct answer is:
$e-1$

Solution:
$\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{1}{n} e^{\frac{r}{n}}=\int_0^1 e^x d x=(e-1)$

Topic: Mathematics, Limits, JEE Main, JEE Main 2004
\(\lim _{n \rightarrow \infty} \frac{1}{n}[(n+1)(n+2)\) \((2 n)]^{\frac{1}{n}}=\)

Options:
A 1
B 0
C \(\frac{2}{e}\)
D \(\frac{4}{e}\)

The correct answer is:
\(\frac{4}{e}\)

Solution:
Let \(P=\lim _{n \rightarrow \infty} \frac{1}{n}[(n+1)(n+2) \ldots \ldots(2 n)]^{1 / n}\)
On applying ' \(\log ^{\prime}\) both sides, we get
\(\log P=\lim _{n \rightarrow \infty} \frac{1}{n}\left[\log \left(\frac{n+1}{n}\right)\left(\frac{n+2}{n}\right) \ldots \ldots\left(\frac{n+n}{n}\right)\right]\)
\(\begin{aligned}
& =\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \log \left(1+\frac{r}{n}\right)=\int_0^1 \log (1+x) d x \\
& =[x \log (1+x)]_0^1-\int_0^1 \frac{x}{1+x} d x \\
& =\log 2-(x-\log (1+x))_0^1 \\
& =\log 2-1+\log 2 \\
& =\log 4-1=\log \left(\frac{4}{e}\right) \Rightarrow P=\frac{4}{e}
\end{aligned}\)
Hence, option (d) is correct.

Topic: Mathematics, Limits, JEE Main
\( \lim _{n \rightarrow \infty} \frac{1^{p}+2^{p}+3^{p}+\ldots+n^{p}}{n^{p+1}} \) is \( (p>-1) \)

Options:
A \( \frac{1}{p+1} \)
B \( \frac{1}{p-1} \)
C \( \frac{1}{p}-\frac{1}{p-1} \)
D \( \frac{1}{p+2} \)

The correct answer is:
\( \frac{1}{p+1} \)

Solution:
Ltn→∞1p+2p+...+npnp×1n

=Ltn→∞1n1np+2np+...+nnp

=Ltn→∞Σ r=1r=n1nrnp

=∫01xpdx xp+1p+101=1p+1

Topic: Mathematics, Definite Integration, JEE Main
\(\lim _{n \rightarrow \infty}\left\{\frac{1}{n+m}+\frac{1}{n+2 m}+\frac{1}{n+3 m}+\ldots+\frac{1}{n+n m}\right\}=\)

Options:
A \(\frac{\log _8(m)}{m}\)
B \(\frac{\log _e(1+m)}{1+m}\)
C \(\frac{\log _e(1+m)}{m}\)
D \(\frac{\log _e(1+m)}{1-m}\)

The correct answer is:
\(\frac{\log _e(1+m)}{m}\)

Solution:
Given,
\(\begin{aligned}
& =\lim _{n \rightarrow \infty} \frac{1}{n}\left(\sum_{k=1}^n \frac{1}{1+m\left(\frac{k}{n}\right)}\right) \\
& =\int_0^1 \frac{1}{1+m x} d x=\frac{1}{m} \int_0^1 \frac{m}{1+m x} d x \\
& =\frac{1}{m}\left[\log _e(1+m x)\right]_0^1=\frac{1}{m}\left[\log _e(1+m)-\log (1)\right] \\
& =\frac{\log _e(1+m)}{m}
\end{aligned}\)

Topic: Mathematics, Limits, JEE Main
\( \lim _{n \rightarrow \infty}\left(\frac{(n+1)^{1 / 3}}{n^{4 / 3}}+\frac{(n+2)^{1 / 3}}{n^{4 / 3}}+\ldots \ldots+\frac{(2 n)^{1 / 3}}{n^{4 / 3}}\right) \) is equal to

Options:
A \( \frac{3}{4}(2)^{4 / 3}-\frac{3}{4} \)
B \( \frac{4}{3}(2)^{3 / 4} \)
C \( \frac{4}{3}(2)^{4 / 3} \)
D \( \frac{3}{4}(2)^{4 / 3}-\frac{4}{3} \)

The correct answer is:
\( \frac{3}{4}(2)^{4 / 3}-\frac{3}{4} \)

Solution:
The given limit can be written as
limn→∞1n1+1n13+1+2n13+….+1+nn13
=limn→∞1n∑r=1n1+rn13
=∫01(1+x)1/3dx
=1+x434301
=34(24/3–1)
=34.24/3-34

Topic: Mathematics, Definite Integration, JEE Main
limn→∞tan∑r=1ntan-111+r+r2 is equal to_______.

Options:

The correct answer is:
1

Solution:
limn→∞tan∑r=1ntan-111+rr+1

=limn→∞tan∑r=1ntan-1r+1-r1+rr+1

=tanlimn→∞∑r=1ntan-1r+1-tan-1r

=tanlimn→∞tan-1n+1-π4

=tanπ4=1

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2021 (24 Feb Shift 1)
limt→011sin2t+21sin2t+31sin2t......n1sin2tsin2t is equal to

Options:
A n2+n
B n
C nn+12
D n2

The correct answer is:
n

Solution:
L=limt→011sin2t+21sin2t+31sin2t......n1sin2tsin2t

=limt→01cosec2t+2cosec2t+3cosec2t......ncosec2tsin2t

=limt→0 nsin2tsin2t1ncosec2t+2ncosec2t+3ncosec2t......1sin2t

=nlimt→0 0+0+0......10=n

Topic: Mathematics, Limits, JEE Main, JEE Main 2023 (24 Jan Shift 1)
limx→0∫0x2sintdtx3 is equal to:

Options:
A 0
B 23
C 32
D 115

The correct answer is:
23

Solution:
limx→0+∫0x2sintdtx3=limx→0+sinx2x3x2

=limx→0+sinxx×23=23

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (24 Feb Shift 1)