datavorous/entrance-exam-dataset huggingface dataset:
The value of \( \lambda \) and \( \mu \) for which the system of equations \( x+y+z=6, x+2 y+3 z=10 \) and \( x+2 y+\lambda z=\mu \) have no solution, are

Options:
A \( \lambda=3, \mu \neq 10 \)
B \( \lambda \neq 3, \mu=10 \)
C \( \lambda \neq 3, \mu \neq 10 \)
D None of these

The correct answer is:
\( \lambda=3, \mu \neq 10 \)

Solution:
Given system of equation is
x+y+z=6
x+2y+3z=10
x+2y+λz=μ
If given system of equation has no solution, then D = 0 and atleast one of the determinants D1, D2, D3≠0
Here D=0
⇒ 11112312λ=0
⇒ 12λ-6-1λ-3+12-2=0
⇒ 2λ-6-λ+3+0=0
⇒ λ-3=0
⇒ λ=3
Also, D1=61110233μ23≠0
⇒ 66-6-130-3μ+120-2μ≠0
⇒ 0-30+3μ+20-2μ≠0
⇒ μ-10≠0
⇒ μ≠10

Topic: Mathematics, Determinants, JEE Main
The value of \( \lambda \), for which the line \( 2 x-\frac{8}{3} \lambda y=-3 \) is a normal to the conic \( x^{2}+\frac{y^{2}}{4}=1 \) is:

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

The correct answer is:
\( \frac{\sqrt{3}}{2} \)

Solution:
Given that the equation of ellipse is

x2+y24=1, where a2=1, b2=4

So, equation of normal in parametric form can be written as

xcosθ -2ysinθ =1-4 …i

Comparing with 2x-83λy=-3

1cosθ=2⇒cosθ=12⇒sinθ=±32

and

-2sinθ=-83λ⇒λ=34sinθ

So, λ = ±32

Topic: Mathematics, Ellipse, JEE Main
The value of \( \lim _{\mathrm{x} \rightarrow 0}\left(\left[\frac{100 \mathrm{x}}{\sin \mathrm{x}}\right]+\left[\frac{99 \sin \mathrm{x}}{\mathrm{x}}\right]\right) \), (where \( [\mathrm{x}] \) represents greatest integral function less than or equal to \( \mathrm{x} \) ) is \( 99 \lambda . \) Then the value of \( \lambda \) is

Options:

The correct answer is:
2

Solution:
We know that,

lim x → 0 sin x x → 1 -
then,

limx→0100xsin x⇒limx→0100sin xx=100

limx→099 sin xx⇒limx→099 sin xx=98
Hence ⇒ lim x→0 [ 100x sinx ]+[ 99sinx x ]=100+98=198
99λ=198
Hence, λ=2

Topic: Mathematics, Limits, JEE Main
The value of

limn→∞1+2-3+4+5-6+…+(3n-2)+(3n-1)-3n2n4+4n+3-n4+5n+4 is

Options:
A 2+12
B 3(2+1)
C 32(2+1)
D 322

The correct answer is:
32(2+1)

Solution:
The given limit limn→∞1+2-3+4+5-6+…+(3n-2)+(3n-1)-3n2n4+4n+3-n4+5n+4 can be expressed as,limn→∞∑r=1n3r-2+3r-1-3r2n4+4n+3-n4+5n+4

=limn→∞∑r=1n3r-12n4+4n+3-n4+5n+4

=limn→∞0+3+6+9+….n terms2n4+4n+3-n4+5n+4.......(1)

We know the sum of the nth term of A.P. is given by

Sn=n22a+n-1d

Using above formula in equation(1) we get,

=limn→∞3n(n-1)22n4+4n+3-n4+5n+4

=limn→∞3n(n-1)22n4+4n+3-n4+5n+4

=limn→∞3(1-1n)22+4n3+3n4-1+5n3+4n4

=3(1-0)22+0+0-1+0+0

=32(2-1)=32(2+1)

Topic: Mathematics, Limits, JEE Main, JEE Main 2023 (25 Jan Shift 1)
The value of limn→∞1n∑j=1n2j-1+8n2j-1+4n is equal to:

Options:
A 5+loge32
B 2-loge23
C 3+2loge23
D 1+2loge32

The correct answer is:
1+2loge32

Solution:
limn→∞1n∑j=1n2jn-1n+82jn-1n+4

∫012x+82x+4dx=∫01dx+∫0142x+4dx

=1+412ln2x+4|01

=1+2ℓn32

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (27 Jul Shift 1)
The value of limn→∞1n∑r=02n-1n2n2+4r2 is:

Options:
A 12tan-12
B tan-14
C 12tan-14
D 14tan-1(4)

The correct answer is:
12tan-14

Solution:
Let

I=limn→∞1n∑r=02n-1n2n2+4r2

I=limn→∞1n∑r=02n-1n2n2+4r2

I=limn→∞1n∑r=02n-111+4rn2

Let rn=x, lower limit x=0 when r=0 and upper limit x=2 when r=2n-1

⇒I=∫0211+4x2dx

⇒I=14∫021122+x2dx

⇒I=14×2tan-12x02=12tan-1(4)

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (26 Aug Shift 1)
The value of limn→∞6tan∑r=1ntan-11r2+3r+3 is equal to

Options:
A 1
B 2
C 3
D 6

The correct answer is:
3

Solution:
Let S=∑r=1ntan-11r2+3r+3

So, Tr=tan-11r2+3r+2+1

=tan-1(r+2)-(r+1)1+(r+1)(r+2)

=tan-1r+2-tan-1r+1

T1=tan-13-tan-12

T2=tan-14-tan-13..
Tn=tan-1n+2-tan-1n+1

⇒∑r=1ntan-11r2+3r+3=tan-1n+2-tan-12

i.e. 6tanlimn→∞∑r=1ntan-11r2+3r+3=6tanlimn→∞∑r=1ntan-1n+2-tan-12

=6tanπ2-tan-12

=6tantan-112=3

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2022 (28 Jun Shift 2)
The value of limn→∞⁡3.2n+1-4.5n+15.2n+7.5n is equal to

Options:
A 35
B -47
C -207
D 0

The correct answer is:
-207

Solution:
limnâââ¡3.2n+1-4.5n+15.2n+7.5n
=limnââ5n6.25n-205n5.25n+7=-207 âµlimnâÂ ââ¡25n=0

Topic: Mathematics, Limits, JEE Main
The value of limn→∞⁡cos⁡x2cos⁡x4cos⁡x8……...cos⁡x2n+1 is equal to

Options:
A xsin⁡x
B sin⁡xx
C 0
D None of these

The correct answer is:
sin⁡xx

Solution:
Let, L=limn→∞⁡cos⁡x2cos⁡x22cos⁡x23…..cos⁡x2n+1
=limn→∞⁡12sin⁡x2n+1cos⁡x2……cos⁡x2n2sin⁡x2n+1cos⁡x2n+1
=limn→∞122sin⁡x2n+1cos⁡x2…2cos⁡x2nsin⁡x2n
=limn→∞⁡12n+1sin⁡x2n+12cos⁡x2sin⁡x2=limn→∞⁡sin⁡x2n+1sin⁡x2n+1
=sin⁡xxlimn→∞⁡x2n+1sin⁡x2n+1=sin⁡xx

Topic: Mathematics, Limits, JEE Main
The value of $\lim _{n \rightarrow \infty}\left[\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\ldots \ldots . . \frac{n}{n^2+n^2}\right]$ is

Options:
A $\frac{\pi}{4}$
B $\log 2$
C zero
D 1

The correct answer is:
$\frac{\pi}{4}$

Solution:
$$
\text { Hints : } \lim _{n \rightarrow \infty}\left[\frac{n}{n^2+1^2}+\frac{n}{n^2+2^2}+\ldots+\frac{n}{n^2+n^2}\right]
$$
$$
=\lim _{n \rightarrow \infty} \sum_{r=1}^n \frac{n}{n^2+r^2}=\lim _{n \rightarrow \infty} \frac{1}{n} \sum_{r=1}^n \frac{1}{1+\left(\frac{r}{n}\right)^2}=\int_0^1 \frac{d x}{1+x^2}=\left[\tan ^{-1} x\right]_0^1=\frac{\pi}{4}
$$

Topic: Mathematics, Limits, JEE Main
The value of lim n → ∞ Σ r = 1 r = 4 n n r 3 r + 4 n 2 is equal to

Options:
A 1 8
B 1 10
C 1 6
D 1 9

The correct answer is:
1 10

Solution:
lim n â â Î£ r = 1 r = 4 n 1 r / n 3 r n + 4 2 Â· 1 n
= â« 0 4 1 x 3 x + 4 2 dx
Put, 3 x + 4 = t
3 2 x dx= dt â 1 x dx = 2 3 dt
when x = 0 then t = 4
when x = 4 then t = 10
= 2 3 â« 4 1 0 1 t 2 dt = 2 3 - 1 t 4 1 0
= - 2 3 1 1 0 - 1 4
= 1 1 0

Topic: Mathematics, Definite Integration, JEE Main
The value of \( \lim _{n \rightarrow \infty} \sum_{k=1}^{n} \frac{k^{\frac{1}{a}}\left\{n^{a-\frac{1}{a}}+k^{a-\frac{1}{a}}\right\}}{n^{a+1}}= \)

Options:
A \( 1 \)
B \( 2 \)
C \( 3 \)
D None of these

The correct answer is:
\( 1 \)

Solution:
limn→∞ ∑k=1nk1ana-1a+ka-1ana+1

=limn→∞ ∑k=1n1n⋅kn1a+kna

=∫01x1a+xadx (applying definite integral as limit of sum)

=x1a+11a+1+ xa+1a+101

=aa+1+1a+1=1

Topic: Mathematics, Definite Integration, JEE Main
The value of limx→0⁡1+6x13-1+4x12x2 is equal to

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

The correct answer is:
-2

Solution:
Using binomial expansion, the given limit becomes
lim x→0 1+2x+ 1 3 1 3 −1 2! ⋅36 x 2 +… − 1+2x+ 1 2 1 2 −1 2! ⋅16 x 2 +… x 2
=limx→0⁡2-4x2+…x2=-2

Topic: Mathematics, Limits, JEE Main
The value of limx→01-cos2x3+cosxxtan4x is equal to

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

The correct answer is:
2

Solution:
Given limx→01-cos 2x3+cos xx tan 4x
=limx→02sin2x3+cosxx tan 4x
=limx→0sin xx24xtan 4x3+cos x×12
=2

Topic: Mathematics, Limits, JEE Main, JEE Main 2013 (07 Apr)
The value of lim x â 0 1 - cos 2 x 3 + cos x x tan 4 x is equal to

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

The correct answer is:
2

Solution:
lim x → 0 1 - cos 2 x 3 + cos x x tan 4 x
= lim x → 0 2 sin 2 x 3 + cos x x tan 4 x
= lim x → 0 sin x x 2 4 x tan 4 x 3 + cos x × 1 2
=2

Topic: Mathematics, Limits, JEE Main
The value of limx→0⁡1-cos3⁡xsin2⁡xcos⁡x is equal to

Options:

The correct answer is:
1.5

Solution:
limxâ0â¡1-cos3â¡xsin2â¡xcosâ¡x
=limxâ0â¡1-cosâ¡x1+cos2â¡x+cosâ¡xsin2â¡xcosâ¡x
=limxâ0â¡2sin2â¡x21+cos2â¡x+cosâ¡x2sinâ¡x2cosâ¡x22cosâ¡x
= lim x→0 1 2 ⋅ 1+ cos 2 x+cosx cos x 2 2 cosx
=1231=32

Topic: Mathematics, Limits, JEE Main
The value of limx→0⁡1-cos⁡4x5+cos⁡xxtan⁡5x is equal to

Options:

The correct answer is:
9.6

Solution:
The value of the limit is
limx→0⁡2sin22x×⁡5+cos⁡xx×tan⁡5x5x×5x
limx→0⁡2sin⁡2x2x2×45×5+cos⁡x×5xtan⁡5x
⇒2×1×45×6×1

Topic: Mathematics, Limits, JEE Main
The value of limx→0⁡1-cos4⁡x5sin2⁡xcos⁡x is equal to

Options:

The correct answer is:
0.4

Solution:
limx→0⁡1-cos4⁡x5sin2⁡xcos⁡x
=limx→0⁡1-cos⁡x1+cos3⁡x+cos2⁡x+cos⁡x5sin2⁡xcos⁡x
=limx→0⁡2sin2⁡x21+cos⁡x+cos2⁡x+cos3⁡x2sin⁡x2cos⁡x225cos⁡x
= lim x→0 1 2 ⋅ 1+cosx+ cos 2 x+ cos 3 x cos x 2 2 5cosx

=25

Topic: Mathematics, Limits, JEE Main
The value of limx→01+sinx-cosx+ln1-xx·tan2x is

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

The correct answer is:
- 1 2

Solution:
limx→01+sinx-cosx+ℓn1-xx·tan2x
=limx→01+sinx-cosx+ℓn1-xx300 [ ∵  lim x→0 x tan x =1 ]
=limx→0cosx+sinx-11-x3x2 00 [using L'hospital rule]
=limx→0-sinx+cosx-1(1-x)26x 00 [Using L'hospital rule again]
=limx→0-cosx-sinx-2(1-x)36 [Using L'hospital rule again]
= - 1 - 2 6 = - 1 2

Topic: Mathematics, Limits, JEE Main
The value of limx→0⁡1+tan⁡x1+sin⁡x2sin⁡x is equal to

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

The correct answer is:
1

Solution:
Let, fx=1+tan⁡x1+sin⁡x
and gx=2sin⁡x
Clearly, fx→1 and gx→∞ as x→0.
∴limx→0⁡1+tan⁡x1+sin⁡x2sin⁡x=elimx→02sin⁡x⁡1+tan⁡x1+sin⁡x-1
using limx→a⁡fxgx=elimx→agx⁡fx-1for 1∞ form
=elimx→0⁡2sin⁡xtan⁡x-sin⁡x1+sin⁡x=elimx→0⁡21-cos⁡xcos⁡x1+sin⁡x=e0=1

Topic: Mathematics, Limits, JEE Main
The value of limxâ0â¡1x181-cosâ¡x33-cosâ¡x66+cosâ¡x33âcosâ¡x66 is Î»2, then the value of 900Î» is equal to (here, Î»>0)

Options:

The correct answer is:
25

Solution:
limx→0⁡1-cos⁡x33-cos⁡x661-cos⁡x33⋅1x18
limx→0⁡1-cos⁡x331-cos⁡x66x18
limx→0⁡1-cos⁡x33x332x69⋅1-cos⁡x66x662⋅x662⋅1x18
limx→0⁡12×x69⋅12⋅x1236⋅1x18=1362
∴λ=136

Topic: Mathematics, Limits, JEE Main
The value of limx→0-⁡43x+1521x21+6x+621x is equal to ______. If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
5

Solution:
limx→0-26x+15⋅21x21+6x+6⋅21x=limx→0-21x25x+1521x225x+6
=limx→0-⁡25x+15225x+6=0+1520+6=156=52

Topic: Mathematics, Limits, JEE Main
The value of limx→0-⁡43x+1521x21+6x+621x is equal to

Options:

The correct answer is:
2.5

Solution:
limx→0-26x+15⋅21x21+6x+6⋅21x=limx→0-21x25x+1521x225x+6
=limx→0-⁡25x+15225x+6=0+1520+6=156=52

Topic: Mathematics, Limits, JEE Main
The value of limxâ0â¡9lnâ¡2-cosâ¡25x5ln2sinâ¡3x+1â¡ is equal to

Options:

The correct answer is:
62.5

Solution:
limx→0⁡9ln2-cos⁡25x5ln2sin⁡3x+1=limx→0⁡9ln1+1-cos⁡25x5ln21+sin⁡3x
=limx→0⁡91-cos⁡25x5sin⁡3x2
⇒limx→0⁡925x2103x2=62510=62.5

Topic: Mathematics, Limits, JEE Main
The value of limx→0+cos-1x-[x]2·sin-1x-[x]2x-x3, where x denotes the greatest integer ≤x is:

Options:
A π
B 0
C π4
D π2

The correct answer is:
π2

Solution:
Given L=limx→0+cos-1x-[x]2·sin-1x-[x]2x-x3

∵ x∈[0,1)⇒x=0

Then L=limx→0+cos-1x1-x2×limx→0+sin-1xx

L=π2×limx→0+11-x2 Apply L'Hospital rule for 00 form

L=π2

Topic: Mathematics, Limits, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The value of limx→0⁡cos⁡x+sin⁡bxax is equal to

Options:
A e-ab
B eab2
C eab
D e-b2a

The correct answer is:
eab

Solution:
Let, L=limx→0⁡cos⁡x+sin⁡bxax=elimx→0⁡axcos⁡x+sin⁡bx-1

(using 1∞ form)
L=elimx→0⁡a-sin⁡x+bcos⁡bx1=eab

Topic: Mathematics, Limits, JEE Main
The value of limx→0cos⁡x+sin⁡x1x is equal to (take e=2.71)

Options:

The correct answer is:
2.71

Solution:
∵limx→a⁡fxϕx=elimx→a⁡ϕxfx-1 as fx→1 & ϕx→∞ as x→a

So, limx→0cos⁡x+sin⁡x1x=elimx→0⁡1xcos⁡x+sin⁡x-1
=elimx→0⁡-sin⁡x+cos⁡x1=e

Topic: Mathematics, Limits, JEE Main
The value of limxâ0â¡logâ¡1+2x5x+limxâ2â¡x4-24x-2 is equal to

Options:

The correct answer is:
32.4

Solution:
limxâ0â¡logâ¡1+2x5x+limxâ0â¡x4-24x-2
= lim x→0 log 1+2x 2x ⋅ 2 5 + lim x→2 4⋅ x 3
=25+32=32.4

Topic: Mathematics, Limits, JEE Main
The value of limx→0log(1+x+x2)+log(1−x+x2)secx−cosx is equal to

Options:
A −1
B 1
C 0
D 2

The correct answer is:
1

Solution:
lim x â 0 log 1 + x + x 2 + log 1 - x + x 2 sec x - cos x
= lim x â 0 log 1 + x 2 2 - x 2 1 - cos 2 x / cos x
= lim x â 0 log 1 + x 2 + x 4 sin x tan x
= lim x â 0 log 1 + x 2 1 + x 2 x 2 1 + x 2 Â· x 2 1 + x 2 Â· 1 sin x x Â· tan x x Â· x 2
= 1 as lim x â 0 log 1 + x x = 1

Topic: Mathematics, Limits, JEE Main
The value of limxâ0â¡sin2â¡3x3+secâ¡x-2 is equal to

Options:

The correct answer is:
72

Solution:
After rationalization, we get,

limx→0⁡sin2⁡3x2+3+sec⁡x3+sec⁡x-4
=limx→0sin23x9x2⋅9x21−cosx2+3+1
=924=72

Topic: Mathematics, Limits, JEE Main
The value of $\lim _{x \rightarrow 0} \frac{1}{x}\left[\tan ^{-1}\left(\frac{x+1}{2 x+1}\right)-\frac{\pi}{4}\right]$ is :

Options:
A
1

B
$-\frac{1}{2}$

C
2

D
0

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

Solution:
$\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)\left[\tan ^{-1}\left(\frac{x+1}{2 x+1}\right)-\frac{\pi}{4}\right]$
$=\lim _{x \rightarrow 0}\left(\frac{1}{x}\right)\left[\tan ^{-1}\left(\frac{x+1}{2 x+1}\right)-\tan ^{-1}(1)\right]$
$=\lim _{x \rightarrow 0}\left(\frac{1}{x}\right) \cdot \tan ^{-1}\left(\frac{\frac{x+1}{2 x+1}-1}{1+\frac{x+1}{2 x+1}}\right)$
$=\lim _{x \rightarrow 0} \frac{1}{x} \cdot \tan ^{-1}\left(\frac{-x}{3 x+2}\right)$
$=-\lim _{x \rightarrow 0}\left[\frac{\tan ^{-1}\left(\frac{x}{3 x+2}\right)}{\frac{x}{3 x+2}} \times \frac{1}{3 x+2}\right]=-\frac{1}{2}$

Topic: Mathematics, Limits, JEE Main, JEE Main 2013 (09 Apr Online)
The value of limxâ0x6000-sin x6000x2sinx6000 is

Options:
A 1000
B 100
C 1100
D 1010

The correct answer is:
1000

Solution:
limx→0⁡x6000-sinx6000x2sinx6000x6000.x600000
limx→0x6000-sinx6000x6002
= -limx→0sinx6000-x6000x6002
=-limx→0sinxx6000-1x2
=-limx→06000sinxx5999xcosx-sinxx22x (By L-Hospital Rule)
=-60002limx→0xcosx-sinxx3

=-3000 limx→0cosx-x sinx-cosx3x2
=1000

Topic: Mathematics, Limits, JEE Main
The value of limxâ0+â¡xcotâ¡x+xlnâ¡x is equal to

Options:
A 1
B 2
C 3
D 0

The correct answer is:
1

Solution:
limx→0+⁡xcot⁡x+limx→0+⁡xln⁡x
=limx→0+⁡xtan⁡x+limx→0+⁡ln⁡x1x
=limx→0+⁡xtan⁡x+limx→0+⁡1x-1x2=1+limx→0+⁡-x
=1+0=1

Topic: Mathematics, Limits, JEE Main
The value of $\lim _{x \rightarrow 0}\left(\frac{e^x-1}{x}\right)$

Options:
A $1 / 2$
B $\infty$
C $1$
D $0$

The correct answer is:
$1$

Solution:
$\lim _{x \rightarrow 0} \frac{e^x-1}{x}=\lim _{x \rightarrow 0} \frac{\left(1+\frac{x}{1!}+\frac{x^2}{2!}+\ldots-1\right)}{x}=1$

Topic: Mathematics, Limits, JEE Main
The value of $\lim _{x \rightarrow 0^{+}} x^m(\log x)^n, m, n \in N$ is

Options:
A $0$
B $\frac{m}{n}$
C $m n$
D None of these

The correct answer is:
$0$

Solution:
$\lim _{x \rightarrow 0+} x^m(\log x)^n=\lim _{x \rightarrow 0+} \frac{(\log x)^n}{x^{-m}}\left(\right.$ Form $\left.\frac{\infty}{\infty}\right)$
$=\lim _{x \rightarrow 0+} \frac{n(\log x)^{(n-1)} \frac{1}{x}}{-m x^{-m-1}}$
(By $L$-Hospital's rule)
$=\lim _{x \rightarrow 0+} \frac{n(\log x)^{n-1}}{-m x^{-m}}$
$\left(\right.$ Form $\left.\frac{\infty}{\infty}\right)$
$=\lim _{x \rightarrow 0+} \frac{n(n-1)(\log x)^{(n-2)} \frac{1}{x}}{(-m)^2 x^{-m-1}}$
(By $L$-Hospital's rule)
$=\lim _{x \rightarrow 0+} \frac{n(n-1)(\log x)^{n-2}}{m^2 x^{-m}} \quad\left(\right.$ Form $\left.\frac{\infty}{\infty}\right)$
...........................
.......................
$=\lim _{x \rightarrow 0+} \frac{n!}{(-m)^n x^{-m}}=0$
(Differentiating $N^r$ and $D^r n$ times).

Topic: Mathematics, Limits, JEE Main
The value of $\lim _{x \rightarrow 0} \frac{\int_0^{x^2} \sec ^2 t d t}{x \sin x}$ is

Options:
A
0

B
3

C
2

D
1

The correct answer is:
1

Solution:
$\operatorname{Lim}_{x \rightarrow 0} \frac{\frac{d}{d x} \int_0^{x^2} \sec ^2 t d t}{\frac{d}{d x}(x \sin x)}=\operatorname{Lim}_{x \rightarrow 0} \frac{\sec ^2 x^2 \cdot 2 x}{\sin x+x \cos x}$
(by L'Hospital rule)
$\operatorname{Lim}_{x \rightarrow 0} \frac{2 \sec ^2 x^2}{\left(\frac{\sin x}{x}+\cos x\right)}=\frac{2 \times 1}{1+1}=1$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2003
The value of limx→0+⁡xx2+xxx is equal to

Options:
A 0
B 1
C 2
D 12

The correct answer is:
1

Solution:
limx→0+⁡xx2=elimx→0+⁡x2ln⁡x=eL (say)
L=limx→0+⁡ln⁡x1x2=limx→0+⁡1x-2x3 {applying L’ hospital rule}
=limx→0+⁡-x22=0
Hence, limx→0+⁡xx2=e0=1

Similarly, limx→0+⁡xx=elimx→0+⁡xlnx=elimx→0+⁡lnx1x

=elimx→0+⁡1x-1x2=elimx→0+⁡-x=e0=1

Hence, limx→0+⁡xxx=0

Therefore, limx→0+⁡xx2+limx→0+⁡xxx=1

Topic: Mathematics, Limits, JEE Main
The value oflimx→1+∫1xt-1dtsin⁡(x-1) is equal to

Options:

The correct answer is:
0

Solution:
Given: lim x→ 1 + ∫ 1 x (t−1) dt sin(x−1)
Using LâHospitalâs rule and leibuitzrule,
l= lim x→ 1 + d dx ∫ 1 x (t−1) dt d dx sin x−1
lim x→ 1 + x−1 cos x−1 = 0 1 =0

Topic: Mathematics, Limits, JEE Main
The value of limx→1-⁡2π-4sin-1⁡x1-x is equal to

Options:
A π
B 12π
C π2
D 2π

The correct answer is:
2π

Solution:
We have, limxâ1-â¡Ï-2sin-1â¡x1-x2
=limxâ1-â¡Ï-2sin-1â¡x1-xÃÏ+2sin-1â¡xÏ+2sin-1â¡xÃ2
=limxâ1-â¡22Ï2-sin-1â¡x1-xÏ+2sin-1â¡x
=limxâ1-â¡22cos-1â¡x1-x⋅12Ï
=limÎ¸â0+â¡22Î¸2sinâ¡Î¸2⋅12Ï [Putting x=cosâ¡Î¸ ]
=4222Ï=2Ï

Topic: Mathematics, Limits, JEE Main
The value of limx→1⁡∑r=110xr-1r2x-1 is equal to

Options:

The correct answer is:
27.5

Solution:
We have, limx→1⁡∑r=110xr-1r2x-1
=12∑r=110r⋅1r-1=1+2+3+…+102=1010+14=552

Topic: Mathematics, Limits, JEE Main
The value of limx→1x2-1sin2πxx4-2x3+2x-1 is equal to:

Options:
A π26
B π23
C π22
D π2

The correct answer is:
π2

Solution:
Given, limx→1x2-1sin2πxx4-2x3+2x-1

=limx→1x2-1sin2πxx2-1x2-2x+1

=limh→0sin2π1+h1+h-12

=limh→0sin2πhh2=limh→0sinπhπh2π2=π2

Topic: Mathematics, Limits, JEE Main, JEE Main 2022 (29 Jun Shift 2)
The value of limx→1⁡x25-2x5+14x-12 is equal to

Options:

The correct answer is:
0.01

Solution:
limx→1⁡x25-2x5+14x-12=limy→1⁡y2-2y+14y5-12
(Putting x5=y; as x→1, y→1)
limy→1⁡y-124y-12y4+y3+y2+y+12
=125×4=0.01

Topic: Mathematics, Limits, JEE Main
The value of limx→π2⁡11cos2⁡x+21cos2⁡x+.………+101cos2⁡xcos2⁡x is equal to

Options:

The correct answer is:
10

Solution:
Let, L=limx→π2⁡11cos2⁡x+21cos2⁡x+.………+101cos2⁡xcos2⁡x

Putting 1cos2⁡x=y, we get,
L=limy→∞⁡1y+2y+.…….+10y1y
=limy→∞10110y+210y+.…….+1010y1y
=100+0+.……+10=10

Topic: Mathematics, Limits, JEE Main
The value of limx→2π⁡1-sec⁡xsec⁡xln⁡sec⁡x is equal to

Options:
A 0
B 1
C 2
D -1

The correct answer is:
-1

Solution:
Let, sec⁡x=t
limt→1⁡1-ttln⁡t

Applying L'Hospital rule, we, get,

limt→1⁡-tt1+ln⁡t1t
=-11=-1

Topic: Mathematics, Limits, JEE Main
The value of limx→2π⁡cos⁡x-cos⁡xcos⁡x1-cos⁡x+ln⁡cos⁡x is equal to

Options:

The correct answer is:
2

Solution:
Let, L=limx→2πcos⁡x-(cos⁡x)cos⁡x1-cos⁡x+ln⁡cos⁡x
Putting cos⁡x=t , we get,
L=limt→1⁡t-tt1-t+ln⁡t
Now by L-Hospital rule
L=limt→11-tt(1+ln⁡t)-1+1t
Again by L-Hospital rule
=limt→1-tt1t-tt1+ln⁡t2-1t2=2

Topic: Mathematics, Limits, JEE Main
The value of limx→2⁡∑r=17xr-2r2rx-2 is equal to

Options:

The correct answer is:
63.5

Solution:
We have, ∑r=17limx→2⁡xr-2r2rx-2
= ∑ r=1 7 r⋅ 2 r−1 2r = 1+ 2 1 + 2 2 +.…..+ 2 6 2 = 127 2

Topic: Mathematics, Limits, JEE Main
The value of limx→∞⁡2x1/2+3x1/3+4x1/4+.....+x1/n2x-31/2+2x-31/3+....+2x-31/n is

Options:
A 2
B 2
C 13
D 0

The correct answer is:
2

Solution:
Given,
limxâââ¡2x1/2+3x1/3+....+nx1/n2x-31/2+2x-31/3+...+2x-31/n
=limhâ0â¡21h1/2+31h1/3+....+n1h1/n1h1/22-3h1/2+1h1/32-3h1/3+.....+1h1/n2-3h1/n
On putting x=1h as xââ , hâ 0
=limhâ0â¡2+3h12 - 13+4h12 - 14+....+nh12 - 1n2-3h1/2+h12 - 132-3h1/3+....+h12 - 1n2-3h1/n
=2+0+0+0+.......21/2+0+0+.......=2.

Topic: Mathematics, Limits, JEE Main
The value of limxâÏ2â¡x3lnsinâ¡x (where, . denotes the greaest integer function)

Options:
A does not exist
B is equal to 1
C is equal to 0
D is equal to -1

The correct answer is:
is equal to 0

Solution:
∵π6<1,
∴π6=0
∴limx→π2x2lnsin⁡x=0

Topic: Mathematics, Limits, JEE Main
The value of lim x→∞ ( 3x−4 3x+2 ) x+1 3 is equal to

Options:
A e −1/3
B e −2/3
C e −1
D e −2

The correct answer is:
e −2/3

Solution:
lim x→∞ ( 3x−4 3x+2 ) x+1 3 = lim x→∞ ( 3x+2−6 3x+2 ) x+1 3 = lim x→∞ ( 1− 6 3x+2 ) x+1 3 = lim x→∞ [ ( 1− 6 3x+2 ) 3x+2 −6 ] −6 3x+2 . x+1 3 = lim x→∞ e −2(x+1) 3x+2 = e −2/3 , { ∵ lim x→∞ −2(x+1) 3x+2 = −2 3 } .

Topic: Mathematics, Limits, JEE Main
The value of limx→∞⁡3x-43x+2x+13 is equal to

Options:
A e-1/3
B e-2/3
C e-1
D e-2

The correct answer is:
e-2/3

Solution:
â¡limxââ3x-43x+2x+13
=â¡limxââ 1+-63x+2x+13
=eâ¡limxââ-63x+2Ãx+13 âµ 1â form
=elimxââ-2x-23x+2=e-2/3 (Applying L' Hospital rule)

Topic: Mathematics, Limits, JEE Main
The value of limx→π4⁡sin⁡2xsec2⁡2x is equal to

Options:
A -12
B 12
C e-12
D e12

The correct answer is:
e-12

Solution:
limx→π4⁡sin⁡2xsec2⁡2x

=elimx→π4⁡sec2⁡2xsin⁡2x-1=elimx→π4⁡sin⁡2x-1cos2⁡2x

Applying L'Hospital Rule, we get,

=elimx→π4⁡2cos2x2cos2x-2sin⁡2x
=elimx→π4⁡-12sin⁡2x=e-12

Topic: Mathematics, Limits, JEE Main
The value of limx→5π4⁡cot3⁡x-tan⁡xcos⁡x+5π4 is equal to

Options:

The correct answer is:
8

Solution:
limx→5π4⁡cot3⁡x-tan⁡xcos⁡x+5π4
=limx→5π41-tan4xcosx+5π4=2limx→5π41-tan2xcosx+5π4
=2limx→5π4cos2x-sin2x-12(cos⁡x-sin⁡x)·1cos2x
=-42limx→5π4cos⁡x+sin⁡x=8

Topic: Mathematics, Limits, JEE Main
The value of limx→π6⁡2cos⁡x+π31-3tan⁡x is equal to

Options:
A 12
B 3
C 34
D 32

The correct answer is:
32

Solution:
Let x=π6-t
Given limit is limt→0⁡2cos⁡π2-t1-3tan⁡π6-t
⇒limt→0⁡2sin⁡t1-31-3tant3+tant=limt→0⁡2sin⁡t3+tant4tant=32

Topic: Mathematics, Limits, JEE Main
The value of limx→∞⁡e21+2xxx2 is equal to

Options:
A e
B e-1
C e12
D e-12

The correct answer is:
e

Solution:
Let, L=limx→∞e21+2xxx2

Put 1x=y

L=limy→0+e21+2y1y12y

L=elimy→0+12yln⁡e21+2y1y=eM (let)
Now, we have,
M=limy→0+12ylne21+2y1y=limy→0+lne2-1yln1+2y2y
=limy→0+2y-ln1+2y2y2
=limy→0+2y-2y-4y22+8y33-16y44+⋯2y2
=limy→0+⁡22-4y3+8y24-......=1
Hence, L=e1=e

Topic: Mathematics, Limits, JEE Main
The value of limx→∞⁡e21+2xxx is equal to

Options:
A e2
B e-1
C e12
D e-12

The correct answer is:
e2

Solution:
Let, L=limx→∞e21+2xxx=limy→0+e21+2y1y1y by putting 1x=y

L=elimy→0+1ylne21+2y1y
Now, we have,
limy→0+1ylne21+2y1y=limy→0+lne2-1yln1+2yy
=limy→0+2y-ln1+2yy2
=limy→0+2y-2y-4y22+8y33-16y44+…y2
=limy→0+42+8y3+16y24-…=2
Hence, L=e2

Topic: Mathematics, Limits, JEE Main
The value of \( \lim _{x \rightarrow 0} \frac{\int_{0}^{x^{2}} \sin \sqrt{t}}{x^{3}} d t \) is

Options:
A \( 0 \)
B \( 2 / 9 \)
C \( 1 / 3 \)
D \( 2 / 3 \)

The correct answer is:
\( 2 / 3 \)

Solution:
Let L=limx→0∫0x2sintx3dt
= lim x → 0 sin x · 2 x 3 x 2 (by Leibnitz rule)
=23limx→0sin xx

=231

=23

Hence, L=23.

Topic: Mathematics, Limits, JEE Main
The value of \( \lim _{x \rightarrow 0}(\cos x+\sin x)^{\frac{1}{x}} \) is equal to (take \( \left.e=2.71\right) \)

Options:

The correct answer is:
2.71

Solution:
∵limx→a⁡fxϕx=elimx→a⁡ϕxfx-1 as fx→1 & ϕx→∞ as x→a

So, limx→0cos⁡x+sin⁡x1x=elimx→0⁡1xcos⁡x+sin⁡x-1
=elimx→0⁡-sin⁡x+cos⁡x1=e

Topic: Mathematics, Limits, JEE Main
The value of \( \lim _{x \rightarrow 1} \frac{\sqrt[5]{x^{2}}-2 \sqrt[5]{x}+1}{4(x-1)^{2}} \) is equal to

Options:

The correct answer is:
0.01

Solution:
limx→1⁡x25-2x5+14x-12=limy→1⁡y2-2y+14y5-12
(Putting x5=y; as x→1, y→1)
limy→1⁡y-124y-12y4+y3+y2+y+12
=125×4=0.01

Topic: Mathematics, Limits, JEE Main
The value of \(\lim _{x \rightarrow 1} \frac{x+x^2+\ldots+x^n-n}{x-1}\) is

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

The correct answer is:
\(\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

Solution:
Hints: \(\operatorname{Lt}_{x \rightarrow 1} \frac{(x-1)+\left(x^2-1\right)+\left(x^3-1\right) \ldots \ldots \ldots \ldots .\left(x^4-1\right)}{x-1}\)
\(=1+2+3 \ldots \ldots \ldots \ldots \ldots \ldots+n=\frac{\mathrm{n}(\mathrm{n}+1)}{2}\)

Topic: Mathematics, Limits, JEE Main
The value of \( \lim _{x \rightarrow \infty} \frac{e^{x+1} \log \left(x^{3} e^{-x}+1\right)}{10 x^{3}} \) is equal to
(Use \( e=2.7 \) )

Options:

The correct answer is:
0.27

Solution:
Given, limx→∞e⋅exlogx3ex+110x3
=limx→∞e logx3ex+1x3ex×110

=e×1×110=0.27 ∵x3ex→0 if x→∞

Topic: Mathematics, Limits, JEE Main
The value of limx→π⁡tan⁡πcos2⁡xsin2⁡2x is equal to

Options:
A 1
B π
C -π4
D π2

The correct answer is:
-π4

Solution:
Let, x=π+t
limt→0⁡tan⁡πcos2⁡π+tsin2⁡2π+2t=limt→0⁡tan⁡πcos2⁡tsin2⁡2t
=limt→0⁡tan⁡π-πsin2⁡tsin2⁡2t=limt→0⁡-tan⁡πsin2⁡t4πsin2⁡tcos2⁡t·π=-π4

Topic: Mathematics, Limits, JEE Main
The value of limx→πtan⁡(πcos2⁡x)sin2⁡x is equal to

Options:
A 1
B π
C -π
D π2

The correct answer is:
-π

Solution:
Let, x=Ï+t
lim t→0 tan π cos 2 (π+t) sin 2 (π+t) ⇒ lim t→0 tan(π cos 2 t) sin 2 t
lim t→0 tan π−π sin 2 t sin 2 t ⇒ lim t→0 −tan(π sin 2 t) π sin 2 t ·π=−π

Topic: Mathematics, Limits, JEE Main
The value of $\lim _{x \rightarrow \infty}\left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2}\right)^{x}$ is

Options:
A $e^{2}$
B $e^{-2}$
C $e^{6}$
D None of these

The correct answer is:
$e^{2}$

Solution:
$\begin{aligned} & \lim _{x \rightarrow \infty}\left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2}\right)^{x} \\ &=\lim _{x \rightarrow \infty}\left(1+\frac{2 x-1}{x^{2}-4 x+2}\right)^{x} \\ &=e^{\lim _{x \rightarrow-\infty}\left(\frac{x(2 x-1)}{x^{2}-4 x+2}\right)}=e^{2} \end{aligned}$

Topic: Mathematics, Limits, JEE Main