datavorous/entrance-exam-dataset huggingface dataset:
The value of \( \int_{0}^{2 \pi}[\sin 2 x(1+\cos 3 x)] d x \), where \( [t] \) denotes the greatest integer function is

Options:
A \( \pi \)
B \( 2 \pi \)
C \( -\pi \)
D \( -2 \pi \)

The correct answer is:
\( -\pi \)

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
The value of \(\int_{-2}^2(x \cos x+\sin x+1) d x\) is

Options:
A 2
B 0
C -2
D 4

The correct answer is:
4

Solution:
Hints: \(\int_{-2}^2(x \cos x+\sin x+1) d x=\int_{-2}^2 d x=4\)

Topic: Mathematics, Definite Integration, JEE Main
The value of integral $\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{x}{1+\sin x} d x$ is

Options:
A
$\frac{\pi}{2}(\sqrt{2}+1)$

B
$\pi(\sqrt{2}-1)$

C
$2 \pi(\sqrt{2}-1)$

D
$\pi \sqrt{2}$

The correct answer is:
$\frac{\pi}{2}(\sqrt{2}+1)$

Solution:
Let $I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \frac{x}{1+\sin x} d x$ also let $K=\frac{x}{1+\sin x}$
Multiplying numerator and denominator by $(1-\sin x)$, we get;
$$
\begin{aligned}
K &=\frac{x(1-\sin x)}{1-(\sin x)^2}=\frac{x(1-\sin x)}{(\cos x)^2} \\
&=x(1-\sin x) \sec ^2 x \\
&=x \sec ^2 x-x \sin x \sec ^2 x=x \sec ^2 x-x \tan \\
x & \sec x
\end{aligned}
$$
Now, $I=\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} x \sec ^2 x d x-\int_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} x \sec x \tan x d x$
$$
=\left[x \tan x-\int \frac{d x}{d x} \tan x d x\right]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}-\left[x \sec x-\int \frac{d x}{d x} \sec x d x\right]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}
$$
$$
\begin{aligned}
&=[x \tan x-\ln |\sec x|]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}} \\
&-[x \sec x-\ln |\sec x+\tan x|]_{\frac{\pi}{4}}^{\frac{3 \pi}{4}}+c \\
&\Rightarrow I=\left\{\left[\frac{3 \pi}{4} \tan \frac{3 \pi}{4}-\ln \left|\frac{3 \pi}{4}\right|\right.\right. \\
&\left.-\left[\frac{3 \pi}{4} \sec \frac{3 \pi}{4}-\ln \left|\sec \frac{3 \pi}{4}+\tan \frac{3 \pi}{4}\right|\right]\right\} \\
&-\left\{\left[\frac{\pi}{4} \tan \frac{\pi}{4}-\ln \left|\frac{\pi}{4}\right|\right.\right. \\
&\left.-\left[\frac{\pi}{4} \sec \frac{\pi}{4}-\ln \left|\sec \frac{\pi}{4}+\tan \frac{\pi}{4}\right|\right]\right\} \\
&=\frac{\pi}{2}(\sqrt{2}+1)
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2018 (15 Apr Shift 2 Online)
The value of \( \int_{-\pi / 2}^{\pi / 2} \frac{d x}{[x]+[\sin x]+4} \), where \( [t] \) denotes the greatest integer less than or equal to \( t \), is

Options:
A \( \frac{3}{20}(4 \pi-3) \)
B \( \frac{3}{10}(4 \pi-3) \)
C \( \frac{1}{12}(7 \pi-5) \)
D \( \frac{1}{12}(7 \pi+5) \)

The correct answer is:
\( \frac{3}{20}(4 \pi-3) \)

Solution:
We have∫-π2π2dx[x]+[sinx]+4

=∫-π20dx[x]+3+∫0π2dx[x]+4

=∫-π2-1dx1+∫-10dx2+∫01dx4+∫1π2dx5
=[x]-π2-1+x2-10+x401+x51π2=-1+π2+0+12+14+π10-15
=-20+10π+10+5+2π-420=12π-920=3204π-3

Topic: Mathematics, Definite Integration, JEE Main
The value of ∑k=14sin2πk5-icos2πk54 is (where i is iota)

Options:

The correct answer is:
1

Solution:
∑k=14sin2πk5-icos2πk5
∑k=14-i2sin2πk5-icos2πk5
=-i∑k=14ei2πk5
=-i-1+ei0+ei2π5+ei4π5+ei6π5+ei8π5 (Sum of roots of the fifth root of unity is zero)
=i
⇒∑k=14(sin2πk5-icos2πk54=i4=1

Topic: Mathematics, Complex Number, JEE Main
The value of ∑k=110sin⁡2πk11-icos⁡2πk11 is (where i = - 1 )

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

The correct answer is:
i

Solution:
We have,
∑k=110sin⁡2πk11-icos⁡2πk11=∑k=110-i2sin⁡2πk11-icos⁡2πk11
=-i ∑k=110cos⁡2πk11+isin⁡2πk11=-i∑k=110ei2πk11
=- i ∑k=010ei2πk11-1
=- i sum of 11th roots of unity - 1
=- i 0 - 1=i.

Topic: Mathematics, Complex Number, JEE Main
The value of $\sum_{k=1}^{10}\left(\sin \frac{2 k \pi}{11}+i \cos \frac{2 k \pi}{11}\right)$ is

Options:
A
i

B
1

C
-1

D
-i

The correct answer is:
-i

Solution:
$\sum_{k=1}^{10}\left(\sin \frac{2 k \pi}{11}+i \cos \frac{2 k \pi}{11}\right)=\sum_{k=1}^{10} \sin \frac{2 k \pi}{11}+i \sum_{k=1}^{10} \cos \frac{2 k \pi}{11}$
$=0+i(-1)=-i$

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2006
The value of ∑k=199ik!+ωk! is (where, i=-1 and ω is non-real cube root of unity)

Options:
A 190+ω
B 192+ω2
C 190+i
D 192+i

The correct answer is:
190+i

Solution:
∑k=199ik!+∑k=199ωk!
∑k=199ik!=i1!+i2!+i3!+i4!+…+i99!
=i-1+i6+1+1+1+…+1
=i-2+96=i+94
∑k=199ωk!=ω1+ω2!+ω3!+ω4!+…+ω99!
=ω+ω2+1+1+1+…+1=96
Sum =i+94+96=i+190

Topic: Mathematics, Complex Number, JEE Main
The value of Kc us 64 at 800 K for the reaction

N2g+3H2g→2 NH3 g

The value of Kc for the following reaction is :

NH3 g→12 N2 g+32 H2 (g)

Options:
A 1/64
B 8
C 1/4
D 1/8

The correct answer is:
1/8

Solution:
N2g+3H2g⇌2NH3g

KC=NH32N2H23=64

For the reaction

NH3g⇌12N2g+32H2g

K'C=N21/2H23/2NH3=1KC=164=18

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2020 (06 Sep Shift 2)
The value of $k$ for which $2 x^2-k x+x+8=0$ has equal and real roots are

Options:
A -9 and -7
B 9 and 7
C -9 and 7
D 9 and -7

The correct answer is:
9 and -7

Solution:
Equation $2 x^2-k x+x+8=0$ has equal and real roots, then $D=b^2-4 a c=0$
$\Rightarrow(1-k)^2-4.2 .8=0 \Rightarrow k^2+1-2 k-64=0$
$\Rightarrow k^2-2 k-63=0 \Rightarrow k=9,-7$

Topic: Mathematics, Quadratic Equation, JEE Main
The value of k for which both the roots of the equation 4x2-20kx+25k2+15k-66=0 are less than 2, lies in

Options:
A 45,2
B 0, 2
C -1,-45
D -∞,-1

The correct answer is:
-∞,-1

Solution:
Let, fx=4x2-20kx+25k2+15k-66=0……………..(i)
Let the roots of fx=0 be α, β
Since α, β are real.
∴ ∆≥0
⇒400k2-4.425k2+15k-66≥0
⇒ -15k+66≥0 ⇒ k≤225…………..ii
We have α, β<2
∴ α+β<4
⇒ --20k4<4 ⇒k<45……………iii
fx=4x-αx-β
∴ f2=42-α2-β=4++=+ve
∴ f2=16-40k+25k2+15k-66>0
⇒25k2-25k-50>0 ⇒ k2-k-2>0
⇒ k+1k-2>0 ⇒ k<-1 or k>2…………iv
Combining ii, iii & iv, we get k∈-∞,-1

Topic: Mathematics, Quadratic Equation, JEE Main
The value of $\mathrm{k}$ for which the equation $(K-2) x^2+8 x+K+4=0$ has both roots real, distinct and negative is

Options:
A
6

B
3

C
4

D
1

The correct answer is:
3

Solution:
$(K-2) x^2+8 x+K+4=0$
If real roots then,
$$
\begin{aligned}
& 8^2-4(K-2)(K+4)>0 \\
\Rightarrow & K^2+2 K-8 < 16 \\
\Rightarrow & K^2+6 K-4 K-24 < 0 \\
\Rightarrow & (K+6)(K-4) < 0 \\
\Rightarrow & -6 < K < 4
\end{aligned}
$$
If both roots are negative then $\alpha \beta$ is $+v e$
$$
\begin{aligned}
& \Rightarrow \frac{K+4}{K-2}>0 \Rightarrow K>-4 \\
& \text { Also, } \frac{K-2}{K+4}>0 \Rightarrow K>2
\end{aligned}
$$
Roots are real so, $-6 < K < 4$
So, 6 and 4 are not correct.
Since, $K>2$, so 1 is also not correct value of $K$.
$$
\therefore K=3
$$

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2012 (07 May Online)
The value of $k$ for which the lines $7 x-8 y+5=0,3 x-4 y+5=0$ and $4 x+5 y+k=0$ are concurrent is given by

Options:
A -45
B 44
C 54
D -54

The correct answer is:
-45

Solution:
The lines are concurrent, if $\left|\begin{array}{ccc}7 & -8 & 5 \\ 3 & -4 & 5 \\ 4 & 5 & k\end{array}\right|=0$
$\Rightarrow 7(-4 k-25)+8(3 k-20)+5(15+16)=0 \Rightarrow k=-45$

Topic: Mathematics, Straight Lines, JEE Main
The value of $k$ for which the planes $3 x-6 y-2 z=7$ and $2 x+y-k z=5$ are perpendicular to each other, is

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
Planes are perpendicular, if $6-6+2 k=0 \Rightarrow k=0$.

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The value of k for which the sum of the squares of the roots of 2x2-2k-2x-k-1=0 is least is

Options:
A 1
B 32
C 2
D 52

The correct answer is:
32

Solution:
Î±+Î²=k-2
Î±Î²=-k+12
Î±2+Î²2=Î±+Î²2-2Î±Î²=k-22+k+1
=k2-4k+4+k+1
=k2-3k+5
⇒Î±2+Î²2=k-322+114
Which is least for k=32

Topic: Mathematics, Quadratic Equation, JEE Main
The value of \( k \) in order that \( f(x)=\sin x-\cos x-k x+b \) decreases for all real values is given by:

Options:
A \( k < 1 \)
B \( k>1 \)
C \( k>\sqrt{2} \)
D \( k < \sqrt{2} \)

The correct answer is:
\( k>\sqrt{2} \)

Solution:
Given:

⇒fx=sinx-cosx-kx+b⇒f'x=cosx+sinx-k

For decreasing function f'x < 0

So, cos x + sin x - k < 0

The maximum value of cos x + sin x is 2.

⇒2-k<0

⇒k>2

Topic: Mathematics, Application of Derivatives, JEE Main
The value of $k \in \mathrm{N}$ for which the integral $I_n=\int_0^1\left(1-x^k\right)^n d x, n \in \mathbb{N}$, satisfies $147 I_{20}=148 I_{21}$ is

Options:
A 14
B 8
C 10
D 7

The correct answer is:
7

Solution:
$\begin{aligned} & I_n=\int_0^1\left(1-x^k\right)^n \cdot 1 d x \\ & I_n=\left(1-x^k\right)^n \cdot x-n k \int_0^1\left(1-x^k\right)^{n-1} \cdot x^{k-1} \cdot d x \\ & I_n=n k \int_0^1\left[\left(1-x^k\right)^n-\left(1-x^k\right)^{n-1}\right] d x \\ & I_n=n k I_n-n k I_n \\ & \frac{I_n}{I_{n-1}}=\frac{n k}{n k+1} \\ & \frac{I_{21}}{I_{20}}=\frac{21 k}{1+21 k} \\ & =\frac{147}{148} \Rightarrow k=7\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (08 Apr Shift 1)
The value of $K_p$ for the equilibrium reaction $\mathrm{N}_2 \mathrm{O}_4(g) \rightleftharpoons 2 \mathrm{NO}_2(g)$ is 2 .
The percentage dissociation of $\mathrm{N}_2 \mathrm{O}_4(g)$ at a pressure of $0.5 \mathrm{~atm}$ is

Options:
A
25

B
88

C
50

D
71

The correct answer is:
71

Solution:
$\begin{array}{llcc}\text { } & \mathrm{N}_2 \mathrm{O}_4(\mathrm{~g}) \rightleftharpoons 2 \mathrm{NO}_2(\mathrm{~g}) \\ & \text { Initial moles } & 1 & 0 \\ & \text { Moles of equil. } & (1-\alpha) & 2 \alpha\end{array}$
( $\alpha=$ degree of dissociation $)$
Total number of moles at equil.
$$
\begin{aligned}
& =(1-\alpha)+2 \alpha \\
& =(1+\alpha) \\
p_{N_2 \mathrm{O}_4} & =\frac{(1-\alpha)}{(1+\alpha)} \times P \\
p_{N O_2} & =\frac{2 \alpha}{(1+\alpha)} \times P \\
K_P & =\frac{\left(p_{N_O}\right)^2}{p_{N_2 O_4}}=\frac{\left(\frac{2 \alpha}{(1+\alpha)} \times P\right)^2}{\left(\frac{1-\alpha}{1+\alpha}\right) \times P}=\frac{4 \alpha^2 P}{1-\alpha^2}
\end{aligned}
$$
$$
\begin{aligned}
& \text { Given, } K_P=2, P=0.5 \mathrm{~atm} \\
& \therefore \quad K_P=\frac{4 \alpha^2 P}{1-\alpha^2} \\
& =\frac{4 \alpha^2 \times 0.5}{1-\alpha^2} \\
& \alpha=0.707 \approx 0.71 \\
& \therefore \quad \text { Percentage dissociation } \\
& =0.71 \times 100=71
\end{aligned}
$$

Topic: Chemistry, Chemical Equilibrium, JEE Main, JEE Main 2012 (19 May Online)
The value of k∈R, for which the following system of linear equations

3x-y+4z=3

x+2y-3z=-2

6x+5y+kz=-3

has infinitely many solutions, is:

Options:
A 3
B -5
C 5
D -3

The correct answer is:
-5

Solution:
We have,

3x-y+4z=3

x+2y-3z=-2

6x+5y+kz=-3

For infinitely many solutions,

A=0

⇒3-1412-365 k=0

⇒3(2k+15)+1k+18+45-12=0

⇒3(2k+15)+k+18-28=0

⇒7k+35=0

⇒k=-5

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (20 Jul Shift 2)
The value of $k$ so that the function $f(x)=\left\{\begin{aligned} k\left(2 x-x^2\right), & \text { when } x \lt 0 \\ \cos x, & \text { when } x \geq 0\end{aligned}\right.$ is continuous at $x=0$, is

Options:
A 1
B 2
C 4
D None of these

The correct answer is:
None of these

Solution:
$f(0^-)=\lim _{x \rightarrow 0^-} k\left(2 x-x^2\right)=0$;
$f(0^+)=\lim _{x \rightarrow 0^+} \cos x=1$
$\therefore f(0)=\cos x=1$
Hence no value of $k$ can make $f(0^-)=1$.

Topic: Mathematics, Continuity and Differentiability, JEE Main
The value of \( k \) such that \( \frac{x-4}{1}=\frac{y-2}{1}=\frac{z-k}{2} \) lies in the plane \( 2 x-4 y+z=7 \), is

Options:
A \( 7 \)
B \( -7 \)
C No real value
D 5

The correct answer is:
\( 7 \)

Solution:
Given, equation of plane is 2x-4y+z=7

Equation of line is x-41=y-21=z-k2

For the equation of line to lie in the plane the points (4,2,k) must lie on the plane.

So, putting the points in the equation of plane,

⇒24-42+k=7⇒8-8+k=7⇒k=7

Topic: Mathematics, Three Dimensional Geometry, JEE Main
The value of $k$ such that the lines $2 x-3 y+k=0, \quad 3 x-4 y-13=0 \quad$ and $8 x-11 y-33=0$ are concurrent, is

Options:
A $20$
B $-7$
C $7$
D $-20$

The correct answer is:
$-7$

Solution:
Since, the lines
$$
\begin{aligned}
& 2 x-3 y+k=0,3 x-4 y-13=0 \\
& \text { and } \quad 8 x-11 y-33=0 \\
& \text { are concurrent } \\
& \therefore \quad\left|\begin{array}{ccc}
2 & -3 & k \\
3 & -4 & -13 \\
8 & -11 & -33
\end{array}\right|=0 \\
& \Rightarrow \quad 2(132-143)+3(-99+104) \\
& +k(-33+32)=0 \\
& \Rightarrow \quad-22+15-k=0 \\
& \Rightarrow \quad k=-7 \\
&
\end{aligned}
$$

Topic: Mathematics, Straight Lines, JEE Main
The value of k which the function fx= 45tan⁡4xtan⁡5x,0<x<π2k+25,x=π2 is continuous at x=π2, is

Options:
A 25
B -25
C 1720
D 35

The correct answer is:
35

Solution:
Given function is continuous at x= π 2
⇒fπ2=fπ+2=fπ-2

⇒k+25=45tan⁡4xtan⁡5x

⇒k+25=45tan2πcot5π2
⇒k+25=450
⇒k+25=1
⇒k=35

Topic: Mathematics, Continuity and Differentiability, JEE Main, JEE Main 2017 (09 Apr Online)
The value of limn→∞12-22+32-42+52....+2n+12n2 is equal to

Options:
A 2
B 4
C 6
D 8

The correct answer is:
2

Solution:
limn→∞12+3+23-2+.......2n+1-2n2n+1+2nn2
=limn→∞1+2+3+4+5+....+2n+1n2

=limn→∞2n+12n+22n2
=limn→∞n22+1n2+2n2n2=2×22=2

Topic: Mathematics, Limits, JEE Main
The value of limn→∞12n+12n+1+12n+2+.....+14n is equal to

Options:
A e2
B ln⁡2
C ln⁡4
D 3ln⁡2

The correct answer is:
ln⁡2

Solution:
limn→∞∑r=02n12n+r=limn→∞ 1n ∑r=02n12+rn
=∫0212+xdx=ln⁡x+202
=ln⁡4-ln⁡2=ln⁡2

Topic: Mathematics, Definite Integration, JEE Main
The value of limn→∞cos⁡xcosx2cosx4...cosx2n is equal to

Options:
A xsin⁡x
B sin⁡xx
C sin⁡2x2x
D 2xsin⁡2x

The correct answer is:
sin⁡2x2x

Solution:
Required limit =limn→∞cosx20cosx21cosx22...cosx2n
=limn→∞12sinx2ncos⁡x...cosx2n-12sinx2ncosx2n
=limn→∞122sinx2ncos⁡x...2cosx2n-1sinx2n-1
=limn→∞12n+1sinx2n2cos⁡xsin⁡x
=limn→∞sin⁡2x2n+1sinx2n
=sin⁡2x2xlimn→∞x2nsinx2n=sin⁡2x2x

Topic: Mathematics, Limits, JEE Main
The value of limn→∞e1nn2+2e2nn2+3e3nn2+…+2e2n is

Options:
A e2-1
B e2+1
C 2e2+1
D 2e2- 1

The correct answer is:
e2+1

Solution:
Let, I=limn→∞ ∑r=12nrn2ern
I=limn→∞ ∑r=12nrnern1n
I=∫02xexdx
I=xex02-∫02exdx
I=2e2-e2-1
I=e2+1

Topic: Mathematics, Definite Integration, JEE Main
The value of limn→∞∑r=1n2r+3r6r is equal to

Options:
A 0
B 1
C 6
D 32

The correct answer is:
32

Solution:
limnâââr=1n2r6r+3r6r⇒limnâââr=1n13r+12r
⇒limnââ131-13n1-13+121-12n1-12
⇒131-023+121-012⇒12+1=32

Topic: Mathematics, Sequences and Series, JEE Main
The value of limnâââr=1n2rn2er2n2 is equal to

Options:
A e
B 2e
C e–2
D e–1

The correct answer is:
e–1

Solution:
limn→∞∑r=1n2rner2n21n=∫012xex2dx=ex201=e-1

Topic: Mathematics, Definite Integration, JEE Main
The value of Î=limnâââr=n2nnr2 is equal to

Options:
A 13
B 12
C π2
D π3

The correct answer is:
12

Solution:
Ι=limn→∞∑r=n2n1r2n2⋅1n
=∫121x2dx=-1x12
=-12--1
=12

Topic: Mathematics, Definite Integration, JEE Main
The value of limn→∞x+22x+32x+.....+n2x12+22+32+....+n2 is equal to (where x represents the greatest integer part of x)

Options:
A x
B 2x
C x2
D x6

The correct answer is:
x

Solution:
Let fx=x+22x+32x+...+n2x12+22+32+...+n2
Now, we have,
fx≤x+22x+32x+...+n2x12+22+32+...+n2=x
and, fx>x-1+22x-1+32x-1+...+n2x-112+22+32+...+n2
=xΣn2-nΣn2=x-6n+12n+1 ∵x-1≤x<x, ∀x∈R
Thus, we have,
x-6n+12n+1<fx≤x
Now, we have,
limn→∞x-6n+12n+1=x & limn→∞x=x
Hence, by Sandwich Theorem, we have
limn→∞fx=x

Topic: Mathematics, Limits, JEE Main
The value of limxâ0 1-cos3sinâ¡xsinâ¡x sinsinâ¡xcossinâ¡x
is equal to

Options:
A 32
B 1
C 0
D 2

The correct answer is:
32

Solution:
Let, sin⁡x=t
⇒limt→0 1-cos3ttsin⁡tcos⁡t
⇒limt→01-cos⁡tt2×tsin⁡t×1+cos⁡t+cos2⁡tcos⁡t
⇒12×1×1+1+11=32

Topic: Mathematics, Limits, JEE Main
The value of limx→01-cos⁡2xx is equal to

Options:
A 12
B -12
C Does not exist
D 1

The correct answer is:
Does not exist

Solution:
limx→01-cos⁡2xx
=limx→01-cos⁡2xx.11+cos⁡2x
=limx→02|sin⁡x|x.11+cos⁡2x
Now, we have
LHL=limx→0-2sin⁡xx.11+cos⁡2x
=-2.12=-1
RHL=limx→02sin⁡xx.11+cos⁡2x
=2.12=1
∵LHL≠RHL
Hence, limit does not exist.

Topic: Mathematics, Limits, JEE Main
The value of limx→0-21x+23x321x+523x is

Options:
A 13
B 15
C 1
D 14

The correct answer is:
13

Solution:
limx→0-21x1+22x21x3+5⋅22x⇒limx→0-1+21x3+522x
=1+03+0=13

Topic: Mathematics, Limits, JEE Main
The value of limx→0-42+3x+5(21x)21+6x+6(21x) is equal to

Options:
A 56
B 8
C 16
D 52

The correct answer is:
56

Solution:
limx→0-24+6x+5.21x21+6x+6.21x=limx→0-21x1625x+521x225x+6
limx→0-1625x+5225x+6=160+520+6=56

Topic: Mathematics, Limits, JEE Main
The value of limx→0costan⁡x-cos⁡x4x4 is equal to

Options:
A -13
B -112
C 12
D 1

The correct answer is:
-112

Solution:
costan⁡x-cos⁡x=2sinx+tan⁡x2sinx-tan⁡x2
⇒limx→0costan⁡x-cos⁡x4x4=limx→02sinx+tan⁡x2sinx-tan⁡tan⁡x24x4
=limx→02sinx+tan⁡x2sinx-tan⁡x24x4x+tan⁡x2x-tan⁡x2x2-tan2x4
=18limx→0x2-tan2xx4
=18limx→0x2-x+x33+215x5+....2x4
=18limx→01x21-1+x23+215x4+....2=-112

Topic: Mathematics, Limits, JEE Main
The value of limx→0e-x22-cos⁡xx3tan⁡x is equal to

Options:
A 14
B 18
C 112
D 116

The correct answer is:
112

Solution:
Using expansion,
limxâ01+-x22+x442!+....-1-x22!+x24!-....x3x+x33+....
limxâ01-1+-x22+x22+x48-x424+....x3x+x33+....
â18-124=112

Topic: Mathematics, Limits, JEE Main
The value of limx→0ex-x-1x-sin⁡xln1+xx6 is equal to

Options:
A 12
B 16
C 112
D 13

The correct answer is:
112

Solution:
Let, L=limx→0ex-x-1x2x-sin⁡xx3ln⁡1+xx

Applying expansion series, we get,
L=limx→01+x+x22!….-x+1x-x-x33!+x-x22+…x6
=12.16.1=112

Topic: Mathematics, Limits, JEE Main
The value of limx→0ln2-cos⁡15xln2sin⁡3x+1 is equal to _______. If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
25

Solution:
limx→0ln2-cos⁡15xln2sin⁡3x+1=limx→0ln1+1-cos⁡15xln21+sin⁡3x
=limx→01-cos⁡15xsin⁡3x2
=limx→015x223x2=12252×9=12.5

Topic: Mathematics, Limits, JEE Main
The value of limx→0ln2-cos⁡15xln2sin⁡3x+1 is equal to

Options:

The correct answer is:
12.5

Solution:
limx→0ln2-cos⁡15xln2sin⁡3x+1=limx→0ln1+1-cos⁡15xln21+sin⁡3x
=limx→01-cos⁡15xsin⁡3x2
=limx→015x223x2=12252×9=12.5

Topic: Mathematics, Limits, JEE Main
The value of limx→0ln10-9cos⁡2xln2sin⁡3x+1 is equal to

Options:

The correct answer is:
2

Solution:
limx→0ln10-9cos⁡2xln2sin⁡3x+1=limx→0ln1+9-9cos⁡2xln21+sin⁡3x
=limx→091-cos⁡2xsin⁡3x2=limx→092x23x2=2

Topic: Mathematics, Limits, JEE Main
The value of limx→0logsin⁡5x+cos⁡5xtan⁡3x is equal to

Options:
A 103
B 203
C 56
D 53

The correct answer is:
53

Solution:
limx→012log1+sin⁡10xtan⁡3x (as, 1+sin10x=(sin5x+cos5x)2)
=limx→012log(1+sin10x)sin⁡10x⋅sin⁡10x10x⋅3xtan3x⋅10x3x
=12×1×1×103=53

Topic: Mathematics, Limits, JEE Main
The value of limxâ0lnâ¡1+2x+4x2+lnâ¡1-2x+4x2secâ¡x-cosâ¡x is equal to

Options:
A -1
B 1
C 0
D 4

The correct answer is:
4

Solution:
limxâ0lnâ¡1+2x+4x2+lnâ¡1-2x+4x2secâ¡x-cosâ¡x
=limxâ0lnâ¡1+4x22-4x21-cos2x/cosâ¡x
=limxâ0lnâ¡1+4x2+16x4sinâ¡xtanâ¡x
=limxâ0lnâ¡1+4x21+4x24x21+4x24x21+4x2â1sinâ¡xxâtanâ¡xxâx2
=4 as limxâ0lnâ¡1+xx=1

Topic: Mathematics, Limits, JEE Main
The value of limxâ0secâ¡x-secâ¡xsec x1-secâ¡x+lnâ¡secâ¡x is equal to

Options:
A 0
B 1
C 2
D -1

The correct answer is:
2

Solution:
Let, sec⁡x=t
limt→1t-tt1-t+ln⁡t=limt→11-tt(1+ln⁡t)-1+1t {by L-hospital rule}
=limt→1-tt1t-tt1+ln⁡t2-1/t2 {again by L-hospital rule}
=-1-1-1=2

Topic: Mathematics, Limits, JEE Main
The value of limx→0sec⁡x+tan⁡x1x is equal to

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

The correct answer is:
e

Solution:
limx→0sec⁡x+tan⁡x1x (1∞ form)
=elimx→01xsec⁡x+tan⁡x-1=elimx→01+sin⁡x-cos⁡xxcos⁡x=elimx→0cos⁡x+sin⁡x-xsin⁡x+cos⁡x by L'Hospital rule
=e1+00+1=e

Topic: Mathematics, Limits, JEE Main
The value of limxâ0sinâ¡x35x is equal to

(where, . represents the greatest integer function)

Options:
A 13
B 0
C 53
D 1

The correct answer is:
53

Solution:
5x-1<5xâ¤5x
sinâ¡x35x-1âhx<sinâ¡x35xâfxâ¤sinâ¡x35xâgx
by sandwich theorem
âµ limxâ0gx=limxâ0hx=53
â´limxâ0fx=53

Topic: Mathematics, Limits, JEE Main
The value of limxâ0tan23x5-4+secx is equal to

Options:
A 25
B -95
C 95
D -365

The correct answer is:
-365

Solution:
After rationalization,
limx→0tan23x5+4+secx5-4+sec⁡x
⇒limx→0tan23x9x2⋅9x2cosx-15+4+1
⇒9-225=-365

Topic: Mathematics, Limits, JEE Main
The value of limx→0xcot4xtan23xcot26x is equal to

Options:
A 0
B 4
C 29
D 1

The correct answer is:
1

Solution:
limx→09x2tan⁡3x219x⋅tan⁡6x236⋅x2⋅4xtan⁡4x36x24x
=1×19×1×1×364=1

Topic: Mathematics, Limits, JEE Main
The value of limx→∞11x+21x+31x+...+101x1010x is

Options:
A 10!
B 10
C 9!
D 0

The correct answer is:
10!

Solution:
limxââ11x+21x+31x+...+101x1010x
=limyâ0+1y+2y+3y+...+10y1010y where, y=1x
=elimyâ0+10y1y+2y+3y+...+10y10-1
=elimyâ0+1y+2y+3y+...+10y-10y
=elimyâ0+1y-1y+2y-1y+3y-1y+...+10y-1y
=elog1+log2+log3+...+log10
=elog1â2â3.....10=10!

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

Options:
A 2π
B 12π
C 4π
D 0

The correct answer is:
0

Solution:
After rationalization
limxâ1-Ï-4tan-1x1-xÏ+4tan-1x=12Ïlimxâ1-Ï-4tan-1x1-x
By L-hospital rule, we get,
12Ïlimxâ1--41+x2-121-x=4Ïlimxâ1-1-x1+x2=0

Topic: Mathematics, Limits, JEE Main
The value of limx→1x25-2x5+1x-12 is equal to

Options:

The correct answer is:
0.04

Solution:
limx→1x25-2x5+1x-12=limx→1y2-2y+1y5-12
Putting x5=y, as x→1, y→1
limy→1y-12y-12y4+y3+y2+y+12
=125

Topic: Mathematics, Limits, JEE Main
The value of limxâ1xtanxx-1 is equal to (where x denotes the fractional part of x)

Options:
A -1
B 0
C 1
D Does not exist

The correct answer is:
Does not exist

Solution:
LHL=limx→1-xtanxx-1
Let, x=1-h, as x→1-, h→0+
⇒LHL=limh→0+1-htan1-h-h
∴LHL=limh→0+1-h-htan1=-∞
Now, RHL=limx→1+xtanxx-1
Let, x=1+h, as x→1+, h→0+
⇒RHL=limh→0+1+htan1+hh
⇒RHL=limh→0+1+htanhh=limh→0+1+h
∴RHL=1+0=1
Since, LHL≠RHL
∴ the limit of the function does not exist at x=1

Topic: Mathematics, Limits, JEE Main
The value of limx→π2x3ln1+cot⁡x is equal to (where, . denotes the greatest 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:
âµÏ6<1
â´Ï6=0
â´limxâÏ2x3lnâ¡1+cotâ¡x=0

Topic: Mathematics, Limits, JEE Main
The value of limx→π32-3sin⁡x-cos⁡x3x-π2 is equal to the reciprocal of the number

Options:

The correct answer is:
9

Solution:
limx→π32-232sin⁡x+12cos⁡x3x-π2
⇒limx→π32-2cosx-π33x-π2
Let, x-π3=t
⇒limt→02-2cos⁡t3t2=limt→021-cos⁡t9t2=29⋅12=19

Topic: Mathematics, Limits, JEE Main
The value of limxââex+1logx3e-x+1sin32x is equal to

(Use e=2.7)

Options:

The correct answer is:
0.34

Solution:
Given, limx→∞e⋅exlogx3ex+1sin⁡2x3
=limx→∞e logx3ex+1x3ex×2x3sin⁡2x3×18

=e×1×1×18=e8 ∵x3ex→0 if x→∞

Topic: Mathematics, Limits, JEE Main
The value of limx→∞ln⁡x22+3x2 is equal to

Options:
A 13
B 23
C 1
D 0

The correct answer is:
0

Solution:
By L-hospital rule,
limxââ2lnâ¡x.1x6x
âlimxââlnâ¡x3x2
Again by L-hospital rule,
limxââ1x6xâlimxââ16x2=0

Topic: Mathematics, Limits, JEE Main
The value of limx→πsin2πcos2xtanπsec2x is equal to

Options:
A 1
B 2
C -2
D 0

The correct answer is:
-2

Solution:
limx→πsin2π1-sin2xtanπ1+tan2x=limx→πsin2π-2πsin2xtanπ+πtan2x
=limx→π-sin2πsin2xtanπtan2x

=limx→π-sin2πsin2x2πsin2x×2πsin2xπtan2x×πtan2xtanπtan2x
=-1×2×1=-2

Topic: Mathematics, Limits, JEE Main
The value oflimx→-∞x2tan1x4x2-x+1 is equal to

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

The correct answer is:
-12

Solution:
limx→-∞xtan1x1x-x4-1x+1x2=1-4=-12

Topic: Mathematics, Limits, JEE Main
The value of limy→01+1+y4-2y4

Options:
A exists and equals 122
B exists and equals 142
C does not exist
D exists and equals 1222+1

The correct answer is:
exists and equals 142

Solution:
To solve this question, we need to apply the concept of rationalization twice.
Given limit is limy→01+1+y4-2y4 ( 00 form)
=limy→01+1+y4-2y4×1+1+y4+21+1+y4+2
=limy→01+y4-1y41+1+y4+2×1+y4+11+y4+1
=limy→0y4y41+1+y4+21+y4+1
=limy→011+1+y4 +21+y4+1
=142.

Topic: Mathematics, Limits, JEE Main, JEE Main 2019 (09 Jan Shift 1)
The value of \( L t\left\{\frac{\int_{0}^{x^{2}} \sec ^{2} t d t}{x \sin x}\right\} \) is

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

The correct answer is:
\( 1 \)

Solution:
Ltx→0 ∫0x2sec2 t dtx sin x 00 form

Apply L'Hospital's Rule, we get

=Ltx→0 sec2x22xxcosx+sinx 00 form

=Ltx→02sec2x2cosx+sinxx=21+1=1

Topic: Mathematics, Definite Integration, JEE Main