datavorous/entrance-exam-dataset huggingface dataset:
The value of ∫12x-1x2-xdx is equal to (where c is the constant of integration)

Options:
A sec-1x-1+c
B sec-12x-1+c
C tan-1x+c
D tan-12x-1+c

The correct answer is:
sec-12x-1+c

Solution:
∫22x-14x2-4xdx=2∫dx2x-12x-12-1
=2×12sec-12x-1+c

Topic: Mathematics, Indefinite Integration, JEE Main
The value of $\left(\frac{1+\sqrt{3} i}{1-\sqrt{3 i}}\right)^{64}+\left(\frac{1-\sqrt{3} i}{1+\sqrt{3} i}\right)^{64}$ is

Options:
A 0
B -1
C 1
D $i$

The correct answer is:
-1

Solution:
We know that, $\omega=\frac{-1+\sqrt{3} i}{2} 1-\sqrt{3} i=-2 \omega$
and $\omega^{2}=\frac{-1-\sqrt{3} i}{2} 1+\sqrt{3} i=-2 \omega^{2}$
Now, $\begin{aligned}\left(\frac{1+\sqrt{3} i}{1-\sqrt{3 i}}\right)^{64}+\left(\frac{1-\sqrt{3} i}{1+\sqrt{3} i}\right)^{64} \\ &=\left(\frac{-2 \omega^{2}}{-2 \omega}\right)^{64}+\left(\frac{-2 \omega}{-2 \omega^{2}}\right)^{64} \end{aligned}$
$=\omega^{64}+\frac{1}{\omega^{64}}=\omega+\omega^{2}$
$$
\left[\because \cdot \omega^{3}=1\right]
$$
$$
\left.=-1 \quad \mid \because 1+\omega+\omega^{2}=0\right]
$$

Topic: Mathematics, Complex Number, JEE Main
The value of $\int_{-\pi}^\pi \frac{\cos ^2 x}{1+a^x} d x, a>0$, is

Options:
A
$a \pi$

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

C
$\frac{\pi}{a}$

D
$2 \pi$

The correct answer is:
$\frac{\pi}{2}$

Solution:
$$
\int_{-\pi}^\pi \frac{\cos ^2 x}{1+a^x} d x=\int_0^\pi \cos ^2 x d x=\frac{\pi}{2} .
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2005
The value of $\int_1^a[x] f^{\prime}(x) d x, a>1$, where $[x]$ denotes the greatest integer not exceeding $x$ is

Options:
A
$\operatorname{af}(a)-\{f(1)+f(2)+\ldots+f([a])\}$

B
$[a] f(a)-\{f(1)+f(2)+\ldots+f([a])\}$

C
$[a] f([a])-\{f(1)+f(2)+\ldots+f(a)\}$

D
$a f([a])-\{f(1)+f(2)+\ldots+f(a)\}$

The correct answer is:
$[a] f(a)-\{f(1)+f(2)+\ldots+f([a])\}$

Solution:

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2006
The value of $\int_{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y$ is :

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

The correct answer is:
$\pi^2$

Solution:
$\int_{-\pi}^\pi \frac{2 y(1+\sin y)}{1+\cos ^2 y} d y$
\(=\underset{\text{(Odd)}}{\int_{-\pi}^\pi \frac{2 y}{1+\cos ^2 y} d y}+\underset{\text{(Even)}}{\int_{-\pi}^\pi \frac{2 y \sin y}{1+\cos ^2 y} d y}\)
$=0+2.2 \int_0^\pi y\left(\frac{\sin y}{1+\cos ^2 y}\right) d y$
$\begin{aligned} & I=4 \int_0^\pi \frac{y \sin y}{1+\cos ^2 y} d y \\ & I=4 \int_0^\pi \frac{(\pi-y) \sin y}{1+\cos ^2 y} d y \\ & 2 I=4 \int_0^\pi \frac{\pi \sin y}{1+\cos ^2 y} d y \\ & I=2 \pi \int_0^\pi \frac{\sin y}{1+\cos ^2 y} d y \\ & =2 \pi\left(-\tan ^{-1}(\cos y)\right)_0^\pi \\ & =-2 \pi\left[\left(-\frac{\pi}{4}\right)-\left(\frac{\pi}{4}\right)\right] \\ & =-2 \pi\left[-\frac{2 \pi}{4}\right]=\pi^2\end{aligned}$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2024 (05 Apr Shift 1)
The value of 1+cosÏ81+cos3Ï81+cos5Ï81+cos7Ï8 is equal to

Options:
A 12
B 14
C 18
D 116

The correct answer is:
18

Solution:
Given expression can be written as
1+cos⁡π81+cos⁡7π81+cos⁡3π81+cos⁡5π8
=1+cosπ81-cosπ81+cos3π81-cos3π8
=1-cos2π81-cos23π8
=sin2π8sin23π8
= sin 2 π 8 · cos 2 π 8
= 1 4 2 sin π 8 cos π 8 2
= 1 4 sin 2 π 4
= 1 8

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of â«1-cosâ¡Î¸3101+cosâ¡Î¸1310dÎ¸ is equal to (where, c is the constant of integration)

Options:
A 58tanâ¡Î¸58+c
B 58tan⁡θ285+c
C 516tan⁡θ285+c
D 58tanâ¡Î¸516+c

The correct answer is:
58tan⁡θ285+c

Solution:
∫1-cos⁡θ3101+cos⁡θ1310dθ=∫2310sin⁡θ23521310cos⁡θ2135dθ
=12∫tan⁡θ235sec2θ2dθ
Let, tan⁡θ2=t ⇒12sec2θ2dθ=dt
=∫t35dt=t8585+c
=58tan⁡θ285+c

Topic: Mathematics, Indefinite Integration, JEE Main
The value of-1+i31-i30 is :

Options:
A 65
B 215 i
C -215
D -215i

The correct answer is:
-215i

Solution:
-1+3i1-i30=2cos2π3+isin2π32cosπ4-isinπ430

=230(cos20π+isin20π)215cos15π2-isin15π2

=215(1+0i)(0+i)=-215i

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2020 (05 Sep Shift 2)
The value of $(1+i)^5(1-i)^7$

Options:
A -64
B -64 i
C 64 i
D 64

The correct answer is:
-64 i

Solution:
$\begin{aligned} & (1+i)^5(1-i)^7 \\ = & (1+i)^5(1-i)^5(1-i)^2=[(1+i)(1+i)(1-i)]^5\left(1-2 i+i^2\right) \\ = & \left(1-i^2\right)^5(-2 i)=(2)^5(-2 i)=-64 i \ldots\left[\because i^2=-1\right]\end{aligned}$

Topic: Mathematics, Complex Number, JEE Main
The value of 1+sin2π9+icos2π91+sin2π9-icos2π93 is

Options:
A -121-i3
B 121-i3
C -123-i
D 123+i

The correct answer is:
-123-i

Solution:
Given,

1+sin2π9+icos2π91+sin2π9-icos2π93

Now let z=sin2π9+icos2π9,

So, z¯=sin2π9-icos2π9=1z

So, 1+sin2π9+icos2π91+sin2π9-icos2π93

=1+z1+z¯3

=1+z1+1z3

=z31+z1+z3

=z3

=sin2π9+icos2π93

=i3cos2π9-isin2π93

=-icos3×2π9-isin3×2π9

=-icos2π3-isin2π3

=-i-12-i32

=-123-i

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2023 (24 Jan Shift 2)
The value of 1+sin2π9+icos2π91+sin2π9−icos2π93 is

Options:
A 12(1-i3)
B 12(3-i)
C -12(3-i)
D −121−i3

The correct answer is:
-12(3-i)

Solution:
1+sin2π9+icos2π91+sin2π9−icos2π93

=1+cos5π18+isin5π181+cos5π18-isin5π183=2cos25π36+i2sin5π36·cos5π362cos25π36-i2sin5π36·cos5π363

=cos5π36+isin5π36cos5π36-isin5π363=cos5π36+isin5π366

=cos6×5π36+isin6×5π36=cos5π6+isin5π6

=-32+i12

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2020 (02 Sep Shift 1)
The value of â = 1sinâ¡3Î¸sin3â¡Î¸2cosâ¡Î¸sinâ¡6Î¸sin3â¡2Î¸4cos2â¡Î¸-1sinâ¡9Î¸sin3â¡3Î¸ is equal to

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

The correct answer is:
0

Solution:
Multiplying C1 by 3 sin θ and dividing ∆ by 3 sin θ, we get,
∆ =13sin⁡θ 3sin⁡θsin⁡3θsin3⁡θ3sin⁡2θsin⁡6θsin3⁡2θ3sin⁡3θsin⁡9θsin3⁡3θ
∵3sin⁡θ 4cos2⁡θ-1=3sin⁡θ3-4sin2⁡θ=3 sin⁡θ-4sin3⁡θ=3sin⁡3θ
Now, applying C1↔C1-4C3

C1 becomes identical with C2

Hence, ∆=0

Topic: Mathematics, Determinants, JEE Main
The value of \( (1+i \sqrt{3})^{6}+(1-i)^{8} \) is

Options:
A \( 16(2-i) \)
B \( 32(3-2 i) \)
C \( 80 \)
D \( 48 \)

The correct answer is:
\( 80 \)

Solution:
We have,

1+i3=212+i32

=2cosπ3+isinπ3

=2eiπ3, ∵cosθ+isinθ=eiθ

Therefore,

1+i36=26ei2π, using De-Moivre's Theorem

=26cos2π+isin2π=26 ...i

Similarly,

1-i8=24e-2iπ, using De-Moivre's Theorem

=24cos2π-isin2π=24 ...ii

Hence, by 1 and 2, we have

1+i36+1-i8=26+24=80.

Topic: Mathematics, Complex Number, JEE Main
The value of 2α+β 0<α, β<π2, satisfying the equation cos⁡αcos⁡βcos⁡α+β=-18 is equal to

Options:
A 56π
B π2
C π
D 7π12

The correct answer is:
π

Solution:
From the given relation,
2cos⁡αcos⁡βcos⁡α+β=-14
cosα+β+cosα-βcosα+β=-14
Let, cosα+β=t
t+cosα-βt=-14
⇒4t2+4tcosα-β+1=0
The roots will be real if D≥0
⇒16cos2α-β-16≥0⇒16cos2α-β≥16
⇒cos2α-β≥1⇒cos2α-β=1⇒α-β=0⇒α=β
Then the above quadratic equation becomes
4t2+4t+1=0⇒2t+12=0⇒t=-12
⇒cosα+β=-12=cos2π3⇒α+β=2π3
Now, α+β=2π3 and α=β⇒α=β=π3

⇒2α+β=π

Topic: Mathematics, Trigonometric Equations, JEE Main
The value of (2⋅1P0−3⋅2P1+4⋅3P2−........ up to 51th term)+( 1!−2!+3!−....... up to 51th term) is equal to

Options:
A 1-5151!
B 1+51!
C 1+52!
D 1

The correct answer is:
1+52!

Solution:
2·P01=2×1!1!=2!

3·P12=3×2!1!=3!

4·P23=4×3!1!=4!

By this observation given question can be written as

=2!-3!+4!-5!+.......-51!+52!+1!-2!+3!-4!+......-50!+51!

=1!+52!=1+52!

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2020 (03 Sep Shift 1)
The value of ∫-π2π211+esinxdx is :

Options:
A π4
B π
C π2
D 3π2

The correct answer is:
π2

Solution:
I=∫-π2π211+esinxdx

⇒I=∫-π2π211+esinπ2-π2-xdx=∫-π2π211+esin-xdx

⇒I=∫-π2π2esinx1+esinxdx

Using the property, ∫abfxdx=∫cbfa+b-xdx

2I=∫-π2π21dx⇒I=12∫-π2π2dx

I=12x-π2π2⇒I=π2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2020 (05 Sep Shift 1)
The value of $\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^2 x}{1+2^x} d x$ is :

Options:
A
$\pi$

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

C
$4 \pi$

D
$\frac{\pi}{4}$

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

Solution:
$\mathrm{I}=\int_{\pi / 2}^{\pi / 2} \frac{\sin ^2 x}{1+2^x} d x$
$$
\Rightarrow \mathrm{I}=\int_{-\pi / 2}^{\pi / 2} \frac{\sin ^2 x}{1+2^{-x}} d x \text {, by replacing } x \text { by }
$$
$$
\begin{aligned}
& \left(\frac{\pi}{2}-\frac{\pi}{2}-x\right) \\
\Rightarrow \quad \mathrm{I} & =\int_{-\pi / 2}^{\pi / 2} \frac{2^x \cdot \sin ^2 x}{1+2^x} d x
\end{aligned}
$$
Adding equations (i) and (ii), we get
$$
\begin{array}{rl}
2 & I=\int_{-\pi / 2}^{\pi / 2} \sin ^2 x d x=\frac{1}{2} \int_{-\pi / 2}^{\pi / 2}(1-\cos 2 x) d x \\
\Rightarrow \quad \mathrm{I} & =\frac{1}{4}\left[x+\frac{\sin 2 x}{2}\right]_{-\pi / 2}^{\pi / 2} \\
& =\frac{1}{4}\left[\left(\frac{\pi}{2}+\frac{\sin \pi}{2}\right)-\left(-\frac{\pi}{2}+\frac{\sin (-\pi)}{2}\right)\right] \\
\Rightarrow \quad \mathrm{I} & =\frac{1}{4}\left[\frac{\pi}{2}+\frac{\pi}{2}\right]=\frac{\pi}{4}
\end{array}
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2013 (23 Apr Online)
The value of ∫-π2π21+sin2x1+πsinxdx is :

Options:
A π2
B 5π2
C 3π2
D 3π4

The correct answer is:
3π4

Solution:
I=∫abfxdx

f(a+b-x)=f(x)

I=∫-π2π21+sin2x1+πsinxdx…i

I=∫-π2π21+sin2x1+π-sinxdx...ii

Add equation i and ii

2I=∫-π2π21+sin2xdx

2I=π+12∫-π2π21-cos2xdx

2I=π+12x-sin2x2-π2π2

2I=π+12π2-0--π2-0

2I=π+12π

2I=3π2

⇒I=3π4

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The value of ∫-223x2-3x-6dx is

Options:

The correct answer is:
19

Solution:
∫-223x2-x-2dx

=3∫-22x2-x-2dx

=3∫-2-1x2-x-2dx+∫-12-x2-x-2dx

=3x33-x22-2x-2-1-x33-x22-2x-12

=37-23

=19

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (25 Feb Shift 2)
The value of ∫-π/2π/2cos2x1+3x dx is:

Options:
A π2
B π4
C 2π
D 4π

The correct answer is:
π4

Solution:
Given I=∫-π/2π/2cos2x1+3xdx ...i

Using ∫abfxdx=∫abfa+b-xdx, we get

I=∫-π/2π/2cos2x1+3-xdx=∫-π/2π/23xcos2x1+3xdx ...ii

Adding i & ii we get,

2I=∫-π/2π/21+3xcos2x1+3xdx

2I=∫-π/2π/2cos2xdx

2I=2∫0π/2cos2xdx

I=∫0π/2cos2xdx

I=12∫0π/21+cos2xdx=12x+sin2x20π/2

I=π4

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (26 Feb Shift 1)
The value of ∫-π/2π/2dxx+sin⁡x + 4, where t denotes the greatest integer less than or equal to t, is

Options:
A 3204π-3
B 3104π-3
C 1127π-5
D 1127π+5

The correct answer is:
3204π-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, JEE Main 2019 (10 Jan Shift 2)
The value of â«Ï2Ï[2sinx]dx is equal to (where [.] represents the greatest integer function)

Options:
A −π
B 5π 3
C â5Ï 3
D −2π

The correct answer is:
−5π 3

Solution:
Let, I= â« Ï 2Ï [2sinx]dx

For Ïâ¤xâ¤7Ï6 or 11Ï6â¤xâ¤2Ï

âââ1â¤2âsinx<0
ââââââ[2sinx]=â1
For 7Ï6â¤x<11Ï6

â2â¤2sinx<â1
ââââââ[2sinx]=â2
â´âââââI= â« Ï 7Ï/6 â1dx+ â« 7Ï/6 11Ï/6 â2dx+ â« 11Ï/6 2Ï â1âdx
=( â 7Ï 6 +Ï )+2( â 11Ï 6 + 7Ï 6 )+( â2Ï+ 11Ï 6 )
=â Ï 6 â 8Ï 6 â Ï 6 =â 10Ï 6 = â5Ï 3

Topic: Mathematics, Definite Integration, JEE Main
The value of $\int_{-2}^3\left|1-x^2\right| d x$ is

Options:
A
$\frac{28}{3}$

B
$\frac{14}{3}$

C
$\frac{7}{3}$

D
$\frac{1}{3}$

The correct answer is:
$\frac{28}{3}$

Solution:
$\int_{-2}^{-1}\left(x^2-1\right) d x+\int_{-1}^1\left(1-x^2\right) d x+\int_1^3\left(x^2-1\right) d x=\frac{x^3}{3}-\left.x\right|_{-2} ^{-1}+x-\left.\frac{x^3}{3}\right|_{-1} ^1+\frac{x^3}{3}-\left.x\right|_1 ^3=\frac{28}{3}$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2004
The value of $2^{1 / 4}, 4^{1 / 8}, 8^{1 / 6}+\ldots \ldots \infty$ is

Options:
A
1

B
2

C
3/2

D
4

The correct answer is:
2

Solution:
No Solution Available

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2002
The value of $\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\ldots .$. is equal to

Options:
A $\frac{2^n-1}{n+1}$
B $n \cdot 2^n$
C $\frac{2^n}{n}$
D $\frac{2^n+1}{n+1}$

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

Solution:
We know that
$\frac{(1+x)^n-(1-x)^n}{2}=C_1 x+C_3 x^3+C_5 x^5+\ldots$
Integrating from $x=0$ to $x=1$, we get
$\begin{aligned} & \frac{1}{2} \int_0^1\left\{(1+x)^n-(1-x)^n\right\} d x \\
& =\int_0^1\left(C_1 x+C_3 x^3+C_5 x^5+\ldots .\right) d x\end{aligned}$
$\Rightarrow \frac{1}{2}\left\{\frac{(1+x)^{n+1}}{n+1}+\frac{(1-x)^{n+1}}{n+1}\right\}_0^1=\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\ldots$
or $\frac{C_1}{2}+\frac{C_3}{4}+\frac{C_5}{6}+\ldots .=\frac{1}{2}\left\{\frac{2^{n+1}-1}{n+1}+\frac{0-1}{n+1}\right\}$
$=\frac{1}{2}\left(\frac{2^{n+1}-2}{n+1}\right)=\frac{2^n-1}{n+1}$

Topic: Mathematics, Binomial Theorem, JEE Main
The value of 2cos-1⁡23-2cos-1⁡6+123 is equal to

Options:
A π3
B π4
C π2
D π6

The correct answer is:
π3

Solution:
2cos-123-2cos-16+123
=2tan-112-2tan-13-21+6=2tan-112-2tan-13-tan-12
=2tan-112+tan-12-2tan-1⁡3
=π-2π3=π3

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of ∫-ππ2cos⁡x1+ex dx is equal to

Options:
A π2
B π
C 0
D 2π

The correct answer is:
0

Solution:
Let, Ι=∫-ππ2cos⁡x1+exdx
Ι=∫0π2cos⁡x1+ex+2cos⁡x1+e-xdx
Ι=∫0π2cos⁡x11+ex+ex1+exdx
Ι=∫0π2cos⁡xdx
Ι=0

Topic: Mathematics, Definite Integration, JEE Main
The value of $\sqrt{2} \int \frac{\sin x d x}{\sin \left(x-\frac{\pi}{4}\right)}$ is

Options:
A
$x+\log \left|\cos \left(x-\frac{\pi}{4}\right)\right|+c$

B
$x-\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|+c$

C
$x+\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|+c$

D
$x-\log \left|\cos \left(x-\frac{\pi}{4}\right)\right|+c$

The correct answer is:
$x+\log \left|\sin \left(x-\frac{\pi}{4}\right)\right|+c$

Solution:
$$
\begin{aligned}
& \sqrt{2} \int \frac{\sin x d x}{\sin \left(x-\frac{\pi}{4}\right)}=\sqrt{2} \int \frac{\sin \left(x-\frac{\pi}{4}+\frac{\pi}{4}\right) d x}{\sin \left(x-\frac{\pi}{4}\right)} \\
& =\sqrt{2} \int\left(\cos \frac{\pi}{4}+\cot \left(x-\frac{\pi}{4}\right) \sin \frac{\pi}{4}\right) d x \\
& =\int d x+\int \cot \left(x-\frac{\pi}{4}\right) d x \\
& =x+\ln \left|\sin \left(x-\frac{\pi}{4}\right)\right|+c .
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2008
The value of $2 \sin ^{-1}\left(\frac{1}{2}\right)+\cot ^{-1}\left(\frac{1}{\sqrt{3}}\right)$ is

Options:
A $\frac{\pi}{4}$
B $\frac{\pi}{3}$
C $\frac{\pi}{6}$
D $\frac{2 \pi}{3}$

The correct answer is:
$\frac{2 \pi}{3}$

Solution:
$2 \sin ^{-1}\left(\frac{1}{2}\right)+\cos ^{-1}\left(\frac{1}{\sqrt{3}}\right)=2 \times \frac{\pi}{6}+\frac{\pi}{3}=\frac{2 \pi}{3}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of 2sin-1⁡45+2sin-1⁡513+2sin-1⁡1665 is equal to

Options:
A 3π2
B π2
C π
D 2π

The correct answer is:
π

Solution:
2sin-1⁡45+sin-1⁡513+sin-1⁡1665
=2sin-1⁡451-5132+5131-452+2sin-1⁡1665
=2sin-1⁡4865+1565+2sin-1⁡1665
=2sin-1⁡6365+sin-1⁡1665
=2cos-1⁡1665+sin-1⁡1665 ∵sin-1⁡x=cos-1⁡1-x2
=π

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
The value of 2sin2â¡Î¸+4cosâ¡Î¸+Î±sinâ¡Î±sinâ¡Î¸+cosâ¡2 (Î±+ Î¸) is

Options:
A cosâ¡Î¸+cosâ¡Î±
B independent of Î¸
C independent of Î±
D independent of both θ and α

The correct answer is:
independent of θ

Solution:
2sin2⁡θ+2cos⁡θ+αcos⁡θ-α-cos⁡θ+α+2cos2⁡θ+α-1
=2sin2⁡θ+2cos2⁡θ-sin2⁡α-1
=2-2sin2⁡α-1=cos2⁡α-sin2⁡α

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
The value of 2sinπ8sin2π8sin3π8sin5π8sin6π8sin7π8 is :

Options:
A 142
B 18
C 182
D 14

The correct answer is:
18

Solution:
⇒2sinπ8sin2π8sin3π8sin5π8sin6π8sin7π8

⇒2sinπ8sin2π8sin3π8sinπ-3π8sinπ-2π8sinπ-π8

⇒2sin2π8sin22π8sin23π8

⇒2sin2π8sin22π8sin23π8

⇒2sin2π8sin2π4sin23π8

⇒sin2π8sin23π8

⇒sin2π8sin2π2-π8

⇒sin2π8cos2π8

⇒142sinπ8cosπ82

14sin2π4=18

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2021 (26 Aug Shift 2)
The value of 2sin12°-sin72° is

Options:
A 51-34
B 1-58
C 31-52
D 31-54

The correct answer is:
31-54

Solution:
To find ⇒2sin12°-sin72

⇒sin12°+sin12°-sin72

⇒ sin12°+2cos42°sin-30° ⇒sin12°-cos42°
⇒ sin12°-sin48° ⇒2cos30°sin−18°

⇒−3sin18°=−35−14 =31−54

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2022 (25 Jun Shift 2)
The value of 3+14+13+14+13+…∞ is equal to

Options:
A 1.5+3
B 2+3
C 3+23
D 4+3

The correct answer is:
1.5+3

Solution:
Let x=3+14+13+14+13+…∞

So, x=3+14+1x=3+14x+1x

⇒x-3=x(4x+1)

⇒(4x+1)(x-3)=x

⇒4x2-12x+x-3=x

⇒4x2-12x-3=0

x=12±(12)2+12×42×4=12±12(16)8

=12±4×238=3±232

x=32±3=1.5±3.

But only positive value is accepted

So, x=1.5+3

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (18 Mar Shift 1)
The value of ∫3636-x23x4dx is equal to

Options:
A π2
B π6
C π3
D π4

The correct answer is:
π3

Solution:
Let, I=∫3636-x23x4dx
Substituting, x=6sin⁡θ
dx=6cos⁡θdθ
I=∫π6π236cos2θ364sin4θ6cos⁡θdθ
I=∫π6π2cos4θsin4θdθ
=∫π6π2cot2θcosec2θ-1dθ
=∫π6π2cot2θcosec2θdθ-∫π6π2cot2θdθ
=-[cot3θ3]π6π2-∫π6π3(cosec2θ-1)dθ
=-130-33+cotθπ6π2+θπ6π2
=3+0-3+π2-π6=π3

Topic: Mathematics, Definite Integration, JEE Main
The value of 4+15+14+15+14+……∞ is:

Options:
A 2+2530
B 2+4530
C 4+4530
D 5+2530

The correct answer is:
2+2530

Solution:
Given y=4+15+14+15+14+……∞⏟y

y=4+15+1y

⇒y-4=y(5y+1)

⇒5y2-20y-4=0

⇒y=20±-202-4·5·-42·5

⇒y=20±48010

∴ y=2-480100→rejected

So, y=2+480100=2+2530

Correct with Option (A).

Topic: Mathematics, Quadratic Equation, JEE Main, JEE Main 2021 (17 Mar Shift 1)
The value of ∫π6π3esec2⁡xsin⁡xcos3⁡xdx is equal to

Options:
A 12e4
B 12e43
C 12e4-e43
D 12e2-1

The correct answer is:
12e4-e43

Solution:
Let, I=∫esec2xsin⁡xcos3⁡xdx
⇒I=∫esec2⁡xtanx·sec2⁡xdx
⇒I=∫e·etan2⁡xtan⁡xsec2dx
Substituting, tan2⁡x=t
2tan⁡xsec2⁡xdx=dt
⇒I=12∫e·etdt
I=e2et+C
I=e2etan2⁡x+C
=esec2⁡x2+C
Ι=12esec2⁡xπ6π3=12e4-e43

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

$6+\log _{\frac{3}{2}}\left(\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}}} \cdots}}\right) \text { is }$

Options:

The correct answer is:
4

Solution:
Let $x=\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}} \sqrt{4-\frac{1}{3 \sqrt{2}}}}} \ldots \ldots$

$\Rightarrow \quad x=\frac{1}{3 \sqrt{2}} \sqrt{4-x}$

$\Rightarrow \quad 3 \sqrt{2} x=\sqrt{4-x}$

$\begin{array}{l}

\Rightarrow \quad 18 x^{2}=4-x \\

\Rightarrow \quad 18 x^{2}+x-4=0 \\

\Rightarrow \quad(9 x-4)(2 x+1)=0 \\

\Rightarrow \quad x=\frac{4}{9} \text { or } x=-\frac{1}{2}

\end{array}$

(Not possible because log is not define for -ve value)

$\begin{aligned}

\therefore \quad & 6+\log _{\frac{3}{2}}\left(\frac{4}{9}\right)=6+\log _{\frac{3}{2}}\left(\frac{3}{2}\right)^{-2} \\

&=6-2=4

\end{aligned}$

Topic: Mathematics, Quadratic Equation, JEE Main
The value of 8π∫0π2cosx2023sinx2023+cosx2023dx is ______.

Options:

The correct answer is:
2

Solution:
Let

I=8π∫0π2cosx2023sinx2023+cosx2023dx ...1

Using ∫0af(x)dx=∫0af(a-x)dx

I=8π∫0π2sinx2023sinx2023+cosx2023dx ...2

Adding 1 and 2, we get

2I=8π∫0π21 dx

⇒2I=8πx0π2

⇒I=2

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The value of 12∫03x2-3x+2dx is ______

Options:

The correct answer is:
22

Solution:
I=∫03x2-3x+2dx

=∫03x-1x-2dx

=∫01x2-3x+2dx-∫12x2-3x+2dx+∫23x2-3x+2dx

=x33-3x22+2x01-x33-3x22+2x12+x33-3x22+2x23

=13-32+2-83-6+4-13-32+2+9-272+6-83-6+4

=23-3+4-163-12+8+9-272+6

=20-1096=116

So,

12I=12×116=22

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2023 (24 Jan Shift 1)
The value of ${ }^{15} C_0^2-{ }^{15} C_1^2+{ }^{15} C_2^2-\ldots .-{ }^{15} C_{15}^2$ is

Options:
A $15$
B $-15$
C $0$
D $51$

The correct answer is:
$0$

Solution:
As we know that
${ }^n C_0-{ }^n C_1^2+{ }^n C_2^2-{ }^n C_3^2+\ldots+(-1)^n \cdot{ }^n C_n^2=0$ (if $n$ is odd)
and in the question $n=15$ (odd).

Topic: Mathematics, Binomial Theorem, JEE Main
The value of \( 0.16^{\log _{2.5}\left(\frac{1}{3}+\frac{1}{3^{2}}+\frac{1}{3^{3}}+\ldots . . \infty\right)} \) is

Options:

The correct answer is:
4.00

Solution:
0.16log2.513+132+133+….∞

=0.16log2.5131-13=425log5212

=52-2log5212=12-2=4

Topic: Mathematics, Sequences and Series, JEE Main
The value of 21C1-10C1+21C2-10C2+21C3-10C3+21C4-10C4+…+21C10-10C10 is

Options:
A 221-211
B 221-210
C 220-29
D 220-210

The correct answer is:
220-210

Solution:
We have,

21C1-10C1+21C2-10C2+21C3-10C3+21C4-10C4+…+21C10-10C10

=21C1+ 21C2+ 21C3+…+ 21C10-10C1+10C2+10C3+…+10C10

=1221C1+21C2+21C3+…+21C10+21C11+…+21C20-210-1

=12221-2-210-1

=220-1-210+1

=220-210.

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2017 (02 Apr)
The value of 364 cos2 9∘-1 4 cos2 27∘-1 4 cos2 81∘-1 4 cos2 243∘-1 is

Options:
A 54
B 18
C 27
D 36

The correct answer is:
36

Solution:
We need to find the value of

364 cos2 9∘-1 4 cos2 27∘-1 4 cos2 81∘-1 4 cos2 243∘-1

=364 cos29∘-1 4 sin29∘-1 4 cos227∘-14 sin227∘-1

=3642cos9∘sin9°2-4cos29°-4sin29°+1 42sin27∘cos9°2-4cos227°-4sin227°+1

=364 sin218∘-4 +1 4 sin254∘-4+1

=364 sin218∘-3 4 sin254∘-3
=365-124-3 5+124-3
=366-25-1246+25-124
=-36252-3616=36

Hence this is the correct option.

Topic: Mathematics, Trigonometric Equations, JEE Main
The value of 364 cos2 9∘-1 4 cos2 27∘-1 4 cos2 81∘-1 4 cos2 243∘-1 is

Options:
A 54
B 18
C 27
D 36

The correct answer is:
36

Solution:
We need to find the value of

364 cos2 9∘-1 4 cos2 27∘-1 4 cos2 81∘-1 4 cos2 243∘-1

=364 cos29∘-1 4 sin29∘-1 4 cos227∘-14 sin227∘-1

=3642cos9∘sin9°2-4cos29°-4sin29°+1 42sin27∘cos9°2-4cos227°-4sin227°+1

=364 sin218∘-4 +1 4 sin254∘-4+1

=364 sin218∘-3 4 sin254∘-3
=365-124-3 5+124-3
=366-25-1246+25-124
=-36252-3616=36

Hence this is the correct option.

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2023 (08 Apr Shift 2)
The value of $\sqrt{42+\sqrt{42+\sqrt{42+\ldots \ldots .}}}$ is equal

Options:
A $7$
B $-6$
C $5$
D $4$

The correct answer is:
$7$

Solution:
Let $y=\sqrt{42+\sqrt{42+\sqrt{42+\ldots}}}$
$$
\Rightarrow \quad y=\sqrt{42+y}
$$
On squaring both sides, we get
$$
\begin{aligned}
& y^2=42+y \\
& \Rightarrow \quad y^2-y-42=0 \\
&
\end{aligned}
$$
$$
\begin{array}{rlrl}
\Rightarrow & & (y-7)(y+6) & =0 \\
\Rightarrow & y =7,-6
\end{array}
$$
Since, $y=-6$ is not satisfied the given equation.
$\therefore$ The required solution is $y=7$.

Topic: Mathematics, Quadratic Equation, JEE Main
The value of ${ }^{50} \mathrm{C}_4+\sum_{\mathrm{r}=1}^6{ }^{56-r} \mathrm{C}_3$ is

Options:
A
${ }^{55} \mathrm{C}_4$

B
${ }^{55} \mathrm{C}_3$

C
${ }^{56} \mathrm{C}_3$

D
${ }^{56} \mathrm{C}_4$

The correct answer is:
${ }^{56} \mathrm{C}_4$

Solution:
$$
\begin{aligned}
& { }^{50} \mathrm{C}_4+\sum_{\mathrm{r}=1}^6{ }^{56-r} \mathrm{C}_3 \\
& \Rightarrow{ }^{50} \mathrm{C}_4+\left[{ }^{55} \mathrm{C}_3+{ }^{54} \mathrm{C}_3+{ }^{53} \mathrm{C}_3+{ }^{52} \mathrm{C}_3+{ }^{51} \mathrm{C}_3+{ }^{50} \mathrm{C}_3\right] \\
& =\left({ }^{50} \mathrm{C}_4+{ }^{50} \mathrm{C}_3\right)+{ }^{51} \mathrm{C}_3+{ }^{52} \mathrm{C}_3+{ }^{53} \mathrm{C}_3+{ }^{54} \mathrm{C}_3+{ }^{55} \mathrm{C}_3 \\
& \Rightarrow\left({ }^{51} \mathrm{C}+{ }^{51} \mathrm{C}\right)+{ }^{52} \mathrm{C}_3+{ }^{53} \mathrm{C}_3+{ }^{54} \mathrm{C}_3+{ }^{55} \mathrm{C}_3 \\
& \Rightarrow{ }^{55} \mathrm{C}_4+{ }^{55} \mathrm{C}_3={ }^{56} \mathrm{C}_4 .
\end{aligned}
$$

Topic: Mathematics, Binomial Theorem, JEE Main, JEE Main 2005
The value of $1000\left[\frac{1}{1 \times 2}+\frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots+\frac{1}{999 \times 1000}\right]$ is

Options:
A 1000
B 999
C 1001
D $\frac{1}{999}$

The correct answer is:
999

Solution:
$1000\left[\frac{1}{1 \times 2} \times \frac{1}{2 \times 3}+\frac{1}{3 \times 4}+\ldots+\frac{1}{999 \times 1000}\right]$
$=1000\left\{\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)\right.$
$\left.+\ldots+\left(\frac{1}{999}-\frac{1}{1000}\right)\right]$
$=1000\left(1-\frac{1}{1000}\right)$
$=1000 \times \frac{999}{1000}$
$=999$

Topic: Mathematics, Sequences and Series, JEE Main
The value of a+1a+2a+21a+2a+3a+31a+3a+4a+41 is

Options:
A 0
B a+2a+3a+4
C -2
D a+1a+2a+3

The correct answer is:
-2

Solution:
R2→R2-R1 and R3→R3-R1

Δ=a+1a+2a+21a+2a+3-a-110a2+7a+12-a2-3a-220

=a2+3a+2a+212a+2104a+1020

=4a+2-4 a-10

=4a+8-4a-10=-2

Topic: Mathematics, Determinants, JEE Main, JEE Main 2021 (26 Feb Shift 1)