datavorous/entrance-exam-dataset huggingface dataset:
If $\mathrm{S}+\mathrm{O}_2 \rightarrow \mathrm{SO}_2 ;(\Delta H=-298.2)$
$\mathrm{SO}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{SO}_3 ;(\Delta H=-98.2)$
$\mathrm{SO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4 ;(\Delta H=-130.2)$
$\mathrm{H}_2+\frac{1}{2} \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O} ;(\Delta \mathrm{H}=-287.3)$
then the enthalpy of formation of $\mathrm{H}_2 \mathrm{SO}_4$ at $298 \mathrm{~K}$ will be

Options:
A $-433.7 \mathrm{~kJ/mole}$
B $-650.3 \mathrm{~kJ/mole}$
C $+320.5 \mathrm{~kJ/mole}$
D $-814.4 \mathrm{~kJ/mole}$

The correct answer is:
$-814.4 \mathrm{~kJ/mole}$

Solution:
$\mathrm{S}+\mathrm{O}_2 \rightarrow \mathrm{SO}_2 ; \Delta \mathrm{H}=-298.2 \mathrm{~kJ} / \mathrm{mole}$
$\mathrm{SO}_2+12 \mathrm{O}_2 \rightarrow \mathrm{SO}_3 \quad \& \mathrm{H}=-98.7 \mathrm{~kJ} / \mathrm{mole}$
$\mathrm{SO}_3+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{H}_2 \mathrm{SO}_4 ; \Delta \mathrm{H}=-130.2 \mathrm{~kJ} / \mathrm{mole}$
$\mathrm{H}_2+12 \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{O} \underset{\sim}{ } \Delta \mathrm{H}=-287.3 \mathrm{~kJ} / \mathrm{mole}$
The formation of $\mathrm{H}_2 \mathrm{SO}_4$
$\mathrm{H}_2+\mathrm{S}+2 \mathrm{O}_2 \rightarrow \mathrm{H}_2 \mathrm{SO}_4$ can be calculated by adding the above four equations
$\Delta \mathrm{H}=-(298.2+98.7+130.2+287.3)=-814.4 \mathrm{~kJ} / \mathrm{mole}$

Topic: Chemistry, Thermodynamics (C), JEE Main
If S=∑r=180rr4+r2+1, then the value of 6481S1000 is _______. If your answer is K, mark 100K as answer after rounding off to nearest integer.

Options:

The correct answer is:
324

Solution:
Tr=12r2+r+1-r2-r+1r2+r+1r2-r+1
=121r2-r+1-1r2+r+1
=121r2-r+1-1r+12-r+1+1
=12fr-fr+1
S=ΣTr=∑r=18012fα-fα+1
=12f1-f2+f2-f3........+f80-f81
=12f1-f81=121-1812-81+1
=121-16481⇒6481S=64802=3240
=6481S1000=32401000=3.24

Topic: Mathematics, Sequences and Series, JEE Main
If S=âr=180rr4+r2+1, then the value of 6481S1000 is

Options:

The correct answer is:
3.24

Solution:
Tr=12r2+r+1-r2-r+1r2+r+1r2-r+1
=121r2-r+1-1r2+r+1
=121r2-r+1-1r+12-r+1+1
=12fr-fr+1
S=ΣTr=∑r=18012fα-fα+1
=12f1-f2+f2-f3........+f80-f81
=12f1-f81=121-1812-81+1
=121-16481⇒6481S=64802=3240
=6481S1000=32401000=3.24

Topic: Mathematics, Sequences and Series, JEE Main
If S=x∈0, 2π :0cos⁡x-sin⁡xsin⁡x0cos⁡xcos⁡xsin⁡x0=0, then ∑x ∈ Stanπ3+x is equal to:

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

The correct answer is:
-4-23

Solution:
Given,0cos⁡x-sin⁡xsin⁡x0cos⁡xcos⁡xsin⁡x0=0

Expanding the determinant,
⇒0(0-cos⁡xsin⁡x)-cos⁡x0-cos2⁡x-sin⁡xsin2⁡x-0=0
⇒ cos3⁡x-sin3⁡x=0
⇒ tan3⁡x=1⇒ tan⁡x=1⇒x=π4,5π4 ⇒S=π4, 5π4

∴∑x ∈ Stanπ3+x
=∑x ∈ S3+tanx1−3 tanx
=23+11-3=-4-23

Topic: Mathematics, Determinants, JEE Main, JEE Main 2017 (08 Apr Online)
If $\mathrm{S}(x)=(1+x)+2(1+x)^2+3(1+x)^3+\cdots+60(1+x)^{60}, x \neq 0$, and $(60)^2 \mathrm{~S}(60)=\mathrm{a}(\mathrm{b})^{\mathrm{b}}+\mathrm{b}$, where $a, b \in N$, then $(a+b)$ equal to ______

Options:

The correct answer is:
3660

Solution:
$\begin{aligned}
S(x)= & (1+x)+2(1+x)^2+3(1+x)^3+. .+60(1+x)^{60} \\
(1+x) S= & (1+x)^2+\ldots \ldots . \quad 59(1+x)^{60}+60(1+x)^{61} \\
& -x S=\frac{(1+x)(1+x)^{60}-1}{x}-60(1+x)^{61}
\end{aligned}$
Put $\mathrm{x}=60$
$-60 \mathrm{~S}=\frac{61\left((61)^{60}-1\right)}{60}-60(61)^{61}$
on solving 3660

Topic: Mathematics, Sequences and Series, JEE Main, JEE Main 2024 (06 Apr Shift 2)
If $S=x^2+y^2+2 x+17 y+4=0$
$S^{\prime}=x^2+y^2+7 x+6 y+11=0$ and
$S^{\prime \prime}=x^2+y^2-x+22 y+3=0$ are three
circles, then the length of tangent from their radical center to $S=0$ is ......... units.

Options:
A $\sqrt{53}$
B $\sqrt{57}$
C $\sqrt{15}$
D $\sqrt{17}$

The correct answer is:
$\sqrt{57}$

Solution:
Given, $S=x^2+y^2+2 x+17 y+4=0$
$S^{\prime \prime}=x^2+y^2+7 x+6 y+11=0$
$S=x^2+y^2-x+22 y+3=0$
$S-S^{\prime}=5 x-11 y+7=0$ ...(i)
$S^{\prime}-S^{\prime \prime}=8 x-16 y+8=0$
$x-2 y+1=0$ ...(ii)
Solving Eqs. (i) and (ii), we get $(3,2)$
$\therefore$ Radical centre of circle is $(3,2)$.
Length of tanget from $(3,2)$ to $S$ is $\sqrt{S_1}$
$=\sqrt{(3)^2+(2)^2+2(3)+17(2)+4}=\sqrt{57}$

Topic: Mathematics, Circle, JEE Main
If S=z∈C:z-iz+2i∈R, then

Options:
A S is a circle in the complex plane
B S contains exactly two elements
C S contains only one element
D S is a straight line in the complex plane

The correct answer is:
S is a straight line in the complex plane

Solution:
Given z-iz+2i∈R

Then argz-iz+2i is 0 or π

⇒S is straight line in complex plane.

Topic: Mathematics, Complex Number, JEE Main, JEE Main 2021 (27 Aug Shift 1)
If $\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a$, then $\frac{d y}{d x}$ is

Options:
A $\frac{y+1}{x-1}$
B $\frac{y-1}{x+1}$
C $\frac{x-1}{y+1}$
D $\frac{x-1}{y-1}$

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

Solution:
We have,
$$
\begin{gathered}
\sec ^{-1}\left(\frac{1+x}{1-y}\right)=a \Rightarrow \frac{1+x}{1-y}=\sec a \\
\Rightarrow \quad \frac{1(1-y)-(1+x)\left(-\frac{d y}{d x}\right)}{(1-y)^{2}}=0 \\
\Rightarrow \quad 1-y+(1+x) \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{y-1}{x+1}
\end{gathered}
$$

Topic: Mathematics, Differentiation, JEE Main
If $\sec \theta=1 \frac{1}{4}$, then $\tan \frac{\theta}{2}=$

Options:
A $\frac{1}{3}$
B $\frac{3}{4}$
C $\frac{1}{4}$
D $\frac{5}{4}$

The correct answer is:
$\frac{1}{3}$

Solution:
Given that
$\begin{aligned}
\sec \theta & =\frac{5}{4} \\
& \sec \theta=\frac{1+\tan ^2(\theta / 2)}{1-\tan ^2(\theta / 2)} \Rightarrow \frac{5}{4}=\frac{1+\tan ^2(\theta / 2)}{1-\tan ^2(\theta / 2)} \\
\Rightarrow & 5-5 \tan ^2(\theta / 2)=4+4 \tan ^2(\theta / 2) \\
& \Rightarrow 9 \tan ^2(\theta / 2)=1 \Rightarrow \tan (\theta / 2)=\frac{1}{3} .
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If ∫sec 2x-1dx=α logecos 2x+β+cos 2x1+cos1βx + constant, then β-α is equal to ______.

Options:

The correct answer is:
1

Solution:
Given,

∫sec 2x-1dx=α logecos 2x+β+cos 2x1+cos1βx+C

Now solving L.H.S we get,

∫sec 2x-1dx=∫1-cos 2xcos 2xdx

⇒∫sec 2x-1dx=2∫sin x2cos2x-1dx

Now let cos x=t ⇒-sin x dx=dt

⇒∫sec 2x-1dx=-2∫dt2t2-1

⇒∫sec 2x-1dx=-ln|2cos x+cos 2x|+C

=-12ln2 cos2x+cos 2x+2cos 2x·2 cos x+C

=-12lncos 2x+12+cos 2x·1+cos 2x+C

Now on comparing with ∫sec 2x-1dx=α logecos 2x+β+cos 2x1+cos1βx+C

We get, β=12,α=-12⇒β-α=1

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2023 (30 Jan Shift 2)
If $\int \sec ^2 x \operatorname{cosec}^4 x d x=-\frac{1}{3} \cot ^3 x+k \tan x$ $-2 \cot x+C$, then $k$ is equal to

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

The correct answer is:
$1$

Solution:
Let $I=\int \sec ^2 x \operatorname{cosec}^4 x d x$
$\begin{aligned} & =\int \frac{1}{\sin ^4 x \cos ^2 x} d x=\int \frac{\sin ^2 x+\cos ^2 x}{\sin ^4 x \cos ^2 x} d x \\ & =\int \frac{d x}{\sin ^2 x \cos ^2 x}+\int \frac{d x}{\sin ^4 x} \\ & \left.=\int \frac{\left(\sin ^2 x+\cos ^2 x\right) d x}{\sin ^2 x \cos ^2 x}+\int \operatorname{cosec}^2 x\right) \\ & =\int\left(\sec ^2 x+\operatorname{cosec}^2 x\right) d x \\ & \quad+\int \operatorname{cosec}^2 x\left(1+\sin ^2 x+\cos ^2 x\right) \\ & =\tan x-\cot x+\int \operatorname{cosec}{ }^2 x d x \\ & =\tan x-\cot x-\cot x-\frac{\cot ^3 x}{3}+C \\ & =-\frac{1}{3} \cot ^3 x+\tan x-2 \cot ^2 x+C \cot ^2 x d x\end{aligned}$
But it given that,
$\begin{aligned} & I=-\frac{1}{3} \cot ^3 x+k \tan x-2 \cot x+C \\ \therefore \quad k & =1\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
If $\sec A+\tan A=p$, then what is the value of $\sin A$ ? $[2008-\mathrm{II}]$

Options:
A $\frac{p^{2}-1}{p^{2}+1}$
B $\frac{p^{2}+1}{p^{2}-1}$
C 1
D None of these

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

Solution:
$\sec A+\tan A=P$
$\Rightarrow \frac{1}{\cos A}+\frac{\sin A}{\cos A}=p$
$\Rightarrow \frac{1+\sin A}{\cos A}=p$
$\Rightarrow \frac{(1+\sin A)^{2}}{\cos ^{2} A}=p^{2}$
$\Rightarrow \frac{(1+\sin A)^{2}}{1-\sin ^{2} A}=p^{2}$
$\Rightarrow \frac{(1+\sin A)^{2}}{(1+\sin A)(1-\sin A)}=p^{2}$

Topic: Mathematics, Trigonometric Equations, JEE Main
If set A=x: tan⁡x=sec⁡x, x∈0,4π and set B=x: sin2⁡x=1, x∈0,4π, then

Options:
A B⊂A
B A=B
C A∩B=B
D nA×B=0

The correct answer is:
nA×B=0

Solution:
For A, sin⁡xcos⁡x=1cos⁡x⇒sin⁡x=1
∴ x=π2, 5π2
but at x=π2, 5π2

tan⁡x, sec⁡θ is not defined
∴ A=ϕ
for B, sin⁡x=1, sin⁡x=-1
x=π2,5π2 or x=3π2,7π2
∴B=π2,3π2,5π2,7π2
∴nA×B=0

Topic: Mathematics, Sets and Relations, JEE Main
If seven women and seven men are to be seated around a circular table such that there is a man on either side of every woman, then the number of seating arrangements is

Options:
A
$6 ! 7 !$

B
$(6 !)^2$

C
$(7 !)^2$

D
7 !

The correct answer is:
$6 ! 7 !$

Solution:
7 women can be arranged around a circular table in 6 ! ways.
Among these 7 men can sit in 7 ! ways. Hence, number of seating arrangement $=7 ! \times 6 !$

Topic: Mathematics, Permutation Combination, JEE Main, JEE Main 2012 (26 May Online)
If sin-1⁡13+sin-1⁡23=sin-1⁡x, then the value of x is

Options:
A 0
B (5-42)9
C (5+42)9
D π2

The correct answer is:
(5+42)9

Solution:
Given that, sin-1â¡13+sin-1â¡23=sin1â¡x
Taking sine on both sides
ââ¡131-49+231-19=â¡x
ââ¡13.53+23.83=â¡x
â â¡5+429=â¡x
â´ x=5+429

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\quad \int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x=f(x)-\log \left(1+x^2\right)$ then $f(x)$ is equal to

Options:
A $2 x \tan ^{-1} x$
B $-2 x \tan ^{-1} x$
C $x \tan ^{-1} x$
D $-x \tan ^{-1} x$

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

Solution:
Let $I=\int \sin ^{-1}\left(\frac{2 x}{1+x^2}\right) d x$
Put $x=\tan \theta \Rightarrow d x=\sec ^2 \theta d \theta$
$\begin{aligned}
\therefore \quad I & =\int \sin ^{-1}(\sin 2 \theta) \cdot \sec ^2 \theta d \theta \\
& =2 \int \theta \sec ^2 \theta d \theta \\
& =2\left[\theta \tan \theta-\int \tan \theta d \theta\right] \\
& =2[\theta \tan \theta+\log \cos \theta]+c \\
& =2\left[x \tan ^{-1} x+\log \frac{1}{\sqrt{1+x^2}}\right]+c \\
& =2 x \tan ^{-1} x-\log \left(1+x^2\right)+c
\end{aligned}$
But $I=f(x)-\log \left(1+x^2\right)+c$
$\begin{array}{cc}
\Rightarrow & f(x)-\log \left(1+x^2\right)+c \\
& =2 x \tan ^{-1} x-\log \left(1+x^2\right)+c \\
& f(x)=2 x \tan ^{-1} x
\end{array}$

Topic: Mathematics, Indefinite Integration, JEE Main
If $\sin ^{-1}(1-x)-2 \sin ^{-1} x=\pi / 2$, then $x$ equals

Options:
A $\left(0,-\frac{1}{2}\right)$
B $\left(\frac{1}{2}, 0\right)$
C $\{0\}$
D $(-1,0)$

The correct answer is:
$\{0\}$

Solution:
$\begin{aligned} & \sin ^{-1}(1-x)-2 \sin ^{-1} x=\frac{\pi}{2} \\ & \Rightarrow \sin ^{-1}(1-x)=\left(\frac{\pi}{2}+2 \sin ^{-1} x\right) \\ & \Rightarrow 1-x=\sin \left(\frac{\pi}{2}+2 \sin ^{-1} x\right) \\ & \Rightarrow 1-x=\sin \frac{\pi}{2} \cos \left(2 \sin ^{-1} x\right)+\cos \frac{\pi}{2} \sin \left(2 \sin ^{-1} x\right) \\ & \Rightarrow 1-x=\cos \left(2 \sin ^{-1} x\right) \\ & \Rightarrow 1-x=\cos \left\{\cos ^{-1}\left(1-2 x\right)\right\} \Rightarrow 2 x^2-x=0 \\ & x=0, x=1 / 2 \text { which does not satisfy the equation } \\ & \therefore x=0 \text { is only the solution. }\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If sin-1⁡2a1+a2+sin-1⁡2b1+b2=2cot-1⁡1x, then x is equal to ∀a,b∈0,1

Options:
A a-b1+ab
B b1+ab
C b1-ab
D a+b1-ab

The correct answer is:
a+b1-ab

Solution:
Since, sin-1â¡2x1+x2=2tan-1â¡x for xâ-1,1
sin-12a1+a2+sin-12b1+b2=2cot-1â¡1x
âtan-1â¡a+tan-1â¡b=tan-1â¡x
âtan-1â¡a+b1-ab=tan-1â¡x
âx=a+b1-ab

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
$\text { If } \sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{2}{3}=\sin ^{-1} x \text {, }$ then $x$ is equal to

Options:
A $0$
B $\frac{\sqrt{5}-4 \sqrt{2}}{9}$
C $\frac{\sqrt{5}+4 \sqrt{2}}{9}$
D $\frac{\pi}{2}$

The correct answer is:
$\frac{\sqrt{5}+4 \sqrt{2}}{9}$

Solution:
$\begin{aligned} & \sin ^{-1} \frac{1}{3}+\sin ^{-1} \frac{2}{3} \\ & =\sin ^{-1}\left[\frac{1}{3} \sqrt{1-\frac{4}{9}}+\frac{2}{3} \sqrt{1-\frac{1}{9}}\right]=\sin ^{-1}\left[\frac{\sqrt{5}+4 \sqrt{2}}{9}\right] \text {, as }\left\{\sin ^{-1}(x)+\sin ^{-1}(y)=\sin ^{-1}\left\{x \sqrt{1-y^2}+y \sqrt{1-x^2}\right\}\right\} \\ & \text {Therefore } \quad x=\frac{\sqrt{5}+4 \sqrt{2}}{9}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If α=sin-1⁡32+sin-1⁡13 and β=cos-1⁡32+cos-1⁡13, then

Options:
A α>β
B α=β
C α<β
D α+β=2π

The correct answer is:
α<β

Solution:
α+β=sin-1⁡32+cos-1⁡32+sin-1⁡13+cos-1⁡13
=π2+π2=π
Also, α=π3+sin-1⁡13<π3+sin-1⁡12
As sin⁡θ is increasing in 0, π2
∴ α<π3+π6=π2
⇒ β>π2>α
⇒ α<β

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\sin ^{-1}\left(\frac{x}{5}\right)+\operatorname{cosec}^{-1}\left(\frac{5}{4}\right)=\frac{\pi}{2}$ then a value of $x$ is

Options:
A
$1$

B
$3$

C
$4$

D
$5$

The correct answer is:
$3$

Solution:
$\sin ^{-1} \frac{x}{5}+\sin ^{-1} \frac{4}{5}=\frac{\pi}{2}$
$\Rightarrow \sin ^{-1} \frac{x}{5}=\cos ^{-1} \frac{4}{5} \Rightarrow \sin ^{-1} \frac{x}{5}=\sin ^{-1} \frac{3}{5}$
$\therefore x=3$.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2007
If $\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2},$ then the value of $x$ is

Options:
A 5
B 4
C 12
D 11

The correct answer is:
5

Solution:
Given,
$\sin ^{-1}\left(\frac{x}{13}\right)+\operatorname{cosec}^{-1}\left(\frac{13}{12}\right)=\frac{\pi}{2} ..(i)$
Let $\operatorname{cosec}^{-1} \frac{13}{12}=y$
Then. $\operatorname{cosec} y=\frac{13}{12} \Rightarrow \sin y=\frac{12}{13}$
$\therefore \cos y=\sqrt{1-\sin ^{2} y}$
$\begin{array}{l}
=\sqrt{1-\left(\frac{12}{13}\right)^{2}}=\sqrt{1-\frac{144}{169}} \\
=\sqrt{\frac{25}{169}}=\frac{5}{13} \Rightarrow y=\cos ^{-1} \frac{5}{13}
\end{array}$
Eq. (i) becomes, $\sin ^{-1}\left(\frac{x}{13}\right)+\cos ^{-1}\left(\frac{5}{13}\right)=\frac{\pi}{2}$
We know that
$\sin ^{-1} \theta+\cos ^{-1} \theta=\frac{\pi}{2}$
$\therefore$ Both angles of Eq. (i) should be same.
$\therefore x=5$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If sin-1α17+cos-145-tan-17736=0,0<α<13, then sin-1sinα+cos-1cosα is equal to

Options:
A π
B 16
C 0
D 16-5π

The correct answer is:
π

Solution:
Given,

sin-1α17+cos-145-tan-17736=0

⇒sin-1α17=tan-17736-cos-145

⇒sin-1α17=tan-17736-tan-134

⇒sin-1α17=tan-17736-341+7736×34

⇒sin-1α17=tan-1815

⇒sin-1α17=sin-1817

⇒α=8

Now solving sin-1sinα+cos-1cosα=sin-1sinα+cos-1cosα

⇒sin-1sin8+cos-1cos8=3π-8+8-2π=π

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2023 (31 Jan Shift 1)
If $\int \sin ^{-1}\left(\sqrt{\frac{x}{a+x}}\right) d x=A(x)+$ constant, then $A(x)=$

Options:
A $\operatorname{atan}^1 \sqrt{\frac{x}{a}}+a x$
B $\frac{1}{\sqrt{a+x}} \tan ^1 \sqrt{\frac{x}{a}}-\sqrt{a x}$
C $(a+x) \tan ^1 \sqrt{x}+a \sqrt{x}$
D $(a+x) \tan ^1 \sqrt{\frac{x}{a}}-\sqrt{a x}$

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

Solution:
Let $I=\int \sin ^2 \sqrt{\frac{x}{a+x}} d x$
Put $x=a \tan ^2 \theta \Rightarrow d x=2 a \tan \theta \sec ^2 \theta d \theta$
$$
\begin{aligned}
& \therefore \quad I=\int \sin \sqrt{\frac{a \tan ^2 \theta}{a+a \tan ^2 \theta}} 2 a \tan \theta \sec ^2 \theta d \theta \\
& =\int \sin ^2 \sqrt{\sin ^2 \theta} \cdot 2 a \tan \theta \sec ^2 \theta d \theta \\
& =2 a \int \theta \tan \theta \sec ^2 \theta d \theta \\
& =2 a\left[\theta \frac{\tan ^2 \theta}{2}-\int \frac{\tan ^2 \theta}{2} \cdot 1 d \theta\right] \\
& =a \theta \tan ^2 \theta-a \int \tan ^2 \theta d \theta \\
& =a \theta \tan ^2 \theta-a(\tan \theta-\theta)+c \\
& =x \tan ^2 \sqrt{\frac{x}{a}}-a \sqrt{\frac{x}{a}}+a \tan ^2 \sqrt{\frac{x}{a}}+c \\
& =(x+a) \tan ^2 \sqrt{\frac{x}{a}}-\sqrt{x a}+\text { constant } \\
& \therefore A(x)=(x+a) \tan ^2 \sqrt{\frac{x}{a}}-\sqrt{x a} \\
&
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main
If \( \sin ^{-1} \sin 17+\cos ^{-1} \cos 27+\tan ^{-1} \tan 37 \) is equal to \( k-\lambda \pi \), then the value of \( k+\lambda \) is

Options:

The correct answer is:
62.00

Solution:
Let E=sin-1sin17+cos-1cos27+tan-1tan37

Since, sin-1sin17=5π-17, cos-1cos27=27-8π

and tan-1tan 37=37-15π

Now, E=5π-17+27-8π+37-15π

⇒E=47-18π

Therefore, k=47, λ=18

Hence, k+λ=18+47=62

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\theta=\sin ^{-1}\left[\sin \left(-600^{\circ}\right)\right]$, then one of the possible value of $\theta$ is

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

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

Solution:
$\theta=\sin ^{-1}\left[\sin \left(-600^{\circ}\right)\right]$
$\begin{aligned} & \Rightarrow \theta=\sin ^{-1}\left[-\left(\sin 240^{\circ}\right)\right]=\sin ^{-1}\left[-\sin \left(180^{\circ}+60^{\circ}\right)\right] \\ & \Rightarrow \theta=\sin ^{-1} \sin 60^{\circ}=\sin ^{-1}\left[\sin \left(\frac{\pi}{3}\right)\right]=\frac{\pi}{3}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If ∫sin-1x1+xdx=Axtan-1x+Bx+C, where C is a constant of integration, then the ordered pair Ax, Bx can be :

Options:
A x-1, x
B x-1, -x
C x+1, x
D x+1, -x

The correct answer is:
x+1, -x

Solution:
I=∫sin-1x1+xdx

=∫tan-1x⏟I1⏟IIdx

=xtan-1x-∫11+x.12x.xdx+C

=xtan-1x-12∫t.2t.dt1+t2+C (putting x=t2⇒dx=2tdt)

=xtan-1x-∫t21+t2dt+C

=xtan-1x-t+tan-1t+C

=xtan-1x-x+tan-1x+C

=x+1tan-1x-x+C

Comparing with Axtan-1x+Bx+C, we get

Ax, Bx=x+1, -x

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2020 (03 Sep Shift 2)
If sin-1x2-cos-1x2=a; 0<x<1, a≠0, then the value of 2x2-1 is

Options:
A cos2aπ
B sin2aπ
C cos4aπ
D sin4aπ

The correct answer is:
sin2aπ

Solution:
sin-1x2-cos-1x2=a

Let sin-1x=t

x=sin t

sin-1sin t2-cos-1sin t2=a

t2-π2-t2=a

t2-π24+t2-πt=a

πt-π24=a

t=aπ+π4

x=sinaπ+π4

2x2-1=2sin2aπ+π4-1

=2sinaπ cosπ4+cosaπ sinπ42-1

=sinaπ+cosaπ2-1

=sin2aπ+cos2aπ+2cosaπsinaπ-1

=2cosaπsinaπ

=sin2aπ

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main, JEE Main 2021 (27 Aug Shift 1)
If $\sin ^{-1}\left(x-\frac{x^2}{2}+\frac{x^3}{4}-\ldots \infty\right)$
$+\cos ^{-1}\left(x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \infty\right)=\frac{\pi}{2}$ and $0 < \mathrm{x} < \sqrt{2}$,
then $x$ is equal to

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

The correct answer is:
$1$

Solution:
Given that, $\sin ^{-1}\left(x-\frac{x^2}{2}+\frac{x^2}{4}-\ldots \infty\right)$
$\cos ^{-1}\left(x-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \infty\right)=\frac{\pi}{2}$
Here, $\quad x-\frac{x^2}{2}+\frac{x^3}{4}-\ldots \infty$
Forms G.P. so its sum can be given as $\frac{x}{1-\frac{x}{2}}$ Similarly, for series
$x^2-\frac{x^4}{2}+\frac{x^6}{4}-\ldots \infty \Rightarrow \text { sum }=\frac{x^2}{1-\frac{x^2}{2}}$
$\begin{array}{ll}\sin ^{-1}\left[\frac{x}{1-\frac{x}{2}}\right]+\cos ^{-1}\left[\frac{x^2}{1-\frac{x^2}{2}}\right]=\frac{\pi}{2} & \\ \sin ^{-1}\left[\frac{2 x}{2-x}\right]+\cos ^{-1}\left[\frac{2 x^2}{2-x^2}\right]=\frac{\pi}{2} & \\ & \quad\left[\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right]\end{array}$
$\begin{aligned} & \frac{2 x}{2-x}=\frac{2 x^2}{2-x^2} \\ & \frac{1}{2-x}=\frac{x}{2-x^2} \\ & \left(2-x^2\right)=x(2-x) \\ & 2-x^2=2 x-x^2 \\ & 2 x=2 \Rightarrow x=1\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\sin ^{-1}\left(\frac{3}{x}\right)+\sin ^{-1}\left(\frac{4}{x}\right)=\frac{\pi}{2}$, then $x$ is equal to

Options:
A $3$
B $5$
C $7$
D $11$

The correct answer is:
$5$

Solution:
Given that,
$\begin{aligned}
& \sin ^{-1}\left(\frac{3}{x}\right)+\sin ^{-1}\left(\frac{4}{x}\right)=\frac{\pi}{2} \\
& \therefore \sin ^{-1}\left(\frac{3}{x}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{4}{x}\right) \\
& \Rightarrow \sin ^{-1}\left(\frac{3}{x}\right)=\cos ^{-1}\left(\frac{4}{x}\right) \\
& \Rightarrow \sin ^{-1}\left(\frac{3}{x}\right)=\sin ^{-1}\left(\frac{\sqrt{x^2-16}}{x}\right) \\
& \Rightarrow \frac{3}{x}=\frac{\sqrt{x^2-16}}{x} \\
& \Rightarrow 9=x^2-16 \Rightarrow x^2=25 \\
& \Rightarrow x= \pm 5 \\
& \Rightarrow x=5
\end{aligned}$
($\because-5$ is not satisfied the given equation)

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If sin-1⁡x4-2sin-1⁡x2+1≤0, then (where, . represents the greatest integral part of x)

Options:
A x∈sin⁡1,sin⁡2∪-1,0
B x∈-sin⁡1,0∪sin⁡1,1
C x∈sin⁡1,1∪-sin⁡1,0
D x∈-sin⁡1,sin⁡2

The correct answer is:
x∈sin⁡1,1∪-sin⁡1,0

Solution:
sin-1⁡x2-12≤0
⇒sin-1⁡x2-1=0⇒sin-1⁡x=1,-1
⇒1≤sin-1⁡x<2 or -1≤sin-1⁡<0
but sin-1⁡x∈-π2,π2
∴1≤sin-1⁡x≤π2 or -1≤sin-1⁡x<0
⇒x∈-sin⁡1,0∪sin⁡1,1

Topic: Mathematics, Functions, JEE Main
If $\sin ^{-1} x=\frac{\pi}{5}$ for some $x \in(-1,1)$, then the value of $\cos ^{-1} x$ is

Options:
A $\frac{3 \pi}{10}$
B $\frac{5 \pi}{10}$
C $\frac{7 \pi}{10}$
D $\frac{9 \pi}{10}$

The correct answer is:
$\frac{3 \pi}{10}$

Solution:
$\begin{aligned} & \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2} \\ & \Rightarrow \cos ^{-1} x=\frac{\pi}{2}-\sin ^{-1} x=\frac{\pi}{2}-\frac{\pi}{5}=\frac{3 \pi}{10}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If sin-1x<cos-1x, then

Options:
A -1≤x<12
B -3≤x<-1
C 12<x≤1
D 1<x<3

The correct answer is:
-1≤x<12

Solution:
Given, sin-1x<cos-1x

⇒cos-1x>π2-cos-1x sin-1x+cos-1x=π2

⇒2cos-1x>π2

⇒cos-1x>π4

⇒x<cosπ4 (As cosθ is decreasing function.)

⇒x<12 ……1

Domain of cos-1θ is -1,1 ……2

Combining equation (1) and (2) we get,

x∈[-1,12)

Topic: Mathematics, Inverse Trigonometric Functions, TS EAMCET, TS EAMCET 2021 (04 Aug Shift 1)
If $\sin ^{-1} x+\cos ^{-1} y=\frac{3 \pi}{10}$, then the value of $\cos ^{-1} x+\sin ^{-1} y$ is

Options:
A $\frac{\pi}{10}$
B $\frac{7\pi}{10}$
C $\frac{9\pi}{10}$
D $\frac{3\pi}{10}$

The correct answer is:
$\frac{7\pi}{10}$

Solution:
$\begin{array}{ll} & \sin ^{-1} x+\cos ^{-1} y=\frac{3 \pi}{10} \\ \therefore \quad & \frac{\pi}{2}-\cos ^{-1} x+\frac{\pi}{2}-\sin ^{-1} y=\frac{3 \pi}{10} \\ \therefore \quad & \pi-\cos ^{-1} x-\sin ^{-1} y=\frac{3 \pi}{10} \\ \therefore \quad & \cos ^{-1} x+\sin ^{-1} y=\pi-\frac{3 \pi}{10}=\frac{7 \pi}{10}\end{array}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x$, then $x \in$ to

Options:
A $\{1,0\}$
B $\{-1,1\}$
C $\left\{0, \frac{1}{2}\right\}$
D $\{2,0\}$

The correct answer is:
$\left\{0, \frac{1}{2}\right\}$

Solution:
We have,
$\begin{aligned}
& \sin ^{-1} x+\sin ^{-1}(1-x)=\cos ^{-1} x \\
& \Rightarrow \sin ^{-1}(1-x)=\frac{\pi}{2}-\sin ^{-1} x-\sin ^{-1} x \\
& =\frac{\pi}{2}-2 \sin ^{-1} x \\
& {\left[\because \cos ^{-1} x+\sin ^{-1} x=\frac{\pi}{2}\right]} \\
& \Rightarrow \sin ^{-1}(1-x)=\sin ^{-1} 1-\sin ^{-1} 2 x \sqrt{1-x^2} \\
& \Rightarrow \sin ^{-1}(1-x)=\sin ^{-1}\left(1 \sqrt{1-4 x^2\left(1-x^2\right)}-0\right) \\
& \Rightarrow \quad(1-x)=\sqrt{1-4 x^4+4 x^2}=\sqrt{\left(1-2 x^2\right)^2} \\
& \Rightarrow \quad(1-x)=1-2 x^2 \\
& \Rightarrow \quad 2 x^2-x=0 \\
& \Rightarrow \quad x(2 x-1)=0 \\
& \Rightarrow \quad x=0, \frac{1}{2} \\
& \therefore \quad x \in\left\{0, \frac{1}{2}\right\} \\
&
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3}$, then $\cos ^{-1} x+\cos ^{-1} y=$

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

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

Solution:
$\begin{aligned} & \sin ^{-1} x+\sin ^{-1} y=\frac{2 \pi}{3} \quad \Rightarrow \frac{\pi}{2}-\cos ^{-1} x+\frac{\pi}{2}-\cos ^{-1} y=\frac{2 \pi}{3} \\ & \Rightarrow \cos ^{-1} x+\cos ^{-1} y=\pi-\frac{2 \pi}{3}=\frac{\pi}{3}\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=\frac{3 \pi}{2},$ then the value of $x^{9}+y^{9}+z^{9}-\frac{1}{x^{9} y^{9} z^{9}}$ is equal to

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
We know that $\left|\sin ^{-1} x\right| \leq \frac{\pi}{2}$
Hence, from the given relation we observe that each of $\sin ^{-1} x, \sin ^{-1} y$ and $\sin ^{-1} x$ will be $\frac{\pi}{2}$
so that $x=y=x=\sin \frac{\pi}{2}=1$
$\begin{aligned} \therefore \quad & x^{9}+y^{9}+z^{9}-\frac{1}{x^{9} y^{9} z^{9}} \\ &=(1)^{9}+(1)^{9}+(1)^{9}-\frac{1}{(1)^{9}(1)^{9}(1)^{9}} \\ &=1+1+1-\frac{1}{1 \times 1 \times 1} \\ &=3-1=2 \end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main