datavorous/entrance-exam-dataset huggingface dataset:
If $\cos x=\tan y, \cot y=\tan z$ and $\cot z=\tan x$, then $\sin x$ equals to

Options:
A $\frac{\sqrt{5}+1}{4}$
B $\frac{\sqrt{5}-1}{4}$
C $\frac{\sqrt{5}+1}{2}$
D $\frac{\sqrt{5}-1}{2}$

The correct answer is:
$\frac{\sqrt{5}-1}{2}$

Solution:
Given, $\cos x=\tan y, \cot y=\tan z$
and $\cot z=\tan x$
$\therefore \quad \cos x=\tan y$
$\Rightarrow \quad \cos x=\frac{1}{\tan z}$
$\Rightarrow \quad \cos x=\cot z$
$\Rightarrow \quad \cos x=\tan x$
$\Rightarrow \quad \cos x=\frac{\sin x}{\cos x}$
$\Rightarrow \cos ^2 x=\sin x$
$\Rightarrow 1-\sin ^2 x=\sin x$
$\Rightarrow \sin ^2 x+\sin x-1=0$
$\begin{aligned} \therefore \quad \sin x & =\frac{-1 \pm \sqrt{1-4 \times(-1)}}{2 \times 1} \\ & =\frac{-1 \pm \sqrt{5}}{2} \\ \therefore \sin x=\frac{\sqrt{5}-1}{2} & \left(\because \frac{-1-\sqrt{5}}{2} < -1\right)\end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
If $\cos (x-y), \cos x, \cos (x+y)$ are three distinct numbers which are in harmonic progression and $\cos x \neq \cos y$, then $1+\cos y$ is equal to

Options:
A $\cos ^2 x$
B $-\cos ^2 x$
C $\cos ^2 x-1$
D $\cos ^2 x-2$

The correct answer is:
$\cos ^2 x$

Solution:
$\cos (x-y), \cos x, \cos (x+y)$ are in HP.
Then, $\cos x=\frac{2 \cos (x-y) \cos (x+y)}{\cos (x+y)+\cos (x-y)}$
$\cos x=\frac{\cos 2 x+\cos 2 y}{2 \cos x \cdot \cos y}$
$\cos x=\frac{2 \cos ^2 x+2 \cos ^2 y-2}{2 \cos x \cdot \cos y}$
$\cos ^2 x \cdot \cos y=\cos ^2 x+\cos ^2 y-1$
$\cos ^2 x(\cos y-1)=\left(\cos ^2 y-1\right)$
$\cos ^2 x(1-\cos y)=\left(1-\cos ^2 y\right)$
$\cos ^2 x\left(2 \sin ^2 \frac{y}{2}\right)=\sin ^2 y$
$\cos ^2 x\left(2 \sin ^2 \frac{y}{2}\right)=\left(2 \sin \frac{y}{2} \cdot \cos \frac{y}{2}\right)^2$
$\cos ^2 x\left(2 \sin ^2 \frac{y}{2}\right)=4 \sin ^2 \frac{y}{2} \cdot \cos ^2 \frac{y}{2}$
$\cos ^2 x=\left(2 \cos ^2 \frac{y}{2}-1\right)+1$
$\cos y+1=\cos ^2 x$
or $\quad 1+\cos y=\cos ^2 x$

Topic: Mathematics, Trigonometric Equations, JEE Main
If $\operatorname{cosec} A+\cot A=\frac{11}{2}$, then $\tan A=$

Options:
A $\frac{21}{22}$
B $\frac{15}{16}$
C $\frac{44}{117}$
D $\frac{117}{43}$

The correct answer is:
$\frac{44}{117}$

Solution:
$\begin{gathered}
\operatorname{cosec} A+\cot A=\frac{11}{2} \Rightarrow \operatorname{cosec} A-\cot A=\frac{2}{11} \\
\text { Therefore } 2 \cot A=\frac{117}{22} \Rightarrow \tan A=\frac{44}{117}
\end{gathered}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $\operatorname{cosec} A+\cot A=\frac{11}{2}$, then $\tan A=$

Options:
A $\frac{21}{22}$
B $\frac{15}{16}$
C $\frac{44}{117}$
D $\frac{117}{43}$

The correct answer is:
$\frac{44}{117}$

Solution:
$\begin{aligned}
& \operatorname{cosec} A+\cot A=\frac{11}{2} \Rightarrow \operatorname{cosec} A-\cot A=\frac{2}{11} \\
& \text { Therefore } 2 \cot A=\frac{117}{22} \Rightarrow \tan A=\frac{44}{117}
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If cosec θ=p+q p-q p≠q, p≠0, then cotπ4+θ2 is equals to:

Options:
A pq
B p q
C q p
D pq

The correct answer is:
q p

Solution:
cosec θ = p + q p - q ⇒ 1 sin θ = p + q p - q
let tan θ 2 = t then 1 + t 2 2t = p + q p - q ∵sin2θ=2tanθ1+tan2θ

Using componendo and dividendo rule, we get
⇒ 1 + t 2 + 2t 1 + t 2 - 2t = p + q + p - q p + q - p + q
⇒ 1 + t 1 - t 2 = p q
⇒ |1-t1+t|  =qp=|cot(π4+θ)| ∵tanπ4+θ=1+tanθ1-tanθ

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2014 (09 Apr Online)
If $\operatorname{cosec} \theta=\frac{p+q}{p-q}$, then $\cot \left(\frac{\pi}{4}+\frac{\theta}{4}\right)$ is equal to

Options:
A $\frac{\sqrt{p}}{q}$
B $\frac{\sqrt{q}}{p}$
C $\sqrt{p q}$
D $p q$

The correct answer is:
$\frac{\sqrt{q}}{p}$

Solution:
We have,
$\begin{array}{r}\frac{1}{\sin \theta}=\frac{p+q}{p-q} \\ \frac{1+\tan ^2 \theta / 2}{2 \tan \theta / 2}=\frac{p+q}{p-q}\end{array}$
Applying componendo and dividendo,
$\begin{aligned} & \frac{1+\tan ^2 \theta / 2+2 \tan \theta / 2}{1+\tan ^2 \theta / 2-2 \tan \theta / 2}=\frac{p+q+p-q}{p+q-p+q} \\ & \Rightarrow \quad \frac{(1+\tan \theta / 2)^2}{(1-\tan \theta / 2)^2}=\frac{2 p}{2 q}=\frac{p}{q} \\ & \Rightarrow \quad \frac{\tan \frac{\pi}{4}+\tan \frac{\theta}{2}}{1-\tan \frac{\pi}{4} \tan \frac{\theta}{2}}=\sqrt{\frac{p}{q}} \\ & \Rightarrow \quad \tan \left(\frac{\theta}{2}+\frac{\pi}{4}\right)=\sqrt{\frac{p}{q}} \\ & \Rightarrow \quad \cot \left(\frac{p}{4}+\frac{\theta}{2}\right)=\sqrt{\frac{p}{q}} \\ & \end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
If \( \cosh (x)=\frac{5}{4} \), then \( \cosh (3 x)= \)

Options:
A \( \frac{61}{16} \)
B \( \frac{63}{16} \)
C \( \frac{65}{16} \)
D \( \frac{61}{63} \)

The correct answer is:
\( \frac{65}{16} \)

Solution:
Given, that coshx=54

and we know that

cosh3x=4cos3hx-3coshx

⇒cosh3x=4543-354=6516

⇒cosh3x=6516

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If cot⁡α+β=0, then sin⁡α+2β is equal to

Options:
A sin⁡α
B cos⁡α
C sin⁡β
D cos⁡2β

The correct answer is:
sin⁡α

Solution:
Given, cot⁡α+β=0 ⇒ cos ⁡α+β=0
⇒α+β=2n+1π2 , n∈I
∴ sin⁡α+2β=sin⁡2α+2β-α
=sin⁡[2n+1π-α]
=sin⁡(2nπ+π-α)
=sin⁡π-α=sin⁡α

Topic: Mathematics, Trigonometric Equations, JEE Main
If $\theta=\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18)$, then $\cot \theta$ is equal to

Options:
A 2
B 3
C 4
D 1

The correct answer is:
3

Solution:
It is given that,
$\begin{aligned}
\theta & =\cot ^{-1}(7)+\cot ^{-1}(8)+\cot ^{-1}(18) \\
& =\tan ^{-1}\left(\frac{1}{7}\right)+\tan ^{-1}\left(\frac{1}{8}\right)+\tan ^{-1}\left(\frac{1}{18}\right) \\
& =\tan ^{-1}\left(\frac{\frac{1}{7}+\frac{1}{8}}{1-\frac{1}{7 \times 8}}\right)+\tan ^{-1} \frac{1}{18} \\
& =\tan ^{-1}\left(\frac{15}{55}\right)+\tan ^{-1}\left(\frac{1}{18}\right)=\tan ^{-1}\left(\frac{\frac{3}{11}+\frac{1}{18}}{1-\frac{3}{11 \times 18}}\right) \\
& =\tan ^{-1}\left(\frac{54+11}{195}\right)=\tan ^{-1}\left(\frac{1}{3}\right) \\
\Rightarrow & \tan \theta=\frac{1}{3} \Rightarrow \cot \theta=3
\end{aligned}$
Hence, option (2) is correct.

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If cotα=1 and secβ=-53, where π<α<3π2 and π2<β<π, then the value of tanα+β and the quadrant in which α+β lies, respectively are

Options:
A -17 and IVth quadrant
B 7 and Ist quadrant
C -7 and IVth quadrant
D 17 and Ist quadrant

The correct answer is:
-17 and IVth quadrant

Solution:
Given,

cotα=1,secβ=-53

So, cosβ=-35,tanβ=-43 and tanα=1

Now using formula tanα+β=tanα+tanβ1-tanαtanβ

⇒tanα+β=1-431+43×1=-17

We know that tanθ is negative in 2nd & 4th quadrant but as π<α<3π2, so α+β will lie in IVth quadrant.

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main, JEE Main 2022 (28 Jun Shift 2)
If cot-1⁡x2-7 cot-1⁡x+10>0, then the range of x will be

Options:
A (-â, cotâ¡2)
B -â, cotâ¡5
C cotâ¡2, cotâ¡5
D cotâ¡2, â

The correct answer is:
cotâ¡2, â

Solution:
cot-1⁡x-5cot-1⁡x-2>0
⇒cot-1⁡x∈-∞, 2∪5, ∞
⇒cot-1⁡x∈0, 2 (Taking intersection with range of cot −1 x )
⇒n∈(cot⁡2, ∞)

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If cot-1â¡x-x22+x34-.â¦â¦+tan-1â¡x2-x42+x64-.â¦â¦=Ï2 for 0<x<2, then x is equal to

Options:
A 0 only
B 1 only
C 0,1 both
D None of these

The correct answer is:
1 only

Solution:
Since tan-1⁡x+cot-1⁡x=π2 for x∈R .
∴x-x22+x34-.…..=x2-x42+x64-.…..
⇒x1+x2=x21+x220<x<2
⇒x2+x=x22+x2⇒2x+x3=2x2+x3
⇒x2=x⇒x=0,1
But x≠0,∴x=1 .

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $\cot A=\frac{11}{60}, \cos B=\frac{7}{25}$ and neither $A$ nor $B$ is in the first quadrant, then $\left(A+\frac{B}{2}\right)$ lies in the quadrant

Options:
A I
B II
C III
D IV

The correct answer is:
I

Solution:
Given, $\cot A=\frac{11}{60}$
$\therefore \tan A=\frac{60}{11}$ and $\cos B=\frac{7}{25}$
$\begin{aligned} & \because A \text { and } B \notin Q_1 \\ & \therefore A \in Q_3 \text { and } B \in Q_4 \\ & \therefore \tan B=-\frac{24}{7} \quad\left(\because \tan B=\text { negative in } Q_4\right) \\ & \Rightarrow \frac{2 \tan \frac{B}{2}}{1-\tan ^2 \frac{B}{2}}=-\frac{24}{7} \\ & \Rightarrow 12 \tan ^2 \frac{B}{2}-7 \tan \frac{B}{2}-12=0 \\ & \Rightarrow\left(3 \tan ^{\frac{B}{2}}-4\right)\left(4 \tan \frac{B}{2}+3\right)=0 \\ & \Rightarrow \tan \frac{B}{2}=\frac{4}{3}(\text { Rejected }) \text { because } B \in Q_4 \text { i.e. } \\ & \frac{3 \pi}{2} < B < 2 \pi \\ & \therefore \quad \frac{3 \pi}{4} < \frac{B}{2} < \pi \\ & \text { i.e. } \quad \frac{B}{2} \in Q_2 \\ & \therefore \quad \tan \frac{B}{2}=\text { negative } \\ & \text { or } \quad \tan \frac{B}{2}=-\frac{3}{4}\end{aligned}$
$\begin{aligned} & \text { Now, } \tan \left(A+\frac{B}{2}\right)=\frac{\tan A+\tan \frac{B}{2}}{1-\tan A \cdot \tan \frac{B}{2}} \\ & =\frac{\frac{60}{11}-\frac{3}{4}}{1-\frac{60}{11}\left(-\frac{3}{4}\right)} \\ & =\frac{\frac{240-33}{44}}{\frac{224}{44}}=\frac{207}{224} \\ & \therefore\left(A+\frac{B}{2}\right) \in Q_1 \\ & \end{aligned}$

Topic: Mathematics, Trigonometric Equations, JEE Main
If $D=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|$ for $x \neq 0, y \neq 0$ then $D$ is

Options:
A
divisible by neither $x$ nor $y$

B
divisible by both $x$ and $y$

C
divisible by $x$ but not $y$

D
divisible by $y$ but not $x$

The correct answer is:
divisible by both $x$ and $y$

Solution:
$D=\left|\begin{array}{ccc}1 & 1 & 1 \\ 1 & 1+x & 1 \\ 1 & 1 & 1+y\end{array}\right|$
$\mathrm{C}_2 \rightarrow \mathrm{C}_2-\mathrm{C}_1$ \& $\mathrm{C}_3 \rightarrow \mathrm{C}_3-\mathrm{C}_1$
$\left|\begin{array}{lll}
1 & 0 & 0 \\
1 & x & 0 \\
1 & 0 & y
\end{array}\right|=x y$.
Hence $D$ is divisible by both $x$ and $y$.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2007
If d1 is the shortest distance between the lines x+1=2 y=-12 z, x=y+2=6 z-6 and d2 is the shortest distance between the lines x-12=y+8-7=z-45,x-12=y-21=z-6-3, then the value of 323 d1 d2 is :

Options:

The correct answer is:
16

Solution:
Let,

L1:x+11=y12=z-112,

And L2:x1=y+21=z-116

Now, given d1= shortest distance between L1 and L2

Now using the formula of shortest distance we get,

d1=a→2-a→1·b→1×b→2b→1×b→2

⇒d1=i^-2j^+k^·i^+12j^-112k^×i^+j^+16k^i^+12j^-112k^×i^+j^+16k^

⇒d1=i^-2j^+k^·i^6-j^4+k^2i^6-j^4+k^2

⇒d1=16+12+1249144

⇒d1=76712=2

Now, let

L3:x-12=y+8-7=z-45, L4:x-12=y-21=z-6-3

Similarly finding d2= shortest distance between L3 & L4 we get,

d2=10j^+2k^·2i^-7j^+5k^×2i^+j^-3k^2i^-7j^+5k^×2i^+j^-3k^

⇒d2=10j^+2k^·16i^+16j^+16k^16i^+16j^+16k^

⇒d2=160+32163

⇒d2=123

Hence, 323 d1 d2=323×2123=16

Topic: Mathematics, Three Dimensional Geometry, JEE Main, JEE Main 2024 (30 Jan Shift 1)
If ∫dθcos2θtan2θ+sec2θ= λtanθ+2logefθ+C where C is a constant of integration, then the ordered pair λ,fθ is equal to:

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

The correct answer is:
-1,1+tanθ

Solution:
∫sec2θ1+tan2θ1-tan2θ+2tanθ1-tan2θdθ

=∫sec2θ1-tan2θ1+tanθ2dθ

=∫sec2θ1-tanθ1+tanθdθ

Let tanθ=t⇒sec2θdθ=dt

=∫1-t1+tdt=∫-1+21+tdt

=-t+2ln1+t+C

=-tanθ+2ln1+tanθ+C

⇒λ=-1 and fθ=1+tanθ

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2020 (09 Jan Shift 2)
If $D, E$ and $F$ are respectively the mid-points of $A B, A C$ and $B C$ in $\triangle A B C$, then $\overrightarrow{\mathbf{B E}}+\overrightarrow{\mathbf{A F}}$ is equal to :

Options:
A $\overrightarrow{\mathbf{D C}}$
B $\frac{1}{2} \overrightarrow{\mathbf{B F}}$
C $2 \overrightarrow{\mathbf{B F}}$
D $\frac{3}{2} \overrightarrow{\mathbf{B F}}$

The correct answer is:
$\overrightarrow{\mathbf{D C}}$

Solution:
Let $A=\overrightarrow{\mathbf{a}}, B=\overrightarrow{\mathbf{b}}, C=\overrightarrow{\mathbf{c}}$
Now, $\quad \overrightarrow{\mathbf{A F}}=\frac{1}{2}(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}})-\overrightarrow{\mathbf{a}}$
$\overrightarrow{\mathbf{B E}}=\frac{1}{2}(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{c}})-\overrightarrow{\mathbf{b}}$
and $\quad \overrightarrow{\mathbf{C D}}=\frac{1}{2}(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})-\overrightarrow{\mathbf{c}}$
$\therefore \quad \overrightarrow{\mathbf{A F}}+\overrightarrow{\mathbf{B E}}=\frac{1}{2}(\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}})-\overrightarrow{\mathbf{a}}+\frac{1}{2}(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{c}})-\overrightarrow{\mathbf{b}}$
$=-\frac{1}{2} \overrightarrow{\mathbf{b}}-\frac{1}{2} \overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{c}}$
$=\overrightarrow{\mathbf{c}}-\frac{1}{2}(\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}})=\overrightarrow{\mathbf{D C}}$

Topic: Mathematics, Vector Algebra, JEE Main
If $\mathrm{D}$ is determinant of order 3 and $\mathrm{D}^{\prime}$ is the determinant obtained by replacing the elements of $D$ by their cofactors, then which one of the following is correct?

Options:
A $\mathrm{D}^{\prime}=\mathrm{D}^{2}$
B $\mathrm{D}^{\prime}=\mathrm{D}^{3}$
C $\mathrm{D}^{\prime}=2 \mathrm{D}^{2}$
D $\mathrm{D}^{\prime}=3 \mathrm{D}^{3}$

The correct answer is:
$\mathrm{D}^{\prime}=\mathrm{D}^{2}$

Solution:
$\mathrm{D}^{\prime}=$ cofactor $\mathrm{D}$ $\Rightarrow\left|\mathrm{D}^{\prime}\right|=\mid$ cofactor $\mathrm{D} \mid$
$\Rightarrow\left|\mathrm{D}^{\prime}\right|=|\mathrm{D}|^{3-1}$
$\Rightarrow\left|\mathrm{D}^{\prime}\right|=|\mathrm{D}|^{2}$
So $\mathrm{D}^{\prime}=\mathrm{D}^{2}$

Topic: Mathematics, Determinants, JEE Main
If $\mathrm{D}$ is the domain and $\mathrm{G}$ is the range of the real valued function $f(x)=\sqrt{\frac{1-x^2}{1+x^2}}$, then $\mathrm{D} \cap \mathrm{G}=$

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

The correct answer is:
$[0,1]$

Solution:
We have $f(x)=\sqrt{\frac{1-x^2}{1+x^2}}$
For domain of $\mathrm{f}(\mathrm{x}), \frac{1-\mathrm{x}^2}{1+\mathrm{x}^2} \geq 0$ $\Rightarrow \mathrm{x} \in[-1,1]$
For range of $\mathrm{f}(\mathrm{x})$
$$
y=\frac{1-x^2}{1+x^2}
$$
On solving we get $\mathrm{y} \leq 1$ but $\mathrm{y} \geq 0$
Hence $\mathrm{D}=[-1,1], \mathrm{G}=[0,1]$ Now $\mathrm{D} \cap \mathrm{G}=[0,1]$

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2022 (19 Jul Shift 2)
If Dr=r158r2359r32510, then the value of -1100∑r=15Dr-375 is equal to

Options:
A 5
B 2
C 9
D 3

The correct answer is:
3

Solution:
∑r=15Dr=∑r=15r158∑r=15r2359∑r=15r32510=15158553592252510
applying C1↔C1-C2, we get,
∑r=15Dr=0158203592002510=-28000
-1100∑r=15Dr-375=2435=3

Topic: Mathematics, Determinants, JEE Main
If ∫dx16-9x2=Asin-1⁡Bx+C then A+B=

Options:
A 94
B 194
C 34
D 1312

The correct answer is:
1312

Solution:
I= ∫ dx 4 2 − ( 3x ) 2
I= 1 3 sin −1 ( 3x 4 )+C
A+B= 1 3 + 3 4 = 4+9 12 = 13 12

Topic: Mathematics, Indefinite Integration, JEE Main
If ∫dxcos3x 2sin2x=tanxA+CtanxB+k, where k is a constant of integration, then A+B+C equals

Options:
A 165
B 2710
C 710
D 215

The correct answer is:
165

Solution:
∫dxcos3x4sinxcosx=∫dx2cos4xtanx
Let, tanx=t2⇒sec2x=1+t4
And sec2xdx=2tdt
∴I= ∫sec4xdx2tanx= ∫sec2x.sec2xdx2tanx
=∫1+t42t dt2t=∫1+t4dt
=t+t55+k, where k is a constant of integration.
= tanx+15tan52x+k

Where, t= tanx.
A=12, B=52, C=15
A+B+C=165

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2016 (09 Apr Online)
If $\frac{d}{d x} G(x)=\frac{e^{\tan x}}{x}, x \in(0, \pi / 2)$, then $\int_{1 / 4}^{1 / 2} \frac{2}{x} \cdot e^{\tan \left(\pi x^2\right)} d x$ is equal to

Options:
A
$G(\pi / 4)-G(\pi / 16)$

B
$2[G(\pi / 4)-G(\pi / 16)]$

C
$\pi[G(1 / 2)-G(1 / 4)]$

D
$G(1 / \sqrt{2})-G(1 / 2)$

The correct answer is:
$G(\pi / 4)-G(\pi / 16)$

Solution:
Let $\frac{d}{d x} G(x)=\frac{e^{\tan x}}{x}, x \in\left(0, \frac{\pi}{2}\right)$
Now,
$$
\begin{aligned}
I & =\int_{1 / 4}^{1 / 2} \frac{2}{x} e^{\tan \pi x^2} \cdot d x \\
& =\int_{1 / 4}^{1 / 2} \frac{2 \pi x}{\pi x^2} e^{\tan \pi x^2} \cdot d x
\end{aligned}
$$
Let $\pi x^2=t \Rightarrow 2 \pi x d x=d t$
When $x=\frac{1}{2}, t=\frac{\pi}{4}$ and $x=\frac{1}{4}, t=\frac{\pi}{16}$
$$
\begin{aligned}
\therefore I & =\int_{\pi / 16}^{\pi / 4} \frac{e^{\tan t}}{t} d t=g(t) \mid \begin{array}{l}
\frac{\pi}{4} \\
\frac{\pi}{16}
\end{array} \\
& =G\left(\frac{\pi}{4}\right)-G\left(\frac{\pi}{16}\right)
\end{aligned}
$$

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2012 (12 May Online)
If $\frac{d y}{d x}+\frac{y}{x}=x^2$, then $2 y(2)-y(1)=$

Options:
A $\frac{11}{4}$
B $\frac{15}{4}$
C $\frac{9}{4}$
D $\frac{13}{4}$

The correct answer is:
$\frac{15}{4}$

Solution:
$\frac{d y}{d x}+\frac{y}{x}=x^2$
On comparing to $\frac{d y}{d x}+P(x) y=Q(x)$
$$
\begin{aligned}
& P(x)=\frac{1}{x}, Q(x)=x^2 \\
& \mathrm{IF}=e^{\int P d x}=e^{\int \frac{1}{x} d x}=e^{\ln x}=x \\
& y \cdot(\mathrm{IF})=\int Q(x)(\mathrm{IF}) d x+C \\
& y \cdot x=\int x^2 x d x+C=\frac{x^4}{4}+C \\
& y=\frac{x^3}{4}+\frac{C}{x} \\
& y(2)=\frac{2^3}{4}+\frac{C}{2}=2+\frac{C}{2} \\
& y(1)=\frac{1}{4}+\frac{C}{1}=\frac{1}{4}+C \\
& 2 y(2)-y(1)=2\left(2+\frac{C}{2}\right)-\left(\frac{1}{4}+C\right) \\
&=4+C-\frac{1}{4}-C=\frac{15}{4}
\end{aligned}
$$

Topic: Mathematics, Differential Equations, JEE Main
If $\frac{d y}{d x}=\frac{y+x \tan \frac{y}{x}}{x}$, then $\sin \left(\frac{y}{x}\right)=$

Options:
A $c x^{2}$
B $c x$
C $c x^{3}$
D $\log x$

The correct answer is:
$c x$

Solution:
We have, $\frac{d y}{d x}=\frac{y+x \tan \frac{y}{x}}{x} \quad \text{...(i)}$
Given, differential equation is in homogeneous form.
$\therefore$ put $y=v x$ in Eq. (i), we get
$v+x \frac{d v}{d x}=v+\tan v \Rightarrow \frac{1}{\tan v} d v=\frac{d x}{x}$
Taking integration on both sides, we get $\log (\sin v)=\log x+\log c$
$$
\Rightarrow \log \frac{\sin v}{x}=\log c \Rightarrow \sin \left(\frac{y}{x}\right)=x c
$$

Topic: Mathematics, Differential Equations, JEE Main
If $\frac{d y}{d x}=y+3>0$ and $y(0)=2$, then $y(\ln 2)$ is equal to

Options:
A
5

B
13

C
-2

D
7

The correct answer is:
7

Solution:
$\frac{d y}{d x}=y+3 \Rightarrow \frac{d y}{y+3}=d x$
$\ln (y+3)=x+c$
$x=0 \Rightarrow y=2$
$\Rightarrow \ln 5=0+c$
$c=\ln 5$
$\ln (y+3)=x+\ln 5$
$y+3=e^{x+\ln 5} \Rightarrow y+3=e^{\ln 2+\ln 5}$
$y+3=10 \Rightarrow y=7$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2011
If â«dxex-1=2tan-1fx+C, (where x>0 and C is the constant of integration) then the range of fx is

Options:
A 0,∞
B 0,∞
C 1,∞
D 1,∞

The correct answer is:
0,∞

Solution:
Le,t ex-1=t2
⇒ex⋅dx=2tdt
∴Ι=∫2tdttex
=∫2dtt2+1
=2tan-1t+C
=2tan-1ex-1+C
∴fx=ex-1
For x∈0,∞;ex∈1,∞
⇒ex-1∈0,∞

Topic: Mathematics, Indefinite Integration, JEE Main
If ∫dxx+12x2+2x+2=Ax+1+Btan-1⁡x+1+C, where A and B are constants and C is the constant of integration, then A-B is equal to

Options:

The correct answer is:
0

Solution:
∫dxx+12x2+2x+2=∫x2+2x+2-x2+2x+1x2+2x+1x2+2x+2dx
=∫dxx+12-∫dxx+12+1
=-1x+1-tan-1⁡x+1+C
So A=B=-1

Hence, A-B=0

Topic: Mathematics, Indefinite Integration, JEE Main
If ∫dxx2-2x+102=Atan-1x-13+fxx2-2x+10+C, then (where C is a constant of integration)

Options:
A A=127 and fx=9x-1
B A=181 and fx=3x-1
C A=154 and fx=9x-12
D A=154 and fx=3x-1

The correct answer is:
A=154 and fx=3x-1

Solution:
We have I=∫dxx2-2x+102=∫dxx-12+322

Let x –1=3tanθ⇒dx=3sec2θdθ

⇒I=∫3sec2θdθ81sec4θ=127∫cos2θdθ

=154∫1+cos2θdθ

=154θ+sin2θ2+C

=154θ+sinθcosθ+C, where C is the constant of integration.

=154tan-1x-13+x-131+x-132+C

=154tan-1x-13+3x-1x2-2x+10+C

Hence, A=154 & fx=3(x-1)

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2019 (10 Apr Shift 1)
If ∫dxx2+x+12=atan-12x+13+b2x+1x2+x+1+C, x>0 where C is the constant of integration, then the value of 9(3a+b) is equal to _________.

Options:

The correct answer is:
15

Solution:
I=∫dxx+122+342

Put x+12=t

=∫dtt2+342

Put t=32tan θ=32∫sec2θ dθ916sec4θ

=439∫1+cos 2θdθ

=439θ+sin 2θ2+C

=439tan-12x+13+32x+13+2x+12+C

=439tan-12x+13+132x+1x2+x+1+C

Hence, 93a+b=15

Topic: Mathematics, Indefinite Integration, JEE Main
If ∫dxx2+x+12=atan-12x+13+b2x+1x2+x+1+C, x>0 where C is the constant of integration, then the value of 9(3a+b) is equal to _________.

Options:

The correct answer is:
15

Solution:
I=∫dxx+122+342

Put x+12=t

=∫dtt2+342

Put t=32tan θ=32∫sec2θ dθ916sec4θ

=439∫1+cos 2θdθ

=439θ+sin 2θ2+C

=439tan-12x+13+32x+13+2x+12+C

=439tan-12x+13+132x+1x2+x+1+C

Hence, 93a+b=15

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2021 (27 Aug Shift 1)
If â«dxx2+x=lnâ¡fx+C (where C is the constant of integration), then the range of y=fx, âxâR--1,0 is

Options:
A R-1
B R-0
C R-0,1
D R-0,-1

The correct answer is:
R-0,1

Solution:
Ι=∫dxxx+1=∫dxx-∫dxx+1
⇒Ι=ln⁡x-ln⁡x+1+C
=ln⁡xx+1+C
i.e. y=fx=xx+1
⇒yx+y=x
⇒x=y1-y
⇒y∈R-0,1 as x≠0

Topic: Mathematics, Indefinite Integration, JEE Main
If ∫dxx31+x623 =xfx1+x613+C, where C is a constant of integration, then the function fx is equal to

Options:
A 3x2
B -12x3
C -16x3
D -12x2

The correct answer is:
-12x3

Solution:
We have, ∫dxx31+x623=xfx1+x613+C

Now, I=∫dxx71x6+123

Put, t=1x6+1⇒dt=-6x7dx

∴I=-16∫dtt23=-12t13+C, where C is the constant of integration=-121x6+113+C=-121+x613x2+C

∴fx=-12x3

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2019 (08 Apr Shift 2)
If Ι=∫dxx3x8+13/4=λ1+x814x2+c (where c is the constant of integration), then the value of λ is equal to

Options:
A 2
B 12
C -2
D -12

The correct answer is:
-12

Solution:
Ι=∫x-91+x-834dx
Let, 1+x-8=t
-8x-9dx=dt
Ι=∫-dt8t3/4
=-18∫t-3/4dt
=-18t1/414+c
=-1+x-81/42+c
=-1+x81/42x2+c
⇒λ=-12

Topic: Mathematics, Indefinite Integration, JEE Main
If â« dx x + x 3 =a x +b x 3 +c x 6 +d ln(x6+1)+e, e being an arbitrary constant, then the value of 20a+b+c+d is

Options:

The correct answer is:
37

Solution:
Let x=u6, dx=6u5du
∫dxx+x3=∫6u5duu3+u2=6∫u3u+1du
=6∫u2-u+1-1u+1du
=2u3-3u2+6u-6lnu+1+e
=2x-3x3+6x6-6lnx6+1+e
∴ a=2, b=-3, c=6, d=-6
∴ 20a+ b+ c+ d=37

Topic: Mathematics, Indefinite Integration, JEE Main
If dydx+2x-y2y-12x-1=0,x,y>0,y1=1, then y2 is equal to

Options:
A 2+log23
B 2+log22
C 2-log-23
D 2-log23

The correct answer is:
2-log23

Solution:
Given, dydx+2x-y2y-12x-1=0 x, y>0, y1=1

Now rearranging and integrating both side of dydx=-2x2y-12y2x-1, we get

⇒∫2y2y-1dy=-∫2x2x-1dx

⇒1ln2∫2yln22y-1dy=-1ln2∫2xln22x-1dx

⇒1ln2ln2y-1=-1ln2ln2x-1+C

At x=1,y=1

Putting this values in above equation we get C=0

So, ln2y-1+ln2x-1=0

⇒2x-12y-1=1

⇒2y-1=12x+1

At x=2

⇒2y=13+1=43

y=log243=log24-log23=2-log23

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (27 Jun Shift 1)
If dy dx=2xy+2y·2x2x+2x+yloge2, y0=0, then for y=1, the value of x lies in the interval :

Options:
A 1, 2
B 12, 1
C 2, 3
D 0, 12

The correct answer is:
1, 2

Solution:
Given differential equation is

dydx=2x·y+2y·2x2x+2x+yloge2

⇒dydx=2xy+2y2x1+2yloge2

⇒∫1+2yloge2y+2ydy=∫dx

⇒∫dy+2yy+2y=∫dx

⇒lny+2y=x+C∵∫f'xfxdx=lnfx+C

Now ∵y0=0

⇒C=0

∴lny+2y=x

Now for y=1 we have

x=ln1+2=ln3∈1, 2

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2021 (31 Aug Shift 2)
If dydx+2ytanx=sinx,0<x<π2 and yπ3=0, then the maximum value of yx is

Options:
A 18
B 34
C 14
D 38

The correct answer is:
18

Solution:
We know that,

dydx+y2tanx=sinx is a linear differential equation so,

I.F=e∫2tanxdx=e2lnsecx=sec2x

The general solution will be

ysec2x=∫sinxsec2xdx+C

⇒ysec2x=secx+C

∵yπ3=0⇒C=-2

Hence the particular solution is

ysec2x=secx-2

⇒y=cosx-2cos2x

⇒y=18-2cosx-142

So, the maximum value of yx is 18

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2022 (26 Jul Shift 1)
If dydx+3cos2xy=1cos2x , x∈-π3,π3, and yπ4=43, then y-π4 equals

Options:
A 13
B 13+e3
C 13+e6
D -43

The correct answer is:
13+e6

Solution:
Given dydx+3sec2xy=sec2x

This is linear differential equation, of the type dydx+Py=Q, where P & Q are functions of x or constants.

Here, P=3sec2x & Q=sec2x

Now, integrating factor I.F.=e∫Pdx

=e∫3sec2x dx=e3tanx

Hence, the solution of the differential equation is yI.F.=∫QI.F.dx+c

⇒y·e3tanx=∫e3tanx·sec2xdx

Let, tanx=t, ⇒sec2xdx=dt

⇒y·e3tanx=∫e3tdt+c

⇒y·e3tanx=e3t3+c

⇒y·e3tanx=e3tanx3+c

⇒y=ce-3tanx+13

Given, yπ4=43⇒

43=ce-3+13

⇒c=e3

Thus, y=e3·e-3tanx+13

Hence, y-π4=e3·e-3tan-π4+13

⇒y-π4=e3·e-3-1+13

=e3·e3+13=e6+13.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2019 (10 Jan Shift 1)
If dydx=xyx2+y2;y1=1; then a value of x satisfying yx=e is:

Options:
A 123e
B e2
C 2e
D 3e

The correct answer is:
3e

Solution:
Put y=vx

dydx=v+xdvdx

v+xdvdx=vx2x2+v2x2

⇒1+v2v3dv=-1xdx

⇒∫1v3+1vdv=∫-1xdx

⇒-121v2+lnv=-lnx+c

⇒-x22y2=-lny+c

When x=1,y=1 then

-12=c

⇒x2=y21+2lny

⇒x2=e23

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (09 Jan Shift 2)