datavorous/entrance-exam-dataset huggingface dataset:
If ∫x5e-x2dx=gxe-x2+c, where c is a constant of integration, then g-1 is equal to

Options:
A -52
B -1
C 1
D -12

The correct answer is:
-52

Solution:
Let I=∫x5e-x2dx

Let, -x2=t⇒xdx=-12dt

∴ I=-12∫t2etdt

By using integrating by parts, i.e. ∫u·vdx=u∫vdx-∫ddxu∫vdxdx+c, we get

I=-12t2∫etdt-∫2t∫etdtdt+c

⇒I=-12t2et-2∫tetdt+c

Again, applying integrating by parts, we get

⇒I=-12t2et-2t∫etdt-∫1·etdt+c

⇒I=-12t2et-2tet-et+c

⇒I=-12t2et-2tet+2et+c

⇒I=-et2t2-2t+2+c

⇒I=-e-x22x4+2x2+2+c

Given, I=∫x5e-x2dx=gxe-x2+c

⇒gx=-12x4+2x2+2

⇒g-1=-12-14+2-12+2

⇒g-1=-52.

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2019 (10 Apr Shift 2)
If $\int \frac{x^4+1}{x^6+1} d x=A \tan ^{-1} x+B \tan ^{-1} x^3+c$, then $(A, B)=$

Options:
A $\left(1, \frac{1}{3}\right)$
B $\left(1, \frac{1}{4}\right)$
C $\left(1, \frac{1}{6}\right)$
D $\left(1, \frac{4}{3}\right)$

The correct answer is:
$\left(1, \frac{1}{3}\right)$

Solution:
We have,
$$
\int \frac{x^4+1}{x^6+1} d x=A \tan ^{-1} x+B \tan ^{-1} x^3+c
$$
Let $I=\int \frac{x^4+1}{x^6+1} d x=\int \frac{x^4+1+x^2-x^2}{\left(x^2\right)^3+(1)^3} d x$
$$
\begin{aligned}
= & \int \frac{\left(x^4-x^2+1\right)+x^2}{\left(x^4+1\right)\left(x^4+1-x^2\right)} d x \\
& {\left[\because a^3+b^3=(a+b)\left(a^2+b^2-a b\right)\right] } \\
= & \int \frac{d x}{x^2+1}+\int \frac{x^2}{\left(x^2+1\right)\left(x^4+1-x^2\right)} d x \\
= & \int \frac{d x}{x^2+1}+\int \frac{x^2}{x^6+1} d x \\
= & \tan ^{-1} x+\int \frac{x^2}{\left(x^3\right)^2+1} d x
\end{aligned}
$$
Put $x^3=t$
$$
\begin{aligned}
\Rightarrow \quad 3 x^2 d x & =d t \Rightarrow x^2 d x=\frac{1}{3} d t \\
\therefore \quad \quad \quad & =\tan ^{-1} x+\frac{1}{3} \int \frac{d t}{t^2+1} \\
& =\tan ^{-2} x+\frac{1}{3} \tan ^{-1}(t)+c \\
& =\tan ^{-2} x+\frac{1}{3} \tan ^{-1}\left(x^3\right)+c
\end{aligned}
$$
On comparing it with Eq.(i), we get $A=1$ and $B=\frac{1}{3}$.

Topic: Mathematics, Indefinite Integration, JEE Main
If x=6 and y=-2 then x-2y=9. The contrapositive of this statement is

Options:
A If x-2y≠9 then x≠6 or y≠-2
B If x-2y≠9 then x≠6 and y≠-2
C If x-2y=9 then x=6 and y=-2
D None of these

The correct answer is:
If x-2y≠9 then x≠6 or y≠-2

Solution:
Let p,q,r be the three statements such that
p:x=6, q:y=-2 and r:x-2y=9
Here given statement is p∧q→r and its contrapositive is ~r→~p∧q
i.e. ∼r→∼p∨∼q
i.e. if x-2y≠9 then x≠6 or y≠-2

Topic: Mathematics, Mathematical Reasoning, JEE Main
If $x=9$ is a chord of contact of the hyperbola $x^2-y^2=9$, then the equation of the tangent at one of the points of contact is

Options:
A $x+\sqrt{3} y+2=0$
B $3 x+2 \sqrt{2} y-3=0$
C $3 x-\sqrt{2} y+6=0$
D $x-\sqrt{3} y+2=0$

The correct answer is:
$3 x+2 \sqrt{2} y-3=0$

Solution:
Given that, $x=9$ is a chord of contact of hyperbola.
$$
\begin{aligned}
& x^2-y^2=9 \\
& \text { put } x=9, \quad 81-y^2=9 \\
& \Rightarrow \quad y^2=72 \\
& \Rightarrow \quad y=6 \sqrt{2} \text { or }-6 \sqrt{2} \\
&
\end{aligned}
$$
$\therefore$ Points are $(9,6 \sqrt{2})$ and $(9,-6 \sqrt{2})$
Now, differentiating Eq. (i) w.r.t. $x$, we get
$$
\begin{aligned}
& 2 x-2 y \frac{d y}{d x}=0 \\
& \Rightarrow \quad \frac{d y}{d x}=\frac{x}{y} \\
& \text { at }(9,6 \sqrt{2}) \quad\left(\frac{d y}{d x}\right)_{(9,6 \sqrt{2})}=\frac{9}{6 \sqrt{2}}=\frac{3}{2 \sqrt{2}} \\
& \text { and at }(9-6 \sqrt{2})\left(\frac{d y}{d x}\right)_{(9,-6 \sqrt{2})}=-\frac{3}{2 \sqrt{2}} \\
&
\end{aligned}
$$
$\therefore$ Equation of tangent at $(9,6 \sqrt{2})$ is
$$
\begin{aligned}
(y-6 \sqrt{2}) & =\frac{3}{2 \sqrt{2}}(x-9) \\
\Rightarrow \quad 2 \sqrt{2} y-24 & =3 x-27 \\
\Rightarrow \quad 3 x-2 \sqrt{2} y-3 & =0
\end{aligned}
$$
and equation of tangent at $(9,-6 \sqrt{2})$ is
$$
\begin{aligned}
(y+6 \sqrt{2}) & =\frac{-3}{2 \sqrt{2}}(x-9) \\
\Rightarrow \quad 2 \sqrt{2} y+24 & =-3 x+27 \\
\Rightarrow \quad 3 x+2 \sqrt{2} y-3 & =0
\end{aligned}
$$

Topic: Mathematics, Hyperbola, JEE Main
If x=33n,n is a positive integral value, then the probability that x will have 3 at its units place is

Options:
A 13
B 14
C 15
D 12

The correct answer is:
14

Solution:
Given that, x=33n
Where, n is a positive integral value.
Here, only four digits may be at the unit place ie, 1, 3, 7, 9.
â´ nS=4
Let E be the event of getting 4 at its units place.
â´ nE=1
â´ P E = n E n S = 1 4

Topic: Mathematics, Probability, JEE Main
If $x=a(1-\cos \theta), \quad y=a(\theta-\sin \theta)$, then $\frac{d^{2} y}{d x^{2}}=$

Options:
A $\frac{\cos ^{2}\left(\frac{\theta}{2}\right)}{2 \operatorname{acosec} \theta}$
B $\frac{\operatorname{cosec} \theta}{2 a \cos ^{2}\left(\frac{\theta}{2}\right)}$
C $\frac{\cos \left(\frac{\theta}{2}\right)}{2 \operatorname{asin} \theta}$
D $\frac{\sin \left(\frac{\theta}{2}\right)}{2 \mathrm{a} \cos \theta}$

The correct answer is:
$\frac{\operatorname{cosec} \theta}{2 a \cos ^{2}\left(\frac{\theta}{2}\right)}$

Solution:
$\frac{d x}{d \theta}=a \sin \theta, \frac{d y}{d \theta}=a(1-\cos \theta)$
$\frac{d y}{d x}=\frac{a(1-\cos \theta)}{a \sin \theta}=\frac{2 \sin ^{2} \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}=\tan \frac{\theta}{2}$
$\frac{d^{2} y}{d x^{2}}=\frac{1}{2} \sec ^{2} \frac{\theta}{2} \cdot \frac{d \theta}{d x}$
$=\frac{1}{2} \sec ^{2} \frac{\theta}{2} \times \frac{1}{a \sin \theta}=\frac{\operatorname{cosec} \theta}{2 \cos ^{2} \frac{\theta}{2}}$

Topic: Mathematics, Differentiation, JEE Main
If $x=a \cos ^{3} \theta$ and $y=a \sin ^{3} \theta$, then $\frac{d y}{d x}$ is

Options:
A $\sqrt[3]{\frac{y}{x}}$
B $\sqrt[3]{\frac{\mathrm{x}}{\mathrm{y}}}$
C $-\sqrt[3]{\frac{x}{y}}$
D $-\sqrt[3]{\frac{\mathrm{y}}{\mathrm{x}}}$

The correct answer is:
$-\sqrt[3]{\frac{\mathrm{y}}{\mathrm{x}}}$

Solution:
If $x=a \cos ^{3} \theta$ and $y=a \sin ^{3} \theta$
$\Rightarrow \quad \frac{d x}{d \theta}=3 a \cos ^{2} \theta(-\sin \theta)$
$\Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{d} \theta}=3 \mathrm{a} \sin ^{2} \theta(\cos \theta)$
$\Rightarrow \quad \frac{d y}{d x}=\frac{d y}{d \theta} \times \frac{d \theta}{d x}$
$=\left(3 a \sin ^{2} \theta \cdot \cos \theta\right) \cdot \frac{1}{-\left(3 a \sin \theta \cdot \cos ^{2} \theta\right)}$
$\frac{d y}{d x}=-\frac{\sin \theta}{\cos \theta}, \frac{d y}{d x}=-\tan \theta$
$\Rightarrow \quad \frac{d y}{d x}=-\frac{(y / a)^{-1 / 3}}{(x / a)^{1 / 3}}$
$$
\begin{aligned}
&=-\left[\frac{y}{a} \times \frac{a}{x}\right]^{1 / 3} \\
\frac{d y}{d x} &=-\left(\frac{y}{x}\right)^{-1 / 3}=-\sqrt[3]{\frac{y}{x}}
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If $x=a\left(\cos t+\log \tan \frac{t}{2}\right), y=a \sin t$, then $\frac{d y}{d x}=$

Options:
A $\tan t$
B $\cot t$
C $-\cot t$
D $\tan t$

The correct answer is:
$\tan t$

Solution:
We have,
$$
x=a\left(\cos t+\log \tan \frac{t}{2}\right)
$$
Now, $\frac{d x}{d t}=a\left\{-\sin t+\frac{1}{\tan \frac{t}{2}} \cdot \sec ^{2} \frac{t}{2} \cdot \frac{1}{2}\right\}$
$$
=a\left\{-\sin t+\frac{1}{\sin t}\right\}
$$
$$
=a\left(\frac{1-\sin ^{2} t}{\sin t}\right)=\frac{a \cos ^{2} t}{\sin t}
$$
and $y=a \sin t$
$$
\frac{d y}{d t}=a \cos t
$$
$\mathrm{So}$,
$$
\begin{aligned}
\frac{d y}{d x} &=\frac{d y / d t}{d x / d t} \\
&=\frac{a(\cos t \sin t)}{a \cos ^{2} t}=\tan t
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If $x=a t^2, y=2 a t$, then $\frac{d^2 y}{d x^2}=$

Options:
A $-\frac{1}{t^2}$
B $\frac{1}{2 a t^3}$
C $-\frac{1}{t^3}$
D $-\frac{1}{2 a t^3}$

The correct answer is:
$-\frac{1}{2 a t^3}$

Solution:
$\begin{aligned} & \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{2 a}{2 a t} \Rightarrow \frac{d y}{d x}=\frac{1}{t}=\frac{2 a}{y} \\ & \Rightarrow y \frac{d y}{d x}=2 a \Rightarrow y \frac{d^2 y}{d x^2}+\left(\frac{d y}{d x}\right)^2=0 \\ & \Rightarrow \frac{d^2 y}{d x^2}=\frac{-(d y / d x)^2}{y}=\frac{-1}{2 a t^3} .\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
If $x=a t+b t^{2},$ where $x$ is in metre $(\mathrm{m})$ and $t$ is in hour $(\mathrm{h})$, then unit of $b$ will be

Options:
A $\frac{\mathrm{m}^{2}}{\mathrm{h}}$
B $m$
C $\frac{m}{h}$
D $\frac{m}{h^{2}}$

The correct answer is:
$\frac{m}{h^{2}}$

Solution:
Given, $x=a t+b t^{2}$
$\begin{array}{l}
{[x]=\left[b t^{2}\right]} \\
\text { Unit of } b=\frac{x}{t^{2}}=\frac{\text { metre }}{(h)^{2}}=\frac{m}{h^{2}}
\end{array}$

Topic: Physics, Units and Dimensions, JEE Main
If $x=a t+b t^2$, where $x$ is the distance travelled by the body in kilometres while $t$ is the time in seconds, then the units of 6 are

Options:
A $\mathrm{km} / \mathrm{s}$
B $\mathrm{km} - \mathrm{s}$
C $\mathrm{km} / \mathrm{s^{2}}$
D $\mathrm{km} - \mathrm{s^{2}}$

The correct answer is:
$\mathrm{km} / \mathrm{s^{2}}$

Solution:
$[x]=\left[b t^2\right] \Rightarrow[b]=\left[x / t^2\right]=k m / s^2$

Topic: Physics, Units and Dimensions, JEE Main
If $x=a(t-\sin t)$ and $y=a(1-\cos t)$, then $\frac{d y}{d x}=$

Options:
A $\tan \left(\frac{t}{2}\right)$
B $-\tan \left(\frac{t}{2}\right)$
C $\cot \left(\frac{t}{2}\right)$
D $-\cot \left(\frac{t}{2}\right)$

The correct answer is:
$\cot \left(\frac{t}{2}\right)$

Solution:
\(\begin{aligned}
\Rightarrow \text { So, } \frac{d y}{d t} & =\frac{d}{d t}(a(1-\cos t))=a \cdot \sin t \\
\frac{d x}{d t} & =\frac{d}{d t}(a(t-\sin t))
\end{aligned}\)
So, \(\frac{d y}{d x}=\frac{a \sin t}{a(1-\cos t)}=a(1-\cos t)\)
\(=\frac{2 \sin (t / 2) \cdot \cos (t / 2)}{2 \sin ^2(t / 2)}=\cot (t / 2)\)

Topic: Mathematics, Differentiation, JEE Main
If x=a, y=b, z=c is a solution of the system of linear equations
x+8y+7z=0
9x+2y+3z=0
x+y+z=0
Such that the point a,b,c lies on the plane x+2y+z=6 , then 2a+b+c equals:

Options:
A 2
B -1
C 1
D 0

The correct answer is:
1

Solution:
x+8y+7z=0 ...i 9x+2y+3z=0 ...ii x+y+z=0 ...iii .....

Solve the equation i &iii

⇒7y+6z=0

Solve the equation ii &iii

⇒7x+z=0
x=λ & y=-6-7λ7 & z=-7λ
x=λ& y=6λ& z=-7λ
⇒λ+12λ−7λ=6⇒6λ=6⇒λ=1 | ⇒2λ+6λ−7λ⇒λ=1
Hence, a=1, b=6, c=-7
∴2a+b+c=1

Topic: Mathematics, Determinants, JEE Main, JEE Main 2017 (09 Apr Online)
If $x$ and $y$ are acute angles such that $\cos x+\cos y=\frac{3}{2}$ and $\sin x+\sin y=\frac{3}{4}$, then $\sin (x+y)$ equals to

Options:
A $\frac{2}{5}$
B $\frac{3}{4}$
C $\frac{3}{5}$
D $\frac{4}{5}$

The correct answer is:
$\frac{4}{5}$

Solution:
Given, $\cos x+\cos y=\frac{3}{2}$
$$
\Rightarrow \quad 2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{3}{2}
$$
and
$$
\begin{aligned}
& \sin x+\sin y=\frac{3}{4} \\
& \Rightarrow \quad 2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)=\frac{3}{4} \\
& \therefore \quad \frac{2 \sin \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}{2 \cos \left(\frac{x+y}{2}\right) \cos \left(\frac{x-y}{2}\right)}=\frac{3 / 4}{3 / 2} \\
& \Rightarrow \tan \left(\frac{x+y}{2}\right)=\frac{1}{2} \\
& \therefore \sin (x+y)=\frac{2 \tan \left(\frac{x+y}{2}\right)}{1+\tan ^2\left(\frac{x+y}{2}\right)} \\
& =\frac{2 \times \frac{1}{2}}{1+\left(\frac{1}{2}\right)^2}=\frac{4}{4+1}=\frac{4}{5} \\
&
\end{aligned}
$$

Topic: Mathematics, Trigonometric Equations, JEE Main
If $x$ and $y$ are digits such that $17 !=3556xy428096000,$ then $x+y$ equals

Options:
A 15
B 6
C 12
D 13

The correct answer is:
15

Solution:
Given, $171=3556 \times 4428096000$
Since, 17! is divisible by $9,$ so sum of the digits $(48+x+y)$ must be divisible by 9
So, $x+y$ can be 15 or 6 . Also, 17! is divisible by $11,$ so $|10+x-y \mid$ must be multiple of 11 or $0 .$ The only possibility is
$$
\begin{array}{l}
|x-y|=1 \\
\therefore \quad x+y=15
\end{array}
$$

Topic: Mathematics, Permutation Combination, JEE Main
If $X$ and $Y$ are the matrices of order $2 \times 2$ each and
$2 X-3 Y=\left[\begin{array}{cc}-7 & 0 \\ 7 & -13\end{array}\right]$ and $3 X+2 Y=\left[\begin{array}{ll}9 & 13 \\ 4 & 13\end{array}\right]$, then what is $Y$ equal to?

Options:
A $\left[\begin{array}{cc}1 & 3 \\ -2 & 1\end{array}\right]$
B $\left[\begin{array}{ll}1 & 3 \\ 2 & 1\end{array}\right]$
C $\left[\begin{array}{cc}3 & 2 \\ -1 & 5\end{array}\right]$
D $\left[\begin{array}{cc}3 & 2 \\ 1 & -5\end{array}\right]$

The correct answer is:
$\left[\begin{array}{cc}3 & 2 \\ -1 & 5\end{array}\right]$

Solution:
Let $X$ and $Y$ be two matrices of order $2 \times 2$ each.
Given, $2 X-3 Y=\left[\begin{array}{cc}-7 & 0 \\ 7 & -13\end{array}\right]$ ...(i)
and $3 X+2 Y=\left[\begin{array}{ll}9 & 13 \\ 4 & 13\end{array}\right]$ ...(ii)
On multiplying Eq. (i) by 3 and Eq. (ii) by 2, we get
$6 X-9 Y=\left[\begin{array}{cc}-21 & 0 \\ 21 & -39\end{array}\right]$ ...(iii)
$6 X+4 Y=\left[\begin{array}{cc}18 & 26 \\ 8 & 26\end{array}\right]$ ...(iv)
On subtracting Eqs. (iii) from (iv), we get
$13 Y=\left[\begin{array}{cc}39 & 26 \\ -13 & 65\end{array}\right] \Rightarrow Y=\left[\begin{array}{cc}3 & 2 \\ -1 & 5\end{array}\right]$

Topic: Mathematics, Matrices, JEE Main
If x and y are the solutions of the equation 5+8sin2x=2y2-8y+21, then the least possible value of x2y3 is

Options:
A 2π2
B 4π2
C 9π2
D π2

The correct answer is:
2π2

Solution:
∵5+8sin2x∈5,13
Also, 2y2-8y+21=2y-22+13≥13
So, equality should hold true if,
5+8sin2x=13 and 2y2-8y+21=13
⇒sinx=±1, y=2
⇒sinx=2n+1π2, y=2
For least value, x=±π2
Hence, x2y3=π24·8=2π2

Topic: Mathematics, Trigonometric Equations, JEE Main
If x and y are the solutions of the equation 12sinâ¡x+5cosâ¡x=2y2-8y+21, then the value of 12cotxy2 is (Given, x<Ï)

Options:

The correct answer is:
5

Solution:
LHS=12sin⁡x+5cos⁡x∈-122+52,122+52
i.e. -13,13 i.e.
maximum value of LHS is 13
RHS=2y2-4y+4+13
=2y-22+13
RHS≥13
Roots of the equation exist if LHS=RHS=13.

RHS=13 when y=2
LHS=13⇒12sin⁡x+5cos⁡x=13
⇒1213sin⁡x+513cos⁡x=1
sinx+α=1, where tan⁡α=512
x+tan-1512=π2⇒x=π2-tan-1512
⇒xy=π-2tan-1512⇒512=tanπ2-xy2=cotxy2
⇒12cotxy2=5

Topic: Mathematics, Trigonometric Equations, JEE Main
If x and y are two distinct integers and n is a natural number then xn-yn is divisible by

Options:
A x2-y2
B x+y
C x-y
D None of these

The correct answer is:
x-y

Solution:
Pn:xn-yn where nâN
Then, P1 :x-y is divisible x-y
P2:x2-y2 is divisible x-y
P3:x3-y3 is divisible x-y
Hence xn-yn is divisible x-y

Topic: Mathematics, Mathematical Induction, JEE Main
If xâ and yâ are two non zero, non-collinear vectors satisfying a-3Î±2+b-4Î±+c-1xâ+a-3Î²2+b-4Î²+c-1yâ+a-3Î³2+b-4Î³+c-1xâÃyâ=0 (where Î±,Î²,Î³ are three distinct numbers), then the value of a2+b2+c24 is equal to ______. If your answer is K, mark 2K as answer after rounding off to nearest integer.

Options:

The correct answer is:
13

Solution:
Since, xâ, yâ and xâÃyâ are 3 non-coplanar vectors
Hence, a-3Î±2+b-4Î±+c-1=0
a-3Î²2+b-4Î²+c-1=0
a-3Î³2+b-4Î³+c-1=0
We can say that a-3x2+b-4x+c-1=0 has roots Î±,Î²,Î³
Since, quadratic has 3 roots so it becomes an identity
âa-3=0, b-4=0, c-1=0
âa=3, b=4, c=1
âa2+b2+c24=264=6.5

Topic: Mathematics, Vector Algebra, JEE Main
If xâ and yâ are two non zero, non-collinear vectors satisfying a-3Î±2+b-4Î±+c-1xâ+a-3Î²2+b-4Î²+c-1yâ+a-3Î³2+b-4Î³+c-1xâÃyâ=0 (where Î±,Î²,Î³ are three distinct numbers), then the value of a2+b2+c24 is equal to

Options:

The correct answer is:
6.5

Solution:
Since, xâ, yâ and xâÃyâ are 3 non-coplanar vectors
Hence, a-3Î±2+b-4Î±+c-1=0
a-3Î²2+b-4Î²+c-1=0
a-3Î³2+b-4Î³+c-1=0
We can say that a-3x2+b-4x+c-1=0 has roots Î±,Î²,Î³
Since, quadratic has 3 roots so it becomes an identity
âa-3=0, b-4=0, c-1=0
âa=3, b=4, c=1
âa2+b2+c24=264=6.5

Topic: Mathematics, Vector Algebra, JEE Main
If xâ and yâ be two non-collinear vectors and a triangle ABC with side lengths a,b,c satisfy the vector equation
20a-15bxâ+15b-12cyâ+12c-20axâÃyâ=0â,
then âABC is

Options:
A an acute angle triangle
B an obtuse angle triangle
C a right angle triangle
D an isosceles triangle.

The correct answer is:
a right angle triangle

Solution:
∵x→,y→ and x→×y→ are non coplanar vectors
20a-15b=0,15b-12c=0,12c-20a=0
⇒a3=b4=c5
⇒∆ABC is a right angle triangle

Topic: Mathematics, Vector Algebra, JEE Main
If x→ and y→ be two non-zero vectors such that x→+y→=x→ and 2x→+λy→ is perpendicular to y→, then the value of λ is ...... .

Options:

The correct answer is:
1

Solution:
x→+y→2=x→2

⇒ y→2+2x→.y→=0 .....(1)

and 2x→+λy→.y→=0s

⇒ λy→2+2x→.y→=0 .....(2)

by (1) and (2) λ=1

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2020 (06 Sep Shift 2)
If $x$ and $y$ represent the number of arrangements of the letters of word ATRAPATRAM such that (i) all A's are together and (ii) no two A's are together respectively, then $x+y$

Options:
A $\frac{10 !}{4 ! 2 ! 2 !}$
B $\frac{7 ! \times 15}{2 ! 2 ! 4 !}$
C $\frac{6 !}{2 ! 2 !} \times 42$
D $\frac{7 !}{2 ! 2 !}+\frac{6 ! \cdot 7 p_4}{2 ! 2 !}$

The correct answer is:
$\frac{6 !}{2 ! 2 !} \times 42$

Solution:
Given, word ATRAPATRAM
Here, $\mathrm{A} \rightarrow 4$ times
$\mathrm{T} \rightarrow 2$ times
$\mathrm{R} \rightarrow 2$ times
$\mathrm{P} \rightarrow 1$ time
$M \rightarrow 1$ time
(i) All A's are together,
Then it will be consider as one letter and remaining 6 letters and l'A's (including 4 A's) will be consider as 7 letters.
$\therefore \text { Number of arrangement }=\frac{7 !}{2 ! 2 !}=x$ $\ldots(\mathrm{i})$
(ii) no two $\mathrm{A}^{\prime}$ s are together.
such that $-\mathrm{T}-\mathrm{R}-\mathrm{P}-\mathrm{T}-\mathrm{R}-\mathrm{M}-$
$\therefore$ Number of arrangement $={ }^7 C_4 \times \frac{6 !}{2 ! 2 !}=y$ $\ldots(\mathrm{ii})$
$\therefore \quad x+y=\frac{7 !}{2 ! 2 !}+{ }^7 C_4 \times \frac{6 !}{2 ! 2 !}$
$=\frac{7 \times 6 !}{2 ! 2 !}+{ }^7 C_3 \times \frac{6 !}{2 ! 2 !}=\frac{6 !}{2 ! 2 !}\left(7+{ }^7 C_3\right)$
$=\frac{6 !}{2 ! 2 !}(7+35)=\frac{6 !}{2 ! 2 !}(42)$

Topic: Mathematics, Permutation Combination, TS EAMCET, TS EAMCET 2020 (10 Sep Shift 2)
If $x=\cos 10^{\circ} \cos 20^{\circ} \cos 40^{\circ}$, then the value of $x$ is

Options:
A $\frac{1}{4} \tan 10^{\circ}$
B $\frac{1}{8} \cot 10^{\circ}$
C $\frac{1}{8} \operatorname{cosec} 0^{\circ}$
D $\frac{1}{8} \sec 10^{\circ}$

The correct answer is:
$\frac{1}{8} \cot 10^{\circ}$

Solution:
$\begin{aligned}
& x=\cos 10^{\circ} \cos 20^{\circ} \cos 40^{\circ} \\
&=\frac{1}{2 \sin 10^{\circ}}\left[2 \sin 10^{\circ} \cos 10^{\circ} \cos 20^{\circ} \cos 40^{\circ}\right] \\
&=\frac{1}{2.2 \sin 10^{\circ}}\left[2 \sin 20^{\circ} \cos 20^{\circ} \cos 40^{\circ}\right] \\
&=\frac{1}{2.4 \sin 10^{\circ}}\left[2 \sin 40^{\circ} \cos 40^{\circ}\right)=\frac{1}{8 \sin 10^{\circ}}\left(\sin 80^{\circ}\right) \\
&=\frac{1}{8 \sin 10^{\circ}} \cos 10^{\circ}=\frac{1}{8} \cot 10^{\circ}
\end{aligned}$

Topic: Mathematics, Trigonometric Ratios & Identities, JEE Main
If $x \cos \theta+y \sin \theta=2$ is perpendicular to the line $x-y=3$, then what is one of the value of $\theta ?$

Options:
A $\pi / 6$
B $\pi / 4$
C $\pi / 2$
D $\pi / 3$

The correct answer is:
$\pi / 4$

Solution:
Consider a line
$x \cos \theta+y \sin \theta=2$
$\Rightarrow y \sin \theta=-x \cos \theta+2$
$\Rightarrow y=-x \frac{\cos \theta}{\sin \theta}+\frac{2}{\sin \theta}$
$\Rightarrow \mathrm{y}=-x \cot \theta+2 \operatorname{cosec} \theta$
On comparing this equation with $y=m x+c$ we get slope of line $x \cos \theta+y \sin \theta=2$ is $-\cot \theta$
Also, we have a line $x-y=3$ $\Rightarrow y=x-3$
slope of line $x-y=3$ is 1 . Since, both the lines are perpendicular to each other Product of their slopes $=-1$ $\Rightarrow(-\cot \theta)(1)=-1$
$\Rightarrow \cot \theta=1=\cot \frac{\pi}{4}$
$\Rightarrow \theta=\frac{\pi}{4}$

Topic: Mathematics, Application of Derivatives, JEE Main
If $x \frac{d y}{d x}=y(\log y-\log x+1)$, then the solution of the equation is

Options:
A
$y \log \left(\frac{x}{y}\right)=c x$

B
$x \log \left(\frac{y}{x}\right)=c y$

C
$\log \left(\frac{y}{x}\right)=c x$

D
$\log \left(\frac{x}{y}\right)=c y$

The correct answer is:
$\log \left(\frac{y}{x}\right)=c x$

Solution:
$$
\begin{aligned}
& \frac{x d y}{d x}=y(\log y-\log x+1) \\
& \frac{d y}{d x}=\frac{y}{x}\left(\log \left(\frac{y}{x}\right)+1\right)
\end{aligned}
$$
Put $y=v x$
$\frac{d y}{d x}=v+\frac{x d v}{d x}$
$\Rightarrow v+\frac{x d v}{d x}=v(\log v+1)$
$\frac{x d v}{d x}=v \log v$
$\Rightarrow \frac{d v}{v \log v}=\frac{d x}{x}$
put $\log v=z$
$\frac{1}{v} d v=d z$
$\Rightarrow \frac{d z}{z}=\frac{d x}{x}$
$\ln z=\ln x+\ln c$
$z=c x$
$\log v=c x$
$\log \left(\frac{y}{x}\right)=c x$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2005
If $x \frac{d y}{d x}+y=x \frac{f(x y)}{f^{\prime}(x y)}$, then $|f(x y)|$ is equal to

Options:
A $\mathrm{Ce}^{\mathrm{x}^2 / 2}$
B $\mathrm{Ce}^{\mathrm{x}^2}$
C $\mathrm{Ce}^{2 x^2}$
D $\mathrm{Ce}^{\mathrm{x}^2 / 3}$

The correct answer is:
$\mathrm{Ce}^{\mathrm{x}^2 / 2}$

Solution:
$x \frac{d y}{d x}+y=x \frac{f(x y)}{f^{\prime}(x y)} \Rightarrow \frac{d(x y)}{d x}=x \frac{f(x y)}{f^{\prime}(x y)} \Rightarrow \frac{f^{\prime}(x y)}{f(x y)} d(x y)=x d x \Rightarrow \ln |f(x y)|=\frac{x^2}{2}+k \Rightarrow f(x y)=C e^{\frac{x^2}{2}}$

Topic: Mathematics, Differential Equations, JEE Main
If $x d y=y(d x+y d y), y \gt 0$ and $y(1)=1$, then $y(-3)$ is equal to

Options:
A 1
B 3
C 5
D -1

The correct answer is:
3

Solution:
$x d y=y(d x+y d y) \Rightarrow \frac{x d y-y d x}{y^2}=d y \Rightarrow-d\left(\frac{x}{y}\right)=d y$
Integrating both sides, we get $\frac{x}{y}+y=c$
$\because y(1)=1 \Rightarrow c=2 ; \therefore \frac{x}{y}+y=2$
For $x=-3$,
$y^2-2 y-3=0 \Rightarrow y=-1$ or $3 \Rightarrow y=3 \quad(\because y\gt0)$

Topic: Mathematics, Differential Equations, JEE Main
If \( x d y=y d x+y^{2} d y \) and \( y(1)=1 \) then find the value of \( y(-3) \)

Options:

The correct answer is:
3.00

Solution:
Given that

x dy=y dx+y2dy

∫ydx-xdyy2=∫-dy

We know that dxy=ydx-xdyy2

xy=-y+c

As, y1=1

1=-1+c

⇒c=2

⇒xy=-y+2

If x=-3

-3=-y2+2y

⇒y2-2y-3=0

⇒y=3

Topic: Mathematics, Differential Equations, JEE Main
If \( \{x\} \) denotes the fractional part of \( x \) then \( \left\{\frac{3^{2 n}}{8}\right\}, n \in N \) is

Options:
A \( \frac{3}{8} \)
B \( \frac{7}{8} \)
C \( \frac{5}{8} \)
D \( \frac{1}{8} \)

The correct answer is:
\( \frac{1}{8} \)

Solution:
32n = 9n= ( 1 + 8)n
⇒32n=nC0 +nC1·8+nC2·82+.....+nCn·8n

⇒32n8=nC0 +nC1·8+nC2·82+.....+nCn·8n8
⇒32n8=18+nC1 +nC2·8 +......+ nCn·8n - 1
⇒32n8=18 + integer quantity

x∈[0, 1) & I=0
⇒ 32n8=18+0

Topic: Mathematics, Binomial Theorem, JEE Main
If \([x]\) denotes the greatest integer function on \(x\), then the number of positive integral divisors of \(\left[(2+\sqrt{3})^5\right]\) is

Options:
A 6
B 4
C 2
D 8

The correct answer is:
4

Solution:
\(\begin{aligned}
& (2+\sqrt{3})^5={ }^5 C_0 \cdot 2^5 \cdot(\sqrt{3})^0+{ }^5 C_1 \cdot 2^4 \cdot(\sqrt{3}) \\
& { }^5 C_2 \cdot 2^3(\sqrt{3})^2+{ }^5 C_3 \cdot 2^2 \cdot(\sqrt{3})^3 \\
& +{ }^5 C_4 \cdot 2 \cdot(\sqrt{3})^4+{ }^5 C_5 \cdot 2^0(\sqrt{3})^5 \\
& =32+80 \sqrt{3}+240+120 \sqrt{3}+90+9 \sqrt{3} \\
& (2+\sqrt{3})^5=362+209 \sqrt{3}=723.99 \\
& \Rightarrow\left[(2+\sqrt{3})^5\right]=723 \Rightarrow 723=3^1 \times 241^1 \\
\end{aligned}\)
Now, positive integral divisors of
\(723=(1+1)(1+1)=4\)

Topic: Mathematics, Functions, JEE Main
If $[x]$ denotes the greatest integer function, then the domain of the function $f(x)=\sqrt{\frac{x-[x]}{\log \left(x^2-x\right)}}$, is

Options:
A $(1, \infty)$
B $(1, \infty)-Z$
C $R-\left[\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right]$
D $\left[\frac{1-\sqrt{5}}{2}, \frac{\sqrt{5}+1}{2}\right]$

The correct answer is:
$R-\left[\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right]$

Solution:
We have,
$$
f(x)=\sqrt{\frac{x-[x]}{\log \left(x^2-x\right)}}
$$
It is defined
$$
\log \left(x^2-x\right)>0 \quad[\because x-[x]=[x] \geq 0, \forall x \in R]
$$
$$
\begin{aligned}
& \Rightarrow x^2-x>1 \Rightarrow x^2-x-1>0 \\
& \because \quad x^2-x-1=0 \Rightarrow x=\frac{1 \pm \sqrt{5}}{2} \\
& \therefore \quad \quad x \in\left(-\infty, \frac{1-\sqrt{5}}{2}\right) \cup\left(\frac{1+\sqrt{5}}{2}, \infty\right) \\
& \therefore \text { Domain of } f(x)=R-\left[\frac{1-\sqrt{5}}{2}, \frac{1+\sqrt{5}}{2}\right]
\end{aligned}
$$

Topic: Mathematics, Functions, TS EAMCET, TS EAMCET 2019 (03 May Shift 1)
If [x] denotes the greatest integer less than or equal to x, then the value of the integral ∫-π/2π/2x-sinxdx is equal to:

Options:
A -π
B π
C 0
D 1

The correct answer is:
-π

Solution:
We have,

I=∫-π/2π/2x-sinxdx …i

⇒I=∫-π2-1x-sinxdx+∫-10x-sinxdx+∫01x-sinxdx+∫1π2x-sinxdx

⇒I=∫-π/2-1-2-sinxdx+∫-10-1-sinxdx+∫010-sinxdx+∫1π21-sinxdx

⇒I=-2∫-π/2-1dx-∫-10dx-1∫01dx+∫1π20·dx

⇒I=-2-1+π2-1-1

⇒I=-π

Topic: Mathematics, Definite Integration, JEE Main, JEE Main 2021 (20 Jul Shift 2)