datavorous/entrance-exam-dataset huggingface dataset:
If x=2nπ+tan-1pq and y=r is a solution of the equation 12sin⁡x+5cos⁡x=2y2-8y+21, then the value of k, such that p2+q2+kr2=15, is equal to

Options:
A 5
B 14
C 314
D -22

The correct answer is:
14

Solution:
12sin⁡x+5cos⁡x=2y2-8y+21
12sin⁡x+5cos⁡x lies between -13,13
Also, 2y2-8y+21=2y2-4y+4+13=2y-22+13
So, possibility of equality is when LHS=RHS=13
RHS=13, when y=2⇒r=2.
12sin⁡x+5cos⁡x=13⇒sin⁡x1213+cos⁡x513=1
⇒sinx+cos-11213=1⇒x+cos-11213=4n+1π2
⇒x=2nπ+π2-cos-11213=2nπ+sin-11213
x=2nπ+tan-1125⇒p=12,q=5
⇒122+52+k⋅22=15⇒k=225-144-254=14

Topic: Mathematics, Trigonometric Equations, JEE Main
If x2=sinθcosθ0-cosθsinθ1sinθcosθ2 then the value of 4x2+xsin3π2+5 is

Options:
A 13-2
B 13+2
C 2-13
D Both (A) and (B)

The correct answer is:
Both (A) and (B)

Solution:
Given, x2=sinθcosθ0-cosθsinθ1sinθcosθ2

On expanding sinθcosθ0-cosθsinθ1sinθcosθ2 along C3, we get

x2=0-1sinθcosθsinθcosθ+2sinθcosθ-cosθsinθ=-1sinθcosθ-sinθcosθ+2sin2θ+cos2θ

=-1×0+2×1

⇒x2=2⇒x=±2

If x=2, then

4x2+xsin3π2+5=4×22-2+5

=8-2+5=13-2

If x=-2, then

4x2+xsin3π2+5=4×-22+2+5

=8+2+5=13+2

Topic: Mathematics, Determinants, JEE Main
If x=2sinθ-sin2θ and y=2cosθ-cos2θ, θ∈0,2π, then d2ydx2 at θ=π is:

Options:
A 34
B -38
C 32
D -34

The correct answer is:
-38

Solution:
dxdθ=2cosθ-2cos2θ

dydθ=-2sinθ+2sin2θ

∴ dydx=sin2θ-sinθcosθ-cos2θ

=2sinθ2.cos3θ22sinθ2.sin3θ2=cot3θ2

d2ydx2=ddθdydxdθdx=-32cosec23θ2.dθdx

⇒d2ydx2=-32cosec23θ22cosθ-cos2θ

⇒d2ydx2θ=π=34-1-1=38

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2020 (09 Jan Shift 2)
If $\pi \leq x \leq 2 \pi$, then $\cos ^{-1}(\cos x)$ is equal to

Options:
A $x$
B $-x$
C $2 \pi+x$
D $2 \pi-x$

The correct answer is:
$2 \pi-x$

Solution:
It is given that $\pi \leq x \leq 2 \pi$
$\begin{aligned}
& \Rightarrow-\pi \geq-x \geq-2 \pi \Rightarrow \pi \geq 2 \pi-x \geq 0 \\
& \Rightarrow \cos ^{-1}\{\cos (2 \pi-x)\}=2 \pi-x
\end{aligned}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If $x, 2 x+2,3 x+3$, are in G.P., then the fourth term is

Options:
A $27$
B $-27$
C $13.5$
D $-13.5$

The correct answer is:
$-13.5$

Solution:
Given that $x, 2 x+2,3 x+3$ are in G.P.
Therefore, $(2 x+2)^2=x(3 x+3) \Rightarrow x^2+5 x+4=0$
$\Rightarrow(x+4)(x+1)=0 \Rightarrow x=-1,-4$
Now first term $a=x$
Second term $a r=2(x+1)$ $\Rightarrow r=\frac{2(x+1)}{x}$
then $4^{\text {th }}$ term $=a r^3$ $=x\left[\frac{2(x+1)}{x}\right]^3=\frac{8}{x^2}(x+1)^3$
Putting $x=-4$
We get, $T_4=\frac{8}{16}(-3)^3=-\frac{27}{2}=-13.5$

Topic: Mathematics, Sequences and Series, JEE Main
If $\left|x^{2}-x-6\right|=x+2$, then the values of $x$ are

Options:
A $-2,2,-4$
B $-2,2,4$
C $3,2,-2$
D $4,4,3$

The correct answer is:
$-2,2,4$

Solution:
$\left|x^{2}-x-6\right|=x+2$, then
Case I : $x^{2}-x-6 < 0$
$\Rightarrow(x-3)(x+2) < 0$
$\Rightarrow-2 < x < 3$
In this case, the equation becomes $x^{2}-x-6=-x-2$
or $x^{2}-4=0$
$\therefore \mathrm{x}=\pm 2$
Clearly, $x=2$ satisfies the domain of the equation in this case. So, $x=2$ is a solution.
Case II : $x^{2}-x-6 \geq 0$
So, $x \leq-2$ or $x \geq 3$
In this case, the equation becomes
$x^{2}-x-6=0=x+2$
i.e., $x^{2}-2 x-8=0$ or $x=-2,4$
Both these values lie in the domain of the equation in this case, so $x=-2,4$ are the roots.
Hence, roots are $x=-2,2,4$.

Topic: Mathematics, Quadratic Equation, VITEEE, VITEEE 2013
If x2+xx+1x-22x2+3x-13x3x-3x2+2x+32x-12x-1=ax-12 , then a is equal to:

Options:
A -24
B 24
C -12
D 12

The correct answer is:
24

Solution:
Put x=1 on both the sides
22-1430611=a-12
⇒23-0-24-0-14-18=a-12
⇒6-8+14=a-12
⇒12=a-12
⇒24=a.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2015 (11 Apr Online)
If $x^{2}+y^{2}=1$, then $\frac{d^{2} x}{d y^{2}}=$

Options:
A $x^{3}$
B $y^{3}$
C $-\frac{1}{x^{3}}$
D $-y^{3}$

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

Solution:
We have $x^{2}+y^{2}=1$
$\begin{array}{l}
2 x+2 y \frac{d y}{d x}=0 \Rightarrow 2 y \frac{d y}{d x}=-2 x \\
\frac{d y}{d x}=-\frac{x}{y} \Rightarrow \frac{d x}{d y}=-\frac{y}{x}
\end{array}$
Differentiating w.r.t. $\mathrm{y}$ we get
$\frac{d^{2} x}{d y^{2}}=\left[\frac{x(1)-y \frac{d x}{d y}}{x^{2}}\right]=-\left[\frac{x-y\left(\frac{-y}{x}\right)}{x^{2}}\right]=-\left[\frac{x^{2}+y^{2}}{x^{3}}\right]$
$=-\frac{1}{x^{3}} \quad \cdots\left[\because x^{2}+y^{2}=1\right.$, given $]$

Topic: Mathematics, Differentiation, JEE Main
If $x^2+y^2=1$ then $\left(y^{\prime}=\frac{d y}{d x}, y^{\prime}=\frac{d^2 y}{d x^2}\right)$

Options:
A $y y^*-2\left(y^{\prime}\right)^2+1=0$
B $y y^{\prime \prime}+\left(y^{\prime}\right)^2+1=0$
C $y y^{\prime \prime}-\left(y^{\prime}\right)^2-1=0$
D $y y^{\prime \prime}+2\left(y^{\prime}\right)^2+1=0$

The correct answer is:
$y y^{\prime \prime}+\left(y^{\prime}\right)^2+1=0$

Solution:
Differentiating w.r.t. x, 2 x+2 y y^{\prime}=0$ or $x+y y^{\prime}=0$
Differentiating again w.r.t. $x, \quad 1+y^{\prime 2}+y y^{\prime \prime}=0$

Topic: Mathematics, Differential Equations, JEE Main
If x2-y2+2hxy+2gx+2fy+c=0 is the locus of a point, which moves such that it is always equidistant from the lines x+2y+7=0 and 2x-y+8=0, then the value of g+c+h-f equals

Options:
A 14
B 6
C 8
D 29

The correct answer is:
14

Solution:
Let P(x,y) be the point whose distance from x+2y+7=0 and 2x-y+8=0 is equal.

⇒x+2y+75=±2x-y+85

⇒x2+4y2+49+4xy+28y+14x=4x2+y2+64-4xy-16y+32x

⇒-3x2+3y2-15+8xy+44y-18x=0

⇒3x2-3y2-8xy+18x-44y+15=0

⇒x2-y2-83xy+6x-443y+5=0

Now, on comparing x2-y2+2hxy+2gx+2fy+c=0 with above equation we get,

⇒h=-43, g=3, f=-223, c=5

⇒g+c+h-f=3+5-43+223

⇒g+c+h-f=14

Topic: Mathematics, Straight Lines, JEE Main, JEE Main 2024 (30 Jan Shift 2)
If x2+y2=a2 and x216+y29=1 intersect at 4 points P,Q,R and S which form a square, then the area (in sq. units) of the square is

Options:
A 14425
B 254
C 152
D 57625

The correct answer is:
57625

Solution:
Let, P=acosâ¡Î¸,asinâ¡Î¸
Then, acosâ¡Î¸,asinâ¡Î¸ lies on x216+y29=1
Also, Î¸=45Â° because P,Q,R,S form a square
âa22Ã16+a22Ã9=1
âa2216+916Ã9=1âa2=2Ã14425
âa=1225âPQ=245

Hence, the area of the square=57625

Topic: Mathematics, Ellipse, JEE Main
If $x^2+y^2+p x+3 y-5=0$ and $x^2+y^2+5 x \quad+p y+7=0$ cut orthogonally, then $p$ is

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

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

Solution:
$2 g_1 g_2+2 f_1 f_2=c_1+c_2$ for orthogonal cut.
$\Rightarrow p\left(\frac{5}{2}\right)+p\left(\frac{3}{2}\right)=-5+7 \Rightarrow p=\frac{1}{2}$

Topic: Mathematics, Circle, JEE Main
If $x^{2} y^{2}=\sin ^{-1} \sqrt{x^{2}+y^{2}}+\cos ^{-1} \sqrt{x^{2}+y^{2}}$, then $\frac{d y}{d x}=$

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

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

Solution:
(D)
We know, $\sin ^{-1} \theta+\cos ^{-1} \theta=\frac{\pi}{2} \Rightarrow x^{2} y^{2}=\frac{\pi}{2}$ Differentiating w.r.t. $\mathrm{x}$
$\begin{array}{l}
x^{2} \cdot 2 y \frac{d y}{d x}+y^{2} \cdot 2 x=0 \Rightarrow x^{2} \cdot 2 y \frac{d y}{d x}=-y^{2} \cdot 2 x \\
\frac{d y}{d x}=\frac{-y^{2} \cdot 2 x}{x^{2} \cdot 2 y}=\frac{-y}{x}
\end{array}$

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If x2+y2+siny=4, then the value of d2ydx2 at the point -2, 0 is :

Options:
A -34
B 4
C -2
D -32

The correct answer is:
-34

Solution:
Given x2+y2+siny=4

Differentiating both sides with respect to x, we get

2x+2y+cosydydx=0

⇒dydx=-2x2y+cos⁡y

At -2, 0, dydx= 41=4

Also, 2y+cosydydx+2x=0

Again differentiating with respect to x, we get

2y+cosyd2ydx2+2-sinydydx2+2=0

At -2, 0, d2ydx2+2-042+2=0

⇒d2ydx2=-34

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2018 (15 Apr)
If $x^2+y^2+\sin y=4$, then the value of $\frac{d^2 y}{d x^2}$ at the point $(-2,0)$ is

Options:
A
$-34$

B
$-32$

C
$-2$

D
4

The correct answer is:
$-34$

Solution:
Given, $x^2+y^2+\sin y=4$
After differentiating the above equation $w$.
r. t. $x$ we get
$$
\begin{aligned}
&2 x+2 y \frac{d y}{d x}+\cos y \frac{d y}{d x}=0 \ldots(1) \\
&\Rightarrow 2 x+(2 y+\cos y) \frac{d y}{d x}=0 \\
&\Rightarrow \frac{d y}{d x}=\frac{-2 x}{2 y+\cos y} \\
&\text { At }(-2,0),\left(\frac{d y}{d x}\right)_{(-2,0)}=\frac{-2 \times-2}{2 \times 0+\cos 0} \\
&\Rightarrow\left(\frac{d y}{d x}\right)_{(-2,0)}=\frac{4}{0+1} \\
&\Rightarrow\left(\frac{d y}{d x}\right)_{(2,0)}=4 \ldots(2)
\end{aligned}
$$
Again differentiating equation (1) w. r. t to $x$, we get
$2+2\left(\frac{d y}{d x}\right)^2+2 y \frac{d^2 y}{d x^2}-\sin y\left(\frac{d y}{d x}\right)^2+\cos \mathrm{y} \frac{d^2 y}{d x^2}=0$
$\Rightarrow 2+(-2 \sin y)\left(\frac{d y}{d x}\right)^2+(2 y+\cos \mathrm{y}) \frac{d^2 y}{d x^2}=0$
$\Rightarrow(2 y+\cos \mathrm{y}) \frac{d^2 y}{d x^2}=-2-(2-\sin y)\left(\frac{d y}{d x}\right)^2$
$\Rightarrow \frac{d^2 y}{d x^2}=\frac{-2-(2-\sin y)\left(\frac{d y}{d x}\right)^2}{2 y+\cos \mathrm{y}}$
So, at $(-2,0)$
$\frac{d^2 y}{d x^2}=\frac{-2-(2-0) \times 4^2}{2 \times 0+1}$
$\Rightarrow \frac{d^2 y}{d x^2}=\frac{-2-2 \times 16}{1}$
$\Rightarrow \frac{d^2 y}{d x^2}=-34$

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2018 (15 Apr Shift 1 Online)
If $x^2+y^2=t-\frac{1}{t}, x^4+y^4=t^2+\frac{1}{t^2}$, then $x^3 y \frac{d y}{d x}=$

Options:
A 1
B 2
C 3
D 4

The correct answer is:
1

Solution:
$\begin{aligned} & x^4+y^4=\left(t-\frac{1}{t}\right)^2+2=\left(x^2+y^2\right)^2+2 \\ & \Rightarrow x^2 y^2=-1 \Rightarrow y^2=-\frac{1}{x^2}\end{aligned}$
Differentiating, we get $2 y \frac{d y}{d x}=\frac{2}{x^3}$ or $x^3 y \frac{d y}{d x}=1$

Topic: Mathematics, Differentiation, JEE Main
If $x-2 y+k=0$ is a tangent to the parabola $\mathrm{y}^2-4 \mathrm{x}-4 \mathrm{y}+8=0$, then the value of $\mathrm{K}$ is

Options:
A $2$
B $\frac{2}{5}$
C $7$
D $-7$

The correct answer is:
$7$

Solution:
$y=\mathrm{m} x+\mathrm{c} \Rightarrow x-2 y+k=0$
$$
\begin{aligned}
& m=\frac{1}{2}, c=\frac{k}{2} \\
& y^2-4 x-4 y+8=0 \\
& (y-2)^2=4(x-1)
\end{aligned}
$$
From (i), $x=2 y-k$
$$
\begin{aligned}
& (y-2)^2=4(2 y-k-1) \\
& y^2-12 y+(8-4 k)=0
\end{aligned}
$$
$\because$ line is tangent
$\therefore$ Above quadratic equation must have equal roots i.e. $\mathrm{D}=0$
$$
(12)^2-4(8-4 k)=0 \Rightarrow k=7
$$

Topic: Mathematics, Parabola, JEE Main
If $x^2 y-x^3 \frac{d y}{d x}=y^4 \cos x$, then $x^3 y$ is equal to

Options:
A $\sin x$
B $2 \sin x+c$
C $-3 \sin x+c$
D $3 \cos x+c$

The correct answer is:
$-3 \sin x+c$

Solution:
Given that
$x^2 y-x^3 \frac{d y}{d x}=y^4 \cos x$
On dividing by $y^4$, we get
$\begin{aligned} \frac{x^2}{y^3}-\frac{x^3}{y^4} \frac{d y}{d x} & =\cos x \\ \Rightarrow \quad \frac{x^3}{y^4} \frac{d y}{d x} & =\frac{x^2}{y^3}-\cos x\end{aligned}$
$\begin{array}{cc}\Rightarrow & \frac{1}{y^4} \frac{d y}{d x}-\frac{1}{x y^3}=-\frac{1}{x^3} \cos x \\ \text { Let } & -\frac{1}{y^3}=t\end{array}$
$\begin{array}{llrl}
\Rightarrow & \frac{1}{y^4} \frac{d y}{d x} & =\frac{1}{3} \cdot \frac{d t}{d x} \\
& \Rightarrow & \frac{1}{3} \cdot \frac{d t}{d x}+\frac{1 t}{x} & =\frac{1}{x^3} \cos x \\
& \Rightarrow & \frac{d t}{d x}+\frac{3}{x} t & =\frac{3}{x^3} \cos x
\end{array}$
This is a linear differential equation in $t$, on comparing with $\frac{d t}{d x}+P t=Q$, we get
$\begin{gathered}
P=\frac{3}{x}, Q=\frac{3}{x^3} \cos x \\
\therefore \quad \text { I.F. }=e^{\int P d x}=e^{\int \frac{3}{x} d x}=e^{\log x^3}=x^3
\end{gathered}$
$\therefore$ Complete solution is
$\begin{aligned}
& t x^3 & =3 \int \frac{x^3}{x^3} \cos x d x+c_1 \\
\Rightarrow & t x^3 & =3 \sin x+c_1 \\
\Rightarrow \quad & -\frac{1}{y^3} x^3 & =3 \sin x+c_1 \\
\Rightarrow \quad & y^{-3} x^3 & =-3 \sin x+c
\end{aligned}$

Topic: Mathematics, Differential Equations, JEE Main
If $x^3+8 x y+y^3=64$, then $\frac{d y}{d x}=$

Options:
A $-\frac{3 x^2+8 y}{8 x+3 y^2}$
B $\frac{3 x^2+8 y}{8 x+3 y^2}$
C $\frac{3 x+8 y^2}{8 x^2+3 y}$
D None of these

The correct answer is:
$-\frac{3 x^2+8 y}{8 x+3 y^2}$

Solution:
$\begin{aligned} & x^3+8 x y+y^3=64 \Rightarrow 3 x^2+8\left(y+x \frac{d y}{d x}\right)+3 y^2 \frac{d y}{d x}=0 \\ & \therefore \frac{d y}{d x}=-\frac{3 x^2+8 y}{8 x+3 y^2} .\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
If x=3cosâ¡t and y=5sinâ¡t, where t is a parameter, then 9d2ydx2 at t=-Ï6 is equal to

Options:

The correct answer is:
40

Solution:
dydx=dydtdxdt=5cos⁡t-3sin⁡t=-53cott
d2ydx2=-53⋅-cosec2t⋅dtdx=531sin2⁡t1-3sin⁡t
=598

Topic: Mathematics, Differentiation, JEE Main
If x3dy+xy·dx=x2dy+2ydx; y2=e and x>1, then y4 is equal to :

Options:
A e2
B 12+e
C 32e
D 32+e

The correct answer is:
32e

Solution:
Given x3dy+xy·dx=x2dy+2ydx

⇒x3-x2dy=2-xydx

⇒dyy=2-xx3-x2dx

Integrating both sides with respect to x, we get

∫dyy=∫2-xx2x-1dx+k ............i

Let 2-xx2x-1=Ax+Bx2+Cx-1

⇒2-x=Axx-1+Bx-1+Cx2

Putting x=0⇒2=-B⇒B=-2

Putting x=1⇒2-1=C⇒C=1

Putting x=2⇒2-2=A21+B1+C22⇒2A+2=0⇒A=-1

From equation i, we get

∫dyy=∫-1x+-2x2+1x-1dx+k

⇒lny=-lnx+2x+lnx-1+k ..........ii

Given y2=e i.e. at x=2, y=e

⇒lne=-ln2+22+ln2-1+k

⇒1=-ln2+1+0+k

⇒k=ln2

Putting in equation ii, we get

lny=-lnx+2x+lnx-1+ln2

Now putting x=4, we get

lny=-ln4+24+ln4-1+ln2

⇒lny=-2ln2+12+ln3+ln2

⇒lny=-ln2+12+ln3

⇒lny=12+ln32

⇒ln2y3=12

⇒2y3=e12

⇒y=32e

⇒y4=32e

Topic: Mathematics, Differential Equations, JEE Main
If x3dy+xy·dx=x2dy+2ydx; y2=e and x>1, then y4 is equal to :

Options:
A e2
B 12+e
C 32e
D 32+e

The correct answer is:
32e

Solution:
Given x3dy+xy·dx=x2dy+2ydx

⇒x3-x2dy=2-xydx

⇒dyy=2-xx3-x2dx

Integrating both sides with respect to x, we get

∫dyy=∫2-xx2x-1dx+k ............i

Let 2-xx2x-1=Ax+Bx2+Cx-1

⇒2-x=Axx-1+Bx-1+Cx2

Putting x=0⇒2=-B⇒B=-2

Putting x=1⇒2-1=C⇒C=1

Putting x=2⇒2-2=A21+B1+C22⇒2A+2=0⇒A=-1

From equation i, we get

∫dyy=∫-1x+-2x2+1x-1dx+k

⇒lny=-lnx+2x+lnx-1+k ..........ii

Given y2=e i.e. at x=2, y=e

⇒lne=-ln2+22+ln2-1+k

⇒1=-ln2+1+0+k

⇒k=ln2

Putting in equation ii, we get

lny=-lnx+2x+lnx-1+ln2

Now putting x=4, we get

lny=-ln4+24+ln4-1+ln2

⇒lny=-2ln2+12+ln3+ln2

⇒lny=-ln2+12+ln3

⇒lny=12+ln32

⇒ln2y3=12

⇒2y3=e12

⇒y=32e

⇒y4=32e

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2020 (03 Sep Shift 2)
If $\int x^3 e^{5 x} d x=\frac{e^{5 x}}{5^4}[f(x)]+C$, then $f(x)$ is equal to

Options:
A $\frac{x^3}{5}-\frac{3 x^2}{5^2}+\frac{6 x}{5^3}-\frac{6}{5^4}$
B $5 x^3-5^2 x^2+5^3 x-6$
C $5^2 x^3-15 x^2+30 x-6$
D $5^3 x^3-75 x^2+30 x-6$

The correct answer is:
$5^3 x^3-75 x^2+30 x-6$

Solution:
We have given that,
$\int x^3 e^{5 x} d x=\frac{e^{5 x}}{5^4}\left[\int(x)\right]+c$
By using the method of intergration by parts, we get
$\begin{aligned}
& \int x^3 e^{5 x} d x=x^3 \int e^{5 x} d x-\int\left\{\frac{d x^3}{d x} \int e^{5 x} d x\right\} d x \\
& =\frac{x^3 e^{5 x}}{5}-\int \frac{3 x^2 e^{5 x}}{5} d x+C
\end{aligned}$
By using the method of integration by parts, we get
$\begin{aligned}
& =\frac{x^3 e^{5 x}}{5}-\frac{3}{5} x^2 \int e^{5 x} d x \\
& +\frac{3}{5} \int\left\{\frac{d x^2}{d x} \int e^{5 x} d x\right\} d x+C \\
& =\frac{x^3 e^{5 x}}{5}-\frac{3 x^3 e^{5 x}}{25}+\frac{6}{25} \int x e^{5 x} d x \\
& =\frac{x^3 e^{5 x}}{5}-\frac{3 x^3 e^{5 x}}{25}+\frac{6 x}{25} \int e^{5 x} d x \\
& -\frac{6}{25} \int\left\{\frac{d x}{d x} \int e^{5 x} d x\right\} d x \\
& =\frac{x^3 e^{5 x}}{5}-\frac{3 x^3 e^{5 x}}{25}+\frac{6 x e^{5 x}}{125}-\frac{6}{125} \int e^{5 x} d x
\end{aligned}$
$\begin{aligned}
& =\frac{x^3 e^{5 x}}{5}-\frac{3 x^3 e^{5 x}}{25}+\frac{6 x e^{5 x}}{125}-\frac{6}{125}+C \\
& =\frac{e^{5 x}}{625}\left[125 x^3-75 x^2+30 x-6\right]+C
\end{aligned}$
Comparing the above equation with
$\begin{aligned}
& \frac{e^{5 x}}{5^4}[f(x)]+C, \text { we get } \\
& f(x)=5^3 x^3-75 x^2+30 x-6
\end{aligned}$

Topic: Mathematics, Indefinite Integration, JEE Main
If x→=3i^-6j^-k^, y→=i^+4j^-3k^ and z→=3i^-4j^-12k^, then the magnitude of the projection of x→×y→ on z→ is

Options:
A 14
B 12
C 15
D 10

The correct answer is:
14

Solution:
Given x→=3i^-6j^-k^, y→=i^+4j^-3k^ and z→=3i^-4j^-12k^.

⇒x→×y→=i^j^k^3-6-114-3

=i^18+4-j^-9+1+k^12+6=22i^+8j^+18k^=211i^+4j^+9k^

Now, we know that the projection of a vector a→ on b→ is defined as a→·b→b→

Hence, the magnitude of projection of x→×y→ on z→ is x→×y→·z→z→

=211i+4j+9k·3i^-4j^-12k^3i^-4j^-12k^

=233-16-10832+42+122

=2×91169=2×9113=14.

Topic: Mathematics, Vector Algebra, JEE Main, JEE Main 2014 (19 Apr Online)
If x=3 is the chord of contact of the circle x2+y2=81, then the equation of the corresponding pair of tangents, is

Options:
A x 2 - 8 y 2 + 5 4 x + 7 2 9 = 0
B x 2 - 8 y 2 - 5 4 x + 7 2 9 = 0
C x 2 - 8 y 2 - 5 4 x - 7 2 9 = 0
D x 2 - 8 y 2 = 7 2 9

The correct answer is:
x 2 - 8 y 2 - 5 4 x + 7 2 9 = 0

Solution:
Put x=3 in the equation of circle to get the point of contact, y=Â±6 2
Hence, equation of the pair of tangents is
3 x + 6 2 y - 8 1 3 x - 6 2 y - 8 1 = 0
x - 2 7 2 - 8 y 2 = 0
x 2 - 54 x - 8 y 2 + 7 2 9 = 0

Topic: Mathematics, Circle, JEE Main
If x=3 tant and y=3 sect, then the value of d2ydx2 at t=π4, is:

Options:
A 16
B 162
C 132
D 322

The correct answer is:
162

Solution:
Given x=3tant and y=3sect

⇒dxdt=3sec2t and dydt=3secttant

⇒dydx=dydtdxdt

⇒dydx=sint

⇒d2ydx2=cost·dtdx=cost ·cos2t3=cos3t3

At t=π4

d2ydx2=162

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2019 (09 Jan Shift 2)
If $\int \frac{3 x+1}{(x-3)(x-5)} d x=\int \frac{-5}{(x-3)} d x+\int \frac{B}{(x-5)} d x$ then the value of $B$ is

Options:
A 3
B 4
C 6
D 8

The correct answer is:
8

Solution:
We have,

$\begin{array}{l}

\frac{3 x+1}{(x-3)(x-5)}=\frac{-5}{x-3}+\frac{\mathrm{B}}{x-5} \\

3 x+1=-5(x-5)+\mathrm{B}(x-3) \\

\text { Put } x=5 \\

3(5)+1=\mathrm{B}(5-3) \\

16=2 \mathrm{~B} \text { or } \mathrm{B}=8

\end{array}$

Topic: Mathematics, Indefinite Integration, JEE Main
If $\int \frac{\sqrt{x}}{\sqrt{x}-\sqrt[3]{x}} d x=x+E x^{5 / 6}+D x^{2 / 3}+C x^{1 / 2}+B x^{1 / 3}+A x^{1 / 6}$ $+\log (\sqrt[6]{x}-1)^6+K$, then $A+B+C+D+E=$

Options:
A $\frac{137}{10}$
B $\frac{129}{10}$
C $\frac{119}{10}$
D $\frac{117}{10}$

The correct answer is:
$\frac{137}{10}$

Solution:
Let $I=\int \frac{\sqrt{x}}{\sqrt{x}-\sqrt[3]{x}} d x$
Put $x=t^6 \Rightarrow d x=6 t^5 d t$
$$
\begin{aligned}
& \therefore I=\int \frac{t^3 \cdot 6 t^5}{t^3-t^2} d t \\
& =6 \int \frac{t^8}{t^2(t-1)} d t=6 \int \frac{t^6}{t-1} d t \\
& =6 \int \frac{\left(t^6-1\right)+1}{t-1} d t=6 \int \frac{\left(t^3-1\right)\left(t^3+1\right)+1}{t-1} d t \\
& =6 \int \frac{(t-1)\left(t^2+t+1\right)\left(t^3+1\right)+1}{t-1} d t \\
& =6 \int\left[t^5+t^2+t^4+t+t^3+1+\frac{1}{t-1}\right] d t \\
& =6 \int\left[t^5+t^4+t^3+t^2+t+1+\frac{1}{t-1}\right] d t \\
& =6\left[\frac{t^6}{6}+\frac{t^5}{5}+\frac{t^4}{4}+\frac{t^3}{3}+\frac{t^2}{2}+t+\log (t-1)\right]+k \\
& =x+\frac{6}{5} x^{5 / 6}+\frac{3}{2} x^{2 / 3}+2 x^{3 / 2}+3 x^{3 / 3} \\
& +6 x^{1 / 6}+\operatorname{Iog}(\sqrt[6]{x-1})^6+k \\
&
\end{aligned}
$$
$$
\begin{aligned}
& \therefore A=6, B=3, C=2 D=\frac{3}{2} B=\frac{6}{5} \\
& \therefore A+B+C+D+B=6+3+2+\frac{3}{2}+\frac{6}{5} \\
& =11+\frac{3}{2}+\frac{6}{5}=\frac{110+15+12}{10}=\frac{137}{10}
\end{aligned}
$$

Topic: Mathematics, Indefinite Integration, JEE Main
If $\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x=A(\mathrm{x})\left(\sqrt{1-x^{2}}\right)^{m}+C,$ for a suitable chosen integer $\mathrm{m}$ and a function $\mathrm{A}(\mathrm{x})$, where $\mathrm{C}$ is a constant of integration, then $(\mathrm{A}(\mathrm{x}))^{\mathrm{m}}$ equals :

Options:
A $\frac{-1}{27 x^{9}}$
B $\frac{-1}{3 x^{3}}$
C $\frac{1}{27 x^{6}}$
D $\frac{1}{9 x^{4}}$

The correct answer is:
$\frac{-1}{27 x^{9}}$

Solution:
$A(x)\left(\sqrt{1-x^{2}}\right)^{m}+C=\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x$

$=\int \frac{\sqrt{\frac{1}{x^{2}}-1}}{x^{3}} d x$

Let $\frac{1}{x^{2}}-1=u^{2}$

$\Rightarrow-\frac{2}{x^{3}}=\frac{2 u d u}{d x}$

$\frac{d x}{x^{3}}=-u d u$

$A(x)\left(\sqrt{1-x^{2}}\right)^{m}+C=\int\left(-u^{2}\right) d u=-\frac{u^{3}}{3}+C$

$\begin{array}{l}

=-\frac{1}{3}\left(\frac{1}{x^{2}}-1\right)^{\frac{3}{2}}+C \\

=-\frac{1}{3} \cdot \frac{1}{x^{3}} \cdot\left(1-x^{2}\right)^{\frac{3}{2}}+C \\

=\frac{-1}{3 x^{3}}\left(\sqrt{1-x^{2}}\right)^{3}+C

\end{array}$

Compare both sides,

$\begin{array}{l}

\Rightarrow A(x)=-\frac{1}{3 x^{3}} \text { and } m=3 \\

\Rightarrow(A(x))^{3}=\frac{-1}{27 x^{9}}

\end{array}$

Topic: Mathematics, Indefinite Integration, JEE Main
If $\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x=A(\mathrm{x})\left(\sqrt{1-x^{2}}\right)^{m}+C,$ for a suitable chosen integer $\mathrm{m}$ and a function $\mathrm{A}(\mathrm{x})$, where $\mathrm{C}$ is a constant of integration, then $(\mathrm{A}(\mathrm{x}))^{\mathrm{m}}$ equals :

Options:
A $\frac{-1}{27 x^{9}}$
B $\frac{-1}{3 x^{3}}$
C $\frac{1}{27 x^{6}}$
D $\frac{1}{9 x^{4}}$

The correct answer is:
$\frac{-1}{27 x^{9}}$

Solution:
$A(x)\left(\sqrt{1-x^{2}}\right)^{m}+C=\int \frac{\sqrt{1-x^{2}}}{x^{4}} d x$

$=\int \frac{\sqrt{\frac{1}{x^{2}}-1}}{x^{3}} d x$

Let $\frac{1}{x^{2}}-1=u^{2}$

$\Rightarrow-\frac{2}{x^{3}}=\frac{2 u d u}{d x}$

$\frac{d x}{x^{3}}=-u d u$

$A(x)\left(\sqrt{1-x^{2}}\right)^{m}+C=\int\left(-u^{2}\right) d u=-\frac{u^{3}}{3}+C$

$\begin{array}{l}

=-\frac{1}{3}\left(\frac{1}{x^{2}}-1\right)^{\frac{3}{2}}+C \\

=-\frac{1}{3} \cdot \frac{1}{x^{3}} \cdot\left(1-x^{2}\right)^{\frac{3}{2}}+C \\

=\frac{-1}{3 x^{3}}\left(\sqrt{1-x^{2}}\right)^{3}+C

\end{array}$

Compare both sides,

$\begin{array}{l}

\Rightarrow A(x)=-\frac{1}{3 x^{3}} \text { and } m=3 \\

\Rightarrow(A(x))^{3}=\frac{-1}{27 x^{9}}

\end{array}$

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2019 (11 Jan Shift 1)
If x-42x2x2xx-42x2x2xx-4=A+Bx x-A2, then the ordered pair (A, B) is equal to

Options:
A 4, 5
B -4,-5
C -4,Â 3
D -4, 5

The correct answer is:
-4, 5

Solution:
x-42x2x2xx-42x2x2xx-4=A+Bx x-A2
Put x=0⇒-4000-4000-4=A3⇒A=-4
x-42x2x2xx-42x2x2xx-4=Bx-4x+42
1-4x2221-4x2221-4x=B-4x1+4x2
Put x→∞⇒122212221=B

On expanding the determinant along the first row, we get B=5.

Topic: Mathematics, Determinants, JEE Main, JEE Main 2018 (08 Apr)
If x-42x2x2xx-42x2x2xx-4=A+Bx x-A2, then the ordered pair (A, B) is equal to

Options:
A 4, 5
B -4,-5
C -4,Â 3
D -4, 5

The correct answer is:
-4, 5

Solution:
x-42x2x2xx-42x2x2xx-4=A+Bx x-A2
Put x=0â-4000-4000-4=A3âA=-4
x-42x2x2xx-42x2x2xx-4=Bx-4x+42
1-4x2221-4x2221-4x=B-4x1+4x2
Put xâââ122212221=BâB=5

Topic: Mathematics, Determinants, JEE Main
If x4+2xi-3x2+yi=3-5i+1+2yi, then the number of ordered pairs x,y is/are equal to ∀x,y∈R and i2=-1

Options:
A 0
B 1
C 2
D 3

The correct answer is:
2

Solution:
x4-3x2-3-1=0 & 2x-y+5-2y=0
x2-4x2+1=0 & 2x-3y+5=0
⇒x=2 or -2 & 2x-3y+5=0
When x=2,y=3 and when x=-2,y=13
⇒x,y=2,3 & -2,13
⇒ Number of ordered pair of x,y is equal to 2

Topic: Mathematics, Complex Number, JEE Main
If X=4n-3n-1 : n∈N and Y=9n-1 : n∈N, where N is the set of natural numbers, then X∪Y is equal to

Options:
A X
B Y
C N
D Y-X

The correct answer is:
Y

Solution:
X=4n-3n-1:n∈N
4n-3n-1=3+1n-3n-1
=nC03o1n+nC1311n-1+.....+nCn3n-3n-1
=1+n3+nC232+.....+3n-3n-1
=nC232+nC333+.....+3n
=32nC2+nC331+nC432+.....+3n-2
=9k, where k is a natural number.
Y=9n-1 : n∈N
∴ X⊆Y
∴ X∪Y=Y

Topic: Mathematics, Sets and Relations, JEE Main, JEE Main 2014 (06 Apr)
If $x=-5+2 \sqrt{-4}$, then the value of $x^4+9 x^3+35 x^2-x+4$ is

Options:
A $80$
B $160$
C $-160$
D $-80$

The correct answer is:
$-160$

Solution:
$x=-5+2 \sqrt{-4}$
$\begin{aligned} & x+5=4 i \\ & (x+5)^2=(4 i)^2 \\ & x^2+10 x+25=-16 \\ & x^2+10 x+41=0\end{aligned}$
Let $x^4+9 x^3+35 x^2-x+4=\left(x^2+10 x+41\right)(p(x))+\mathrm{R}$
$\begin{aligned} & x^2+10 x+41=0 \\ & x^2+9 x^3+35 x^2-x+4=\mathrm{R}\end{aligned}$
i.e. Remainder when $x^4+9 x^3+35 x^2-x+4$ is divided by $x^2+10 x+41$
$R=-160$
$x^4+9 x^3+35 x^2-x+4=-160$

Topic: Mathematics, Complex Number, AP EAMCET, AP EAMCET 2022 (05 Jul Shift 2)
If ∫x5e-4x3dx=148e-4x3fx+C, where C is a constant of integration, then fx is equal to

Options:
A -4x3-1
B -2x3+1
C -2x3-1
D 4x3+1

The correct answer is:
-4x3-1

Solution:
We have, I=∫x5e-4x3dx
Let, -4x3=t⇒-12x2dx=dt
⇒I=∫-t4et-dt12=148∫tetdt
=148tet-∫etdt=t-1et48+C=-4x3-1e-4x348+C,
where C is the constant of integration.

Hence, fx=-4x3-1

Topic: Mathematics, Indefinite Integration, JEE Main, JEE Main 2019 (10 Jan Shift 2)