datavorous/entrance-exam-dataset huggingface dataset:
If y=m1x+c1 and y=m2x+c2, m1≠m2 are two common tangents of circle x2+y2=2 and parabola y2=x, then the value of 8 m1 m2 is equal to

Options:
A 32-4
B 62-4
C -5+62
D 3+42

The correct answer is:
32-4

Solution:
Equation of the tangent to the parabola can be written as y=mx+14m or y-mx-14m=0

Now, the perpendicular distance from 0,0 to the tangent is equal to the radius of the given circle, soOM=2⇒-14m1+m2=2

⇒216m21+m2=1 ⇒32m4+32m2-1=0

⇒ m2=-25+2316+22×25 ignoring the negative sign as it is square function.

Now 8m1m2=8m2

=-25+2316+223=32-4

Topic: Mathematics, Parabola, JEE Main, JEE Main 2022 (25 Jun Shift 1)
If y=mx+4 is a tangent to both the parabolas, y2=4x and x2=2by, then b is equal to

Options:
A -32
B -64
C -128
D 128

The correct answer is:
-128

Solution:
Equation of tangent to y2=4ax in slope Form is y=mx+am.

Comparing with given equation of tangent and parabola we get 1m=4⇒m=14.

Hence, line y=x4+4 is tangent to x2=2by also.

Solving both, we get x2=2bx4+4⇒2x2-bx-16b=0 ..(1).

Since the line is a tangent, so 1 must have equal roots.

⇒-b2-4·2·-16b=0

⇒b2+128b=0⇒b=0,-128

But b can't be zero.

So, b=-128

Topic: Mathematics, Parabola, JEE Main, JEE Main 2020 (07 Jan Shift 1)
If y=mx+5 is a tangent to x3y3=ax3+by3 at point 1,2, then the value of a is equal to

Options:
A 95
B 165
C 94
D 187

The correct answer is:
165

Solution:
P1,2 lies on the curve and line, hence
L→2=m+5⇒m=-3
Curve→8=a+8b
Also, dydx at P1,2 for the curve should be m=-3,
i.e. 3x2y3+x3.3y2y'=3ax2+3by2y'
⇒24+12-3=3a+12b-3
⇒-12=3a-36b
⇒a-12b=-4 & a+8b=8
-20b=-12
⇒b=35 and a=165

Topic: Mathematics, Application of Derivatives, JEE Main
If $y=m x+c$ is tangent on the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$, then the value of $c$ is

Options:
A 0
B $3 / m$
C $\pm \sqrt{9 m^2+4}$
D $\pm 3 \sqrt{1+m^2}$

The correct answer is:
$\pm \sqrt{9 m^2+4}$

Solution:
Here, $a=3, b=2 . \therefore$ By formula, $c^2=b^2+a^2 m^2$
$\therefore c^2=4+9 m^2 ; \therefore c= \pm \sqrt{9 m^2+4} \text {. }$

Topic: Mathematics, Ellipse, JEE Main
If y=mx+c is the normal at a point on the parabola y2=8x whose focal distance is 8 units, then c is equal to:

Options:
A 83
B 103
C 23
D 163

The correct answer is:
103

Solution:
Let the point on the parabola be P( a t 2 , 2at ) where a=2.
Given distance of this point from focus be 8 units
⇒ at2-a2+4a2t2=64
⇒ at2+1=8
⇒ t2+1=4 ⇒ t2=3
⇒ t= ±3
Since y=mx+c is normal to the parabola
⇒c=2at+a t 3
⇒c=103

Topic: Mathematics, Parabola, JEE Main, JEE Main 2017 (09 Apr Online)
If y=sin-12x1+x2, then dydx is

Options:
A -21+x2 for all x
B 21+x2 for all x<1
C 21+x2 for x>1
D None of these

The correct answer is:
21+x2 for all x<1

Solution:
dydx=11-4x21+x22ddx2x1+x2
âdydx=1+x21-x2221+x2-4x21+x22
âdydx=21+x21-x21-x21+x22
âdydx=21-x21-x211+x2
âdydx=-21+x2,if x>121+x2,if x<1

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

Options:
A 0
B 1
C 2
D 3

The correct answer is:
0

Solution:
$\begin{aligned} & y=\sec ^{-1}\left(\frac{x+1}{x-1}\right)+\sin ^{-1}\left(\frac{x-1}{x+1}\right) \\ & \text { or } y=\cos ^{-1} \frac{x-1}{x+1}+\sin ^{-1}\left(\frac{x-1}{x+1}\right) \\ & \therefore y=\frac{\pi}{2} \Rightarrow \frac{d y}{d x}=0 \quad\left(\because \sin ^{-1} x+\cos ^{-1} x=\frac{\pi}{2}\right)\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
If y=sectan-1x, then dydx at x=1 is equal to

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

The correct answer is:
1 2

Solution:
Given, y=sectan-1x

Differentiating both sides w.r.t. x, we get,

dydx=sectan-1xtantan-1x·11+x2

At x=1, we have, tan-1x=π4

dydx=secπ4×tanπ41+12=22=12

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2013 (07 Apr)
If $y \sec x+\tan x+x^2 y=0$, then $\frac{d y}{d x}=$

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

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

Solution:
$\begin{aligned} & y \sec x+\tan x+x^2 y=0 \\ & \Rightarrow \sec x \frac{d y}{d x}+y \sec x \tan x+\sec ^2 x+2 x y+x^2 \frac{d y}{d x}=0 \\ & \Rightarrow \frac{d y}{d x}=-\frac{2 x y+\sec ^2 x+y \sec x \tan x}{x^2+\sec x} .\end{aligned}$

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

Options:
A $\frac{4 x}{1-x^2} \cdot \cos \left(\frac{1+x^2}{1-x^2}\right)$
B $\frac{x}{\left(1-x^2\right)^2} \cdot \cos \left(\frac{1+x^2}{1-x^2}\right)$
C $\frac{x}{\left(1-x^2\right)} \cdot \cos \left(\frac{1+x^2}{1-x^2}\right)$
D $\frac{4 x}{\left(1-x^2\right)^2} \cdot \cos \left(\frac{1+x^2}{1-x^2}\right)$

The correct answer is:
$\frac{4 x}{\left(1-x^2\right)^2} \cdot \cos \left(\frac{1+x^2}{1-x^2}\right)$

Solution:
The given equation is $y=\sin \left(\frac{1+x^2}{1-x^2}\right)$.
Differentiate both sides of the equation with respect to $x$.
$\begin{aligned} & \therefore \frac{\mathrm{d}(y)}{\mathrm{d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\left[\sin \left(\frac{1+x^2}{1-x^2}\right)\right] \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos \left(\frac{1+x^2}{1-x^2}\right) \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left[\frac{1+x^2}{1-x^2}\right] \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos \left(\frac{1+x^2}{1-x^2}\right) .\frac{\left(1-x^2\right) \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left(1+x^2\right)-\left(1+x^2\right) \cdot \frac{\mathrm{d}}{\mathrm{d} x}\left(1-x^2\right)}{\left(1-x^2\right)^2}\end{aligned}$
$\begin{aligned} {\left[\because \frac{\mathrm{d}}{\mathrm{d x}}\left(\frac{\mathrm{u}}{\mathrm{v}}\right)=\frac{\mathrm{v} \cdot \frac{\mathrm{du}}{\mathrm{dx}}-\mathrm{u} \cdot \frac{\mathrm{dv}}{\mathrm{dx}}}{\mathrm{v}^2}\right]}\end{aligned}$
$\begin{aligned} & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos \left(\frac{1+x^2}{1-x^2}\right) \cdot \frac{\left(1-x^2\right)(2 x)-\left(1+x^2\right)(-2 x)}{\left(1-x^2\right)^2} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos \left(\frac{1+x^2}{1-x^2}\right) \cdot \frac{2 x-2 x^3+2 x+2 x^3}{\left(1-x^2\right)^2} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\cos \left(\frac{1+x^2}{1-x^2}\right) \cdot \frac{4 x}{\left(1-x^2\right)^2} \\ & \Rightarrow \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{4 x}{\left(1-x^2\right)^2} \cos \left(\frac{1+x^2}{1-x^2}\right) \\ & \text { Therefore, } \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{4 x}{\left(1-x^2\right)^2} \cos \left(\frac{1+x^2}{1-x^2}\right)\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
If $y=\sin p x$ and $y_n$ is the $n^{\text {th }}$ derivative of $y$, then $\left|\begin{array}{lll}y & y_1 & y_2 \\ y_3 & y_4 & y_5 \\ y_6 & y_7 & y_8\end{array}\right|$ is equal to

Options:
A 1
B 0
C -1
D None of these

The correct answer is:
0

Solution:
$\begin{aligned} D & =\left|\begin{array}{ccc}\sin p x & p \cos p x & -p^2 \sin p x \\ -p^3 \cos p x & p^4 \sin p x & p^5 \cos p x \\ -p^6 \sin p x & -p^7 \cos p x & p^8 \sin p x\end{array}\right| \\ & =p^9\left|\begin{array}{ccc}\sin p x & p \cos p x & -p^2 \sin p x \\ -\cos p x & p \sin p x & p^2 \cos p x \\ -\sin p x & -p \cos p x & p^2 \sin p x\end{array}\right| \\ & =-p^9\left|\begin{array}{ccc}\sin p x & p \cos p x & -p^2 \sin p x \\ \cos p x & p \sin p x & p^2 \cos p x \\ \sin p x & p \cos p x & -p^2 \sin p x\end{array}\right|=0 .\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
If y=sin(sinx) and y''+f(x)·y'+g(x)·y=0, then f(x)·g(x)=

Options:
A 12sin(2x)
B 12cos(2x)
C sin(2x)
D cos(2x)

The correct answer is:
12sin(2x)

Solution:
Since y=sin(sinx) ...i

Differentiating w.r.t. x by chain rule

y'=cossinx.cosx ...ii

Again differentiate w.r.t. x

⇒y''=-sinsinx×cos2x-sinx×cossinx

⇒y''=-sinsinx×cos2x-sinx×cossinx.cosxcosx

by equation i and ii

⇒y''=-cos2x.y-tanx.y'

⇒y''+tanx.y'+cos2x.y=0

Since we are given that y''+f(x)·y'+g(x)·y=0

then by comparison fx=tanx, gx=cos2x

Now fx.gx=tanx.cos2x

=sinx.cosx.22

Since 2sinA.cosA=sin2A

=sin2x2.

Topic: Mathematics, Differentiation, JEE Main
If $y=\sin (\sqrt{\sin x+\cos x})$, then $\frac{d y}{d x}=$

Options:
A $\frac{1}{2} \frac{\cos \sqrt{\sin x+\cos x}}{\sqrt{\sin x+\cos x}}$
B $\frac{\cos \sqrt{\sin x+\cos x}}{\sqrt{\sin x+\cos x}}$
C $\frac{1}{2} \frac{\cos \sqrt{\sin x+\cos x}}{\sqrt{\sin x+\cos x}} \cdot(\cos x-\sin x)$
D None of these

The correct answer is:
$\frac{1}{2} \frac{\cos \sqrt{\sin x+\cos x}}{\sqrt{\sin x+\cos x}} \cdot(\cos x-\sin x)$

Solution:
$\begin{aligned} & y=\sin (\sqrt{\sin x+\cos x}) \\ & \frac{d y}{d x}=\frac{1}{2} \frac{\cos (\sqrt{\sin x+\cos x})}{\sqrt{\sin x+\cos x}}(\cos x-\sin x)\end{aligned}$

Topic: Mathematics, Differentiation, JEE Main
If y=tan-111+x+x2+tan-11x2+3x+3+tan-11x2+5x+7+....+ upto 2n terms ∀x≥0, then y0 is

Options:
A tan-1n
B tan-12n
C 2tan-1n
D 0

The correct answer is:
tan-12n

Solution:
y=tan-111+x+x2+tan-11x2+3x+3+...+2n terms
=tan-1x+1-x1+x1+x+tan-1x+2-x+11+x+1x+2+......2n terms
=tan-1x+1-tan-1x+tan-1x+2-tan-1x+1+...+tan-1x+2n-tan-1x+2n-1
=tan-1x+2n-tan-1x
y0=tan-12n

Topic: Mathematics, Inverse Trigonometric Functions, JEE Main
If y=tan-11x2+x+1+tan-11x2+3x+3 +tan-11x2+5x+7+tan-11x2+7x+13, x>0 and dydxx=0=-k1+k, then the value of k is

Options:

The correct answer is:
16

Solution:
y=tan-11x2+x+1+tan-11x2+3x+3
+tan-1(1x2+5x+7)+tan-11x2+7x+13
=tan-11xx+1+1+tan-11x+2x+1+1+tan-11x+3x+2+1c+tan-11x+4x+3+1
=tan-1x+1-x1+x+1x+tan-1x+2-x+11+x+2x+1 +tan-1x+3-x+21+x+3x+2+tan-1x+4-x+31+x+4x+3
=tan-1x+1-tan-1x+tan-1x+2 -tan-1x+1+tan-1x+3-tan-1x+2+tan-1x+4-tan-1x+3
⇒y=tan-1x+4-tan-1x
⇒dydx=11+(x+4)2-11+x2
⇒dydxx=0=11+16-1
=-1617
=-161+16
⇒k=16

Topic: Mathematics, Differentiation, JEE Main
If y=tan-1secx-tanx, then the value of dydx is

Options:

The correct answer is:
-0.5

Solution:
y=tan-1secx-tanx
y=tan-11-sinxcosx
⇒y=tan-1cosx2-sinx22cos2x2-sin2x2
⇒y=tan-1cosx2-sinx2cosx2+sinx2
⇒y=tan-11-tanx21+tanx2
⇒y=tan-1tanπ4-x2
⇒y=π4-x2
Differentiating with respect to x, we get
⇒dydx=-12=-0.5

Topic: Mathematics, Differentiation, JEE Main
If y=tan-1x1+6x2+tan-12x-12x+1 âx>0, then dydx is equal to

Options:
A 31+9x2
B 11+6x2
C 11+6x2+11+x2
D 31+6x2

The correct answer is:
31+9x2

Solution:
y=tan-13x-2x1+3x⋅2x+tan-12x-11+2x⋅1
y=tan-13x-tan-12x+tan-12x-tan-11
y=tan-13x-π4⇒dydx=31+9x2

Topic: Mathematics, Differentiation, JEE Main
If $y=\tan ^{-1} \frac{4 x}{1+5 x^{2}}+\tan ^{-1} \frac{2+3 x}{3-2 x}$, then $\frac{d y}{d x}=$

Options:
A $\frac{1}{1+25 x^{2}}+\frac{2}{1+x^{2}}$
B $\frac{5}{1+25 x^{2}}+\frac{2}{1+x^{2}}$
C $\frac{5}{1+25 x^{2}}$
D $\frac{1}{1+25 x^{2}}$

The correct answer is:
$\frac{5}{1+25 x^{2}}$

Solution:
$$
\begin{aligned}
y &=\tan ^{-1} \frac{4 x}{1+5 x^{2}}+\tan ^{-1} \frac{2+3 x}{3-2 x} \\
&=\tan ^{-1} \frac{5 x-x}{1+5 x \cdot x}+\tan ^{-1} \frac{\frac{2}{3}+x}{1-\frac{2}{3} \cdot x} \\
&=\tan ^{-1} 5 x-\tan ^{-1} x+\tan ^{-1} \frac{2}{3}+\tan ^{-1} x \\
\Rightarrow & \frac{d y}{d x}=\frac{5}{1+25 x^{2}}
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If $y=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)$, then $\left(1+a^2 x^2\right){y^{\prime}}^{\prime}+2 a^2 x y^{\prime}$ is equal to

Options:
A $-2 a^2$
B $\frac{1}{(2 x+1)^2}$
$a^2$
C $2 a^2$
D $0$

The correct answer is:
$0$

Solution:
$$
\text { Given, } y=\tan ^{-1}\left(\frac{\sqrt{1+a^2 x^2}-1}{a x}\right)
$$
Put $a x=\tan \theta$
$$
\begin{aligned}
\therefore \quad y & =\tan ^{-1}\left(\frac{\sqrt{1+\tan ^2 \theta}-1}{\tan \theta}\right) \\
& =\tan ^{-1}\left(\frac{\sec \theta-1}{\tan \theta}\right) \\
& =\tan ^{-1}\left(\frac{1-\cos \theta}{\sin \theta}\right) \\
& =\tan ^{-1}\left(\frac{2 \sin ^2 \frac{\theta}{2}}{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right) \\
& =\tan ^{-1}\left(\tan ^{\frac{\theta}{2}}\right) \\
& =\frac{\theta}{2}=\frac{1}{2} \tan ^{-1} a x \\
\therefore \quad y & =\frac{1}{2} \tan ^{-1} a x
\end{aligned}
$$
On differentiating w.r.t. $x$, we get
$$
y^{\prime}=\frac{1}{2\left(1+a^2 x^2\right)}
$$
Again, differentiating both side we get
$$
\begin{aligned}
y^{\prime \prime} & =-\frac{1}{2} \frac{\left(a^2 2 x\right)}{\left(1+a^2 x^2\right)^2} \\
\Rightarrow\left(1+a^2 x^2\right) y^{\prime \prime} & =-\frac{a^2 x}{1+a^2 x^2} \\
& =-a^2 x\left(2 y^{\prime}\right) \quad[\because \text { from Eq. (i) }] \\
\Rightarrow\left(1+a^2 x^2\right) y^{\prime \prime} & +2 a^2 x y^{\prime}=0
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If y=tan-1secx3-tanx3,π2<x3<3π2, then

Options:
A xy''+2y'=0
B x2y''-6y+3π2=0
C x2y''-6y+3π=0
D xy''-4y'=0

The correct answer is:
x2y''-6y+3π2=0

Solution:
Given,

y=tan-1secx3-tanx3

=tan-11-sinx3cosx3

=tan-11-cosπ2-x3sinπ2-x3

=tan-1tanπ4-x32

Since π4-x32∈-π2,0 as π2<x3<3π2

So, y=π4-x32

Now differentiating we get,

y'=-3x22,y''=-3x

Now putting the value of x in term of y'' in 4y=π-2x3

We get,

4y=π-2x2-y''3

12y=3π+2x2y''

x2y''-6y+3π2=0

Topic: Mathematics, Differentiation, JEE Main
If y=tan-1secx3-tanx3,π2<x3<3π2, then

Options:
A xy''+2y'=0
B x2y''-6y+3π2=0
C x2y''-6y+3π=0
D xy''-4y'=0

The correct answer is:
x2y''-6y+3π2=0

Solution:
Given,

y=tan-1secx3-tanx3

=tan-11-sinx3cosx3

=tan-11-cosπ2-x3sinπ2-x3

=tan-1tanπ4-x32

Since π4-x32∈-π2,0 as π2<x3<3π2

So, y=π4-x32

Now differentiating we get,

y'=-3x22,y''=-3x

Now putting the value of x in term of y'' in 4y=π-2x3

We get,

4y=π-2x2-y''3

12y=3π+2x2y''

x2y''-6y+3π2=0

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2022 (24 Jun Shift 2)
If $y=\tan ^{-1} \sqrt{\mathrm{x}^{2}-1}$, then the ratio $\frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx} \mathrm{x}^{2}}: \frac{\mathrm{dy}}{\mathrm{dx}}$ is

Options:
A $\frac{x\left(x^{2}-1\right)}{1+2 x^{2}}$
B $\frac{1-2 x^{2}}{x\left(x^{2}-1\right)}$
C $\frac{1+2 x^{2}}{x\left(x^{2}+1\right)}$
D $\frac{x\left(x^{2}+1\right)}{1-2 x^{2}}$

The correct answer is:
$\frac{1-2 x^{2}}{x\left(x^{2}-1\right)}$

Solution:
$\mathrm{y}=\tan ^{-1} \sqrt{\mathrm{x}^{2}-1}$,
Put $\left\{\begin{array}{l}\mathrm{x}=\sec \theta \\ \mathrm{dx}=\sec \theta \cdot \tan \theta \mathrm{d} \theta\end{array}\right.$
$\mathrm{y}=\tan ^{-1} \sqrt{\sec ^{2} \theta-1}=\tan ^{-1}(\tan \theta)=\theta$
$=\sec ^{-1} \mathrm{x}$
$\Rightarrow \quad \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\sec ^{-1} \mathrm{x}\right)=\frac{1}{\mathrm{x} \sqrt{\mathrm{x}^{2}-1}}$
$\begin{aligned} \frac{\mathrm{d}^{2} \mathrm{y}}{\mathrm{dx}} &=\frac{1}{\mathrm{x}} \cdot \frac{-1}{2} \frac{1}{\left(\mathrm{x}^{2}-1\right)^{3 / 2}}(2 \mathrm{x})-\frac{1}{\mathrm{x}^{2}} \cdot \frac{1}{\sqrt{\mathrm{x}^{2}-1}} \\ \quad=-\frac{1}{\left(\mathrm{x}^{2}-1\right)^{3 / 2}}-\frac{1}{\mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)^{1 / 2}} \\=-\frac{1}{\mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)^{3 / 2}}\left(\mathrm{x}^{2}+\mathrm{x}^{2}-1\right) \\ \quad=-\frac{\left(2 \mathrm{x}^{2}-1\right)}{\mathrm{x}^{2}\left(\mathrm{x}^{2}-1\right)^{3 / 2}} \end{aligned}$
$$
\begin{aligned}
&\text { Now, } \frac{d^{2} y}{d x^{2}}: \frac{d y}{d x}=-\frac{\left(2 x^{2}-1\right)}{x^{2}\left(x^{2}-1\right)^{3 / 2}}: \frac{1}{x\left(x^{2}-1\right)^{1 / 2}} \\
&\frac{d^{2} y}{d x^{2}}: \frac{d y}{d x}=\left(1-2 x^{2}\right): x\left(x^{2}-1\right) \\
&\text { or } \quad\left(\frac{d^{2} y}{\frac{d x^{2}}{d y}}\right)=\frac{\left(1-2 x^{2}\right)}{x\left(x^{2}-1\right)}
\end{aligned}
$$

Topic: Mathematics, Differentiation, JEE Main
If y=tan⁡x-sin⁡x, then the value of dydx at x=5π4 is

Options:
A 22+12
B 22-12
C 2+12
D 2-12

The correct answer is:
22-12

Solution:
In the neighbourhood of x=5π4, sin⁡x<0
∴ sin⁡x=-sin⁡x
Now, y=tan⁡x-(-sin⁡x)=tan⁡x+sin⁡x
Now, in the neighbourhood of x=5π4
tan⁡x+sin⁡x≃1+-12, which is positive
∴ y=tan⁡x+sin⁡x
Hence, dydx=sec2⁡x+cos⁡x
Putting x=5π4, we get,

dydx=-22-12=22-12

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

Options:
A
$\mathrm{n}^2 \mathrm{y}$

B
$-n^2 y$

C
$-y$

D
$2 x^2 y$

The correct answer is:
$\mathrm{n}^2 \mathrm{y}$

Solution:

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2002
If y=x+1x2−xxx+x+x+1153cos2x−5cos3x, then 96y'π6 is equal to:

Options:

The correct answer is:
105

Solution:
Given:y=x+1x2-xxx+x+x+1153cos5x-5cos3x

⇒y=x+1x2-xx-1xx+x+xx-1+1153cos5x-5cos3x

⇒y=x+1x2-xx-1xx+x+1x-1+1153cos5x-5cos3x

⇒y=x+1x2-xx-1xx3-1+1153cos5x-5cos3x

⇒y=x+1xx3-1x-1xx3-1+1153cos5x-5cos3x

⇒y=x+1x-1+1153cos5x-5cos3x

⇒y=x-1+1153cos5x-5cos3x

⇒dydx=1+11515cos4x-sinx+15cos2xsinx

⇒dydx=1+cos4x-sinx+cos2xsinx

⇒y'π6=1+cos4π6-sinπ6+cos2π6sinπ6

⇒y'π6=1+3424-12+3222×12

⇒y'π6=1-932+38

⇒y'π6=32-9+1232

⇒y'π6=3532

⇒96y'π6=35×3

⇒96y'π6=105

Topic: Mathematics, Differentiation, JEE Main
If y=x+1x2−xxx+x+x+1153cos2x−5cos3x, then 96y'π6 is equal to:

Options:

The correct answer is:
105

Solution:
Given:y=x+1x2-xxx+x+x+1153cos5x-5cos3x

⇒y=x+1x2-xx-1xx+x+xx-1+1153cos5x-5cos3x

⇒y=x+1x2-xx-1xx+x+1x-1+1153cos5x-5cos3x

⇒y=x+1x2-xx-1xx3-1+1153cos5x-5cos3x

⇒y=x+1xx3-1x-1xx3-1+1153cos5x-5cos3x

⇒y=x+1x-1+1153cos5x-5cos3x

⇒y=x-1+1153cos5x-5cos3x

⇒dydx=1+11515cos4x-sinx+15cos2xsinx

⇒dydx=1+cos4x-sinx+cos2xsinx

⇒y'π6=1+cos4π6-sinπ6+cos2π6sinπ6

⇒y'π6=1+3424-12+3222×12

⇒y'π6=1-932+38

⇒y'π6=32-9+1232

⇒y'π6=3532

⇒96y'π6=35×3

⇒96y'π6=105

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2024 (01 Feb Shift 2)
If y=x+c touches the ellipse 3x2+4y2=12 at the point P, then the value of the length OP (where O is the origin) is equal to

Options:
A 3 units
B 7 units
C 57 units
D 75 units

The correct answer is:
57 units

Solution:
If y=x+c touches x24+y23=1
then, c=Â±41+3
=Â±7
Let's assume, c=7 and y=x+7 touches x24+y23=1 at h,k.
then, xh4+yk3=1 is same as y-x=7
âh4-1=k31=17
âh,k=-47,37=P
Hence, OP=167+97=57 units

Topic: Mathematics, Ellipse, JEE Main
If yx is a solution of dydx-xy1 + x=11 + x and y0=-1, then the value of y2 is

Options:
A -12
B -13
C -14
D -15

The correct answer is:
-13

Solution:
I.F.=e-∫x1+xdx=e-∫x+1-1x+1dx=e-x+ln⁡x+1
= e-x·|(x+1)|
y·x+1e–x=∫x+1x+1e-xdx+c
y·|(x+1)|e–x=-| x+1 |x+1e-x+c
⇒ y=-1x+1+cex| x+1 |
x=0, y=-1⇒ -1=-1+c⇒ c=0
⇒ y=-11+x
y2=-13

Topic: Mathematics, Differential Equations, JEE Main
If $y(x)$ is the solution of the differential equation $\frac{d y}{d x}+\left(\frac{2 x+1}{x}\right) y=e^{-2 x}, x>0,$ where $y(1)=\frac{1}{2} e^{-2},$ then:

Options:
A $y\left(\log _{e} 2\right)=\log _{e} 4$
B $y\left(\log _{e} 2\right)=\frac{\log _{e} 2}{4}$
C $y(x)$ is decreasing in $\left(\frac{1}{2}, 1\right)$
D $y(x)$ is decreasing in (0,1)

The correct answer is:
$y(x)$ is decreasing in $\left(\frac{1}{2}, 1\right)$

Solution:
Given differential equation is,

$\frac{d y}{d x}+\left(2+\frac{1}{x}\right) y=e^{-2 x}, x>0$

$\mathrm{IF}=e^{\int\left(2+\frac{1}{x}\right) d x}=e^{2 x+\ln x}=x e^{2 x}$

Complete solution is given by

$y(x) \cdot x e^{2 x}=\int x e^{2 x} \cdot e^{-2 x} d x+c$

$=\int x d x+c$

$y(x) \cdot e^{2 x} \cdot x=\frac{x^{2}}{2}+c$

Given, $y(1)=\frac{1}{2} e^{-2}$

$\therefore \quad \frac{1}{2} e^{-2} \cdot e^{2} \cdot 1=\frac{1}{2}+c \Rightarrow c=0$

$\therefore \quad y(x)=\frac{x^{2}}{2} \cdot \frac{e^{-2 x}}{x}$

$y(x)=\frac{x}{2} \cdot e^{-2 x}$

Differentiate both sides with respect to $x$

$y^{\prime}(x)=\frac{e^{-2 x}}{2}(1-2 x) < 0 \forall x \in\left(\frac{1}{2}, 1\right)$

Hence, $y(x)$ is decreasing in $\left(\frac{1}{2}, 1\right)$

Topic: Mathematics, Differential Equations, JEE Main
If $y(x)$ is the solution of the differential equation $\frac{d y}{d x}+\left(\frac{2 x+1}{x}\right) y=e^{-2 x}, x>0,$ where $y(1)=\frac{1}{2} e^{-2},$ then:

Options:
A $y\left(\log _{e} 2\right)=\log _{e} 4$
B $y\left(\log _{e} 2\right)=\frac{\log _{e} 2}{4}$
C $y(x)$ is decreasing in $\left(\frac{1}{2}, 1\right)$
D $y(x)$ is decreasing in (0,1)

The correct answer is:
$y(x)$ is decreasing in $\left(\frac{1}{2}, 1\right)$

Solution:
Given differential equation is,

$\frac{d y}{d x}+\left(2+\frac{1}{x}\right) y=e^{-2 x}, x>0$

$\mathrm{IF}=e^{\int\left(2+\frac{1}{x}\right) d x}=e^{2 x+\ln x}=x e^{2 x}$

Complete solution is given by

$y(x) \cdot x e^{2 x}=\int x e^{2 x} \cdot e^{-2 x} d x+c$

$=\int x d x+c$

$y(x) \cdot e^{2 x} \cdot x=\frac{x^{2}}{2}+c$

Given, $y(1)=\frac{1}{2} e^{-2}$

$\therefore \quad \frac{1}{2} e^{-2} \cdot e^{2} \cdot 1=\frac{1}{2}+c \Rightarrow c=0$

$\therefore \quad y(x)=\frac{x^{2}}{2} \cdot \frac{e^{-2 x}}{x}$

$y(x)=\frac{x}{2} \cdot e^{-2 x}$

Differentiate both sides with respect to $x$

$y^{\prime}(x)=\frac{e^{-2 x}}{2}(1-2 x) < 0 \forall x \in\left(\frac{1}{2}, 1\right)$

Hence, $y(x)$ is decreasing in $\left(\frac{1}{2}, 1\right)$

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2019 (11 Jan Shift 1)
If y (x) is the solution of the differential equation x+2dydx=x2+4x-9, x ≠ -2 and y0=0, then y(-4) is equal to

Options:
A -1
B 1
C 0
D 2

The correct answer is:
0

Solution:
x+2dydx=x2+4x+4-13
dydx= x+22x+2- 13(x+2)
dy= x+2-13x+2dx
y=x22+2x-13loge⁡x+2+C
If x=0 then y=0
0= 0+0-13 log2+C
C =13log⁡(2)
If x= -4, then y
y=162-8-13log⁡-2+13log⁡|2|
y=0
Hence, answer is 0.

Topic: Mathematics, Differential Equations, JEE Main, JEE Main 2015 (10 Apr Online)
If $y=x \sin x$, then

Options:
A $\frac{1}{y} \frac{d y}{d x}=\frac{1}{x}+\cot x$
B $\frac{d y}{d x}=\frac{1}{x}+\cot x$
C $\frac{1}{y} \frac{d y}{d x}=\frac{1}{x}-\cot x$
D None of these

The correct answer is:
$\frac{1}{y} \frac{d y}{d x}=\frac{1}{x}+\cot x$

Solution:
$y=x \sin x \Rightarrow \frac{d y}{d x}=x \cos x+\sin x \Rightarrow \frac{1}{y} \frac{d y}{d x}=\cot x+\frac{1}{x}$

Topic: Mathematics, Differentiation, JEE Main
If $y=x+\frac{1}{x}$, then

Options:
A $x^2 \frac{d y}{d x}+x y=0$
B $x^2 \frac{d y}{d x}+x y+2=0$
C $x^2 \frac{d y}{d x}-x y+2=0$
D None of these

The correct answer is:
$x^2 \frac{d y}{d x}-x y+2=0$

Solution:
$y=x+\frac{1}{x} \Rightarrow \frac{d y}{d x}=1-\frac{1}{x^2}$
Therefore,
$x^2 \cdot \frac{d y}{d x}-x y+2=x^2\left(1-\frac{1}{x^2}\right)-x\left(x+\frac{1}{x}\right)+2=0$

Topic: Mathematics, Differentiation, JEE Main
If y=x+x2-115+x-x2-115, then x2-1d2ydx2+xdydx is equal to

Options:
A 224 y2
B 125Â y
C 225 y
D 225Â y2

The correct answer is:
225 y

Solution:
y=x+x2-115+x-x2-115

dydx=15x+x2-1141+xx2-1+15x-x2-1141-xx2-1

dydx=15x+x2-114x+x2-1x2-1+15x-x2-114x-x2-1x2-1

dydx=15x+x2-115-x-x2-115x2-1
⇒ x2-1 .dydx=15x+x2-115-x-x2-115
⇒ xx2-1 .dydx+ x2-1d2ydx2=15×15yx2-1
⇒ xdydx+x2-1d2y dx2=15x2-1×15yx2-1=225y

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2017 (08 Apr Online)
If yx=xx,x>0, then y"2-2y'2 is equal to :

Options:
A 8loge2-2
B 4loge2+2
C 4loge22-2
D 4loge22+2

The correct answer is:
4loge22-2

Solution:
Given, y=xx

⇒lny=xlnx

Differentiate both sides w.r.t. x

1y.y'=1+lnx

⇒y'=y1+lnx .......(1)

⇒y'=xx1+lnx

at x=2⇒y=4

So, y'(2)=41+ln2 .......(2)

Now differentiate the equation 1,

y"=y'1+lnx+yx

⇒y"(2)=y'1+ln2+2

⇒y"(2)-y'(2)=y'(ln2)+2

⇒y"(2)-2y'(2)=ln2-1y'(2)+2

=4(ln2-1)(ln2+1)+2

=4ln22-2

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2023 (01 Feb Shift 2)
If y=y(x) is an implicit function of x such that loge(x+y)=4xy, then d2y dx2 at x=0 is equal to

Options:

The correct answer is:
40

Solution:
Given:

loge(x+y)=4xy

When x=0, then y=1

logex+y=4xy

⇒x+y=e4xy

Now differentiate w.r.t. x

1+y'=e4xy4y+4xy' …i

At (0,1)⇒y'(0)+1=4⇒y'(0)=3

Now, again differentiate equation (i), we get

y"=e4xy4y+4xy′2+e4xy4y'+4y'+4xy"

At 0,1

y"(0)=1(4×1+0)2+1(4×3+4×3+0)

⇒y"(0)=16+24=40

⇒y"(0)=40

Topic: Mathematics, Differentiation, JEE Main
If y=y(x) is an implicit function of x such that loge(x+y)=4xy, then d2y dx2 at x=0 is equal to

Options:

The correct answer is:
40

Solution:
Given:

loge(x+y)=4xy

When x=0, then y=1

logex+y=4xy

⇒x+y=e4xy

Now differentiate w.r.t. x

1+y'=e4xy4y+4xy' …i

At (0,1)⇒y'(0)+1=4⇒y'(0)=3

Now, again differentiate equation (i), we get

y"=e4xy4y+4xy′2+e4xy4y'+4y'+4xy"

At 0,1

y"(0)=1(4×1+0)2+1(4×3+4×3+0)

⇒y"(0)=16+24=40

⇒y"(0)=40

Topic: Mathematics, Differentiation, JEE Main, JEE Main 2021 (26 Aug Shift 1)