Differentiation — Page 3 | JEE Mathematics

Q21

For x > 1, if (2x)2y = 4e2x

-

2y, then (1 + loge 2x)2

{{dy} \over {dx}}

is equal to :

A

{{x\,{{\log }_e}2x - {{\log }_e}2} \over x}

B loge 2x

C x loge 2x

D

{{x\,{{\log }_e}2x + {{\log }_e}2} \over x}

Correct Answer

Option A

Solution

(2x)2y = 4e2x-2y 2y

\ell

n2x =

\ell

n4 + 2x $-$ 2y y =

{{x + \ell n2} \over {1 + \ell n2x}}

y ' =

{{\left( {1 + \ell n2x} \right) - \left( {x + \ell n2} \right){1 \over x}} \over {{{\left( {1 + \ell n2x} \right)}^2}}}

y '

{\left( {1 + \ell n2x} \right)^2} = \left[ {{{x\ell n2x - \ell n2} \over x}} \right]

Q22

Let ƒ and g be differentiable functions on R such that fog is the identity function. If for some a, b

\in

R, g'(a) = 5 and g(a) = b, then ƒ'(b) is equal to :

A 1

B 5

C

{2 \over 5}

D

{1 \over 5}

Correct Answer

Option D

Solution

Given the function composition f(g(x)) is the identity function, it means f(g(x)) = x for all x. $\Rightarrow$ ƒ'(g(x)) g'(x) = 1 put x = a $\Rightarrow$ ƒ'(b) g'(a) = 1 $\Rightarrow$ ƒ'(b) =

{1 \over 5}

Q23

The derivative of

{\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)

with respect to

{\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right)

at x =

{1 \over 2}

is :

A

{{2\sqrt 3 } \over 3}

B

{{2\sqrt 3 } \over 5}

C

{{\sqrt 3 } \over {10}}

D

{{\sqrt 3 } \over {12}}

Correct Answer

Option C

Solution

Let f =

{\tan ^{ - 1}}\left( {{{\sqrt {1 + {x^2}} - 1} \over x}} \right)

Put x = tan $\theta$ $\Rightarrow$ $\theta$ = tan–1 x f =

{\tan ^{ - 1}}\left( {{{\sec \theta - 1} \over {\tan \theta }}} \right)

$\Rightarrow$ f =

{\tan ^{ - 1}}\left( {{{1 - \cos \theta } \over {\sin \theta }}} \right)

=

{\theta \over 2}

$\Rightarrow$ f =

{{{{\tan }^{ - 1}}x} \over 2}

$\therefore$

{{df} \over {dx}}

=

{1 \over {2\left( {1 + {x^2}} \right)}}

....(1) Let g =

{\tan ^{ - 1}}\left( {{{2x\sqrt {1 - {x^2}} } \over {1 - 2{x^2}}}} \right)

Put x = sin $\theta$ $\Rightarrow$ $\theta$ = sin–1 x $\Rightarrow$ g =

{\tan ^{ - 1}}\left( {{{2\sin \theta \cos \theta } \over {1 - 2{{\sin }^2}\theta }}} \right)

$\Rightarrow$ g = tan–1 (tan 2 $\theta$ ) = 2 $\theta$ $\Rightarrow$ g = 2sin-1 x $\Rightarrow$

{{dg} \over {dx}}

=

{2 \over {\sqrt {1 - {x^2}} }}

...(2) Using (i) and (ii), $\therefore$

{{df} \over {dg}}

=

{1 \over {2\left( {1 + {x^2}} \right)}}{{\sqrt {1 - {x^2}} } \over 2}

At x =

{1 \over 2}

,

{\left( {{{df} \over {dg}}} \right)_{x = {1 \over 2}}}

=

{{\sqrt 3 } \over {10}}

Q24

If

\left( {a + \sqrt 2 b\cos x} \right)\left( {a - \sqrt 2 b\cos y} \right) = {a^2} - {b^2}

where a > b > 0, then

{{dx} \over {dy}}\,\,at\left( {{\pi \over 4},{\pi \over 4}} \right)

is :

A

{{a - 2b} \over {a + 2b}}

B

{{a - b} \over {a + b}}

C

{{a + b} \over {a - b}}

D

{{2a + b} \over {2a - b}}

Correct Answer

Option C

Solution

(a + \sqrt 2 b\cos x)(a - \sqrt 2 b\cos y) = {a^2} - {b^2}

\Rightarrow {a^2} - \sqrt 2 ab\cos y + \sqrt 2 ab\cos x - 2{b^2}\cos x\cos y = {a^2} - {b^2}

Differentiating both sides :

0 - \sqrt 2 ab\left( { - \sin y{{dy} \over {dx}}} \right) + \sqrt 2 ab( - \sin x)

- 2{b^2}\left[ {\cos x\left( { - \sin y{{dy} \over {dx}}} \right) + \cos y( - \sin x)} \right] = 0

At

\left( {{\pi \over 4},{\pi \over 4}} \right)

:

ab{{dy} \over {dx}} - ab - 2{b^2}\left( { - {1 \over 2}{{dy} \over {dx}} - {1 \over 2}} \right) = 0

\Rightarrow {{dx} \over {dy}} = {{ab + {b^2}} \over {ab - {b^2}}} = {{a + b} \over {a - b}}

; a, b > 0

Q25

If y2 + loge (cos2x) = y,

x \in \left( { - {\pi \over 2},{\pi \over 2}} \right)

, then :

A |y''(0)| = 2

B |y'(0)| + |y''(0)| = 3

C y''(0) = 0

D |y'(0)| + |y"(0)| = 1

Correct Answer

Option A

Solution

Given y2 + loge (cos2x) = y .....(

1) Put x = 0, we get y2 + loge (1) = y $\Rightarrow$ y2 = y $\Rightarrow$ y = 0, 1 Differentiating (1) we get 2yy' +

{1 \over {\cos x}}\left( { - \sin x} \right)

= y' $\Rightarrow$ 2yy' - 2tanx = y' ....(

2) From (2) when x = 0, y = 0 then y'(0) = 0 From (2) when x = 0, y = 1 then 2y' = y' $\Rightarrow$ y'(0) = 0 Again differentiating (2) we get 2(y')2 + 2yy'' – 2sec2x = y'' from (2) when x = 0, y = 0, y’(0) = 0 then y”(0) = -2 Also from (2) when x = 0, y = 1, y’(0) = 0 then y”(0) = 2 $\therefore$ |y''(0)| = 2

Q26

If

x = 2\sin \theta - \sin 2\theta

and

y = 2\cos \theta - \cos 2\theta

,

\theta \in \left[ {0,2\pi } \right]

, then

{{{d^2}y} \over {d{x^2}}}

at

\theta

=

\pi

is :

A

{3 \over 8}

B

{3 \over 2}

C

{3 \over 4}

D -

{3 \over 4}

Correct Answer

Option A

Solution

x = 2\sin \theta - \sin 2\theta

$\Rightarrow$

{{dx} \over {d\theta }}

=

2\cos \theta - 2\cos 2\theta

y = 2\cos \theta - \cos 2\theta

$\Rightarrow$

{{dy} \over {d\theta }}

= –2sin $\theta$ + 2sin2 $\theta$

{{dy} \over {dx}} = {{{{dy} \over {d\theta }}} \over {{{dx} \over {d\theta }}}}

=

{{\sin 2\theta - \sin \theta } \over {\cos \theta - \cos 2\theta }}

This expression is simplified by recognizing that $\sin 2 \theta = 2\sin \theta \cos \theta$ and $\cos 2 \theta = 2\cos^2 \theta -1$ , leading to the expression:

\frac{dy}{dx} = \frac{2 \sin \frac{\theta}{2} \cdot \cos \frac{3 \theta}{2}}{2 \sin \frac{\theta}{2} \cdot \sin \frac{3 \theta}{2}}=\cot \frac{3 \theta}{2}

This is the rate of change of y with respect to x, as a function of θ.

Next, we differentiate this function with respect to θ to find $\dfrac{d^2 y}{dx^2}$ , yielding :

\frac{d^2 y}{dx^2}=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \frac{d \theta}{d x}=-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2} \cdot \frac{d \theta}{d x}

Here, $\dfrac{d \theta}{d x}$ is the reciprocal of $\dfrac{dx}{d\theta}$ , so the equation becomes :

\frac{d^2 y}{dx^2}=\frac{-\frac{3}{2} \operatorname{cosec}^2 \frac{3 \theta}{2}}{2\left(\cos \theta-\cos2 \theta\right)}

Finally, we evaluate this expression at $\theta = \pi$ , yielding :

\frac{d^2 y}{dx^2}(\pi)=\frac{-3}{4(-1-1)}=\frac{3}{8}

Therefore, the correct answer is (A) $\dfrac{3}{8}$ .

Alternate Method : First, let's find the derivatives of x and y with respect to θ : 1)

\frac{dx}{d\theta} = 2\cos \theta - 2\cos 2\theta

2)

\frac{dy}{d\theta} = -2\sin \theta + 2\sin 2\theta

We know that

\frac{dy}{dx} = \frac{\frac{dy}{d\theta}}{\frac{dx}{d\theta}}

So, we substitute 1) and 2) into this equation : We have

\frac{dy}{dx} = \frac{-2\sin \theta + 2\sin 2\theta}{2\cos \theta - 2\cos 2\theta} = \frac{\sin 2\theta - \sin \theta}{\cos \theta - \cos 2\theta}

For simplification, let's denote the numerator as

N = \sin 2\theta - \sin \theta

and the denominator as

D = \cos \theta - \cos 2\theta

. We have to compute

\frac{d}{d\theta}(\frac{dy}{dx})

which is

\frac{d}{d\theta}(\frac{N}{D})

. We can use the quotient rule for differentiation, which states that if we have a function of the form

\frac{u}{v}

, then its derivative is given by

\frac{vu' - uv'}{v^2}

. So here,

N' = 2\cos 2\theta - \cos \theta

and

D' = -\sin \theta + 2\sin 2\theta

. Applying the quotient rule :

\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{D N' - N D'}{D^2}

Substituting the expressions for

N'

,

D'

,

N

, and

D

we get :

\frac{d}{d\theta}(\frac{dy}{dx}) = \frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}

Now

\frac{d^2 y}{d x^2}=\frac{d}{d x}\left(\frac{d y}{d x}\right)=\frac{d}{d \theta}\left(\frac{d y}{d x}\right) \times \frac{d \theta}{d x}

=

\frac{(\cos \theta - \cos 2\theta)(2\cos 2\theta - \cos \theta) - (\sin 2\theta - \sin \theta)(-\sin \theta + 2\sin 2\theta)}{(\cos \theta - \cos 2\theta)^2}

\times \frac{1}{(2 \cos \theta-2 \cos 2 \theta)}

\begin{aligned} \left.\therefore \frac{d^2 y}{d x^2}\right|_{\theta=\pi} & =\frac{(-1-1)(2+1)-(0-0)(-0+0)}{2(-1-1)^3} \\\\ & =\frac{-2 \times 3}{-2 \times 8}=\frac{3}{8} \end{aligned}

Q27

Let ƒ(x) = (sin(tan–1x) + sin(cot–1x))2 – 1, |x| > 1. If

{{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)

and

y\left( {\sqrt 3 } \right) = {\pi \over 6}

, then y(

{ - \sqrt 3 }

) is equal to :

A

{{5\pi } \over 6}

B

- {\pi \over 6}

C

{\pi \over 3}

D

{{2\pi } \over 3}

Correct Answer

Option B

Solution

Given ƒ(x) = (sin(tan–1x) + sin(cot–1x))2 – 1 = (sin(tan–1x) + sin(

{\pi \over 2}

- tan–1x))2 – 1 = (sin(tan–1x) + cos(tan–1x))2 – 1 = sin2(tan–1x) + cos2(tan–1x) + 2sin(tan–1x)cos(tan–1x) + 1 = 1 + sin(2tan–1x) - 1 = sin(2tan–1x) Also given

{{dy} \over {dx}} = {1 \over 2}{d \over {dx}}\left( {{{\sin }^{ - 1}}\left( {f\left( x \right)} \right)} \right)

Integrating both sides we get y =

{1 \over 2}

sin-1 (f(x)) + C =

{1 \over 2}

sin-1 (sin(2tan–1x)) + C Given

y\left( {\sqrt 3 } \right) = {\pi \over 6}

mean x =

\sqrt 3

and y =

{\pi \over 6}

$\therefore$

{\pi \over 6}

=

{1 \over 2}

sin-1 (sin(2tan–1

\sqrt 3

)) + C $\Rightarrow$

{\pi \over 6}

=

{1 \over 2}

sin-1 (sin(2 $\times$

{\pi \over 3}

)) + C $\Rightarrow$

{\pi \over 6}

=

{1 \over 2}

sin-1 (

{{\sqrt 3 } \over 2}

) + C $\Rightarrow$

{\pi \over 6}

=

{1 \over 2}

$\times$

{\pi \over 3}

+ C $\Rightarrow$ C = 0 Now y(

{ - \sqrt 3 }

) means when x =

{ - \sqrt 3 }

then find y. y =

{1 \over 2}

sin-1 (sin(2tan–1x)) =

{1 \over 2}

sin-1 (sin(2tan–1(

{ - \sqrt 3 }

))) =

{1 \over 2}

sin-1 (sin(-2tan–1(

{ \sqrt 3 }

))) =

{1 \over 2}

sin-1 (sin(-2 $\times$

{\pi \over 3}

)) =

{1 \over 2}

sin-1 (-sin(2 $\times$

{\pi \over 3}

)) =

{1 \over 2}

sin-1 (-

{{\sqrt 3 } \over 2}

) =

{1 \over 2}

$\times$ -sin-1 (

{{\sqrt 3 } \over 2}

) =

{1 \over 2}

$\times$ -

{\pi \over 3}

= -

{\pi \over 6}

Q28

Let y = y(x) be a function of x satisfying

y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}}

where k is a constant and

y\left( {{1 \over 2}} \right) = - {1 \over 4}

. Then

{{dy} \over {dx}}

at x =

{1 \over 2}

, is equal to :

A

{2 \over {\sqrt 5 }}

B

- {{\sqrt 5 } \over 2}

C

{{\sqrt 5 } \over 2}

D

- {{\sqrt 5 } \over 4}

Correct Answer

Option B

Solution

y\sqrt {1 - {x^2}} = k - x\sqrt {1 - {y^2}}

....(1) On differentiating both side of eq. (1) w.r.t. x we get,

{{dy} \over {dx}}\sqrt {1 - {x^2}} - y{{2x} \over {2\sqrt {1 - {x^2}} }}

= 0 -

\sqrt {1 - {y^2}} + {{xy} \over {\sqrt {1 - {y^2}} }}{{dy} \over {dx}}

Put x =

{1 \over 2}

and y =

- {1 \over 4}

, we get

{{dy} \over {dx}}{{\sqrt 3 } \over 2} - \left( { - {1 \over 4}} \right){{{1 \over 2}} \over {{{\sqrt 3 } \over 2}}}

=

- {{\sqrt {15} } \over 4} + {{ - {1 \over 8}} \over {{{\sqrt {15} } \over 4}}}.{{dy} \over {dx}}

$\therefore$

{{dy} \over {dx}} = - {{\sqrt 5 } \over 2}

Q29

If

y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}} ,\alpha \in \left( {{{3\pi } \over 4},\pi } \right)

{{dy} \over {d\alpha }}\,\,at\,\alpha = {{5\pi } \over 6}is

:

A 4

B -4

C

{4 \over 3}

D -

{1 \over 4}

Correct Answer

Option A

Solution

y\left( \alpha \right) = \sqrt {2\left( {{{\tan \alpha + \cot \alpha } \over {1 + {{\tan }^2}\alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}}

=

\sqrt {2\left( {{{1 + {{\tan }^2}\alpha } \over {\tan \alpha \left( {1 + {{\tan }^2}\alpha } \right)}}} \right) + {1 \over {{{\sin }^2}\alpha }}}

=

\sqrt {2\left( {{{\cos \alpha } \over {\sin \alpha }}} \right) + {1 \over {{{\sin }^2}\alpha }}}

=

\sqrt {{{\sin 2\alpha + 1} \over {{{\sin }^2}\alpha }}}

=

{{\left| {\sin \alpha + \cos \alpha } \right|} \over {\left| {\sin \alpha } \right|}}

At $\alpha$ =

{{5\pi } \over 6}

{\left| {\sin \alpha + \cos \alpha } \right|}

= -(sin $\alpha$ + cos $\alpha$ ) and |sin $\alpha$ | = sin $\alpha$ $\therefore$ y( $\alpha$ ) =

{{ - \left( {\sin \alpha + \cos \alpha } \right)} \over {\sin \alpha }}

= -1 - cot $\alpha$ $\therefore$

{{dy} \over {d\alpha }}

= cosec2 $\alpha$ So

{{{dy} \over {d\alpha }}}

at $\alpha$ =

{{5\pi } \over 6}

, = cosec2

{{5\pi } \over 6}

= 4

Q30

Let xk + yk = ak, (a, k > 0 ) and

{{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0

, then k is:

A

{1 \over 3}

B

{2 \over 3}

C

{4 \over 3}

D

{3 \over 2}

Correct Answer

Option B

Solution

xk + yk = ak $\Rightarrow$ kxk - 1 + kyk - 1

{{{dy} \over {dx}}}

= 0 $\Rightarrow$

{{{dy} \over {dx}} + {{\left( {{x \over y}} \right)}^{k - 1}}}

= 0 ...(1) Given

{{dy} \over {dx}} + {\left( {{y \over x}} \right)^{{1 \over 3}}} = 0

...(2) Comparing (1) and (2), we get k - 1 =

- {1 \over 3}

$\Rightarrow$ k =

{2 \over 3}