Differentiation — Page 2 | JEE Mathematics

Q11

If for

x \in \left( {0,{1 \over 4}} \right)

, the derivatives of

{\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)

is

\sqrt x .g\left( x \right)

, then

g\left( x \right)

equals

A

{{{3x\sqrt x } \over {1 - 9{x^3}}}}

B

{{{3x} \over {1 - 9{x^3}}}}

C

{{3 \over {1 + 9{x^3}}}}

D

{{9 \over {1 + 9{x^3}}}}

Correct Answer

Option D

Solution

Let y =

{\tan ^{ - 1}}\left( {{{6x\sqrt x } \over {1 - 9{x^3}}}} \right)

=

{\tan ^{ - 1}}\left[ {{{2.\left( {3{x^{{3 \over 2}}}} \right)} \over {1 - {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}} \right]

= 2

{\tan ^{ - 1}}\left( {3{x^{{3 \over 2}}}} \right)

$\therefore$

{{dy} \over {dx}} = 2.{1 \over {1 + {{\left( {3{x^{{3 \over 2}}}} \right)}^2}}}.3 \times {3 \over 2}{\left( x \right)^{{1 \over 2}}}

=

{9 \over {1 + 9{x^3}}}.\sqrt x

$\therefore$ g(x) =

{9 \over {1 + 9{x^3}}}

Q12

Let f be a polynomial function such that f (3x) = f ' (x) . f '' (x), for all x

\in

R. Then :

A f (2) + f ' (2) = 28

B f '' (2)

-

f ' (2) = 0

C f '' (2)

-

f (2) = 4

D f (2)

-

f ' (2) + f '' (2) = 10

Correct Answer

Option B

Solution

Let

f(x) = {a_0}{x^n} + {a_1}{x^{n - 1}} + {a_2}{x^{n - 1}} + \,\,....\,\, + {a_{n - 1}}x + {a_n}

f'(x) = {a_0}n{x^{n - 1}} + {a_1}(n - 1){x^{n - 2}} + \,\,.....\,\, + {a_{n - 1}}

f''(x) = {a_0}n(n - 1){x^{n - 2}} + {a_1}(n - 1)(n - 2){x^{n - 3}} + \,\,....\,\, + {a_{n - 2}}

Now,

f(3x) = {3^n}{a_0}{x^n} + {3^{n - 1}}{a_1}{x^{n - 1}} + {3^{n - 2}}{a_2}{x^{n - 2}} + \,\,....\,\, + 3{a_{n - 1}} + {a_n}

f'(x)\,.\,f''(x) = [{a_0}n{x^{n - 1}} + {a_1}(n - 1){x^{n - 2}} + \,\,....\,\, + {a_{n - 1}}]

[{a_0}n(n - 1){x^{n - 2}} + {a_1}(n - 1)(n - 2){x^{n - 3}} + \,\,....\,\, + {a_{n - 2}}]

Comparing highest powers of x, we get

{3^n}{a_0}{x_n} = a_0^2(n - 1){x^{n - 1 + n - 2}} = a_0^2{n^2}(n - 1){x^{2n - 3}}

Therefore,

2n - 3 = n

$\Rightarrow$ n = 3 and

{3^n}{a_0} = a_0^2{n^2}(n - 1)

\Rightarrow {a_0} = 27 = {3 \over 2}

Therefore,

f(x) = {3 \over 2}{x^3} + {a_1}{x^2} + {a_2}x + {a_3}

f'(x) = {9 \over 2}{x^2} + 2{a_1}x + {a_2}

f''(x) = 9x + 2{a_1}

f(3x) = {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3}

Now,

f(3x) = f'(x)\,.\,f''(x)

\Rightarrow {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3}

\Rightarrow \left( {{9 \over 2}{x^2} + 2{a_1}x + {a_2}} \right)(9x + 2{a_1})

\Rightarrow {{81} \over 2}{x^3} + 9{a_1}{x^2} + 3{a_2}x + {a_3} = {{81} \over 2}{x^3} + [9{a_1} + 18{a_1}]{x^2} + [4a_1^2 + 9{a_2}]x + 2{a_1}{a_2}

Comparing the coefficients, we get

9{a_1} = 27{a_1}

\Rightarrow {a_1} = 0,\,3{a_2} = 4a_1^2 + 9{a_2} = 9{a_2}

\Rightarrow {a_2} = 0

Therefore,

f(x) = {3 \over 2}{x^3}

f'(x) = {9 \over 2}{x^2}

f''(x) = 9x

Hence,

f''(2) - f'(x) = 18 - 18 = 0

Q13

If y =

{\left[ {x + \sqrt {{x^2} - 1} } \right]^{15}} + {\left[ {x - \sqrt {{x^2} - 1} } \right]^{15}},

then (x2

-

1)

{{{d^2}y} \over {d{x^2}}} + x{{dy} \over {dx}}

is equal to :

A 125 y

B 124 y2

C 225 y2

D 225 y

Correct Answer

Option D

Solution

The given equation is

y = {({x^2} + \sqrt {{x^2} - 1} )^{15}} + {(x - \sqrt {{x^2} - 1} )^{15}}

Differentiating w.r.t. x, we get

{{dy} \over {dx}} = 15{(x + \sqrt {{x^2} - 1} )^{14}}\left( {1 + {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right) + 15{(x - \sqrt {{x^2} - 1} )^{14}}\left( {1 - {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right)

Here, we have used the standard differentiatials

{d \over {dx}}{x^n} = n\,{x^{n - 1}}

That is,

{d \over {dx}}(\sqrt {f(x)} ) = {1 \over {2\sqrt {f(x)} }} \times {d \over {dx}}(f(x))

Therefore

{{dy} \over {dx}} = {{15{{(x + \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} + x)} \over {\sqrt {{x^2} - 1} }} + {{15{{(x - \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} - x)} \over {\sqrt {{x^2} - 1} }}

\Rightarrow \sqrt {{x^2} - 1} {{dy} \over {dx}} = 15{(x + \sqrt {{x^2} - 1} )^{15}} - 15{(x - \sqrt {{x^2} - 1} )^{15}}

Differentiating w.r.t. x, we get

{{1(2x)} \over {2\sqrt {{x^2} + 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}} = 15 \times 15{(x + \sqrt {{x^2} - 1} )^{14}}\left( {1 + {{1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right) - 15 \times 15{(x - \sqrt {{x^2} - 1} )^{14}}\left( {{{1 - 1(2x)} \over {2\sqrt {{x^2} - 1} }}} \right)

\Rightarrow {x \over {\sqrt {{x^2} - 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}} = 225{(x + \sqrt {{x^2} - 1} )^{14}}{{(\sqrt {{x^2} - 1} + x)} \over {\sqrt {{x^2} - 1} }} - {{225{{(x - \sqrt {{x^2} - 1} )}^{14}}(\sqrt {{x^2} - 1} - x)} \over {\sqrt {{x^2} - 1} }}

\Rightarrow \sqrt {{x^2} - 1} \left[ {{x \over {\sqrt {{x^2} - 1} }}{{dy} \over {dx}} + \sqrt {{x^2} - 1} {{{d^2}y} \over {d{x^2}}}} \right] = 225{(x + \sqrt {{x^2} - 1} )^{15}} + 225{(x - \sqrt {{x^2} - 1} )^{15}}

\Rightarrow x{{dy} \over {dx}} + ({x^2} - 1){{{d^2}y} \over {d{x^2}}} = 225\left[ {{{(x + \sqrt {{x^2} - 1} )}^{15}} + {{(x - \sqrt {{x^2} - 1} )}^{15}}} \right]

Substituting

{(x + \sqrt {{x^2} - 1} )^{15}} + {(x - \sqrt {{x^2} - 1} )^{15}} = y

, we get

({x^2} - 1){{{d^2}y} \over {d{x^2}}} + {{x\,dy} \over {dx}} = 225y

Q14

If

x = \sqrt {{2^{\cos e{c^{ - 1}}}}}

and

y = \sqrt {{2^{se{c^{ - 1}}t}}} \,\,\left( {\left| t \right| \ge 1} \right),

then

{{dy} \over {dx}}

is equal to :

A

{y \over x}

B

{x \over y}

C

-

{y \over x}

D

-

{x \over y}

Correct Answer

Option C

Solution

x =

\sqrt {{2^{\cos e{c^{ - 1}}t}}}

\therefore\,\,\,\,

{{dx} \over {dt}}

=

{1 \over {2\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}

$\times$ (

{2^{\cos e{c^{ - 1}}t}}\,.\,\log 2

) $\times$

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

{{dy} \over {dt}}

=

{1 \over {2\sqrt {{2^{{{\sec }^{ - 1}}t}}} }}

$\times$

\left( {{2^{{{\sec }^{ - 1}}t}}\log 2} \right)

$\times$

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

\therefore\,\,\,\,

{{dy} \over {dx}}

=

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

=

{{ - \sqrt {{2^{\cos e{c^{ - 1}}t}}} } \over {\sqrt {{2^{\cos e{c^{ - 1}}t}}} }}

$\times$

{{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}

=

- \sqrt {{{{2^{{{\sec }^{ - 1}}t}}} \over {{2^{\cos e{c^{ - 1}}t}}}}}

= $-$

{y \over x}

Q15

If

f\left( x \right) = \left| \begin{array}{lll}{\cos x} & x & 1 \\ {2\sin x} & {{x^2}} & {2x} \\ {\tan x} & x & 1 \end{array} \right|,

then

\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}

A does not exist.

B exists and is equal to 2.

C existsand is equal to 0.

D exists and is equal to

-

2.

Correct Answer

Option D

Solution

Given,

f\left( x \right) = \left| \begin{array}{lll}{\cos x} & x & 1 \\ {2\sin x} & {{x^2}} & {2x} \\ {\tan x} & x & 1 \end{array} \right|

= cosx(x2 - 2x2) - x(2 sinx - 2x tanx) + (2x sinx - x2 tanx) = x2 (tanx - cosx) $\therefore$

{f^{'}}(x)

= 2x (tanx - cosx) + x2(sec2x + sinx) $\therefore$

\mathop {\lim }\limits_{x \to 0} {{f'\left( x \right)} \over x}

=

\mathop {\lim }\limits_{x \to o} {{2x(\tan x - \cos x) + {x^2}({{\sec }^2}x + \sin x)} \over x}

=

\mathop {\lim }\limits_{x \to o} \,\,2(\tan x - \cos x) + x({\sec ^2}x + \sin x)

= 2 (0-1) + 0 = -2

Q16

The derivative of

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

, with respect to

{x \over 2}

, where

\left( {x \in \left( {0,{\pi \over 2}} \right)} \right)

is :

A 1

B 2

C

{2 \over 3}

D

{1 \over 2}

Correct Answer

Option B

Solution

y = {\tan ^{ - 1}}\left( {{{\tan x - 1} \over {\tan x + 1}}} \right)

$\Rightarrow$

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

$\Rightarrow$

- {\tan ^{ - 1}}\left( {\tan \left( {{\pi \over 4} - x} \right)} \right)

$\Rightarrow$

- \left( {{\pi \over 4} - x} \right)

$\Rightarrow$

{{dy} \over {dx}} = 1

Now if differentiation of

{x \over 2}

w.r.t is

{1 \over 2}

$\Rightarrow$ So differentiation of y w.r.t

{x \over 2}

is

1 \over {1 \over 2}

= 2

Q17

If ey + xy = e, the ordered pair

\left( {{{dy} \over {dx}},{{{d^2}y} \over {d{x^2}}}} \right)

at x = 0 is equal to :

A

\left( {{1 \over e}, - {1 \over {{e^2}}}} \right)

B

\left( { - {1 \over e},{1 \over {{e^2}}}} \right)

C

\left( { - {1 \over e}, - {1 \over {{e^2}}}} \right)

D

\left( {{1 \over e},{1 \over {{e^2}}}} \right)

Correct Answer

Option B

Solution

y = 1 $\Rightarrow$ x = 0

{e^y}{{dy} \over {dx}} + x{{dy} \over {dx}} + y = 0

\Rightarrow e{{dy} \over {dx}} + 1 = 0 \Rightarrow {{dy} \over {dx}} = - {1 \over e}

\Rightarrow {e^y}{{{d^2}y} \over {d{x^2}}} + {e^y}{\left( {{{dy} \over {dx}}} \right)^2} + x{{{d^2}y} \over {d{x^2}}} + 2{{dy} \over {dx}} = 0

x = 0, y = 1

\Rightarrow e{{{d^2}y} \over {d{x^2}}} + e{\left( { - {1 \over e}} \right)^2} + 0 + 2\left( { - {1 \over e}} \right) = 0

\Rightarrow {{{d^2}y} \over {d{x^2}}} = {1 \over {{e^2}}}

Q18

Let f(x) = loge(sin x), (0 < x <

\pi

) and g(x) = sin–1 (e–x ), (x

\ge

0). If

\alpha

is a positive real number such that a = (fog)'(

\alpha

) and b = (fog)(

\alpha

), then :

A a

\alpha

2 + b

\alpha

- a = -2

\alpha

2

B a

\alpha

2 + b

\alpha

+ a = 0

C a

\alpha

2 - b

\alpha

- a = 0

D a

\alpha

2 - b

\alpha

- a = 1

Correct Answer

Option D

Solution

f(x) = ln(sin x), g(x) = sin–1 (e–x) f(g(x)) = ln(sin(sin–1 e–x)) = -x f(g( $\alpha$ )) = – $\alpha$ = b As f(g(x)) = – x $\therefore$ (f(g(x)))' = – 1 $\Rightarrow$ (f(g( $\alpha$ )))' = – 1 = a $\therefore$ b = – $\alpha$ , a = – 1 $\therefore$ a $\alpha$ 2 - b $\alpha$ - a = - $\alpha$ 2 + $\alpha$ 2 + 1 = 1

Q19

If ƒ(1) = 1, ƒ'(1) = 3, then the derivative of ƒ(ƒ(ƒ(x))) + (ƒ(x))2 at x = 1 is :

A 33

B 12

C 9

D 15

Correct Answer

Option A

Solution

Given ƒ(1) = 1, ƒ'(1) = 3 Let y = ƒ(ƒ(ƒ(x))) + (ƒ(x))2 On differentiating both sides with respect to x we get,

{{dy} \over {dx}}

= ƒ'(ƒ(ƒ(x))).ƒ'(ƒ(x)).ƒ'(x) + 2ƒ(x).ƒ'(x) Now at x = 1,

{{dy} \over {dx}}

= ƒ'(ƒ(ƒ(1))).ƒ'(ƒ(1)).ƒ'(1) + 2ƒ(1).ƒ'(1) = ƒ'(ƒ(1)).ƒ'(1).ƒ'(1) + 2.1.ƒ'(1) = ƒ'(1).ƒ'(1).ƒ'(1) + 2.1.ƒ'(1) = 3 $\times$ 3 $\times$ 3 + 2 $\times$ 3 = 33

Q20

If

2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}

, x

\in

\left( {0,{\pi \over 2}} \right)

then

dy \over dx

is equal to:

A

2x - {\pi \over 3}

B

{\pi \over 6} - x

C

{\pi \over 3} - x

D

x - {\pi \over 6}

Correct Answer

Option D

Solution

2y = {\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 \cos x + \sin x} \over {\cos x - \sqrt 3 \sin x}}} \right)} \right)^2}

$\Rightarrow$ 2y =

{\left( {{{\cot }^{ - 1}}\left( {{{\sqrt 3 + \tan x} \over {1 - \sqrt 3 \tan x}}} \right)} \right)^2}

$\Rightarrow$ 2y =

{\left( {{{\cot }^{ - 1}}\left( {{{\tan {\pi \over 3} + \tan x} \over {1 - \tan {\pi \over 3}\tan x}}} \right)} \right)^2}

$\Rightarrow$ 2y =

{\left( {{{\cot }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}

$\Rightarrow$ 2y =

{\left( {{\pi \over 2} - {{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)} \right)^2}

As x

\in

\left( {0,{\pi \over 2}} \right)

then

{{{\tan }^{ - 1}}\tan \left( {{\pi \over 3} + x} \right)}

=

{\left( {{\pi \over 3} + x} \right)}

$\Rightarrow$ 2y =

{\left( {{\pi \over 2} - \left( {{\pi \over 3} + x} \right)} \right)^2}

$\Rightarrow$ 2y =

{\left( {{\pi \over 6} - x} \right)^2}

$\therefore$

2{{dy} \over {dx}} = 2\left( {{\pi \over 6} - x} \right)\left( { - 1} \right)

$\Rightarrow$

{{dy} \over {dx}} = \left( {x - {\pi \over 6}} \right)