JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

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Analyzing +4 | -1

Observe the following statements:

I. If

f\left( x \right) = a{x^{41}} + b{x^{ - 40}}

, then

\frac{{f''\left( x \right)}}{{f\left( x \right)}} = 1640{x^{ - 2}}

II.

\frac{d}{{dx}}\left\{ {{{\tan }^{ - 1}}\left( {\frac{{2x}}{{1 - {x^2}}}} \right)} \right\} = \frac{1}{{1 + {x^2}}}

Which of the following is correct?

View answer and explanation

Correct option: A

I is true, but II is false

Option A (I is true, but II is false) is correct because statement I is verified to be true through differentiation, while statement II is false due to the range of the inverse tangent function.

More Questions from DIFFERENTIATION

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f(x)=|x|^{|\tan x|}

, then

f^{\prime}\left(\frac{-\pi}{6}\right)

is equal to Let

x

and

y

be two real numbers such that

x>0

and

x y=1

. The minimum value of

x+y

is Let

f(x)=[x] \sin (\pi x)

where

[x]

denotes the integer part of

x

. Then at

x=k, k

being an integer, the left derivative of

f(x)

is If

y = {\sec ^{ - 1}}\left[ {\mathrm{cosec}x} \right] + \mathrm{cose}{\mathrm{c}^{ - 1}}\left[ {\sec x} \right] + {\sin ^{ - 1}}[\cos x] + {\cos ^{ - 1}}[\sin x],

then

\frac{{dy}}{{dx}}

is equal to

\frac{d}{d x} \sin ^{-1}\left(3 x-4 x^{3}\right)

is equal to If

z = y + f\left( v \right),

where

v = \left( {\frac{x}{y}} \right),

then

v\frac{{\partial z}}{{\partial x}} + \frac{{\partial z}}{{\partial y}}