Let $f:(-1,1) \rightarrow R$ be a differentiable function with $f(0)=1$ and $f^{\prime}(0)=-1.$ If $g(x)=(f(n f(x)-n))^n$, where $n$ is a natural number and $g^{\prime}(0)= 4,$ then $n$ equals

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Application +4 | -1

Let

f:(-1,1) \rightarrow R

be a differentiable function with

f(0)=1

and

f^{\prime}(0)=-1.

g(x)=(f(n f(x)-n))^n

, where

n

is a natural number and

g^{\prime}(0)= 4,

then

n

equals

View answer and explanation

Correct option: A

Option A is correct because solving for n in the given conditions leads to n = 2 after applying the chain rule and substituting the given values.

Practice All Chapter MCQs →

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

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

x

y

x>0

x y=1

x+y

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

[x]

x

x=k, k

f(x)

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],

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

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

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

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

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