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JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Easy

1 of 1

Analyzing +4 | -1

Let

f(x y)=f(x) f(y)

for all

x>0, y>0

and

f(1+x)=1+x\{1+g(x)\}

, where

\quad \lim\limits _{x \rightarrow 0} g(x)=0

, then

\int \frac{f(x)}{f^{\prime}(x)} d x

, is

View answer and explanation

Correct option: A

\frac{x^2}{2}+C

Option A (

\frac{x^2}{2}+C

) is correct because after finding

f(x) = x

, the integral

\int \frac{f(x)}{f^{\prime}(x)} dx

simplifies to

\int x dx

, which equals

\frac{x^2}{2}+C

.

More Questions from DIFFERENTIATION

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

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

is equal to Let

x

y

be two real numbers such that

x>0

x y=1

. The minimum value of

x+y

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

[x]

denotes the integer part of

x

x=k, k

being an integer, the left derivative of

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}}