If $f\left( x \right) = x\left( {\sqrt x + \sqrt {x + 1} } \right)$, then

$f\left( x \right)$ is continuous but not differentiable at $x = 0$

$f\left( x \right)$ is differentiable at $x = 0$

JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Analyzing +4 | -1

f\left( x \right) = x\left( {\sqrt x + \sqrt {x + 1} } \right)

, then

View answer and explanation

Correct option: C

f\left( x \right)

is not differentiable at

x = 0

Option C is correct because the function is not continuous at x=0 due to the square root of x, and therefore not differentiable.

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}}