JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Hard

1 of 1

Analyzing +4 | -1

Function

f\left( x \right) = \left| {x - 1} \right| + \left| {x - 2} \right|,x \in R

View answer and explanation

Correct option: B

Except

x = 1

and

x = 2

differentiable everywhere in

R

Option B (Except x = 1 and x = 2 differentiable everywhere in R) is correct because the function is not differentiable at x=1 and x=2 due to the absolute value functions.

More Questions from DIFFERENTIATION

Practice All Chapter MCQs →

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}}