JEE MATHEMATICS MCQ: DIFFERENTIATION

1 Medium

1 of 1

Application +4 | -1

f\left( x \right) = \left\{ {\begin{array}{cccccccccccccccccccc} {a{x^2} - b,\left| x \right| < 1}\\ {\frac{1}{{\left| x \right|}},\left| x \right| \ge 1} \end{array}} \right.

is differentiable at

x = 1

, then

View answer and explanation

Correct option: B

a = - \frac{1}{2},b = - \frac{3}{2}

Option B (

a = - \frac{{1}}{{2}},b = - \frac{{3}}{{2}}

) is correct because it satisfies the conditions for differentiability at x=1, ensuring both continuity and equal derivatives from both sides.

More Questions from DIFFERENTIATION

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}}