QJEEM21Maec08: JEE Main 2021 Mathematics - Class 12 JEEM21Maec08 Solution: If f ( x ) = { 1

Question

Verified Solution • 2025–26 Session moderate

If

f (x) = {\begin{matrix} \frac{1}{| x |} & ; | x | \geq 1 \\ a x^{2} + b & ; | x | < 1 \end{matrix}

is differentiable at every point of the domain, then the values of a and b are respectively :

Accepted Answer

f (x) = {\begin{matrix} \frac{1}{| x |}, & | x | \geq 1 \\ a x^{2} + b, & | x | < 1 \end{matrix}

= {\begin{matrix} - \frac{1}{x}; & x \leq - 1 \\ a x^{2} + b; & - 1 < x < 1 \\ \frac{1}{x}; & x \geq 1 \end{matrix}

As f(x) is differentiable so it is also continuous,

at x = 1,

\mathop {\lim }\limits_{x \to {1^ - }} f(x) = \mathop {\lim }\limits_{x \to {1^ + }} f(x)

\Rightarrow a + b = {1 \over 1}

\Rightarrow a + b = 1

...... (1)

As f(x) is differentiable, so at x = 1

L.H.D. = R.H.D.

\Rightarrow 2ax = - {1 \over {{x^2}}}

\Rightarrow 2a = - 1

\Rightarrow a = - {1 \over 2}

From (1),

b = {3 \over 2}

Step-by-Step Verified Solution