If $f(x)=\left\{\begin{array}{cl}\frac{1}{|x|} & ;|x| \geq 1 \\ a x^{2}+b & ;|x|<1\end{array}\right.$ is differentiable at every point of the domain, then the values of $a$ and $b$ are respectively ;

(A) $-\frac{1}{2}, \frac{3}{2}$

(B) $\frac{1}{2},-\frac{3}{2}$

(C) $\frac{1}{2}, \frac{1}{2}$

(D) $\frac{5}{2},-\frac{3}{2}$

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Continuity and Differentiability

f (x) = {\frac{1}{∣ x ∣} a x^{2} + b; ∣ x ∣ \geq 1; ∣ x ∣ < 1

is differentiable at every point of the domain, then the values of

a

and

b

are respectively ;

Answer: (A)

Both branches are even, so matching at

x = 1

settles the constants. Continuity there requires

a (1)^{2} + b = \frac{1}{1} = 1, a + b = 1

For

x \geq 1

the rule is

\frac{1}{x}

, whose derivative

- \frac{1}{x ^{2}}

equals

- 1

x = 1

, while for

∣ x ∣ < 1

the derivative is

2 a x

, equal to

2 a

there. Matching the two slopes gives

2 a = - 1

, so

a = - \frac{1}{2}

and

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

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Single correct option