Question

If f(x) = { 1/|x| if |x| ≥ 1 ; a x² + b if |x| $< 1$ } is differentiable at every point of the domain, then the values of a and b are respectively :

• A. 1/2, 1/2
• B. -1/2, 3/2
• C. 5/2, -3/2
• D. 1/2, -3/2

Hint:

Differentiability implies continuity. Always check both conditions when solving for two unknowns.

Answer 1

The Correct Option is B

Step 1: For differentiability, the function must be continuous at $x=1$. $LHL = \lim_{x \to 1^-} (ax^2 + b) = a+b$. $RHL = \lim_{x \to 1^+} (1/x) = 1$. So, $a+b = 1$.
Step 2: Derivatives must be equal at $x=1$. $f'(x) = 2ax$ for $x<1$. $f'(x) = -1/x^2$ for $x>1$. At $x=1$: $2a(1) = -1/(1)^2 \implies 2a = -1 \implies a = -1/2$.
Step 3: Substitute $a$ in continuity eq: $-1/2 + b = 1 \implies b = 3/2$.