Question

If a function \[ f(x)= \begin{cases} \dfrac{a}{|x|}, & x\le -1 \text{ or } x\ge 1,[6pt] \\ x^2+b, & -1<x<1, \end{cases} \] is differentiable on \(\mathbb{R}\), then \(a+b=\)

• A. \(3\)
• B. \(-2\)
• C. \(-5\)
• D. \(2\)

Hint:

Whenever a piecewise function is said to be differentiable, first apply continuity and then equate the left-hand and right-hand derivatives. Missing the continuity condition is the most common mistake.

Answer 1

The Correct Option is C

Concept: A function is differentiable at a point only if:
• It is continuous at that point.
• The left-hand derivative equals the right-hand derivative. Since the function changes its definition at \(x=-1\) and \(x=1\), differentiability must be checked at both points.

Step 1: Continuity at \(x=1\). From the left side, \[ \lim_{x\to1^-}(x^2+b) = 1+b. \] From the right side, \[ \lim_{x\to1^+}\frac{a}{|x|} = a. \] Continuity gives \[ 1+b=a. \] \[ a-b=1. \]

Step 2: Differentiate both branches. For \(-1<x<1\), \[ f(x)=x^2+b. \] Thus, \[ f'(x)=2x. \] At \(x=1\), \[ f'_-(1)=2. \] For \(x>1\), \[ f(x)=\frac{a}{x}. \] Hence, \[ f'(x) = -\frac{a}{x^2}. \] Therefore, \[ f'_+(1) = -a. \] Differentiability gives \[ 2=-a. \] Thus, \[ a=-2. \]

Step 3: Determine \(b\). Using \[ 1+b=a, \] we get \[ 1+b=-2. \] Hence, \[ b=-3. \]

Step 4: Calculate \(a+b\). \[ a+b = (-2)+(-3) = -5. \] Therefore, \[ \boxed{-5}. \]