a + bx, for x ≤ 0 }
where a and b are constants, is differentiable at x = 0,
then a + b is equal to
For the piecewise function f(x) = { sin 2x, x > 0; a + bx, x ≤ 0 } to be differentiable at x = 0, both continuity and equal derivative limits from left and right must hold at that point.
Continuity Condition
The function requires f(0⁺) = f(0⁻) for continuity at x=0.
Limit as x → 0⁺ of sin(2x) = sin(0) = 0
f(0) from left piece = a + b(0) = a
Thus, a = 0
Differentiability Condition
Derivatives must match: f'(0⁺) = f'(0⁻)
Right derivative: lim (h→0⁺) [sin(2h) – sin(0)] / h = lim [sin(2h)/h] = lim 2 cos(2h) = 2 cos(0) = 2
Left derivative: derivative of a + bx is b, so f'(0⁻) = b
Thus, b = 2
Correct Answer
With a = 0 and b = 2, a + b = 2, matching option (C).
Option Analysis
| Option | a + b Value | Why Incorrect/Correct |
|---|---|---|
| (A) 0 | a = 0, b = 0 | Fails differentiability: left derivative b=0 ≠ right derivative 2 |
| (B) 1 | e.g., a=0, b=1 | Slope mismatch: b=1 ≠ 2 |
| (C) 2 | a=0, b=2 | Satisfies both continuity (a=0) and differentiability (b=2) |
| (D) 3 | e.g., a=1, b=2 | Fails continuity: a=1 ≠ 0 |
Key Takeaway
This ensures the tangent line is unique at x=0, making f differentiable there. Both conditions must be satisfied simultaneously.
