Question

If the derivative of the function \(f(x) = \{\begin{array}{cc}ax^2+b & \text{if }x < -1 \\ bx^2+ax+4 & \text{if }x\geq -1\end{array}\) is continuous everywhere then

• A. \(a = 2,b = 3\)
• B. \(a = 3,b = 2\)
• C. \(a = -2,b = 3\)
• D. \(a = -3,b = -2\)

Hint:

Match function values and derivative values at x = -1.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For the derivative to exist and be continuous, \(f\) must be continuous at \(-1\), differentiable at \(-1\), and \(f'\) must have equal left and right limits there.

Step 2: Continuity of f at x = -1:
Left value: \(a(1) + b = a + b\). Right value: \(b(1) + a(-1) + 4 = b - a + 4\).
\[ a + b = b - a + 4 \Rightarrow a = 2 \]

Step 3: Matching derivatives:
Left derivative: \(2ax\) at \(-1\) gives \(-2a\). Right derivative: \(2bx + a\) at \(-1\) gives \(-2b + a\).
\[ -2a = -2b + a \Rightarrow 3a = 2b \Rightarrow b = 3 \]

Step 4: Check:
With \(a = 2, b = 3\): left value \(5\), right value \(3 - 2 + 4 = 5\). Left slope \(-4\), right slope \(-6 + 2 = -4\). Both agree. Option (A).

Final Answer:
Continuity gives a = 2 and matching slopes gives b = 3. \[ \boxed{\text{(A) }a=2,\ b=3} \]