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: Test the options by substitution:
Check option (A): $a=2, b=3$. Value from the left: $2 + 3 = 5$. From the right: $3 - 2 + 4 = 5$. Equal.

Step 2: Derivative test:
Left slope at $-1$: $2\cdot2\cdot(-1) = -4$. Right slope: $2\cdot3\cdot(-1) + 2 = -4$. Equal.

Step 3: Other options:
Option (B) $a=3,b=2$: values $5$ and $2-3+4 = 3$, not equal. Option (C) $a=-2,b=3$ gives $1$ and $9$. Option (D) $a=-3,b=-2$ gives $-5$ and $-2+3+4=5$. Only (A) works.

Final Answer:
The constants are a = 2 and b = 3. \[ \boxed{\text{(A) }a=2,\ b=3} \]