Question

If Rolle's theorem holds for the function \(f(x)=x^{2}-px+10\), \(x\in [0,4]\) with \(c=2\), then \(p=\)

• A. \(4\)
• B. \(2\)
• C. \(-4\)
• D. \(-2\)

Hint:

Rolle's theorem needs f(0) = f(4) and f'(c) = 0; either condition gives p.

Answer 1

The Correct Option is A

Step 1: Use f'(c) = 0
$f'(x)=2x-p$. At $c=2$: $4-p=0$.

Step 2: Solve
This gives $p=4$.

Step 3: Check
With $p=4$: $f(0)=10$ and $f(4)=16-16+10=10$, so Rolle's conditions hold and option (A) is correct.

Final Answer:
Hence $p=4$. \[ \boxed{\text{(A) }4} \]