Question:

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

Show Hint

Rolle's theorem needs f(0) = f(4) and f'(c) = 0; either condition gives p.
Updated On: Nov 12, 2026
  • \(4\)
  • \(2\)
  • \(-4\)
  • \(-2\)
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The Correct Option is A

Solution and Explanation

Step 1: Understanding the Concept
Rolle's theorem requires \(f(0)=f(4)\) and some \(c\) with \(f'(c)=0\).

Step 2: Use f(0) = f(4)
\(f(0)=10\) and \(f(4)=16-4p+10=26-4p\).
Setting \(26-4p=10\) gives \(p=4\).

Step 3: Verify c = 2
\(f'(x)=2x-p\), so \(f'(2)=4-p=0\) gives \(p=4\). Both conditions agree.

Final Answer:
The value of \(p\) is 4. \[ \boxed{\text{(A) }4} \]
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