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if rolle s theorem holds for the function f x x 2
Question:
If Rolle's theorem holds for the function \(f(x)=x^{2}-px+10\), \(x\in [0,4]\) with \(c=2\), then \(p=\)
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Rolle's theorem needs f(0) = f(4) and f'(c) = 0; either condition gives p.
MHT CET - 2026
MHT CET
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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