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If `f(x)` satisfies the condition of Rolles theorem in [1, 2] then `int_1^2f^(prime)(x)dx` is equal to (A) 1 (B) 3 (C) 0 (D) none of these

A

3

B

0

C

1

D

2

Text Solution

Verified by Experts

The correct Answer is:
B

It is givne that f(x) is continuous on [1,2] differentiable on (1,2) and f(2) =r(1)
`:.underset(1)overset(2)intf'(x)dx=[f(x)]_(1)^(2)=f(2)-f(1)=0`
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