A) 3
B) 0
C) 1
D) 2
Correct Answer: B
Solution :
\[\int_{1}^{2}{{f}'(x)dx=[f(x)]_{1}^{2}}=f(2)-f(1)=0\] (\[\because \]\[f(x)\]satisfies the conditions of Rolle?s theorem, \ \[f(2)=f(1)]\]) .You need to login to perform this action.
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