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Question

Mathematics Question on Continuity and differentiability

The value of C in Rolle's theorem where π2-\frac{π}{2}<C<π2\frac{π}{2} and f(x)=cosxf(x)=cos x on [π2,π2][-\frac{π}{2},\frac{π}{2}] is equal to :

A

0

B

π

C

π2\frac{π}{2}

D

π4\frac{π}{4}

Answer

0

Explanation

Solution

The correct option is(A):0