JEE MATHEMATICS MCQ: DIFFERENTIATION

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Application +4 | -1

Let

f\left(\frac{x+y}{2}\right)=\frac{f(x)+f(y)}{2}

for all real

x

and

y

for

f^{\prime}(0)=-1, f(0)=1

, then

f^{\prime}(2)=

View answer and explanation

Correct option: C

-1

Option C (-1) is correct because the given functional equation implies that

f'(x)

is constant, and since

f'(0) = -1

, then

f'(2) = -1

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