Let non constant differentiable|Applications Derivatives

JEE MAIN 2024

30-01-2024 S1

Question

Let $g: R \rightarrow R$ be a non constant twice differentiable such that $g^{\prime}\left(\frac{1}{2}\right)=g^{\prime}\left(\frac{3}{2}\right)$. If a real valued function $f$ is defined as $f(x)=\frac{1}{2}[g(x)+g(2-x)]$, then

Select the correct option:

f' (x)=0 for atleast two x in (0,2)

f' (x)=0 for exactly one x in (0,1)

f' (x)=0 for no x in (0,1)

f' (3/2)+f^' (1/2)=1

✓ Correct! Well done.

✗ Incorrect. Try again or view the solution.

Solution

Question Tags

JEE Main

Mathematics

Easy

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