Let g : R ? R be a non constant twice differentiable function such that g ' ( 1 2 ) = g ' ( 3 2 ) . If a real valued function f is defined as f ( x ) = 1 2 [ g ( x ) + g ( 2 - x ) ] , then [2024]

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Solution

Q.
Let $g : R \to R$ be a non constant twice differentiable function such that $g' (\frac{1}{2}) = g' (\frac{3}{2}) .$ If a real valued function $f$ is defined as $f (x) = \frac{1}{2} [g (x) + g (2 - x)],$ then [2024]

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

2 $f' (\frac{3}{2}) + f' (\frac{1}{2}) = 1$

3 $f'' (x) = 0$ for at least two $x$ in (0, 2)

4 $f'' (x) = 0$ for exactly one $x$ in (0, 1)

Ans.
(3)

$We have, f (x) = \frac{1}{2} [g (x) + g (2 - x)]$

$So f' (x) = \frac{1}{2} (g' (x) + g' (2 - x) (- 1)) = \frac{1}{2} (g' (x) - g' (2 - x))$

$\Rightarrow f' (\frac{1}{2}) = \frac{1}{2} (g' (\frac{1}{2}) - g' (2 - \frac{1}{2}))$

$= \frac{1}{2} (g' (\frac{1}{2}) - g' (\frac{3}{2})) = 0$ $[∵ g' (\frac{1}{2}) = g' (\frac{3}{2})]$

$Similarly f' (\frac{3}{2}) = \frac{1}{2} (g' (\frac{3}{2}) - g' (\frac{1}{2})) = 0$

$Now, f' (x) = 0 when x = \frac{1}{2} and \frac{3}{2}$

$So, f'' (x) has at least two roots in (0, 2) .$ $[By Rolle's theorem]$

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