Let g ( x ) = 3 f ( x 3 ) + f ( 3 - x ) and f ' ' ( x ) > 0 for all x ? ( 0 , 3 ) . If g is decreasing in and increasing in, then is: [2024]

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Solution

Q.
$Let g (x) = 3 f (\frac{x}{3}) + f (3 - x) and f'' (x) > 0 for all x \in (0, 3) .$

$If g is decreasing in (0, α) and increasing in (α, 3), then 8 α is:$ [2024]

1 20

2 18

3 0

4 24

Ans.
(2)

$Given, g (x) = 3 f (\frac{x}{3}) + f (3 - x)$ ...(i)

Differentiating (i) w.r.t. $x,$ we get

$g' (x) = 3 f' (\frac{x}{3}) \cdot \frac{1}{3} + f' (3 - x) (- 1)$

$\Rightarrow g' (x) = f' (\frac{x}{3}) - f' (3 - x)$

When $g (x)$ is decreasing, $g' (x) < 0,$

$\Rightarrow f' (\frac{x}{3}) < f' (3 - x) \Rightarrow \frac{x}{3} < (3 - x) \Rightarrow x < \frac{9}{4}$

Therefore, $α = \frac{9}{4};$ So, $8 α = 8 \times \frac{9}{4} = 18$

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