Let g ( x ) = f ( x ) + f ( 1 - x ) and f ' ' ( x ) 0 , ? x ( 0 , 1 ) . If g is decreasing in the interval and increasing in the interval ( 1 ) , then tan - 1 ( 2 ) + tan - 1 ( 1 ) + tan - 1 ( + 1 ) is equal to [2023]

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Solution

Q.
Let $g (x) = f (x) + f (1 - x)$ and $f'' (x) > 0, x \in (0, 1)$ . If $g$ is decreasing in the interval $(0, α)$ and increasing in the interval $(α, 1)$ , then $\tan^{- 1} (2 α) + \tan^{- 1} (\frac{1}{α}) + \tan^{- 1} (\frac{α + 1}{α})$ is equal to [2023]

1 $\frac{3 π}{2}$

2 $\frac{3 π}{4}$

3 $π$

4 $\frac{5 π}{4}$

Ans.
(3)

Given, $g (x) = f (x) + f (1 - x)$ and $f'' (x) > 0, x \in (0, 1)$

$∴$ $g' (x) = f' (x) + f' (1 - x) (- 1) = f' (x) - f' (1 - x)$

$and g'' (x) = f'' (x) + f'' (1 - x)$

At $x = \frac{1}{2}$ , $g' (\frac{1}{2}) = f' (\frac{1}{2}) - f' (1 - \frac{1}{2}) = 0 and g'' (x) > 0$

$∴$ $g (x) is concave upward for α = \frac{1}{2} .$

Now, $\tan^{- 1} (2 α) + \tan^{- 1} (\frac{1}{α}) + \tan^{- 1} (\frac{α + 1}{α})$

$= \tan^{- 1} (1) + \tan^{- 1} (2) + \tan^{- 1} (3)$

$= \frac{π}{4} + \tan^{- 1} (\frac{2 + 3}{1 - (2 \times 3)}) = \frac{π}{4} + \tan^{- 1} (- 1) = \frac{π}{4} + (π - \frac{π}{4}) = π$

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