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Let $f(x)=\sin x+\cos x+\tan x+\arcsin x+\arccos x+\arctan x$ . If $M$ and $m$ are maximum and minimum values of $f(x)$ then their arithmetic mean is equal to?

My Approach:

$\arcsin x + \arccos x =\frac{\pi}{2}$, so the equation becomes $\sin x + \cos x + \tan x + \frac{\pi}{2} + \arctan x$ . After this I differentiated this equation but did not came to any conclusion . Also I tried to solve the equation to make something relevant but the equation did not simplified . Please help...

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  • $\begingroup$ Maximum and minimum over what set? All of the functions listed are bounded except for $\tan x$, which is unbounded over $(-\pi/2,\pi/2)$. $\endgroup$
    – Joey Zou
    Sep 1, 2016 at 18:07
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    $\begingroup$ @JoeyZou Obviously, on a subset of $[-1,1]$. $\endgroup$
    – Jean-Claude Arbaut
    Sep 1, 2016 at 18:08
  • $\begingroup$ Since there is a inverse function so domain can be [-1,1]. $\endgroup$
    – saladi
    Sep 1, 2016 at 18:14

1 Answer 1

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Hint: We have $f(x) = \sin x + \cos x + \tan x + \frac{\pi}{2} + \arctan x$ on $[-1,1]$. Now $$ f'(x) = \cos x - \sin x + \sec^2 x + \frac{1}{1+x^2}. $$ On $[-1,1]$, we have $\cos x > 0$, $-\sin x > -1$, $\sec^2x\ge 1$, and $\frac{1}{1+x^2}>0$, so for all $x\in[-1,1]$ we have $$ f'(x) > 0 - 1 + 1 + 0 = 0. $$ Thus $f$ is strictly increasing. What can you say about where it attains its maximum and minimum?

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  • $\begingroup$ Since the function is always increasing hence , the minimum and maximum values will be at x = -1 and x=1 , so their arithmetic mean will be π/2 + cos (1) . Am I correct.... $\endgroup$
    – saladi
    Sep 1, 2016 at 19:12
  • $\begingroup$ @saladi looks good to me! $\endgroup$
    – Joey Zou
    Sep 1, 2016 at 19:55

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