2
$\begingroup$

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...

$\endgroup$
  • $\begingroup$ 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 '16 at 18:07
  • 1
    $\begingroup$ @JoeyZou Obviously, on a subset of $[-1,1]$. $\endgroup$ – Stop hurting Monica Sep 1 '16 at 18:08
  • $\begingroup$ Since there is a inverse function so domain can be [-1,1]. $\endgroup$ – saladi Sep 1 '16 at 18:14
3
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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 '16 at 19:12
  • $\begingroup$ @saladi looks good to me! $\endgroup$ – Joey Zou Sep 1 '16 at 19:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.