Let $f (x) = sin x + cos x + tan x + {sin}^{- 1} x + {cos}^{- 1} x + {tan}^{- 1} x$ . If 'M' and 'm' are maximum and minimum values of f(x), then their arithmetic mean of them is equal to?

\frac{π}{2} + cos 1

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\frac{π}{2} + sin 1

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\frac{π}{4} + tan 1 + cos 1

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\frac{π}{4} + tan 1 + sin 1

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Solution

The correct option is A $\frac{π}{2} + cos 1$
Given,
$f (x) = s i n x + c o s x + t a n x + s i n^{- 1} x + c o s^{- 1} x + t a n^{- 1} x$
x

$f (x) = s i n x + c o s x + t a n x + \frac{π}{2} + t a m^{- 1} x$

Differentiate with respect to x
$f^{'} (x) = c o s x - s i n x + s e c^{2} x + \frac{1}{1 + x^{2}}$
F'(x) is always increasing sinc f'(x)>0

The ranfe of f(x) is given by f(1) and f(-1)
$f (- 1) = - s i n 1 + c o s 1 - t a n 1 - \frac{π}{2} + π - \frac{π}{4}$
= $\frac{π}{4} + c o s 1 - s i n 1 - t a n 1$

$f (1) = s i n 1 + c o s 1 + t a n 1 + \frac{π}{2} + \frac{π}{4}$
= $\frac{3 π}{4} + s i n 1 + c o s 1 + t a n 1$

$∴$ A.M.= $\frac{m + M}{2}$
= $\frac{\frac{π}{4} + c o s 1 - s i n 1 - t a n 1 + \frac{3 π}{4} + c o s 1 + s i n 1 + t a n 1}{2}$

= $\frac{π + 2 c o s 1}{2}$

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Similar questions

	List I		List II
A.	Range of $f (x) = {sin}^{- 1} x + {cos}^{- 1} x + {cot}^{- 1} x$ is	1.	$[0, \frac{π}{2}) \cup (\frac{π}{2}, π]$
B.	Range of $f (x) = cot - 1 x + {tan}^{- 1} x + c o s e c^{- 1} x$ is	2.	$[\frac{π}{2}, \frac{3 π}{2}]$
C.	Range of $f (x) = {cot}^{- 1} x + {tan}^{- 1} x + {cos}^{- 1} x$ is	3.	${0, π}$
D.	Range of $f (x) = {sec}^{- 1} x + c o s e c^{- 1} x + {sin}^{- 1} x$ is	4.	$(\frac{π}{2}, \frac{3 π}{2})$