The range of the function $f (x) = (s i n^{- 1} x + t a n^{- 1} x + s e c^{- 1} x)$ is

(\frac{π}{4}, \frac{3 π}{4})

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[\frac{π}{4}, \frac{3 π}{4}]

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{\frac{π}{4}, \frac{3 π}{4}}

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n o n e o f t h e s e

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Solution

The correct option is B ${\frac{π}{4}, \frac{3 π}{4}}$
The domain of ${sin}^{- 1} (x)$ is $[- 1, 1]$
And the domain of $s e c^{- 1} (x)$ is $(- \infty, - 1] \cup [1, \infty)$

Therefore
The above function is valid for $x = - 1$ and $x = 1$ .

Substituting, we get

$f (- 1) = \frac{- π}{2} - \frac{π}{4} + π$
$= π - \frac{3 π}{4}$
$= \frac{π}{4}$ .
$f (1) = \frac{π}{2} + \frac{π}{4} + 0$
$= \frac{3 π}{4}$

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f (x) = t a n^{- 1} x + s i n^{- 1} x + c o s^{- 1} x

{x \in R : | cos x | \geq sin x} \cap [0, \frac{3 π}{2}] =

	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.	$(\frac{π}{2}, \frac{3 π}{2})$

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