Range of ${sin}^{- 1} x - {cos}^{- 1} x$ is

[\frac{- 3 π}{2}, \frac{π}{2}]

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[\frac{- 5 π}{3}, \frac{π}{3}]

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[\frac{- 3 π}{2}, π]

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[0, π]

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Solution

The correct option is A $[\frac{- 3 π}{2}, \frac{π}{2}]$
range of $s i n^{- 1} x ϵ [- \frac{π}{2}, \frac{π}{2}]$
range of $c o s^{- 1} x ϵ [0, π]$
$∴ f (x) = s i n^{- 1} x - c o s^{- 1} x = s i n^{- 1} x + c o s^{- 1} x - 2 c o s^{- 1} x$
$= \frac{π}{2} - 2 c o s^{- 1} x [∵ s i n^{- 1} x + c o s^{- 1} x = \frac{π}{2}] i d e n t i t y$
range of $f (x) ϵ [\frac{π}{2} - 2 (π), \frac{π}{2} - 2 (0)]$
$∴ f (x) ϵ [- \frac{3 π}{2}, \frac{π}{2}]$

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Similar questions

s i n^{- 1} x - c o s^{- 1} x

If $s i n^{- 1} x + s i n^{- 1} y = \frac{π}{2}$ ,then $c o s^{- 1} x + c o s^{- 1} y$ is equal to

	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.	$(\frac{π}{2}, \frac{3 π}{2})$

α \leq 2 {sin}^{- 1} x + {cos}^{- 1} x \leq β,

Solveis equal to

(A) (B). (C) (D)

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