The range of $f (x) = {sin}^{- 1} x - {cos}^{- 1} x$ is

[0, π]

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[\frac{- 3 π}{2}, \frac{π}{2}]

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[\frac{- π}{2}, \frac{π}{2}]

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[- π, π]

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Solution

The correct option is B $[\frac{- 3 π}{2}, \frac{π}{2}]$
It is known that
${sin}^{- 1} x + {cos}^{- 1} x = \frac{π}{2}$
${cos}^{- 1} x = \frac{π}{2} - {sin}^{- 1} x$
Therefore
${sin}^{- 1} x - {cos}^{- 1} x$
$= {sin}^{- 1} x - (\frac{π}{2} - {sin}^{- 1} x)$
$= 2 {sin}^{- 1} x - \frac{π}{2}$
Now the range of ${sin}^{- 1} x$ is $[\frac{- π}{2}, \frac{π}{2}]$
Therefore, the range of $2 {sin}^{- 1} x$ is $[- π, π]$
Hence the range of $2 {sin}^{- 1} x - \frac{π}{2}$ will be
$[- π - \frac{π}{2}, π - \frac{π}{2}]$
$= [\frac{- 3 π}{2}, \frac{π}{2}]$

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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.	$(\frac{π}{2}, \frac{3 π}{2})$

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