Find the range of ${sin}^{- 1} x$ .

0 \leq y \leq π

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- \frac{π}{2} \leq y \leq \frac{π}{2}

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0

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- \frac{π}{2} < y < \frac{π}{2}

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Solution

The correct option is B $- \frac{π}{2} \leq y \leq \frac{π}{2}$
It is given that ${sin}^{- 1} x = y$ .
We know that the range of the principal valueof ${sin}^{- 1}$ is $[- \frac{π}{2}, \frac{π}{2}]$ .
Therefore, $- \frac{π}{2} \leq y \leq \frac{π}{2}$

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Similar questions

If sin^-1x = y then the range of y is:

Find the value of if sin⁻¹ x = y, then

(A) (B)

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

{sec}^{- 1} x

	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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