The range of sin -1 x-cos -1 x isA-B- -0 -2-C- -3 -2- -D- -0- -2-

The correct option is A [ 3π/2, π/2 ]sin^ 1 cos^ 1x=π/2 cos^ 1x cos^ 1x=π/2 2cos^ 1x 0≤ cos^ 1x≤π⇒ 0 ≤ 2 cos^ 1x ≤ 2π⇒ 2π≤ 2cos^ 1 x≤0⇒ 3π/2≤π/2 2cos^ 1x≤π/ ...

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Standard IX

Mathematics

Using Monotonicity to Find the Range of a Function

The range of ...

Question

The range of $s i n^{- 1} x - c o s^{- 1} x$ is

[\frac{- 3 π}{2}, \frac{π}{2}]

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[\frac{- 5 π}{2}, \frac{π}{3}]

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[\frac{- 3 π}{2}, π]

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[0, \frac{π}{2}]

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Solution

The correct option is A $[\frac{- 3 π}{2}, \frac{π}{2}]$
$s i n^{- 1} - c o s^{- 1} x = \frac{π}{2} - c o s^{- 1} x - c o s^{- 1} x = \frac{π}{2} - 2 c o s^{- 1} x$
$0 \leq c o s^{- 1} x \leq π \Rightarrow 0 \leq 2 c o s^{- 1} x \leq 2 π \Rightarrow - 2 π \leq - 2 c o s^{- 1} x \leq 0 \Rightarrow \frac{- 3 π}{2} \leq \frac{π}{2} - 2 c o s^{- 1} x \leq \frac{π}{2}$
$∴$ Range $= [\frac{- 3 π}{2}, \frac{π}{2}]$

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Similar questions

Q. The range of

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

Q. Match the column

	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.	$(\frac{π}{2}, \frac{3 π}{2})$

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

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

Q. Assertion :Statement 1 Range of

f (x) = {tan}^{- 1} x + {sin}^{- 1} x + {cos}^{- 1} x

(0, π)

Reason: Statement 2

f (x) = {tan}^{- 1} x + {sin}^{- 1} x + {cos}^{- 1} x = \frac{π}{2} + {tan}^{- 1} x

for

x ϵ (- 1, 1]

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The range of sin−1x−cos−1x is

The correct option is A [−3π2,π2]sin−1−cos−1x=π2−cos−1x−cos−1x=π2−2cos−1x 0≤cos−1x≤π⇒0≤2cos−1x≤2π⇒−2π≤−2cos−1x≤0⇒−3π2≤π2−2cos−1x≤π2 ∴ Range =[−3π2,π2]

The range of $s i n^{- 1} x - c o s^{- 1} x$ is