If [sin^ - 1 x] ^ 2 - 2 [ sin^ - 1 x] + 1 ≤ 0 (where, [.] represents the greatest integral part of x ), then

Correct answer: C. x ∈ [ 1, 1] ([sin^ - 1 x] - 1) ^ 2 ≤ 0 ⇒ [sin^ - 1 x] - 1 = 0 ⇒ [sin^ - 1 x] = 1 ⇒ 1 ≤ sin^ - 1 x ∴ 1 ≤ sin^ - 1 x ≤ (π)/(2) ⇒ 1 ≤ x ≤ sin(π)/(2)

JEE Main

Exam: JEE Main
Subject: Mathematics
Chapter: Inverse Trigonometric Functions

JEE Main · Mathematics · Inverse Trigonometric Functions

If [sin^ - 1 x] ^ 2 - 2 [ sin^ - 1 x] + 1 ≤ 0 (where, [.]...

{ \left[sin^{- 1}x\right]}^{2} - 2 \left[

sin^{- 1}x\right] + 1 ≤ 0

(where, \left[.\right] represents the greatest integral part of

), then

Options

A $x \in \left[\sin 1,\sin 2\right]$
B $x \in \left[-\sin 1,\sin 1\right]$
C $x \in \left[\sin 1, 1\right]$
D $x \in \left[-\sin 1,\sin 2\right]$

Correct answer

C. $x \in \left[\sin 1, 1\right]$

Step-by-step solution

{\left(\left[sin^{- 1}x\right] - 1\right)}^{2} \le 0

⇒ \left[sin^{- 1}x\right] - 1 = 0 ⇒ \left[sin^{- 1}x\right] = 1

⇒ 1 \le sin^{- 1}x < 2

but,

sin^{- 1}x \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]<br/>

∴ 1 ≤ sin^{- 1}x

\le \dfrac{\pi}{2}<br/>

⇒ \sin 1 ≤ x ≤ sin\dfrac{\pi}{2}

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