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NTA Abhyas 2020
Exam
NTA Abhyas
Year
2020
Subject
Mathematics
Chapter
Inverse Trigonometric Functions

NTA Abhyas 2020 · Mathematics · Inverse Trigonometric Functions

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

If [sin−1x]2−2{ \left[sin^{- 1}x\right]}^{2} - 2\left[sin^{- 1}x\right] + 1 ≤ 0(where,[.](where, \left[.\right]represents the greatest integral part of xx), then

Options

  1. Ax∈[sin⁡1,sin⁡2]x \in \left[\sin 1,\sin 2\right]
  2. Bx∈[−sin⁡1,sin⁡1]x \in \left[-\sin 1,\sin 1\right]
  3. Cx∈[sin⁡1,1]x \in \left[\sin 1, 1\right]
  4. Dx∈[−sin⁡1,sin⁡2]x \in \left[-\sin 1,\sin 2\right]

Correct answer

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

Step-by-step solution

([sin−1x]−1)2≤0{\left(\left[sin^{- 1}x\right] - 1\right)}^{2} \le 0
⇒[sin−1x]−1=0⇒[sin−1x]=1⇒ \left[sin^{- 1}x\right] - 1 = 0 ⇒ \left[sin^{- 1}x\right] = 1
⇒1≤sin−1x<2⇒ 1 \le sin^{- 1}x < 2 but, sin−1x∈[−π2,π2]<br/>sin^{- 1}x \in \left[- \dfrac{\pi}{2}, \dfrac{\pi}{2}\right]<br />∴ 1 ≤ sin^{- 1}x ≤π2<br/>\le \dfrac{\pi}{2}<br />⇒ \sin 1 ≤ x ≤ sinπ2sin\dfrac{\pi}{2}
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