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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question
lim⁡x→π/2sin⁡xcos⁡−1[14(3sin⁡x−sin⁡3x)]\displaystyle\lim_{x\to\pi/2}\dfrac{\sin x}{\cos^{-1}\left[\dfrac{1}{4}(3\sin x-\sin 3x)\right]}, where [⋅][\cdot] is GIF
A2π\dfrac{2}{\pi}correct
B11
C4π\dfrac{4}{\pi}
DDoes not exist
Solution
Step 1: Use the identity sin⁡3x=3sin⁡x−4sin⁡3x\sin 3x=3\sin x-4\sin^{3}x:
14(3sin⁡x−sin⁡3x)=14(3sin⁡x−3sin⁡x+4sin⁡3x)=sin⁡3x\dfrac{1}{4}(3\sin x-\sin 3x)=\dfrac{1}{4}(3\sin x-3\sin x+4\sin^{3}x)=\sin^{3}x
 Step 2: As x→π/2x\to\pi/2: sin⁡x→1\sin x\to 1, so sin⁡3x→1−\sin^{3}x\to 1^{-} (slightly less than 11). Step 3: Therefore [sin⁡3x]=0[\sin^{3}x]=0 for xx close to but not equal to π/2\pi/2. Step 4: The expression becomes:
lim⁡x→π/2sin⁡xcos⁡−1(0)=1π/2=2π\lim_{x\to\pi/2}\dfrac{\sin x}{\cos^{-1}(0)}=\dfrac{1}{\pi/2}=\dfrac{2}{\pi}
 Correct answer: (1) 2π\dfrac{2}{\pi}
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