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Question: Let $f(x) = \frac{sin^{-1}(1-\{x\})\times cos^{-1}(1-\{x\})}{\sqrt{2\{x\} \times (1-\{x\})}}$, where...

Let f(x)=sin−1(1−{x})×cos−1(1−{x})2{x}×(1−{x})f(x) = \frac{sin^{-1}(1-\{x\})\times cos^{-1}(1-\{x\})}{\sqrt{2\{x\} \times (1-\{x\})}}, where {x}\{x\} denote the fractional part of xx.

R=lim⁡x→0+f(x)R = \lim_{x\to 0^+}f(x) is equal to:

A

π2\frac{\pi}{2}

B

π22\frac{\pi}{2\sqrt{2}}

C

π2\frac{\pi}{\sqrt{2}}

D

2π\sqrt{2\pi}

Answer

π2\frac{\pi}{2}

Explanation

Solution

As x→0+x \to 0^+, {x}=x\{x\}=x. R=lim⁡x→0+sin⁡−1(1−x)cos⁡−1(1−x)2x(1−x)R = \lim_{x\to 0^+} \frac{\sin^{-1}(1-x)\cos^{-1}(1-x)}{\sqrt{2x(1-x)}}. Let 1−x=cos⁡θ1-x = \cos\theta. As x→0+x \to 0^+, θ→0+\theta \to 0^+. R=lim⁡θ→0+sin⁡−1(cos⁡θ)θ2(1−cos⁡θ)cos⁡θ=lim⁡θ→0+(π/2−θ)θ2(2sin⁡2(θ/2))cos⁡θR = \lim_{\theta\to 0^+} \frac{\sin^{-1}(\cos\theta)\theta}{\sqrt{2(1-\cos\theta)\cos\theta}} = \lim_{\theta\to 0^+} \frac{(\pi/2-\theta)\theta}{\sqrt{2(2\sin^2(\theta/2))\cos\theta}}. R=lim⁡θ→0+(π/2−θ)θ2sin⁡(θ/2)cos⁡θR = \lim_{\theta\to 0^+} \frac{(\pi/2-\theta)\theta}{2\sin(\theta/2)\sqrt{\cos\theta}}. Using sin⁡(θ/2)≈θ/2\sin(\theta/2) \approx \theta/2 and cos⁡θ≈1\cos\theta \approx 1 for small θ\theta. R=lim⁡θ→0+(π/2−θ)θ2(θ/2)=lim⁡θ→0+(π/2−θ)=π/2R = \lim_{\theta\to 0^+} \frac{(\pi/2-\theta)\theta}{2(\theta/2)} = \lim_{\theta\to 0^+} (\pi/2-\theta) = \pi/2.