Let x denote the fractional part of x and f(x)= cos^-1(1-x^2) sin^-1(1-x) x-x^3,x≠ 0. If L andR respectively denotes the left hand limit and the…

Given: $f(x)=\frac{{\mathrm{cos}}^{-1}(1-{{x}}^{2}){\mathrm{sin}}^{-1}(1-{x})}{{x}-{{x}}^{3}}$f(x)=x−x3cos−1(1−x2)sin−1(1−x)​

$$\Rightarrow \mathrm{RHL}=\underset{x\rightarrow {0}^{+}}{\mathrm{lim}}f(x)=\underset{h\rightarrow 0}{\mathrm{lim}}f(0+h)$$⇒RHL=x→0+lim​f(x)=h→0lim​f(0+h)

$$\Rightarrow \mathrm{RHL}=\underset{h\rightarrow 0}{\mathrm{lim}}f(h)$$⇒RHL=h→0lim​f(h)

$$\Rightarrow \mathrm{RHL}=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{{\mathrm{cos}}^{-1}(1-{h}^{2}){\mathrm{sin}}^{-1}(1-h)}{h(1-{h}^{2})}$$⇒RHL=h→0lim​h(1−h2)cos−1(1−h2)sin−1(1−h)​

$$\Rightarrow \mathrm{RHL}=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{{\mathrm{cos}}^{-1}(1-{h}^{2})}{h}(\frac{{\mathrm{sin}}^{-1}1}{1})$$⇒RHL=h→0lim​hcos−1(1−h2)​(1sin−11​)

Let ${\mathrm{cos}}^{-1}(1-{h}^{2})=\theta \Rightarrow \mathrm{cos}\theta =1-{h}^{2}$cos−1(1−h2)=θ⇒cosθ=1−h2

$$\Rightarrow \mathrm{RHL}=\frac{\pi }{2}\underset{\theta \rightarrow 0}{\mathrm{lim}}\frac{\theta }{\sqrt{1-\mathrm{cos}\theta }}$$⇒RHL=2π​θ→0lim​1−cosθ​θ​

$$\Rightarrow \mathrm{RHL}=\frac{\pi }{2}\underset{\theta \rightarrow 0}{\mathrm{lim}}\frac{1}{\sqrt{\frac{1-\mathrm{cos}\theta }{{\theta }^{2}}}}$$⇒RHL=2π​θ→0lim​θ21−cosθ​​1​

$$\Rightarrow \mathrm{RHL}=\frac{\pi }{2}\frac{1}{\sqrt{\frac{1}{2}}}$$⇒RHL=2π​21​​1​

$$\Rightarrow \mathrm{RHL}=\frac{\pi }{\sqrt{2}}$$⇒RHL=2​π​

$$\Rightarrow R=\frac{\pi }{\sqrt{2}}$$⇒R=2​π​

Now finding left hand limit,

$$\Rightarrow \mathrm{LHL}=\underset{x\rightarrow {0}^{-}}{\mathrm{lim}}f(x)$$⇒LHL=x→0−lim​f(x)

$$\Rightarrow \mathrm{LHL}=\underset{h\rightarrow 0}{\mathrm{lim}}f(-h)$$⇒LHL=h→0lim​f(−h)

$$\Rightarrow \mathrm{LHL}=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{{\mathrm{cos}}^{-1}(1-{{-h}}^{2}){\mathrm{sin}}^{-1}(1-{-h})}{{-h}-{{-h}}^{3}}$$⇒LHL=h→0lim​−h−−h3cos−1(1−−h2)sin−1(1−−h)​

$$\Rightarrow \mathrm{LHL}=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{{\mathrm{cos}}^{-1}(1-{(-h+1)}^{2}){\mathrm{sin}}^{-1}(1-(-h+1))}{(-h+1)-{(-h+1)}^{3}}$$⇒LHL=h→0lim​(−h+1)−(−h+1)3cos−1(1−(−h+1)2)sin−1(1−(−h+1))​

$$\Rightarrow \mathrm{LHL}=\underset{h\rightarrow 0}{\mathrm{lim}}\frac{{\mathrm{cos}}^{-1}(-{h}^{2}+2h){\mathrm{sin}}^{-1}h}{(1-h)(1-{(1-h)}^{2})}$$⇒LHL=h→0lim​(1−h)(1−(1−h)2)cos−1(−h2+2h)sin−1h​

$$\Rightarrow \mathrm{LHL}=\underset{h\rightarrow 0}{\mathrm{lim}}(\frac{\pi }{2})\frac{{\mathrm{sin}}^{-1}h}{(1-{(1-h)}^{2})}$$⇒LHL=h→0lim​(2π​)(1−(1−h)2)sin−1h​

$$\Rightarrow \mathrm{LHL}=\frac{\pi }{2}\underset{h\rightarrow 0}{\mathrm{lim}}(\frac{{\mathrm{sin}}^{-1}h}{-{h}^{2}+2h})$$⇒LHL=2π​h→0lim​(−h2+2hsin−1h​)

$$\Rightarrow \mathrm{LHL}=\frac{\pi }{2}\underset{h\rightarrow 0}{\mathrm{lim}}(\frac{{\mathrm{sin}}^{-1}h}{h})(\frac{1}{-h+2})$$⇒LHL=2π​h→0lim​(hsin−1h​)(−h+21​)

$$\Rightarrow \mathrm{LHL}=\frac{\pi }{4}$$⇒LHL=4π​

$$\Rightarrow L=\frac{\pi }{4}$$⇒L=4π​

$$\Rightarrow \frac{32}{{\pi }^{2}}({L}^{2}+{R}^{2})=\frac{32}{{\pi }^{2}}(\frac{{\pi }^{2}}{2}+\frac{{\pi }^{2}}{16})$$⇒π232​(L2+R2)=π232​(2π2​+16π2​)

$$\Rightarrow \frac{32}{{\pi }^{2}}({L}^{2}+{R}^{2})=16+2$$⇒π232​(L2+R2)=16+2

$$\Rightarrow \frac{32}{{\pi }^{2}}({L}^{2}+{R}^{2})=18$$⇒π232​(L2+R2)=18