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QJEEM24Ma1468: JEE Main 2024 Mathematics - Class 12 JEEM24Ma1468 Solution: Let denote the fractional part of and . If and respectively…

Let {x}\{x\} denote the fractional part of xx and f(x)=cos⁡−1(1−{x}2)sin⁡−1(1−{x}){x}−{x}3,x≠0f(x)=\frac{\cos ^{-1}\left(1-\{x\}^2\right) \sin ^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, x \neq 0 . If L\mathrm{L} and R\mathrm{R} respectively denotes the left hand limit and the right hand limit of f(x)f(x) at x=0x=0 , then 32π2( L2+R2)\frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\right) is equal to ___________.

Step-by-Step Verified Solution

Finding right hand limit

R = lim⁡x→0+f(x)=lim⁡h→0f(0+h)=lim⁡h→0f(h)\begin{gathered}\lim _{x \rightarrow 0^{+}} f(x)=\lim _{h \rightarrow 0} f(0+h)=\lim _{h \rightarrow 0} f(h)\end{gathered}

amp;=lim⁡h→0cos⁡−1(1−h2)sin⁡−1(1−h)h(1−h2)amp;=lim⁡h→0cos⁡−1(1−h2)h(sin⁡−111)\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-h^2\right) \sin ^{-1}(1-h)}{h\left(1-h^2\right)} \\\\ & =\lim _{\mathrm{h} \rightarrow 0} \frac{\cos ^{-1}\left(1-\mathrm{h}^2\right)}{\mathrm{h}}\left(\frac{\sin ^{-1} 1}{1}\right)\end{aligned}

Let cos⁡−1(1−h2)=θ⇒cos⁡θ=1−h2\cos ^{-1}\left(1-h^2\right)=\theta \Rightarrow \cos \theta=1-h^2

amp;=π2lim⁡θ→0θ1−cos⁡θamp;=π2lim⁡θ→011−cos⁡θθ2amp;=π211/2amp;∴R=π2\begin{aligned}& =\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{\theta}{\sqrt{1-\cos \theta}} \\\\& =\frac{\pi}{2} \lim _{\theta \rightarrow 0} \frac{1}{\sqrt{\frac{1-\cos \theta}{\theta^2}}} \\\\& =\frac{\pi}{2} \frac{1}{\sqrt{1 / 2}} \\\\& \therefore \mathrm{R}=\frac{\pi}{\sqrt{2}}\end{aligned}

Now finding left hand limit

amp;L=lim⁡x→0−f(x)=lim⁡h→0f(−h)\begin{aligned}& L=\lim _{x \rightarrow 0^{-}} f(x) =\lim _{h \rightarrow 0} f(-h)\end{aligned}

amp;=lim⁡h→0cos⁡−1(1−{−h}2)sin⁡−1(1−{−h}){−h}−{−h}3amp;=lim⁡h→0cos⁡−1(1−(−h+1)2)sin⁡−1(1−(−h+1))(−h+1)−(−h+1)3amp;=lim⁡h→0cos⁡−1(−h2+2h)sin⁡−1h(1−h)(1−(1−h)2)\begin{aligned} & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-\{-h\}^2\right) \sin ^{-1}(1-\{-h\})}{\{-h\}-\{-h\}^3} \\\\ & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(1-(-h+1)^2\right) \sin ^{-1}(1-(-h+1))}{(-h+1)-(-h+1)^3} \\\\ & =\lim _{h \rightarrow 0} \frac{\cos ^{-1}\left(-h^2+2 h\right) \sin ^{-1} h}{(1-h)\left(1-(1-h)^2\right)}\end{aligned}

amp;=lim⁡h→0(π2)sin⁡−1h(1−(1−h)2)amp;=π2lim⁡h→0(sin⁡−1h−h2+2h)amp;=π2lim⁡h→0(sin⁡−1hh)(1−h+2)amp;L=π4\begin{aligned} & =\lim _{h \rightarrow 0}\left(\frac{\pi}{2}\right) \frac{\sin ^{-1} h}{\left(1-(1-h)^2\right)} \\\\ & =\frac{\pi}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{-1} h}{-h^2+2 h}\right) \\\\ & =\frac{\pi}{2} \lim _{h \rightarrow 0}\left(\frac{\sin ^{-1} h}{h}\right)\left(\frac{1}{-h+2}\right) \\\\ & \quad L=\frac{\pi}{4}\end{aligned}

amp;32π2( L2+R2)=32π2(π22+π216)amp;=16+2amp;=18\begin{aligned} & \frac{32}{\pi^2}\left(\mathrm{~L}^2+\mathrm{R}^2\right)=\frac{32}{\pi^2}\left(\frac{\pi^2}{2}+\frac{\pi^2}{16}\right) \\\\ & =16+2 \\\\ & =18\end{aligned}

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