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JEE Main 20241 FebruaryMorning Shift
Exam
JEE Main
Year
2024
Date
1 February
Shift
Morning Shift
Subject
Mathematics
Chapter
Limits

JEE Main 2024 · 1 February · Morning Shift · Mathematics · Limits

Let \ x\ denote the fractional part of x and f(x) = ^ - 1 (1 - \ x\...

Let {x}\left\{x\right\} denote the fractional part of xandf(x)=cos⁡−1(1−{x}2)sin⁡−1(1−{x}){x}−{x}3,x≠0x and f\left(x\right) = \dfrac{\cos^{- 1}\left(1 - {\left\{x\right\}}^{2}\right)\sin^{- 1}\left(1 - \left\{x\right\}\right)}{\left\{x\right\} - {\left\{x\right\}}^{3}}, x \ne 0. If LandRL\text{and}R respectively denotes the left hand limit and the right hand limit of f(x)f\left(x\right) at x=0x = 0, then $\dfrac{32}{{\pi}^{2}}\left({L}^{2} + {R}^{2}\right) is equal to __________.

Correct answer

0

Step-by-step solution

Given: f(x)=cos⁡−1(1−{x}2)sin⁡−1(1−{x}){x}−{x}3f\left(x\right) = \dfrac{\cos^{- 1}\left(1 - {\left\{x\right\}}^{2}\right)\sin^{- 1}\left(1 - \left\{x\right\}\right)}{\left\{x\right\} - {\left\{x\right\}}^{3}}

⇒RHL=lim⁡x→0+f(x)=lim⁡h→0f(0+h)⇒ RHL = \lim_{x \to {0}^{+}}f\left(x\right) = \lim_{h \to 0}f\left(0 + h\right)

⇒RHL=lim⁡h→0f(h)⇒ RHL = \lim_{h \to 0}f\left(h\right)

⇒RHL=lim⁡h→0cos⁡−1(1−h2)sin⁡−1(1−h)h(1−h2)⇒ RHL = \lim_{h \to 0}\dfrac{\cos^{- 1}\left(1 - {h}^{2}\right)\sin^{- 1}\left(1 - h\right)}{h\left(1 - {h}^{2}\right)}

⇒RHL=lim⁡h→0cos⁡−1(1−h2)h(sin⁡−111)⇒ RHL = \lim_{h \to 0}\dfrac{\cos^{- 1}\left(1 - {h}^{2}\right)}{h}\left(\dfrac{\sin^{- 1}1}{1}\right)

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

⇒RHL=π2lim⁡θ→0θ1−cosθ⇒ RHL = \dfrac{\pi}{2}\lim_{\theta \to 0}\dfrac{\theta}{\sqrt{1 - cos\theta}}

⇒RHL=π2lim⁡θ→011−cosθθ2⇒ RHL = \dfrac{\pi}{2}\lim_{\theta \to 0}\dfrac{1}{\sqrt{\dfrac{1 - cos\theta}{{\theta}^{2}}}}

⇒RHL=π2112⇒ RHL = \dfrac{\pi}{2}\dfrac{1}{\sqrt{\dfrac{1}{2}}}

⇒RHL=π2⇒ RHL = \dfrac{\pi}{\sqrt{2}}

⇒R=π2⇒ R = \dfrac{\pi}{\sqrt{2}}

Now finding left hand limit,

⇒LHL=lim⁡x→0−f(x)⇒ LHL = \lim_{x \to {0}^{-}}f\left(x\right)

⇒LHL=lim⁡h→0f(−h)⇒ LHL = \lim_{h \to 0}f\left(- h\right)

⇒LHL=lim⁡h→0cos⁡−1(1−{−h}2)sin⁡−1(1−{−h}){−h}−{−h}3⇒ LHL = \lim_{h \to 0}\dfrac{\cos^{- 1}\left(1 - {\left\{- h\right\}}^{2}\right)\sin^{- 1}\left(1 - \left\{- h\right\}\right)}{\left\{- h\right\} - {\left\{- h\right\}}^{3}}

⇒LHL=lim⁡h→0cos⁡−1(1−(−h+1)2)sin⁡−1(1−(−h+1))(−h+1)−(−h+1)3⇒ LHL = \lim_{h \to 0}\dfrac{\cos^{- 1}\left(1 - {\left(- h + 1\right)}^{2}\right)\sin^{- 1}\left(1 - \left(- h + 1\right)\right)}{\left(- h + 1\right) - {\left(- h + 1\right)}^{3}}

⇒LHL=lim⁡h→0cos⁡−1(−h2+2h)sin⁡−1h(1−h)(1−(1−h)2)⇒ LHL = \lim_{h \to 0}\dfrac{\cos^{- 1}\left(- {h}^{2} + 2h\right)\sin^{- 1}h}{\left(1 - h\right)\left(1 - {\left(1 - h\right)}^{2}\right)}

⇒LHL=lim⁡h→0(π2)sin⁡−1h(1−(1−h)2)⇒ LHL = \lim_{h \to 0}\left(\dfrac{\pi}{2}\right)\dfrac{\sin^{- 1}h}{\left(1 - {\left(1 - h\right)}^{2}\right)}

⇒LHL=π2lim⁡h→0(sin⁡−1h−h2+2h)⇒ LHL = \dfrac{\pi}{2}\lim_{h \to 0}\left(\dfrac{\sin^{- 1}h}{- {h}^{2} + 2h}\right)

⇒LHL=π2lim⁡h→0(sin⁡−1hh)(1−h+2)⇒ LHL = \dfrac{\pi}{2}\lim_{h \to 0}\left(\dfrac{\sin^{- 1}h}{h}\right)\left(\dfrac{1}{- h + 2}\right)

⇒LHL=π4⇒ LHL = \dfrac{\pi}{4}

⇒L=π4⇒ L = \dfrac{\pi}{4}

⇒32π2(L2+R2)=32π2(π22+π216)⇒ \dfrac{32}{{\pi}^{2}}\left({L}^{2} + {R}^{2}\right) = \dfrac{32}{{\pi}^{2}}\left(\dfrac{{\pi}^{2}}{2} + \dfrac{{\pi}^{2}}{16}\right)

⇒32π2(L2+R2)=16+2⇒ \dfrac{32}{{\pi}^{2}}\left({L}^{2} + {R}^{2}\right) = 16 + 2

⇒32π2(L2+R2)=18⇒ \dfrac{32}{{\pi}^{2}}\left({L}^{2} + {R}^{2}\right) = 18

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