Let { x } denote the fractional part of x and If L and R respectively denotes the left hand limit and the right hand limit of f ( x ) at x = 0 , then is equal to _____. [2024]

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Solution

Q.
Let ${x}$ denote the fractional part of $x$ and $f (x) = \frac{\cos^{- 1} (1 - {x}^{2}) \sin^{- 1} (1 - {x})}{{x} - {x}^{3}}, x \neq 0 .$

If $L$ and $R$ respectively denotes the left hand limit and the right hand limit of $f (x)$ at $x = 0,$ then $\frac{32}{π^{2}} (L^{2} + R^{2})$ is equal to _____. [2024]

Ans.
(18)

$We have, f (x) = \frac{\cos^{- 1} (1 - {x}^{2}) \sin^{- 1} (1 - {x})}{{x} - {x}^{3}}$

$R = \lim_{h \to 0^{+}} f (x) = \lim_{h \to 0} f (0 + h)$

$= \lim_{h \to 0^{+}} f (h) = \lim_{h \to 0} \frac{\cos^{- 1} (1 - h^{2})}{h} (\frac{\sin^{- 1} 1}{1})$

$Put \cos^{- 1} (1 - h^{2}) = θ \Rightarrow \cos θ = 1 - h^{2}$

$= \frac{π}{2} \lim_{θ \to 0} \frac{θ}{\sqrt{1 - \cos θ}} = \frac{π}{2} \lim_{θ \to 0} \frac{1}{\sqrt{\frac{1 - \cos θ}{θ^{2}}}} = \frac{π}{2} \frac{1}{\sqrt{1 / 2}} = \frac{π}{\sqrt{2}}$

Also, $L = \lim_{x \to 0^{-}} f (x) = \lim_{h \to 0} f (- h)$

$= \lim_{h \to 0} \frac{\cos^{- 1} (1 - {- h}^{2}) \sin^{- 1} (1 - {- h})}{{- h} - {- h}^{3}}$

$= \lim_{h \to 0} \frac{\cos^{- 1} (1 - {(- h + 1)}^{2}) \sin^{- 1} (1 - (- h + 1))}{(- h + 1) - {(- h + 1)}^{3}}$

$= \lim_{h \to 0} \frac{\cos^{- 1} (- h^{2} + 2 h) \sin^{- 1} h}{(1 - h) (1 - {(1 - h)}^{2})}$

$= \lim_{h \to 0} (\frac{π}{2}) \frac{\sin^{- 1} h}{(1 - {(1 - h)}^{2})} = \frac{π}{2} \lim_{h \to 0} (\frac{\sin^{- 1} h}{- h^{2} + 2 h})$

$= \frac{π}{2} \lim_{h \to 0} (\frac{\sin^{- 1} h}{h}) (\frac{1}{- h + 2}) = \frac{π}{4}$

$∴$ $\frac{32}{π^{2}} (L^{2} + R^{2}) = \frac{32}{π^{2}} (\frac{π^{2}}{2} + \frac{π^{2}}{16}) = 16 + 2 = 18$

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