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Step-by-Step Solution
Step 1: Understand the Function and Fractional Part Behavior
The function is
f(x) = \dfrac{\cos^{-1}\bigl(1 - \{x\}^2 \bigr)\,\sin^{-1}\bigl(1 - \{x\}\bigr)}{\{x\} - \{x\}^3}, \quad x \neq 0,
where \{x\} is the fractional part of x . Recall the behavior of \{x\} as x \to 0 :
As x \to 0^{+} , x is a small positive number and so \{\,x\} = x \to 0.
As x \to 0^{-} , x is a small negative number. In that case, \lfloor x \rfloor = -1 , so \{\,x\} = x - \lfloor x \rfloor = x + 1 \to 1.
We need to find the left-hand limit L = \lim\limits_{x \to 0^-} f(x) and the right-hand limit R = \lim\limits_{x \to 0^+} f(x) , then evaluate
\dfrac{32}{\pi^2}\bigl(L^2 + R^2\bigr).
Step 2: Determine the Left-Hand Limit ( x \to 0^{-} )
When x \to 0^{-} , we have \{x\} = 1 + x . Let t = \{x\} \to 1 . To handle small deviations, set t = 1 + n , where n \to 0^{-} . Rewrite the expression in terms of t :
f(x)
= \dfrac{\cos^{-1}\!\bigl(1 - t^2\bigr)\,\sin^{-1}\!\bigl(1 - t\bigr)}{t - t^3}.
Substitute t = 1 + n :
1 - t^2 = 1 - (1 + n)^2 = 1 - [1 + 2n + n^2] = -\,2n - n^2.
1 - t = 1 - (1 + n) = -\,n.
t - t^3 = (1 + n) - (1 + n)^3
= (1 + n) - [1 + 3n + 3n^2 + n^3]
= -\,2n - 3n^2 - n^3.
For small n , we use approximations:
\sin^{-1}(-n) \approx -\,n.
\cos^{-1}(1 - t^2) \approx \tfrac{\pi}{2} \text{ for small } |1 - t^2|.
Focusing on the leading term in n :
\cos^{-1}\!\bigl(1 - t^2\bigr)\sin^{-1}\!\bigl(1 - t\bigr) \;\approx\; \bigl(\tfrac{\pi}{2}\bigr)\,(-n) \;=\; -\tfrac{\pi}{2}\,n.
Meanwhile, t - t^3 \approx -2n for very small n . Hence,
f(x) \;=\; \dfrac{\cos^{-1}\!\bigl(1 - t^2\bigr)\,\sin^{-1}\!\bigl(1 - t\bigr)}{t - t^3}
\;\approx\; \dfrac{-\tfrac{\pi}{2}\,n}{-\,2n}
= \dfrac{\pi}{4}.
Therefore,
L = \lim_{x \to 0^-} f(x) \;=\; \dfrac{\pi}{4}.
Step 3: Determine the Right-Hand Limit ( x \to 0^{+} )
When x \to 0^{+} , \{\,x\} = x . Then
f(x) = \dfrac{\cos^{-1}\bigl(1 - x^2\bigr)\,\sin^{-1}\bigl(1 - x\bigr)}{x - x^3}.
For very small x , x - x^3 \approx x . We approximate each factor:
For \cos^{-1}(1 - x^2) : Near x = 0 , set \eta = x^2 . A known expansion is \cos^{-1}(1 - \eta) \approx \sqrt{2\,\eta} for small \eta . Thus
\cos^{-1}(1 - x^2) \approx \sqrt{2}\,x.
For \sin^{-1}(1 - x) : Near x = 0 , 1 - x \approx 1 , so \sin^{-1}(1) = \tfrac{\pi}{2} . A small deviation suggests
\sin^{-1}(1 - x) \approx \tfrac{\pi}{2} - \sqrt{2\,x}.
Multiplying these approximations gives:
\cos^{-1}(1 - x^2)\,\sin^{-1}(1 - x)
\;\approx\; \bigl(\sqrt{2}\,x\bigr)\,\Bigl(\tfrac{\pi}{2} - \sqrt{2\,x}\Bigr)
\;=\; \tfrac{\pi}{2}\,\sqrt{2}\,x \;-\; 2\,x^{3/2}.
Dividing by x - x^3 \approx x , we obtain:
f(x)
\approx \dfrac{\tfrac{\pi}{2}\,\sqrt{2}\,x - 2\,x^{3/2}}{x}
= \tfrac{\pi}{2}\,\sqrt{2} \;-\; 2\,\sqrt{x}.
As x \to 0^{+} , \sqrt{x} \to 0 , so
R = \lim_{x \to 0^+} f(x) = \tfrac{\pi}{2}\,\sqrt{2}.
Step 4: Compute the Required Expression
We found:
L = \dfrac{\pi}{4},
\quad
R = \dfrac{\pi}{2}\,\sqrt{2}.
Then,
L^2 = \Bigl(\dfrac{\pi}{4}\Bigr)^2 = \dfrac{\pi^2}{16},
\quad
R^2 = \Bigl(\dfrac{\pi}{2}\,\sqrt{2}\Bigr)^2 = \dfrac{\pi^2}{2}.
Adding these:
L^2 + R^2
= \dfrac{\pi^2}{16} + \dfrac{\pi^2}{2}
= \dfrac{\pi^2}{16} + \dfrac{8\,\pi^2}{16}
= \dfrac{9\,\pi^2}{16}.
Finally, we evaluate
\dfrac{32}{\pi^2}\,\bigl(L^2 + R^2\bigr)
= \dfrac{32}{\pi^2} \times \dfrac{9\,\pi^2}{16}
= \dfrac{32 \times 9}{16}
= 18.
Final Answer
\displaystyle 18
