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One-sided limits at 0 of inverse trig functions of the fractional part

This type of question, asked 3 times: 2021 Paper 6, 2021 Paper 7, 1 Feb 2024. All 3 questions below, oldest first.

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  1. Question 1

    2021 Paper 6

    63
    Let α∈R\alpha\in\mathbb{R} be such that the function f(x)={cos⁡−1(1−{x}2) sin⁡−1(1−{x}){x}−{x}3,x≠0α,x=0f(x)=\begin{cases} \dfrac{\cos^{-1}(1-\{x\}^2)\,\sin^{-1}(1-\{x\})}{\{x\}-\{x\}^3}, & x\neq 0 \\ \alpha, & x=0 \end{cases} is continuous at x=0x=0, where {x}=x−[x]\{x\}=x-[x] and [x][x] is the greatest integer less than or equal to xx. Then:

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  2. Question 2

    2021 Paper 7

    80
    The value of lim⁡x→0+cos⁡−1(x−[x]2)⋅sin⁡−1(x−[x]2)x−x3\lim_{x\rightarrow0^{+}} \frac{\cos^{- 1}\left( x- \lbrack x\rbrack^{2} \right)\cdot\sin^{- 1}\left( x- \lbrack x\rbrack^{2} \right)}{x-x^{3}}, where [x] denotes the greatest integer ≤x\leq x is:

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  3. Question 3

    1 Feb 2024

    85
    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}\left( 1 - \{ x\} \right)}{\{ x\} - \{ x\}^{3}},x\neq 0. If LL and RR 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( L^{2}+R^{2} \right) is equal to