Consider a function f: [-1, 1] to R where f(x | vedclass

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(B) Given $f(x) = \sum_{k=0}^{n} \alpha_{2k+1} (\sin^{-1} x)^{2k+1} - \cot^{-1} x$. Taking the derivative with respect to $x$: $f'(x) = \frac{1}{\sqrt

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one-one and into

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(B) Given $f(x) = \sum_{k=0}^{n} \alpha_{2k+1} (\sin^{-1} x)^{2k+1} - \cot^{-1} x$. Taking the derivative with respect to $x$: $f'(x) = \frac{1}{\sqrt{1-x^2}} \left( \alpha_1 + 3\alpha_3 (\sin^{-1} x)^2 + \dots + (2n+1)\alpha_{2n+1} (\sin^{-1} x)^{2n} \right) + \frac{1}{1+x^2}$. Since $\alpha_i > 0$ and the terms $(\sin^{-1} x)^{2k}$ are non-negative,$f'(x) > 0$ for all $x \in (-1, 1)$. Thus,$f(x)$ is a strictly increasing function,which implies it is one-one. The domain is $[-1, 1]$,so the range is $[f(-1), f(1)]$. $f(-1) = \alpha_1(-\pi/2) + \dots + \alpha_{2n+1}(-\pi/2)^{2n+1} - (3\pi/4)$ and $f(1) = \alpha_1(\pi/2) + \dots + \alpha_{2n+1}(\pi/2)^{2n+1} - (\pi/4)$. Since the range $[f(-1), f(1)]$ is a subset of $R$ but not equal to $R$,the function is into. Therefore,$f(x)$ is one-one and into.

Consider a function $f: [-1, 1] \to R$ where $f(x) = \alpha_1 \sin^{-1} x + \alpha_3 (\sin^{-1} x)^3 + \dots + \alpha_{2n+1} (\sin^{-1} x)^{2n+1} - \cot^{-1} x$,where $\alpha_i$ are positive constants and $n \in N < 100$. Then $f(x)$ is:

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