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If $\mathrm{f}(\mathrm{x})=\frac{\mathrm{a}^{2}-1}{\mathrm{a}^{2}+1} \mathrm{x}^{3}-3 \mathrm{x}+5$ is a decreasing
function of $\mathrm{x}$ in $\mathbf{R}$, then the set of possible values of a (independent of $\mathrm{x}$ ) is
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function of $\mathrm{x}$ in $\mathbf{R}$, then the set of possible values of a (independent of $\mathrm{x}$ ) is
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The correct answer is:
[-1,1]
$f^{\prime}(x)=3\left(\frac{a^{2}-1}{a^{2}+1}\right) x^{2}-3$
$\mathrm{f}^{\prime}(\mathrm{x})<0$ for all $\mathrm{x}$ if $\mathrm{a}^{2}-1 \leq 0 \Rightarrow-1 \leq \mathrm{a} \leq 1$
$\mathrm{f}^{\prime}(\mathrm{x})<0$ for all $\mathrm{x}$ if $\mathrm{a}^{2}-1 \leq 0 \Rightarrow-1 \leq \mathrm{a} \leq 1$
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