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(C) Let $f(x) = \cos \left(\sin ^{-1}\left(x \cot \left(	an ^{-1}\left(\cos \left(\sin ^{-1} xight)ight)ight)ight)ight) = k$.First,$\cos(\s

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$12$

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(C) Let $f(x) = \cos \left(\sin ^{-1}\left(x \cot \left(	an ^{-1}\left(\cos \left(\sin ^{-1} xight)ight)ight)ight)ight) = k$.First,$\cos(\sin^{-1} x) = \sqrt{1-x^2}$.Then,$	an^{-1}(\sqrt{1-x^2}) = 	heta \implies 	an 	heta = \sqrt{1-x^2} \implies \cot 	heta = \frac{1}{\sqrt{1-x^2}}$.Substituting this back,we get $\cos(\sin^{-1}(\frac{x}{\sqrt{1-x^2}})) = k$.Let $\sin^{-1}(\frac{x}{\sqrt{1-x^2}}) = \phi \implies \sin \phi = \frac{x}{\sqrt{1-x^2}} \implies \cos \phi = \sqrt{1 - \frac{x^2}{1-x^2}} = \sqrt{\frac{1-2x^2}{1-x^2}}$.So,$\sqrt{\frac{1-2x^2}{1-x^2}} = k \implies \frac{1-2x^2}{1-x^2} = k^2 \implies 1-2x^2 = k^2 - k^2x^2 \implies x^2(k^2-2) = k^2-1 \implies x^2 = \frac{k^2-1}{k^2-2}$.Since $x^2 = \alpha^2 = \beta^2$,we have $\alpha^2 = \beta^2 = \frac{k^2-1}{k^2-2}$.Given the roots of $x^2 - bx - 5 = 0$ are $\frac{1}{\alpha^2} + \frac{1}{\beta^2}$ and $\frac{\alpha}{\beta}$.Note that $\frac{\alpha}{\beta}$ can be $1$ or $-1$. Since $x^2 = \frac{k^2-1}{k^2-2}$,$\alpha = \pm \sqrt{\frac{k^2-1}{k^2-2}}$. If $\alpha = \beta$,$\frac{\alpha}{\beta} = 1$,but the equation $x^2-bx-5=0$ has roots $\frac{2}{\alpha^2}$ and $1$. Product $= \frac{2}{\alpha^2} = -5$,which is impossible as $\alpha^2 > 0$. Thus $\alpha = -\beta$,so $\frac{\alpha}{\beta} = -1$.Roots are $\frac{2}{\alpha^2}$ and $-1$. Product: $\frac{2}{\alpha^2}(-1) = -5 \implies \frac{2}{\alpha^2} = 5 \implies \alpha^2 = \frac{2}{5}$.Sum of roots: $\frac{2}{\alpha^2} - 1 = b \implies 5 - 1 = 4 = b$.From $\alpha^2 = \frac{k^2-1}{k^2-2} = \frac{2}{5} \implies 5k^2 - 5 = 2k^2 - 4 \implies 3k^2 = 1 \implies k^2 = \frac{1}{3}$.Finally,$\frac{b}{k^2} = \frac{4}{1/3} = 12$.

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