Question

For $k\in R$, let the solutions of the equation $\cos \left({\sin }^{-1}\left(x\cot \left({\tan }^{-1}\left(\cos \left({\sin }^{-1}x\right)\right)\right)\right)\right)=k,0<|x|<\frac{1}{\sqrt{2}}$ be $\alpha$ and $\beta$, where the inverse trigonometric functions take only principal values. If the solutions of the equation $x^2-bx-5=0$ are $\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}$ and $\frac{\alpha }{\beta }$, then $\frac{b}{k^2}$ is equal to ______

Answer 1

Answer: 12

$\cos \left({\sin }^{-1}x\right)=\cos \left({\cos }^{-1}\sqrt{1-x^2}\right)=\sqrt{1-x^2}$
$\cot \left({\tan }^{-1}\sqrt{1-x^2}\right)=\cot {\cot }^{-1}\left(\sqrt{\frac{1}{\sqrt{1-x^2}}}\right)=\frac{1}{\sqrt{1-x^2}}$
$\Rightarrow \cos \left({\sin }^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\right)=\frac{\sqrt{1-2x^2}}{\sqrt{1-x^2}}$
$\Rightarrow \frac{\sqrt{1-2x^2}}{\sqrt{1-x^2}}=k$
$\Rightarrow 1-2x^2=k^2\left(1-x^2\right)$
$\Rightarrow \left(k^2-2\right)x^2=k^2-1$
$x^2=\frac{k^2-1}{k^2-2}$
$\alpha =\sqrt{\frac{k^2-1}{k^2-2}}\Rightarrow {\alpha }^2=\frac{k^2-1}{k^2-2}$
$\beta =\sqrt{\frac{k^2-1}{k^2-2}}\Rightarrow {\beta }^2=\frac{k^2-1}{k^2-2}$
$\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}=2\left(\frac{k^2-2}{k^2-1}\right)\&\frac{\alpha }{\beta }=-1$
Sum of roots $=\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}+\frac{\alpha }{\beta }=b$
$\Rightarrow \frac{2\left(k^2-2\right)}{k^2-1}-1=b\dots .(1)$
Product of roots $=\left(\frac{1}{{\alpha }^2}+\frac{1}{{\beta }^2}\right)\frac{\alpha }{\beta }=-5$
$\Rightarrow \frac{2\left(k^2-2\right)}{k^2-1}(-1)=-5$
$\Rightarrow 2k^2-4=5k^2-5$
$\Rightarrow 3k^2=1\Rightarrow k^2=\frac{1}{3}\dots .\text{ Put in (1) }$
$\Rightarrow b=\frac{2\left(k^2-2\right)}{k^2-1}-1=5-1=4$
$\frac{b}{k^2}=\frac{4}{\frac{1}{3}}=12$