For kR, let the solutions of the equation (^-1(x(^-1((^-1x)))))=k, 0

Question

For $k\in\mathbb{R}$, let the solutions of the equation $\cos\left(\sin^{-1}\left(x\cot\left(	an^{-1}(\cos(\sin^{-1}x))ight)ight)ight)=k$, $0<x<\dfrac{1}{\sqrt2}$, 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 $\dfrac{1}{\alpha^2}+\dfrac{1}{\beta^2}$ and $\dfrac{\alpha}{\beta}$, then $\dfrac{b}{k^2}$ is equal to \_\_\_\_\_\_\_\_.

Answer: 12

Accepted Answer

Let $\theta=\sin^{-1}x$, so $\cos\theta=\sqrt{1-x^2}$, and $\tan^{-1}(\cos\theta)$ has cotangent $\dfrac{1}{\sqrt{1-x^2}}$... working through the composition gives $\cos\left(\sin^{-1}\left(\dfrac{x}{\sqrt{1-x^2}}\right)\right)=\dfrac{\sqrt{1-2x^2}}{\sqrt{1-x^2}}=k$.

$$\Rightarrow\frac{1-2x^2}{1-x^2}=k^2\ \Rightarrow\ x^2=\frac{k^2-1}{k^2-2}.$$

The two roots $\alpha,\beta$ of this (for the given range of $x$) satisfy

$$\frac{1}{\alpha^2}+\frac{1}{\beta^2}=\frac{2(k^2-2)}{k^2-1},\qquad \frac{\alpha}{\beta}=-1.$$

By Vieta's on $x^2-bx-5=0$: product of roots $=-5$, so

$$\frac{2(k^2-2)}{k^2-1}\cdot(-1)=-5\ \Rightarrow\ k^2=\frac13,$$

and $b=$ sum of roots $=\dfrac{2(k^2-2)}{k^2-1}-1=4$. Hence

$$\frac{b}{k^2}=\frac{4}{1/3}=12.$$

Solutions · 1