The set of all values of k for which (^-1x)^3+(^-1x)^3=k^3, xR, has a solution, is the int…

Question

The set of all values of $k$ for which $(	an^{-1}x)^3+(\cot^{-1}x)^3=k\pi^3$, $x\in\mathbb{R}$, has a solution, is the interval :

Accepted Answer

Let $A=\tan^{-1}x\in\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, $B=\cot^{-1}x$, with $A+B=\dfrac{\pi}{2}$. Then

$$A^3+B^3=(A+B)\left[(A+B)^2-3AB\right]=\frac{\pi}{2}\left[\frac{\pi^2}{4}-3AB\right],\qquad AB=A\left(\frac{\pi}{2}-A\right).$$

As $A$ ranges over $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$, $AB$ ranges over $\left(-\dfrac{\pi^2}{2},\dfrac{\pi^2}{16}\right]$ (maximum at $A=\dfrac{\pi}{4}$).

$$k=\frac{1}{2\pi^2}\left[\frac{\pi^2}{4}-3AB\right]\cdot\frac{1}{\pi}\cdot\pi = \frac18-\frac{3AB}{2\pi^2}.$$

- At $AB=\dfrac{\pi^2}{16}$: $k=\dfrac18-\dfrac{3}{32}=\dfrac{1}{32}$.
- As $AB\to-\dfrac{\pi^2}{2}$: $k\to\dfrac18+\dfrac34=\dfrac78$ (not attained, since $A\to-\frac\pi2$ is excluded).

So $k\in\left[\dfrac{1}{32},\dfrac78\right)$.

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