The set of all values of k for which (tan^{-1 | vedclass

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(A) Let $f(x) = (	an^{-1} x)^3 + (\cot^{-1} x)^3$. We know that $	an^{-1} x + \cot^{-1} x = \frac{\pi}{2}$. Let $u = 	an^{-1} x$. Then $\cot^{-1} x

Vedclass · Accepted Answer

$[\frac{1}{32}, \frac{7}{8})$

Vedclass · Answer

(A) Let $f(x) = (	an^{-1} x)^3 + (\cot^{-1} x)^3$. We know that $	an^{-1} x + \cot^{-1} x = \frac{\pi}{2}$. Let $u = 	an^{-1} x$. Then $\cot^{-1} x = \frac{\pi}{2} - u$. Since $x \in \mathbb{R}$,$u \in (-\frac{\pi}{2}, \frac{\pi}{2})$. Now,$f(x) = u^3 + (\frac{\pi}{2} - u)^3 = u^3 + \frac{\pi^3}{8} - \frac{3\pi^2}{4}u + \frac{3\pi}{2}u^2 - u^3 = \frac{3\pi}{2}u^2 - \frac{3\pi^2}{4}u + \frac{\pi^3}{8}$. Completing the square: $f(x) = \frac{3\pi}{2}(u^2 - \frac{\pi}{2}u) + \frac{\pi^3}{8} = \frac{3\pi}{2}(u - \frac{\pi}{4})^2 - \frac{3\pi}{2}(\frac{\pi^2}{16}) + \frac{\pi^3}{8} = \frac{3\pi}{2}(u - \frac{\pi}{4})^2 + \frac{\pi^3}{32}$. Since $u \in (-\frac{\pi}{2}, \frac{\pi}{2})$,the range of $(u - \frac{\pi}{4})^2$ is $[0, (-\frac{\pi}{2} - \frac{\pi}{4})^2) = [0, \frac{9\pi^2}{16})$. Thus,the range of $f(x)$ is $[\frac{\pi^3}{32}, \frac{3\pi}{2}(\frac{9\pi^2}{16}) + \frac{\pi^3}{32}) = [\frac{\pi^3}{32}, \frac{27\pi^3}{32} + \frac{\pi^3}{32}) = [\frac{\pi^3}{32}, \frac{28\pi^3}{32}) = [\frac{\pi^3}{32}, \frac{7\pi^3}{8})$. Given $f(x) = k \pi^3$,we have $k \in [\frac{1}{32}, \frac{7}{8})$.

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

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