JEE Main 2022 — Mathematics Trigonometry

Let $T={({\mathrm{tan}}^{-1}x)}^{3}+{({\mathrm{cot}}^{-1}x)}^{3}$T=(tan−1x)3+(cot−1x)3

$$={({\mathrm{tan}}^{-1}x+{\mathrm{cot}}^{-1}x)}^{3}-3{\mathrm{tan}}^{-1}x\cdot {\mathrm{cot}}^{-1}x({\mathrm{tan}}^{-1}x+{\mathrm{cot}}^{-1}x)$$=(tan−1x+cot−1x)3−3tan−1x⋅cot−1x(tan−1x+cot−1x)

$=\frac{{\pi }^{3}}{8}-\frac{3\pi }{2}{\mathrm{tan}}^{-1}x(\frac{\pi }{2}-{\mathrm{tan}}^{-1}x)$=8π3​−23π​tan−1x(2π​−tan−1x) $(\because {\mathrm{tan}}^{-1}x+{\mathrm{cot}}^{-1}x=\frac{\pi }{2})$(∵tan−1x+cot−1x=2π​)

$$=\frac{3\pi }{2}{({\mathrm{tan}}^{-1}x)}^{2}-\frac{3{\pi }^{2}}{4}{\mathrm{tan}}^{-1}x+\frac{{\pi }^{3}}{8}$$=23π​(tan−1x)2−43π2​tan−1x+8π3​

$$T=\frac{3\pi }{2}{({\mathrm{tan}}^{-1}x-\frac{\pi }{4})}^{2}+\frac{{\pi }^{3}}{32}$$T=23π​(tan−1x−4π​)2+32π3​

We know than ${\mathrm{tan}}^{-1}x\in (-\frac{\pi }{2},\frac{\pi }{2})$tan−1x∈(−2π​,2π​)

Minimum value of $T$T is at ${\mathrm{tan}}^{-1}x=\frac{\pi }{4}$tan−1x=4π​ and maximum value at ${\mathrm{tan}}^{-1}x=-\frac{\pi }{2}$tan−1x=−2π​

$$\Rightarrow \frac{{\pi }^{3}}{32}\leq T<\frac{7}{8}{\pi }^{3}$$⇒32π3​≤T<87​π3

$$=\frac{{\pi }^{3}}{32}\leq k{\pi }^{3}<\frac{7}{8}{\pi }^{3}$$=32π3​≤kπ3<87​π3

$$\Rightarrow \frac{1}{32}\leq k<\frac{7}{8}$$⇒321​≤k<87​