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Question Number 146838 by bobhans last updated on 16/Jul/21

$$\mathrm{Given}{4}^{\mathrm{x}}+4^{-\mathrm{x}}-2^{2-\mathrm{x}}+2^{2+\mathrm{x}}-7=0,\mathrm{x}>0$$ $$\mathrm{find}{2}^{\mathrm{x}}+2^{-\mathrm{x}}.$$ $$$$ $$\mathrm{if}\mathrm{x}\in [-\frac{\pi }{6},0]\mathrm{then}\mathrm{minimum}\mathrm{value}\mathrm{of}$$ $$\mathrm{function}\mathrm{f}\left(\mathrm{x}\right)=\cot (\mathrm{x}+\frac{\pi }{3})-\tan (\frac{2\pi }{3}-\mathrm{x})$$ $$\mathrm{when}\mathrm{x}=?$$

Answered by EDWIN88 last updated on 16/Jul/21

$$\left(1\right)\mathrm{let}{2}^{\mathrm{x}}=\mathrm{p}\wedge 2^{-\mathrm{x}}=\mathrm{q}\Rightarrow \mathrm{pq}=1$$ $$\Rightarrow {\mathrm{p}}^2+{\mathrm{q}}^2-4\mathrm{q}+4\mathrm{p}-7=0$$ $$\Rightarrow {(\mathrm{p}-\mathrm{q})}^2+2\mathrm{pq}+4(\mathrm{p}-\mathrm{q})-7=0$$ $$\Rightarrow {(\mathrm{p}-\mathrm{q})}^2+4(\mathrm{p}-\mathrm{q})-5=0$$ $$\Rightarrow (\mathrm{p}-\mathrm{q}+5)(\mathrm{p}-\mathrm{q}-1)=0$$ $$\{\begin{array}{l}\mathrm{p}-\mathrm{q}=-5\\ \mathrm{p}-\mathrm{q}=1\end{array}$$ $$\because \mathrm{p}+\mathrm{q}=\sqrt{{(\mathrm{p}+\mathrm{q})}^2}=\sqrt{{\mathrm{p}}^2+{\mathrm{q}}^2+2\mathrm{pq}}$$ $$=\sqrt{{(\mathrm{p}-\mathrm{q})}^2+4\mathrm{pq}}$$ $$=\sqrt{25+4}\mathrm{or}\sqrt{1+4}$$ $$=\sqrt{29}\mathrm{or}\sqrt{5}$$

Commented byRasheed.Sindhi last updated on 16/Jul/21

$$\mathcal{N}ice!$$

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