Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

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 by Rasheed.Sindhi last updated on 16/Jul/21

$$\mathcal{N}ice!$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com