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Question Number 146838 by bobhans last updated on 16/Jul/21
Given 4^x +4^(−x) −2^(2−x) +2^(2+x) −7=0 ,x>0 find 2^x +2^(−x) . if x∈[ −(π/6),0 ] then minimum value of function f(x)=cot (x+(π/3))−tan (((2π)/3)−x) when x = ?
$$\mathrm{Given}\:\mathrm{4}^{\mathrm{x}} +\mathrm{4}^{−\mathrm{x}} −\mathrm{2}^{\mathrm{2}−\mathrm{x}} +\mathrm{2}^{\mathrm{2}+\mathrm{x}} −\mathrm{7}=\mathrm{0}\:,\mathrm{x}>\mathrm{0} \\ $$$$\:\mathrm{find}\:\mathrm{2}^{\mathrm{x}} +\mathrm{2}^{−\mathrm{x}} . \\ $$$$\: \\ $$$$\:\mathrm{if}\:\mathrm{x}\in\left[\:−\frac{\pi}{\mathrm{6}},\mathrm{0}\:\right]\:\mathrm{then}\:\mathrm{minimum}\:\mathrm{value}\:\mathrm{of}\: \\ $$$$\mathrm{function}\:\mathrm{f}\left(\mathrm{x}\right)=\mathrm{cot}\:\left(\mathrm{x}+\frac{\pi}{\mathrm{3}}\right)−\mathrm{tan}\:\left(\frac{\mathrm{2}\pi}{\mathrm{3}}−\mathrm{x}\right)\: \\ $$$$\mathrm{when}\:\mathrm{x}\:=\:?\: \\ $$
Answered by EDWIN88 last updated on 16/Jul/21
(1) let 2^x = p ∧ 2^(−x) = q⇒pq=1 ⇒ p^2 +q^2 −4q+4p−7=0 ⇒(p−q)^2 +2pq+4(p−q)−7=0 ⇒(p−q)^2 +4(p−q)−5=0 ⇒(p−q+5)(p−q−1)=0 { ((p−q=−5)),((p−q=1)) :} ∵ p+q = (√((p+q)^2 )) =(√(p^2 +q^2 +2pq)) = (√((p−q)^2 +4pq)) = (√(25+4)) or (√(1+4)) = (√(29)) or (√5)
$$\left(\mathrm{1}\right)\:\mathrm{let}\:\mathrm{2}^{\mathrm{x}} \:=\:\mathrm{p}\:\wedge\:\mathrm{2}^{−\mathrm{x}} =\:\mathrm{q}\Rightarrow\mathrm{pq}=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} −\mathrm{4q}+\mathrm{4p}−\mathrm{7}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{p}−\mathrm{q}\right)^{\mathrm{2}} +\mathrm{2pq}+\mathrm{4}\left(\mathrm{p}−\mathrm{q}\right)−\mathrm{7}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{p}−\mathrm{q}\right)^{\mathrm{2}} +\mathrm{4}\left(\mathrm{p}−\mathrm{q}\right)−\mathrm{5}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{p}−\mathrm{q}+\mathrm{5}\right)\left(\mathrm{p}−\mathrm{q}−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\:\:\begin{cases}{\mathrm{p}−\mathrm{q}=−\mathrm{5}}\\{\mathrm{p}−\mathrm{q}=\mathrm{1}}\end{cases} \\ $$$$\because\:\mathrm{p}+\mathrm{q}\:=\:\sqrt{\left(\mathrm{p}+\mathrm{q}\right)^{\mathrm{2}} }\:=\sqrt{\mathrm{p}^{\mathrm{2}} +\mathrm{q}^{\mathrm{2}} +\mathrm{2pq}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\sqrt{\left(\mathrm{p}−\mathrm{q}\right)^{\mathrm{2}} +\mathrm{4pq}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\sqrt{\mathrm{25}+\mathrm{4}}\:\mathrm{or}\:\sqrt{\mathrm{1}+\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\sqrt{\mathrm{29}}\:\mathrm{or}\:\sqrt{\mathrm{5}}\: \\ $$
Commented by Rasheed.Sindhi last updated on 16/Jul/21
Nice!
$$\mathcal{N}{ice}! \\ $$

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