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Question Number 128942 by bramlexs22 last updated on 11/Jan/21

$$\mathrm{Given}:3xf\left(\frac{1}{x}\right)+f\left(x\right)=2x+2$$$$andf\left(3\right),\mathrm{f}\left(9\right),\mathrm{f}\left(\mathrm{a}\right)\mathrm{three}\mathrm{first}$$$$\mathrm{term}\mathrm{in}\mathrm{AP}\mathrm{respectively}.\mathrm{Find}$$$$\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{a}?$$

Answered by liberty last updated on 11/Jan/21

$$\left(1\right)3\mathrm{xf}\left(\frac{1}{\mathrm{x}}\right)+\mathrm{f}\left(\mathrm{x}\right)=2\mathrm{x}+2$$$$\left(2\right)\frac{3}{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\frac{2}{\mathrm{x}}+2$$$$3\mathrm{f}\left(\mathrm{x}\right)+\mathrm{xf}\left(\frac{1}{\mathrm{x}}\right)=2+2\mathrm{x}$$$$9\mathrm{f}\left(\mathrm{x}\right)+3\mathrm{xf}\left(\frac{1}{\mathrm{x}}\right)=6\mathrm{x}+6$$$$\mathrm{substract}\left(2\right)\mathrm{\&}\left(1\right)$$$$8\mathrm{f}\left(\mathrm{x}\right)=4\mathrm{x}+4;\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{x}+1}{2}$$$$\mathrm{then}\{\begin{array}{l}\mathrm{f}\left(3\right)=2;\mathrm{f}\left(9\right)=5\\ \mathrm{f}\left(\mathrm{a}\right)=\frac{\mathrm{a}+1}{2}\end{array}$$$$\mathrm{since}\mathrm{f}\left(3\right),\mathrm{f}\left(9\right),\mathrm{f}\left(\mathrm{a}\right)\mathrm{in}\mathrm{AP}$$$$\iff 2\mathrm{f}\left(9\right)=\mathrm{f}\left(3\right)+\mathrm{f}\left(\mathrm{a}\right)$$$$\frac{\mathrm{a}+1}{2}=10-2=8$$$$\mathrm{a}+1=16\Rightarrow \mathrm{a}=15$$

Answered by mathmax by abdo last updated on 11/Jan/21

$$3\mathrm{xf}\left(\frac{1}{\mathrm{x}}\right)+\mathrm{f}\left(\mathrm{x}\right)=2\mathrm{x}+2\Rightarrow \frac{3}{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\frac{2}{\mathrm{x}}+2\Rightarrow \{\begin{array}{l}\mathrm{f}\left(\mathrm{x}\right)+3\mathrm{xf}\left(\frac{1}{\mathrm{x}}\right)=2\mathrm{x}+2\\ \frac{3}{\mathrm{x}}\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\frac{2}{\mathrm{x}}+2\end{array}$$$${\mathrm{\Delta }}_{\mathrm{s}}=\left|\begin{array}{c}13\mathrm{x}\\ \frac{3}{\mathrm{x}}1\end{array}\right|=1-9=-8\ne 0\Rightarrow \mathrm{f}\left(\mathrm{x}\right)=\frac{\left|\begin{array}{c}2\mathrm{x}+23\mathrm{x}\\ \frac{2}{\mathrm{x}}+21\end{array}\right|}{-8}$$$$=-\frac{1}{8}(2\mathrm{x}+2-3\mathrm{x}(\frac{2}{\mathrm{x}}+2))=-\frac{1}{8}(2\mathrm{x}+2-6-6\mathrm{x})=\frac{4\mathrm{x}+4}{8}=\frac{\mathrm{x}+1}{2}$$$$\mathrm{f}\left(3\right),\mathrm{f}\left(9\right)\mathrm{and}\mathrm{f}\left(\mathrm{a}\right)\mathrm{are}\mathrm{in}\mathrm{a}.\mathrm{p}\Rightarrow \mathrm{f}\left(3\right)+\mathrm{f}\left(\mathrm{a}\right)=2\mathrm{f}\left(9\right)\Rightarrow$$$$2+\frac{\mathrm{a}+1}{2}=10\Rightarrow 4+\mathrm{a}+1=20\Rightarrow \mathrm{a}=20-5=15$$$$$$

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