f-2-x-f-x-2-2-f-2-x-stands-for-f-x-2-f-x-If-possible-solve-stepwise-

Question Number 1902 by Rasheed Soomro last updated on 23/Oct/15

$${f}^{\:\mathrm{2}} \left({x}\right)−{f}\left({x}^{\mathrm{2}} \right)=\mathrm{2}\:,\:{f}^{\:\mathrm{2}} \left({x}\right)\:{stands}\:{for}\:\left[{f}\left({x}\right)\right]^{\mathrm{2}} \\ $$$${f}\left({x}\right)=? \\ $$$$\left({If}\:{possible}\:{solve}\:{stepwise}\right) \\ $$

Commented by 123456 last updated on 23/Oct/15

$$\mathrm{hehe}…\:\mathrm{sorry},\:\mathrm{i}\:\mathrm{make}\:\mathrm{a}\:\mathrm{great}\:\mathrm{confusion}. \\ $$

Commented by Rasheed Soomro last updated on 23/Oct/15

$${How}\:{you}\:{supposed}\:{that} \\ $$$${f}^{\mathrm{2}} \left(\mathrm{0}\right)={f}\left(\mathrm{0}^{\mathrm{2}} \right)\:\:{and}\:\:{f}^{\mathrm{2}} \left(\mathrm{1}\right)={f}\left(\mathrm{1}^{\mathrm{2}} \right)\:? \\ $$

Commented by 123456 last updated on 23/Oct/15

$${f}^{\mathrm{2}} \left(\mathrm{0}\right)−{f}\left(\mathrm{0}^{\mathrm{2}} \right)=\mathrm{2} \\ $$$${f}^{\mathrm{2}} \left(\mathrm{0}\right)−{f}\left(\mathrm{0}\right)=\mathrm{2} \\ $$$${x}={f}\left(\mathrm{0}\right) \\ $$$${x}^{\mathrm{2}} −{x}=\mathrm{2} \\ $$$$\Delta=\mathrm{1}+\mathrm{8}=\mathrm{9} \\ $$$${x}=\frac{\mathrm{1}\pm\mathrm{3}}{\mathrm{2}} \\ $$$${x}=−\mathrm{1}\vee{x}=\mathrm{2} \\ $$$${f}\left(\mathrm{0}\right)=−\mathrm{1}\vee{f}\left(\mathrm{0}\right)=\mathrm{2} \\ $$$$\mathrm{by}\:\mathrm{same}\:\mathrm{way} \\ $$$${f}\left(\mathrm{1}\right)=−\mathrm{1}\vee{f}\left(\mathrm{1}\right)=\mathrm{2} \\ $$

Commented by Rasheed Soomro last updated on 24/Oct/15

$${Good}\:{Try}\:{Sir}!\:{Actually}\:{if}\:{it}\:{is}\:{necessary}\:{to}\:{make}\:{clear}\:{then}\: \\ $$$${really}\:{x}\neq\mathrm{0}. \\ $$

Answered by prakash jain last updated on 24/Oct/15

$${f}\left({x}\right)={a}^{{k}\mathrm{ln}\:{x}} +{a}^{−{k}\mathrm{ln}\:{x}} \\ $$$$\left[{f}\left({x}\right)\right]^{\mathrm{2}} ={a}^{\mathrm{2}{k}\mathrm{ln}\:{x}} +{a}^{−\mathrm{2}{k}\mathrm{ln}\:{x}} +\mathrm{2} \\ $$$${f}\left({x}^{\mathrm{2}} \right)={a}^{\mathrm{2}{k}\mathrm{ln}\:{x}} +{a}^{−\mathrm{2}{k}\mathrm{ln}\:{x}} \\ $$$$\left[{f}\left({x}\right)\right]^{\mathrm{2}} −{f}\left({x}^{\mathrm{2}} \right)=\mathrm{2} \\ $$

Commented by prakash jain last updated on 24/Oct/15

$${f}\left({x}\right)={x}^{{a}} +{x}^{−{a}} \\ $$

Commented by Rasheed Soomro last updated on 25/Oct/15

$$\mathcal{E}{xcellent}\:\mathcal{S}{ir}! \\ $$

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