Question Number 1902 by Rasheed Soomro last updated on 23/Oct/15
![f^( 2) (x)−f(x^2 )=2 , f^( 2) (x) stands for [f(x)]^2 f(x)=? (If possible solve stepwise)](https://www.tinkutara.com/question/Q1902.png)
$${f}^{\:\mathrm{2}} \left({x}\right)−{f}\left({x}^{\mathrm{2}} \right)=\mathrm{2}\:,\:{f}^{\:\mathrm{2}} \left({x}\right)\:{stands}\:{for}\:\left[{f}\left({x}\right)\right]^{\mathrm{2}} \\ $$$${f}\left({x}\right)=? \\ $$$$\left({If}\:{possible}\:{solve}\:{stepwise}\right) \\ $$
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(x)=a^(kln x) +a^(−kln x) [f(x)]^2 =a^(2kln x) +a^(−2kln x) +2 f(x^2 )=a^(2kln x) +a^(−2kln x) [f(x)]^2 −f(x^2 )=2](https://www.tinkutara.com/question/Q1926.png)
$${f}\left({x}\right)={a}^{{k}\mathrm{ln}\:{x}} +{a}^{−{k}\mathrm{ln}\:{x}} \\ $$$$\left[{f}\left({x}\right)\right]^{\mathrm{2}} ={a}^{\mathrm{2}{k}\mathrm{ln}\:{x}} +{a}^{−\mathrm{2}{k}\mathrm{ln}\:{x}} +\mathrm{2} \\ $$$${f}\left({x}^{\mathrm{2}} \right)={a}^{\mathrm{2}{k}\mathrm{ln}\:{x}} +{a}^{−\mathrm{2}{k}\mathrm{ln}\:{x}} \\ $$$$\left[{f}\left({x}\right)\right]^{\mathrm{2}} −{f}\left({x}^{\mathrm{2}} \right)=\mathrm{2} \\ $$
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
$$\mathcal{E}{xcellent}\:\mathcal{S}{ir}! \\ $$