f-R-R-g-R-R-f-x-determinant-x-g-x-g-x-x-g-x-determinant-f-x-x-x-f-x-

FacebookTweetPin

Question Number 316 by 123456 last updated on 25/Jan/15

$$f:\mathbb{R}\to \mathbb{R}$$$$g:\mathbb{R}\to \mathbb{R}$$$$f\left(x\right)=\left|\begin{array}{cc}x& g\left(x\right)\\ g(-x)& -x\end{array}\right|$$$$g\left(x\right)=\left|\begin{array}{cc}f\left(x\right)& x\\ -x& f(-x)\end{array}\right|$$

Answered by prakash jain last updated on 21/Dec/14

$$f\left(x\right)=-x^2-g\left(x\right)g(-x)\dots \left(i\right)$$$$g\left(x\right)=f\left(x\right)f(-x)+x^2\dots \left(ii\right)$$$$f(-x)=f\left(x\right)$$$$g(-x)=g\left(x\right)$$$$f\left(x\right)=-x^2-{\left[g\left(x\right)\right]}^2$$$$g\left(x\right)=x^2+{\left[f\left(x\right)\right]}^2$$$$f\left(x\right)+g\left(x\right)={\left[f\left(x\right)\right]}^2-{\left[g\left(x\right)\right]}^2=[f\left(x\right)+g\left(x\right)][f\left(x\right)-g\left(x\right))]$$$$f\left(x\right)-g\left(x\right)=1$$$$f\left(x\right)=1+g\left(x\right)$$$$1+g\left(x\right)=-x^2-{\left[g\left(x\right)\right]}^2$$$${\left[g\left(x\right)\right]}^2+g\left(x\right)+(1+x^2)=0$$$$g\left(x\right)=\frac{-1\pm \sqrt{1-4(1+x^2)}}{2}=\frac{-1\pm \sqrt{-x^2-3}}{2}$$$$f\left(x\right)=1+g\left(x\right)=\frac{1\pm \sqrt{-x^2-3}}{2}$$$$\mathrm{The}\mathrm{solution}\mathrm{pair}\mathrm{meets}\mathrm{all}\mathrm{condition}\mathrm{except}$$$$f\left(x\right)\mathrm{and}g\left(x\right)\mathrm{are}\mathrm{not}\mathrm{real}\mathrm{for}\mathrm{any}\mathrm{real}x.$$$$\mathrm{So}\mathrm{given}\mathrm{conditons}\mathrm{has}\mathrm{no}\mathrm{solution}.$$

Terms of Service
Privacy Policy
Contact: info@tinkutara.com

FacebookTweetPin