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Question Number 168721 by mathlove last updated on 16/Apr/22

Answered by Rasheed.Sindhi last updated on 16/Apr/22

$$x^{2}f\left(x\right)-2f\left(1/x\right)=g\left(x\right)..........\left(i\right)$$$${(-x)}^{2}f(-x)-2f(-1/x)=g(-x)$$$$x^{2}f\left(x\right)-2f\left(1/x\right)=-g\left(x\right)......\left(ii\right)$$$$\left(i\right)+\left(ii\right):2x^{2}f\left(x\right)-4f\left(1/x\right)=0$$$$x^{2}f\left(x\right)-2f\left(\frac{1}{x}\right)=0...........\left(iii\right)$$$${\left(\frac{1}{x}\right)}^{2}f\left(\frac{1}{x}\right)-2f\left(x\right)=0.......\left(iv\right)$$$$\left(iii\right)\Rightarrow f\left(\frac{1}{x}\right)=\frac{x^{2}f\left(x\right)}{2}$$$$\left(iv\right)\Rightarrow \frac{1}{x^2}\left(\frac{x^{2}f\left(x\right)}{2}\right)-2f\left(x\right)=0$$$$\frac{f\left(x\right)}{2}-2f\left(x\right)=0$$$$f\left(x\right)-4f\left(x\right)=0$$$$f\left(x\right)=0$$$$f\left(5\right)=0$$

Commented by mathlove last updated on 16/Apr/22

$$thankssir$$

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