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Question Number 86160 by jagoll last updated on 27/Mar/20

$$\mathrm{If}(\mathrm{x}-1)\mathrm{f}\left(\mathrm{x}\right)+\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)=\frac{1}{\mathrm{x}-1}$$$$\mathrm{find}\mathrm{f}\left(\mathrm{x}\right)$$

Commented by john santu last updated on 27/Mar/20

$$replacexby\frac{1}{x}$$$$\Rightarrow (\frac{1}{x}-1)f\left(\frac{1}{x}\right)+f\left(x\right)=\frac{1}{\frac{1}{x}-1}$$$$\left(\frac{1-x}{x}\right)f\left(\frac{1}{x}\right)+f\left(x\right)=\frac{x}{1-x}\left[\times x\right]$$$$(1-x)f\left(\frac{1}{x}\right)+xf\left(x\right)=\frac{x^2}{1-x}\to \left(ii\right)$$$$multiplyeq\left(i\right)by(1-x)$$$$(1-x)f\left(\frac{1}{x}\right)-{(1-x)}^{2}f\left(x\right)=-1\leftarrow \left(i\right)$$$$\left(ii\right)-\left(i\right)$$$$(x+{(1-x)}^2)f\left(x\right)=\frac{x^2}{1-x}+1$$$$(x^2-x+1)f\left(x\right)=\frac{x^2-x+1}{1-x}$$$$f\left(x\right)=\frac{1}{1-x}$$$$$$

Commented by jagoll last updated on 27/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{mr}\mathrm{john}\mathrm{\&}\mathrm{tanmay}$$

Commented by mathmax by abdo last updated on 27/Mar/20

$$(x-1)f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{1}{x-1}\Rightarrow (\frac{1}{x}-1)f\left(\frac{1}{x}\right)+f\left(x\right)=\frac{1}{\frac{1}{x}-1}=\frac{x}{1-x}$$$$wegetthesystem\{\begin{array}{l}(x-1)f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{1}{x-1}\\ f\left(x\right)+\frac{1-x}{x}f\left(\frac{1}{x}\right)=\frac{x}{1-x}\end{array}$$$${\mathrm{\Delta }}_s=\left|\begin{array}{c}x-11\\ 1\frac{1-x}{x}\end{array}\right|=\frac{(x-1)(1-x)}{x}-1=\frac{x-x^2-1+x-x}{x}$$$$=\frac{-x^2+x-1}{x}\Rightarrow f\left(x\right)=\frac{\mathrm{\Delta }f\left(x\right)}{\mathrm{\Delta }}$$$$=\frac{\left|\begin{array}{c}\frac{1}{x-1}1\\ \frac{x}{1-x}\frac{1-x}{x}\end{array}\right|}{\frac{-x^2+x-1}{x}}=\frac{-\frac{1}{x}-\frac{x}{1-x}}{\frac{-x^2+x-1}{x}}=\frac{-1+x-x^2}{x(1-x)}\times \frac{x}{-x^2+x-1}$$$$=\frac{1}{1-x}\Rightarrow f\left(x\right)=\frac{1}{1-x}$$$$$$$$$$

Answered by TANMAY PANACEA. last updated on 27/Mar/20

$$(x-1)f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{1}{x-1}$$$$replacingxby\frac{1}{x}$$$$(\frac{1}{x}-1)f\left(\frac{1}{x}\right)+f\left(x\right)=\frac{1}{\frac{1}{x}-1}$$$$f\left(x\right)=\frac{x}{1-x}-\frac{1-x}{x}f\left(\frac{1}{x}\right)$$$$puttingthevalueoff\left(x\right)ingiveneqn$$$$[(x-1)f\left(x\right)+f\left(\frac{1}{x}\right)=\frac{1}{1-x}]$$$$soweget$$$$(x-1)[\frac{x}{1-x}-\frac{1-x}{x}f\left(\frac{1}{x}\right)]+f\left(\frac{1}{x}\right)=\frac{1}{x-1}$$$$-x+\frac{{(x-1)}^2}{x}f\left(\frac{1}{x}\right)+f\left(\frac{1}{x}\right)=\frac{1}{x-1}$$$$f\left(\frac{1}{x}\right)[1+\frac{x^2-2x+1}{x}]=\frac{1}{x-1}+x$$$$f\left(\frac{1}{x}\right)\left[\frac{x+x^2-2x+1}{x}\right]=\frac{1+x^2-x}{x-1}$$$$f\left(\frac{1}{x}\right)=\frac{x}{x-1}=\frac{1}{1-\frac{1}{x}}$$$$f\left(x\right)=\frac{1}{1-x}$$

Commented by jagoll last updated on 27/Mar/20

$$\mathrm{thank}\mathrm{you}\mathrm{mister}$$

Commented by Serlea last updated on 27/Mar/20

$$$$$$\mathrm{I}\mathrm{think}\mathrm{Something}\mathrm{is}\mathrm{wrong}\mathrm{Sir}$$$$\mathrm{Line}\mathrm{number}4:\mathrm{How}\mathrm{did}\mathrm{u}\mathrm{get}\mathrm{ur}\mathrm{Second}\mathrm{f}\left(\frac{1}{\mathrm{x}}\right)$$$$2)\mathrm{Why}{\mathrm{didn}}^{\prime }\mathrm{t}\mathrm{you}\mathrm{multiply}\mathrm{the}\mathrm{both}\mathrm{side}\mathrm{by}(\mathrm{x}-1)\mathrm{only}\mathrm{the}$$$$\mathrm{left}\mathrm{hand}\mathrm{side}\mathrm{of}\mathrm{the}\mathrm{equation}.$$$$$$

Commented by john santu last updated on 27/Mar/20

$$yessir.$$

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