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Question Number 206746 by sonukgindia last updated on 23/Apr/24

Answered by A5T last updated on 23/Apr/24

$$f\left(2\right)+2f(-1)=2...\left(i\right)$$$$f(-1)+2f\left(\frac{1}{2}\right)=-1...\left(ii\right)$$$$f\left(\frac{1}{2}\right)+2f\left(2\right)=\frac{1}{2}...\left(iii\right)$$$$2\times \left(iii\right)-\left(ii\right):4f\left(2\right)-f(-1)=2...\left(iv\right)$$$$2\times \left(iv\right)+\left(i\right):9f\left(2\right)=6\Rightarrow f\left(2\right)=\frac{2}{3}$$

Answered by mr W last updated on 24/Apr/24

$$f\left(x\right)+2f\left(\frac{1}{1-x}\right)=x...\left(i\right)$$$$f\left(\frac{1}{1-x}\right)+2f\left(\frac{x-1}{x}\right)=\frac{1}{1-x}...\left(ii\right)$$$$f\left(\frac{x-1}{x}\right)+2f\left(x\right)=\frac{x-1}{x}...\left(iii\right)$$$$\left(iii\right)into\left(ii\right):$$$$f\left(\frac{1}{1-x}\right)+2[\frac{x-1}{x}-2f\left(x\right)]=\frac{1}{1-x}$$$$f\left(\frac{1}{1-x}\right)=4f\left(x\right)+\frac{1}{1-x}-\frac{2(x-1)}{x}$$$$thisinto\left(i\right):$$$$f\left(x\right)+2[4f\left(x\right)+\frac{1}{1-x}-\frac{2(x-1)}{x}]=x$$$$\Rightarrow f\left(x\right)=\frac{1}{9}[x-\frac{2}{1-x}+\frac{4(x-1)}{x}]$$$$f\left(2\right)=\frac{1}{9}[2-\frac{2}{1-2}+\frac{4(2-1)}{2}]=\frac{2}{3}✓$$

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