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Question Number 184594 by CrispyXYZ last updated on 09/Jan/23

$$\mathrm{If}f\left(x\right)=\ln \mid a+\frac{1}{1-x}\mid +b\mathrm{is}\mathrm{an}\mathrm{odd}\mathrm{function},$$$$\mathrm{then}\mathrm{find}\mathrm{the}\mathrm{value}\mathrm{of}a,b.$$

Answered by mr W last updated on 09/Jan/23

$$f\left(x\right)=-f(-x)$$$$\ln \mid a+\frac{1}{1-x}\mid +b=-\ln \mid a+\frac{1}{1+x}\mid -b$$$$\ln \mid (a+\frac{1}{1-x})(a+\frac{1}{1+x})\mid +2b=0$$$$\ln \mid \left(\frac{a+1-ax}{1-x}\right)\left(\frac{a+1+ax}{1+x}\right)\mid +2b=0$$$$\ln \mid \left(\frac{{(a+1)}^2-{\left(ax\right)}^2}{1-x^2}\right)\mid +2b=0$$$$\ln \mid a^2\left(\frac{{(1+\frac{1}{a})}^2-x^2}{1-x^2}\right)\mid +2b=0$$$$suchthatthisholdsforanyx,$$$$1+\frac{1}{a}=-1\Rightarrow a=-\frac{1}{2}$$$$\ln \mid \frac{1}{4}\mid +2b=0\Rightarrow -2\ln 2+2b\Rightarrow b=\ln 2$$$$f\left(x\right)=\ln \mid \frac{1}{1-x}-\frac{1}{2}\mid +\ln 2=\ln \mid \frac{1+x}{1-x}\mid is$$$$oddfunction.$$

Commented by mr W last updated on 09/Jan/23

Commented by CrispyXYZ last updated on 09/Jan/23

$$\mathrm{Thanks}\mathrm{sir}!\mathrm{Great}\mathrm{solution}!$$

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