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

If f(x)=ln∣a+(1/(1−x))∣+b is an odd function, then find the value of a, b.

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

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

f(x)=−f(−x) ln ∣a+(1/(1−x))∣+b=−ln ∣a+(1/(1+x))∣−b ln ∣(a+(1/(1−x)))(a+(1/(1+x)))∣+2b=0 ln ∣(((a+1−ax)/(1−x)))(((a+1+ax)/(1+x)))∣+2b=0 ln ∣((((a+1)^2 −(ax)^2 )/(1−x^2 )))∣+2b=0 ln ∣a^2 ((((1+(1/a))^2 −x^2 )/(1−x^2 )))∣+2b=0 such that this holds for any x, 1+(1/a)=−1 ⇒a=−(1/2) ln ∣(1/4)∣+2b=0 ⇒−2 ln 2+2b ⇒b=ln 2 f(x)=ln ∣(1/(1−x))−(1/2)∣+ln 2=ln ∣((1+x)/(1−x))∣ is odd function.

$${f}\left({x}\right)=−{f}\left(−{x}\right) \\ $$$$\mathrm{ln}\:\mid{a}+\frac{\mathrm{1}}{\mathrm{1}−{x}}\mid+{b}=−\mathrm{ln}\:\mid{a}+\frac{\mathrm{1}}{\mathrm{1}+{x}}\mid−{b} \\ $$$$\mathrm{ln}\:\mid\left({a}+\frac{\mathrm{1}}{\mathrm{1}−{x}}\right)\left({a}+\frac{\mathrm{1}}{\mathrm{1}+{x}}\right)\mid+\mathrm{2}{b}=\mathrm{0} \\ $$$$\mathrm{ln}\:\mid\left(\frac{{a}+\mathrm{1}−{ax}}{\mathrm{1}−{x}}\right)\left(\frac{{a}+\mathrm{1}+{ax}}{\mathrm{1}+{x}}\right)\mid+\mathrm{2}{b}=\mathrm{0} \\ $$$$\mathrm{ln}\:\mid\left(\frac{\left({a}+\mathrm{1}\right)^{\mathrm{2}} −\left({ax}\right)^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }\right)\mid+\mathrm{2}{b}=\mathrm{0} \\ $$$$\mathrm{ln}\:\mid{a}^{\mathrm{2}} \left(\frac{\left(\mathrm{1}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} −{x}^{\mathrm{2}} }{\mathrm{1}−{x}^{\mathrm{2}} }\right)\mid+\mathrm{2}{b}=\mathrm{0} \\ $$$${such}\:{that}\:{this}\:{holds}\:{for}\:{any}\:{x}, \\ $$$$\mathrm{1}+\frac{\mathrm{1}}{{a}}=−\mathrm{1}\:\Rightarrow{a}=−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{ln}\:\mid\frac{\mathrm{1}}{\mathrm{4}}\mid+\mathrm{2}{b}=\mathrm{0}\:\Rightarrow−\mathrm{2}\:\mathrm{ln}\:\mathrm{2}+\mathrm{2}{b}\:\Rightarrow{b}=\mathrm{ln}\:\mathrm{2} \\ $$$${f}\left({x}\right)=\mathrm{ln}\:\mid\frac{\mathrm{1}}{\mathrm{1}−{x}}−\frac{\mathrm{1}}{\mathrm{2}}\mid+\mathrm{ln}\:\mathrm{2}=\mathrm{ln}\:\mid\frac{\mathrm{1}+{x}}{\mathrm{1}−{x}}\mid\:{is}\: \\ $$$${odd}\:{function}. \\ $$

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

Commented by CrispyXYZ last updated on 09/Jan/23

Thanks sir! Great solution!

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

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