If f(x)=(4 x+x^4)/(1+4 x^3) and g(x)=In ((1+x)/(1-x)), then what is the value of f^ ^ g ( e-1 e+1 )equal to?

Correct answer: B. 1 aligned & f(x)= 4 x+x^4 1+4 x^3 , g(x)= ( 1+x 1-x ) & g( e-1 e+1 )= ( 1+( e-1 e+1 ) 1-( e-1 e+1 ) )= ( e+1+e-1 e+1-e+1 ) aligned aligned & = ((2 )/(2))= e=1 & fo g( e-1 e+1 )=f(1)=(4(1)+(1)^4)/(1+4(1)^3)=(4+1)/(1+4)=(5)/(5)=1 aligned

NDA 2017

Exam: NDA
Year: 2017
Subject: Mathematics
Chapter: Functions

NDA 2017 · Mathematics · Functions

If f(x)=(4 x+x^4)/(1+4 x^3) and g(x)=In ((1+x)/(1-x)), then what is...

f(x)=\frac{4 x+x^4}{1+4 x^3}

and

g(x)=\operatorname{In} \left(\frac{1+x}{1-x}\right), then what is the value of

f^{{}^{\circ}}

\mathrm{g} \left(\frac{\mathrm{e}-1}{\mathrm{e}+1}\right)equal to?

Options

A $2$
B $1$
C $0$
D $\frac{1}{2}$

Correct answer

B. $1$

Step-by-step solution

\begin{aligned} & \mathrm{f}(\mathrm{x})=\frac{4 \mathrm{x}+\mathrm{x}^4}{1+4 \mathrm{x}^3}, \mathrm{ g}(\mathrm{x})=\ln \left(\frac{1+\mathrm{x}}{1-\mathrm{x}}\right) \\ & \mathrm{g}\left(\frac{\mathrm{e}-1}{\mathrm{e}+1}\right)=\ln \left(\frac{1+\left(\frac{\mathrm{e}-1}{\mathrm{e}+1}\right)}{1-\left(\frac{\mathrm{e}-1}{\mathrm{e}+1}\right)}\right)=\ln \left(\frac{\mathrm{e}+1+\mathrm{e}-1}{\mathrm{e}+1-\mathrm{e}+1}\right)\end{aligned}

\begin{aligned} & =\ln \left(\frac{2 \ell}{2}\right)=\ln \mathrm{e}=1 \\ & \text{fo } \mathrm{g}\left(\frac{\mathrm{e}-1}{\mathrm{e}+1}\right)=\mathrm{f}(1)=\frac{4(1)+(1)^4}{1+4(1)^3}=\frac{4+1}{1+4}=\frac{5}{5}=1\end{aligned}

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