Question

QNO. 15 SOLUTION IS REQUIRED

Q15. The function ƒ (x) = $\frac{x+1}{x^3+1}$ can be written as the sum of an even function g(x) and odd function h(x). Then the value of |g (0)| is _____________

Answer 1

$$\begin{gathered} Letf\left(x\right)=g\left(x\right)+h\left(x\right) \\ \frac{x+1}{x^3+1}=g\left(x\right)+h\left(x\right)\_\_\_\_\_\_\_\_\_\_\left(1\right) \\ f\left(-x\right)=g\left(-x\right)+h\left(-x\right) \\ Sinceg\left(x\right)isevenandh\left(x\right)isodd \\ f\left(-x\right)=g\left(x\right)-h\left(x\right) \\ \frac{-x+1}{-x^3+1}=g\left(x\right)-h\left(x\right)\_\_\_\_\_\_\_\_\_\_\_\left(2\right) \\ addingequation1and2 \\ \frac{x+1}{x^3+1}+\frac{-x+1}{-x^3+1}=2g\left(x\right) \\ \frac{x+1}{x^3+1}+\frac{x-1}{x^3-1}=2g\left(x\right) \\ g\left(x\right)=\frac{1}{2}\left(\frac{x+1}{x^3+1}+\frac{x-1}{x^3-1}\right) \\ g\left(0\right)=\frac{1}{2}\left(\frac{1}{1}+\frac{-1}{-1}\right) \\ g\left(0\right)=\frac{1}{2}\left(2\right)=1 \\ \left|g\left(0\right)\right|=1 \end{gathered}$$