91Ó°ÊÓ

The two given functions are $In\left|1-x\right|$and$\left(1+x\right)\left(\sin x+\cos x\right)$.

A function $\mathit{f}\left(x\right)$is an even function when$\mathit{f}\left(-x\right)\mathbf{=}\mathit{f}\left(x\right)$, where as a function $\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$is an odd function when$\mathit{f}\mathbf{(}\mathbf{-}\mathbf{x}\mathbf{)}\mathbf{=}\mathit{f}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$.

The two given functions are $In\left|1-x\right|$and$\left(1+x\right)\left(\sin x+\cos x\right)$.

(a) $In\left|1-x\right|$

Express the function $In\left|1-x\right|$as the sum of an even function.

$$\begin{gathered} \left(f_e\left(x\right)\right)=\frac{1}{2}\left(f\left(x\right)+f\left(-x\right)\right) \\ \left(f_e\left(x\right)\right)=\left(\frac{In\left|1-x\right|+In\left|1+x\right|}{2}\right) \end{gathered}$$

Express the functionas the sum of an odd function.

$$\begin{gathered} \left(f_e\left(x\right)\right)=\frac{1}{2}\left(f\left(x\right)+f\left(-x\right)\right) \\ \left(f_e\left(x\right)\right)=\left(\frac{In\left|1-x\right|+In\left|1+x\right|}{2}\right) \end{gathered}$$

The sum of the even and odd function is expressed as:

$$\begin{gathered} f\left(x\right)=f_e\left(x\right)+f_0\left(x\right) \\ =\left(\frac{In\left|1-x\right|+In\left|1+x\right|}{2}\right)+\left(\frac{In\left|1-x\right|-In\left|1+x\right|}{2}\right) \\ =\frac{1}{2}In\left(1-x^2\right)+\frac{1}{2}In\left(\frac{1-x}{1+x}\right) \end{gathered}$$

Hence, the sum of the even and odd function is, $\frac{1}{2}In\left(1-x^2\right)+\frac{1}{2}In\left(\frac{1-x}{1+x}\right)$.

(b) $\left(1+x\right)\left(\sin x+\cos x\right)$

Express the function$$" width="9" height="19" role="math">

$$\begin{gathered} \left(f_0\left(x\right)\right)=\frac{1}{2}\left(f\left(x\right)-f\left(-x\right)\right) \\ \left(f_0\left(x\right)\right)=\left(\frac{\left(1+x\right)\left(\cos x+\sin x\right)-\left(1-x\right)\left(\cos x-\sin x\right)}{2}\right) \end{gathered}$$

Simplify further as shown below.

$\left(f_0\left(x\right)\right)=\sin x+x\cos x$

Express the functionas the sum of an odd function.

$$\begin{gathered} \left(f_0\left(x\right)\right)=\frac{1}{2}\left(f\left(x\right)-f\left(-x\right)\right) \\ \left(f_0\left(x\right)\right)=\left(\frac{\left(1+x\right)\left(\cos x+\sin x\right)-\left(1-x\right)\left(\cos x-\sin x\right)}{2}\right) \end{gathered}$$

Simplify further as shown below.

$\left(f_0\left(x\right)\right)=\sin x+x\cos x$

The sum of the even and odd function is expressed as:

$$\begin{gathered} f\left(x\right)=f_e\left(x\right)+f_0\left(x\right) \\ =\cos x+x\sin x+\sin x+x\cos x \\ =\left(1+x\right)\left(\cos x+\sin x\right) \end{gathered}$$

Hence, the sum of the even and odd function is, $\left(1+x\right)\left(\cos x+\sin x\right)$.

Therefore, the function $In\left|1-x\right|$can be expressed in the form of the sum of an even function $\left(f_e\left(x\right)\right)$and an odd function $\left(f_0\left(x\right)\right)$as $\left(\frac{In\left|1-x\right|+In\left|1+x\right|}{2}\right)$and $\left(\frac{In\left|1-x\right|+In\left|1+x\right|}{2}\right)$respectively. Similarly, the function $\left(1+x\right)\left(\sin x+\cos x\right)$can be expressed in the form of the sum of an even function $\left(f_e\left(x\right)\right)$and an odd function $\left(f_0\left(x\right)\right)$as $\cos x+x\sin x$and $\sin x+x\cos x$respectively.