$f\left(x\right):\left[\mathrm{0,2}\right]\to R$ and $f^{\text{'}\text{'}}\left(x\right)> 0$ for $x\in \left[\mathrm{0,2}\right]$
    $\Rightarrow f^{\mathrm{\text{'}}}\left(x\right)$ is increasing for $x\in \left[\mathrm{0,2}\right]$
Now, $\varphi \left(x\right)=f\left(x\right)+f(2-x)$
    $\Rightarrow {\varphi }^{\text{'}}\left(x\right)=f^{\text{'}}\left(x\right)-f\text{'}(2-x)$
For $x\in \left[\mathrm{0,1}\right)$ , $x < 2 – x$  

    $\Rightarrow f\text{'}\left(x\right)<f\text{'}(2-x)\Rightarrow \varphi \text{'}\left(x\right)<0$
For $x\in (1, 2]$ , $x > 2 – x$

    $\Rightarrow f^{\text{'}}\left(x\right)>f^{\text{'}}\left(2-x\right)\Rightarrow {\varphi }^{\text{'}}\left(x\right)>0$
Hence, $\varphi$ is decreasing on $\left(0,1\right)$ and increasing on $\left(1,2\right)$.