A function s(x) is defined as $\mathrm{g}\left(\mathrm{x}\right)=\frac{1}{4}\mathrm{f}\left(2{\mathrm{x}}^2-1\right)+\frac{1}{2}\mathrm{f}\left(1-{\mathrm{x}}^2\right)\text{ and }{\mathrm{f}}^{\mathrm{\prime }}\left(\mathrm{x}\right)$ is an increasing function. Then g(x) is increasing in the interval

${\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{x}\right)={\mathrm{xf}}^{\mathrm{\prime }}\left(2{\mathrm{x}}^2-1\right)-{\mathrm{xf}}^{\mathrm{\prime }}\left(1-{\mathrm{x}}^2\right)=\mathrm{x}\left(\mathrm{f}\left(2{\mathrm{x}}^2-1\right)-\mathrm{f}\left(1-{\mathrm{x}}^2\right)\right) \Rightarrow {\mathrm{g}}^{\mathrm{\prime }}\left(\mathrm{x}\right)>0$
If $\mathrm{x}>0,2{\mathrm{x}}^2-1>1-{\mathrm{x}}^2$ (as f ' is an increasing function)
or    $3{\mathrm{x}}^2>2\text{ or }\mathrm{x}\in \left(-\mathrm{\infty },-\sqrt{\frac{2}{3}}\right)\cup \left(\sqrt{\frac{2}{3}},\mathrm{\infty }\right)$
or    $\mathrm{x}\in \left(\sqrt{\frac{2}{3}},\mathrm{\infty }\right)$
If $\mathrm{x}<0,2{\mathrm{x}}^2-1<1-{\mathrm{x}}^2$
or    $3{\mathrm{x}}^2<2\text{ or }\mathrm{x}\in \left(-\sqrt{\frac{2}{3}},\sqrt{\frac{2}{3}}\right)\text{ or }\mathrm{x}\in \left(-\sqrt{\frac{2}{3}},0\right)$