The interval in which $f\left(x\right)=2x^2–\log x$ increases is

$$\begin{gathered} \mathrm{f}\left(\mathrm{x}\right)=2{\mathrm{x}}^2-\mathrm{logx} \\ {\mathrm{f}}^{\mathrm{I}}\left(\mathrm{x}\right)=4\mathrm{x}-1/\mathrm{x} \\ =\frac{4{\mathrm{x}}^2-1}{\mathrm{x}} \end{gathered}$$

$\mathrm{f}\left(\mathrm{x}\right)$ is increasing 

$$\begin{gathered} \Rightarrow {\mathrm{f}}^{\mathrm{I}}\left(\mathrm{x}\right)>0 \\ 4{\mathrm{x}}^2-1/\mathrm{x}>0 \\ \mathrm{x}\left(4{\mathrm{x}}^2-1\right)>0 \\ \mathrm{x}\left(2\mathrm{x}-1\right)\left(2\mathrm{x}+1\right)>0 \end{gathered}$$

$$\begin{gathered} \mathrm{x}\in \left(-1/2,0\right)\cup \left(1/2,\infty \right) \\ f\left(x\right)=2x^2–\log x is defined ifx>0 \\ \Rightarrow f increases in \left(1/2,\infty \right) \end{gathered}$$