Question

Let $f\left(x\right)=2x^2-\log \left|x\right|,x\ne 0.$ Then $f\left(x\right)$ is

• A. monotonically increasing on $(-\frac{1}{2},0)\cup (\frac{1}{2},+\infty )$
• B. monotonically decreasing on $(-\frac{1}{2},0)\cup (\frac{1}{2},+\infty )$
• C. monotonically increasing on $(-\infty -\frac{1}{2})\cup (0,\frac{1}{2})$
• D. monotonically decreasing on $(-\infty ,-\frac{1}{2})\cup (0,\frac{1}{2})$

Answer 1

Answer: (A), (D)

The correct options are
A monotonically increasing on $(-\frac{1}{2},0)\cup (\frac{1}{2},+\infty )$
D monotonically decreasing on $(-\infty ,-\frac{1}{2})\cup (0,\frac{1}{2})$

For calculating increasing or decreasing function, have to calculate $f^{\prime }\left(x\right)$
$f^{\prime }\left(x\right)=4x-\frac{1}{x}=0$
$x=\pm \frac{1}{2}$ and at $x=0$ function will not define.
So $f^{\prime }\left(x\right)>0$ in interval $(\frac{-1}{2},0)\bigcup (\frac{1}{2},\infty )$, thus in this interval function will monotonically increasing.
$f^{\prime }\left(x\right)<0$ in interval $(-\infty ,\frac{-1}{2})\bigcup (0,\frac{1}{2})$, thus in this interval function will monotonically decreasing.
Hence, options 'A' and 'D' are correct.