Given, $f(x)=x+\frac{1}{2}=\frac{2 x+1}{2}$
$\therefore$
$f(2 x)=2 x+\frac{1}{2}$
$\Rightarrow$
$f(2 x)=\frac{4 x+1}{2}$
and $f(4 x)=4 x+\frac{1}{2} \Rightarrow f(4 x)=\frac{8 x+1}{2}$
since, $f(x), f(2 x)$ and $f(4 x)$ are in $\mathrm{HP}$. $\frac{1}{f(x)} \cdot \frac{1}{f(2 x)}$ and $\frac{1}{f(4 x)}$ are in AP.
$\Rightarrow \quad \frac{1}{f(2 x)}=\frac{\frac{1}{f(x)}+\frac{1}{f(4 x)}}{2}$
$\Rightarrow \quad \frac{2}{4 x+1}=\frac{\frac{2}{2 x+1}+\frac{2}{8 x+1}}{2}$
$\Rightarrow \quad-\frac{2}{4 x+1}=\frac{10 x+2}{(2 x+1)(8 x+1)}$
$\Rightarrow(2 x+1)(8 x+1)=(5 x+1)(4 x+1)$
$\Rightarrow 16 x^{2}+10 x+1=20 x^{2}+9 x+1$
$\Rightarrow \quad 4 x^{2}-x=0 \Rightarrow x(4 x-1)=0$
$\Rightarrow \quad x=\frac{1}{4} \quad[\because x \neq 0$
Hence, one real value of $x$ for which the thrce unequal terms are in HP.