Answer: B
For $f(x)$ to be maximum
$f'(x) = 4x - 2 = 0 \\ \implies x = \frac{1}{2}$
So at $x = \frac{1}{2}, f(x)$ is an extremum (either maximum or minimum).
$f(2) = 2(2)^2 - 2(2) + 6 = 8 - 4 + 6 = 10$
$f\left(\frac{1}{2}\right) = 2{\frac{1}{2}}^2 - 2\frac{1}{2} + 6 = 5.5$, so $x = \frac{1}{2}$ is a mimimum.
$f(0) = 6$.
So, the maximum value is at $x = 2$ which is $10$ as there are no other extremum for the given function.