For the given values of initial velocity of projection and angle of inclination of the plane, the maximum range for a projectile projected upwards will be obtained, if the angle of projection is A. $$\alpha = \frac{\pi }{4} – \frac{\beta }{2}$$ B. $$\alpha = \frac{\pi }{2} + \frac{\beta }{2}$$ C. $$\alpha = \frac{\beta }{2} – \frac{\pi }{2}$$ D. $$\alpha = \frac{\pi }{4} – \frac{\beta }{4}$$ E. $$\alpha = \frac{\pi }{2} – \frac{\beta }{2}$$

The correct answer is $\boxed{\alpha = \frac{\pi}{4} – \frac{\beta}{2}}$.

The range of a projectile is the maximum horizontal distance that it travels. It is given by the formula

$$R = \frac{v_0^2 \sin(2\alpha)}{g}$$

where $v_0$ is the initial velocity, $\alpha$ is the angle of projection, and $g$ is the acceleration due to gravity.

The maximum range occurs when $\sin(2\alpha) = 1$, which means that $\alpha = \frac{\pi}{4} – \frac{\beta}{2}$, where $\beta$ is any integer multiple of $\pi$.

The other options are incorrect because they do not result in a maximum range. For example, if $\alpha = \frac{\pi}{2} + \frac{\beta}{2}$, then $\sin(2\alpha) = 0$, which means that there is no horizontal displacement.