Keeping the velocity of projectile constant, the angle of projection is increased from $0^o$ to ${90}^o$, then the horizontal range of the projectile

The correct option is C increases upto ${45}^o$ and decreases afterwards.
Horizontal Range, $R=\frac{u^2\sin 2\theta }{2g}$
$R\propto \sin 2\theta$
$\sin 2\theta$ increases till $2\theta ={90}^0$
i.e $\theta ={45}^0$
Hence, Range $R$ increases till ${45}^0$ then decreases.