Opening of flowers in the cymose inflorescence is

If $x = \frac{1 - t^2}{1 + t^2}$ and $y = \frac{2at}{1 + t^2}$,then $\frac{dy}{dx} = $

$A$ projectile is projected with velocity $k v_e$ in the vertically upward direction from the ground into space. ($v_e$ is the escape velocity and $k < 1$). If air resistance is considered to be negligible,then the maximum height from the centre of the Earth to which it can go will be: ($R =$ radius of Earth)

$\lim _{x \rightarrow \pi / 6} \left[ \frac{3 \sin x - \sqrt{3} \cos x}{6 x - \pi} \right]$ is equal to :

If $0 < x < 1$,then $\sqrt{1+x^2}\left[\left\{x \cos \left(\cot ^{-1} x\right)+\sin \left(\cot ^{-1} x\right)\right\}^2-1\right]^{1 / 2}$ is equal to

The solution of the differential equation $\frac{dy}{dx} = \frac{x-2y+1}{2x-4y}$ is