The equation of motion of the particle is $\mathrm{s}=\mathrm{at}^2+\mathrm{bt}+\mathrm{c}$. If the displacement after 1 second is 20 m , velocity after 2 seconds is $30 \mathrm{~m} /$ seconds and the acceleration is $10 \mathrm{~m} /$ seconds $^2$, then

If $y=\tan ^{-1}\left[\frac{4 \sin 2 x}{\cos 2 x-6 \sin ^2 x}\right]$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=0$ is

If $x=a \sin 2 t(1+\cos 2 t), y=b \cos 2 t(1-\cos 2 t)$ then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ is equal to

The contrapositive of the statement $\sim p \vee(q \wedge \sim r)$ is