If $x^2 y^2=\sin ^{-1} x+\cos ^{-1} x$, then $\frac{\mathrm{d} y}{\mathrm{~d} x}$ at $x=1$ and $y=2$ is

If $\overline{\mathrm{a}}=\hat{\mathrm{j}}-\hat{\mathrm{k}}$ and $\overline{\mathrm{c}}=\hat{\mathrm{i}}-\hat{\mathrm{j}}-\hat{\mathrm{k}}$, then the vector $\overline{\mathrm{b}}$ satisfying $\overline{\mathrm{a}} \times \overline{\mathrm{b}}+\overline{\mathrm{c}}=\overline{0}$ and $\overline{\mathrm{a}} \cdot \overline{\mathrm{b}}=3$ is

The expression $((p \wedge q) \vee(p \vee \sim q)) \wedge(\sim p \wedge \sim q)$ is equivalent to

If $y=4 x-5$ is a tangent to the curve $y^2=p x^3+q$ at $(2,3)$, then the values of $p$ and $q$ are respectively