Let $\bar{u}, \bar{v}, \bar{w}$ be the vectors such that $|\overline{\mathrm{u}}|=1,|\overline{\mathrm{v}}|=2,|\overline{\mathrm{w}}|=3$. If the projection $\overline{\mathrm{v}}$ along $\overline{\mathrm{u}}$ is equal to that of $\overline{\mathrm{w}}$ along $\overline{\mathrm{u}}$ and the vectors $\overline{\mathrm{v}}, \overline{\mathrm{w}}$ are perpendicular to each other then $|\overline{\mathrm{u}}-\overline{\mathrm{v}}+\overline{\mathrm{w}}|$ equals

The area enclosed between the curves $y^2=4 x$ and $y=|x|$ is

If $\tan \mathrm{A}=\frac{1}{\sqrt{x\left(x^2+x+1\right)}}, \tan \mathrm{B}=\frac{\sqrt{x}}{\sqrt{x^2+x+1}}$ and $\tan \mathrm{C}=\sqrt{x^{-1}+x^{-2}+x^{-3}}$ then

Let X be a discrete random variable. The probability distribution of X is given below

$$ \begin{array}{|c|c|c|c|} \hline \mathrm{X} & 30 & 10 & -10 \\ \hline \mathrm{P}(\mathrm{X}) & \frac{1}{5} & \mathrm{~A} & \mathrm{~B} \\ \hline \end{array} $$

and $\mathrm{E}(\mathrm{X})=4$, then the value of AB is equal to