The first derivative of the function $\left(\cos ^{-1}\left(\sin \sqrt{\frac{1+x}{2}}\right)+x^x\right)$ with respect to $x$ at $x=1$ is

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