Let $$\int_\limits0^x \sqrt{1-\left(y^{\prime}(t)\right)^2} d t=\int_0^x y(t) d t, 0 \leq x \leq 3, y \geq 0, y(0)=0$$. Then at $$x=2, y^{\prime \prime}+y+1$$ is equal to

Two vertices of a triangle $$\mathrm{ABC}$$ are $$\mathrm{A}(3,-1)$$ and $$\mathrm{B}(-2,3)$$, and its orthocentre is $$\mathrm{P}(1,1)$$. If the coordinates of the point $$\mathrm{C}$$ are $$(\alpha, \beta)$$ and the centre of the of the circle circumscribing the triangle $$\mathrm{PAB}$$ is $$(\mathrm{h}, \mathrm{k})$$, then the value of $$(\alpha+\beta)+2(\mathrm{~h}+\mathrm{k})$$ equals

The integral $$\int_\limits{1 / 4}^{3 / 4} \cos \left(2 \cot ^{-1} \sqrt{\frac{1-x}{1+x}}\right) d x$$ is equal to

Let $$\alpha, \beta ; \alpha>\beta$$, be the roots of the equation $$x^2-\sqrt{2} x-\sqrt{3}=0$$. Let $$\mathrm{P}_n=\alpha^n-\beta^n, n \in \mathrm{N}$$. Then $$(11 \sqrt{3}-10 \sqrt{2}) \mathrm{P}_{10}+(11 \sqrt{2}+10) \mathrm{P}_{11}-11 \mathrm{P}_{12}$$ is equal to