Let the equation of the circle, which touches $x$-axis at the point $(a, 0), a>0$ and cuts off an intercept of length $b$ on $y-a x i s$ be $x^2+y^2-\alpha x+\beta y+\gamma=0$. If the circle lies below $x-a x i s$, then the ordered pair $\left(2 a, b^2\right)$ is equal to

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function defined by $f(x)=(2+3 a) x^2+\left(\frac{a+2}{a-1}\right) x+b, a \neq 1$. If $f(x+y)=f(x)+f(\mathrm{y})+1-\frac{2}{7} x \mathrm{y}$, then the value of $28 \sum\limits_{i=1}^5|f(i)|$ is

Let $O$ be the origin, the point $A$ be $z_1=\sqrt{3}+2 \sqrt{2} i$, the point $B\left(z_2\right)$ be such that $\sqrt{3}\left|z_2\right|=\left|z_1\right|$ and $\arg \left(z_2\right)=\arg \left(z_1\right)+\frac{\pi}{6}$. Then

The area (in sq. units) of the region $\left\{(x, \mathrm{y}): 0 \leq \mathrm{y} \leq 2|x|+1,0 \leq \mathrm{y} \leq x^2+1,|x| \leq 3\right\}$ is