Let the point p , p + 1 lie inside the region E = x , y : 3 - x ≤ y ≤ 9 - x 2 , 0 ≤ x…

Given,

The point $\left(p,p+1\right)$ lie inside the region $E=\left\{\left(x,y\right):3-x\le y\le \sqrt{9-x^2},0\le x\le 3\right\}$,

And the set of all values of $p$ is the interval $\left(a,b\right),$ 

Now by observation $\left(p,p+1\right)$ lies on $y-x=1$,

Also point $\left(p,p+1\right)$ lies on $x+y=3$

So, solving $y-x=1$ and $x+y=3$ we get, $P\left(1,2\right)$

Again given point $\left(p,p+1\right)$ lies on $x^2+y^2=9$

So, solving $y-x=1$ and $x^2+y^2=9$, we get

$x^2+(1+x{)}^2=9$

$\Rightarrow 2x^2+2x-8=0$

$\Rightarrow x^2+x-4=0.........\left(\mathrm{i}\right)$

$\Rightarrow x=\frac{-1\pm \sqrt{1+4·4}}{2}=\frac{\sqrt{17}-1}{2}$

Hence, $p\in \left(1,\frac{\sqrt{17}-1}{2}\right)$

Now on comparing with $p\in \left(a,b\right)$ we get,

$\Rightarrow a=1, b^2+b=4$   [using $\left(\mathrm{i}\right)$]

Hence, the value of $b^2+b-a^2=4-1=3$