Let the point (p, p + 1) lie inside the region E = \ (x, y): 3 - 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),$</p><p>Now by observation \left(p, p + 1\right)$ lies on $y - x = 1$,

Also point \left(p, p + 1\right)$lies on$x + y = 3$</p><p>So, solving$y - x = 1$and$x + y = 3$we get,$P\left(1, 2\right)$</p><p>Again given point \left(p, p + 1\right)$ lies on ${x}^{2} + {y}^{2} = 9$

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

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

$⇒ 2{x}^{2} + 2x - 8 = 0$

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

$⇒ x = \dfrac{- 1 \pm \sqrt{1 + 4 \cdot 4}}{2} = \dfrac{\sqrt{17} - 1}{2}$

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

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

$⇒ a = 1, {b}^{2} + b = 4$ [using \left(i\right)$]</p><p>Hence, the value of${b}^{2} + b - {a}^{2} = 4 - 1 = 3