Let the point (p,p+1) lie inside the region E=(x,y):3-x y √9-x^2,0 x 3.

Question

Let the point $(p,p+1)$ lie inside the region $E={(x,y):3-x\le y\le \sqrt{9-{x}^{2}},0\le x\le 3}$. If the set of all values of $p$ is the interval $(a,b)$, then ${b}^{2}+b-{a}^{2}$ is equal to \_\_\_\_\_\_\_\_\_\_\_.

Answer: 3

Accepted Answer

Given region,

$E={(x,y):3-x\le y\le \sqrt{9-{x}^{2}},0\le x\le 3}$

Since, point $(p,p+1)$ lie on line $y=x+1$

$	herefore $ Point of intersection of $y=x+1$ and $y=3-x$

i.e.,
$x+1=3-x$

$\Rightarrow$ $2x=2\Rightarrow x=1$

and $y=2$

and point of intersection of $y=x+1$ and

$y=\sqrt{9-{x}^{2}}$

i.e., $(x+1{)}^{2}=9-{x}^{2}$

$\begin{gathered} \Rightarrow {x}^{2}+1+2x=9-{x}^{2} \ \Rightarrow 2{x}^{2}+2x-8=0 \ \Rightarrow {x}^{2}+x-4=0 \ \Rightarrow x=\frac{-1\pm \sqrt{1+4(1)(4)}}{2} \ \Rightarrow x=\frac{-1\pm \sqrt{17}}{2} \end{gathered}$

$\begin{gathered} \Rightarrow \quad  x=\frac{-1+\sqrt{17}}{2}, \quad  x\in [0,3]) \ 	herefore  \quad  p\in (1,\frac{-1+\sqrt{17}}{2}) \end{gathered}$

$\Rightarrow a=1,b=\frac{-1+\sqrt{17}}{2}$

$\begin{gathered} 	herefore {b}^{2}+b-a \ =\frac{1+17-2\sqrt{17}}{4}+\frac{(-1+\sqrt{17})}{2}-1 \ =\frac{18-2\sqrt{17}-2+2\sqrt{17}-4}{4} \ =\frac{12}{4}=3 \end{gathered}$

Solutions · 1