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

Given region, E=\(x, y): 3-x y 9-x^2, 0 x 3\ Since, point (p, p+1) lie on line y=x+1 Point of intersection of y=x+1 and y=3-x i.e., x+1=3-x 2 x=2 x=1 and y=2 and point of intersection of y=x+1 and y=9-x^2 i.e., (x+1)^2=9-x^2 $arraylc & x^2+1+2 x=

Let the point p p1 lie inside the region Ex y 3x leq y leq 9: null (Mathematics)

Given region,

E=\left\{(x, y): 3-x \leq y \leq \sqrt{9-x^2}, 0 \leq x \leq 3\right\}

Since, point

(p, p+1)

lie on line

y=x+1

\therefore

Point of intersection of

y=x+1

and

y=3-x

i.e.,

x+1=3-x

\Rightarrow

2 x=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{array}{lc} &\Rightarrow x^2+1+2 x=9-x^2 \\\\ &\Rightarrow 2 x^2+2 x-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{array}

\begin{array}{lll}\Rightarrow & x=\frac{-1+\sqrt{17}}{2}, & \text { (Since, } x \in[0,3]) \\\\ \therefore & p \in \left(1, \frac{-1+\sqrt{17}}{2}\right)\end{array}

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

\begin{aligned} & \therefore 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{aligned}

Explanation

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