1e:I[80107,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1f:I[76763,["/_next/static/chunks/235ad55241b6443f.js","/_next/static/chunks/8337ab1f037b6654.js","/_next/static/chunks/c9e933b5fd1a0afc.js","/_next/static/chunks/848a330faa2461ac.js","/_next/static/chunks/9f2942e368a262d2.js","/_next/static/chunks/92f9e11244990470.js","/_next/static/chunks/a8b867fbb1dfca1a.js","/_next/static/chunks/1100f0390a901fa7.js","/_next/static/chunks/bbe6bd9d3c7b0355.js","/_next/static/chunks/a82195c3e5de7dd4.js"],"default"] 1d:T432,{"@context":"https://schema.org","@graph":[{"@type":"BreadcrumbList","itemListElement":[{"@type":"ListItem","position":1,"name":"Home","item":"https://www.divinejee.com"},{"@type":"ListItem","position":4,"name":"Let the point p p1 lie inside the region Ex y 3x leq y leq 9","item":"https://www.divinejee.com/question/let-the-point-p-p-1-lie-inside-the-region-e-x-y-3x-leq-y-leq-9x2-0-leq-x-leq-3-if-the-set"}]},{"@type":"QAPage","mainEntity":{"@type":"Question","name":"Question Let the point p p1 lie inside the region Ex y 3x leq y leq 9 on Topic","text":"

Let the point $(p, p+1)$ lie inside the region $E=\\left\\{(x, y): 3-x \\leq y \\leq \\sqrt{9-x^{2}}, 0 \\leq x \\leq 3\\right\\}$. If the set of all values...","answerCount":1,"upvoteCount":0,"datePublished":"2026-01-16","author":{"@type":"Organization","name":"DivineJEE"},"acceptedAnswer":{"@type":"Answer","text":"Check the detailed solution on the page.","upvoteCount":0,"url":"https://www.divinejee.com/question/let-the-point-p-p-1-lie-inside-the-region-e-x-y-3x-leq-y-leq-9x2-0-leq-x-leq-3-if-the-set"}}}]}20:T446,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., $(x+1)^2=9-x^2$

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

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

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

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