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QJEEM23Ma694e: JEE Main 2023 Mathematics - Class 11 JEEM23Ma694e Solution: Let the point lie inside the region . If the set of all values…

Let the point (p,p+1)(p, p+1) lie inside the region E={(x,y):3−x≤y≤9−x2,0≤x≤3}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 of p\mathrm{p} is the interval (a,b)(a, b) , then b2+b−a2b^{2}+b-a^{2} is equal to ___________.

Step-by-Step Verified Solution

Given region,

E={(x,y):3−x≤y≤9−x2,0≤x≤3}E=\left\{(x, y): 3-x \leq y \leq \sqrt{9-x^2}, 0 \leq x \leq 3\right\}

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

∴\therefore Point of intersection of y=x+1y=x+1 and y=3−xy=3-x

i.e., x+1=3−xx+1=3-x

⇒\Rightarrow 2x=2⇒x=12 x=2 \Rightarrow x=1

and y=2y=2

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

y=9−x2y=\sqrt{9-x^2}

i.e., (x+1)2=9−x2(x+1)^2=9-x^2

amp;⇒x2+1+2x=9−x2amp;⇒2x2+2x−8=0amp;⇒x2+x−4=0amp;⇒x=−1±1+4(1)(4)2amp;⇒x=−1±172\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}

⇒amp;x=−1+172,amp; (Since, x∈[0,3])∴amp;p∈(1,−1+172)\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}

⇒a=1,b=−1+172\Rightarrow a=1, b=\frac{-1+\sqrt{17}}{2}

amp;∴b2+b−aamp;=1+17−2174+(−1+17)2−1amp;=18−217−2+217−44amp;=124=3\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}

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