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 _____ . [2023]

Question

Let the point $(p, p + 1)$ lie inside the region $E = {(x, y) : 3 - x \leq y \leq \sqrt{9 - x^{2}}, 0 \leq x \leq 3} .$ If the set of all values of $p$ is the interval $(a, b),$ then $b^{2} + b - a^{2}$ is equal to _____ . [2023]

Smart Achievers · Accepted Answer

(3)

$Given, 3 - x \leq y \leq \sqrt{9 - x^{2}}$

$3 - x \leq y \Rightarrow x + y - 3 \geq 0 \dots (i)$

$y \leq \sqrt{9 - x^{2}} \Rightarrow x^{2} + y^{2} \leq 9$

$Now, (p, p + 1) lies inside the region E .$ $∴$ $p + p + 1 - 3 \geq 0$

$\Rightarrow p \geq 1 and p^{2} + {(p + 1)}^{2} \leq 9$

$\Rightarrow 2 p^{2} + 2 p - 8 \leq 0 \Rightarrow p^{2} + p - 4 \leq 0$

$\Rightarrow p \in (- \frac{(1 + \sqrt{17})}{2}, \frac{\sqrt{17} - 1}{2})$

$∴$ $p \in (1, \frac{\sqrt{17} - 1}{2}) (as p \geq 1)$

$a = 1, b = \frac{\sqrt{17} - 1}{2} \Rightarrow b^{2} + b - a^{2} = 3$

Solution

Q.
Let the point $(p, p + 1)$ lie inside the region $E = {(x, y) : 3 - x \leq y \leq \sqrt{9 - x^{2}}, 0 \leq x \leq 3} .$ If the set of all values of $p$ is the interval $(a, b),$ then $b^{2} + b - a^{2}$ is equal to _____ . [2023]

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Solution

Q. Let the point (p, p+1) lie inside the region E={(x,y):3-x≤y≤9-x2, 0≤x≤3}. If the set of all values of p is the interval (a,b), then b2+b-a2 is equal to _____ . [2023]

Ans. (3) Given, 3-x≤y≤9-x2 3-x≤y⇒x+y-3≥0 ⋯(i) y≤9-x2⇒x2+y2≤9 Now, (p, p+1) lies inside the region E. ∴ p+p+1-3≥0 ⇒p≥1 and p2+(p+1)2≤9 ⇒2p2+2p-8≤0 ⇒ p2+p-4≤0 ⇒p∈(-(1+17)2,17-12) ∴ p∈(1,17-12) (as p≥1) a=1, b=17-12 ⇒ b2+b-a2=3

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Q.
Let the point $(p, p + 1)$ lie inside the region $E = {(x, y) : 3 - x \leq y \leq \sqrt{9 - x^{2}}, 0 \leq x \leq 3} .$ If the set of all values of $p$ is the interval $(a, b),$ then $b^{2} + b - a^{2}$ is equal to _____ . [2023]