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: The region E is defined by 3 - x y 9 - x^2, 0 x 3 and the point is A = (p, p+1). Find: The value of b^2 + b - a^2 where the set of all values of p is (a,b). For A to lie above the line L: y = 3 - x, p + p + 1 - 3 0 so 2p 2 which gives p 1. For A to lie below the semicircle S: y = 9 - x^2, p^2 + (p+1)^2 - 9 < 0. Now simplify: aligned p^2 + p^2 + 2p + 1 - 9 &< 0 2p^2 + 2p - 8 &< 0 p^2 + p - 4 &< 0 aligned The roots of p^2 + p - 4 = 0 are p = -1 172. Hence, p^2 + p - 4 < 0 for -1- 172 < p < -1+ 172. From the region, we also need 0 < p < 3, and from the line condition we already have p 1. Therefore, combining all conditions, 1 < p < 17-12. So, a = 1, b = 17-12. Now evaluate the required expression: b^2 + b - a^2 = 3. Therefore, the final answer is 3.

Let the point (p, p + 1) lie inside the region E = (x,y): 3 - x y… — Solution | JEE Mathematics

NVAMediumJEE 2023Basics: Distance, Section Formula, Locus

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

Answer

Correct answer:3

Step-by-step solution

Standard Method

Given: The region $E$ is defined by

3 - x \leq y \leq \sqrt{9 - x^2}, \quad 0 \leq x \leq 3

and the point is $A = (p, p+1)$ .

Find: The value of $b^2 + b - a^2$ where the set of all values of $p$ is $(a,b)$ .

For $A$ to lie above the line $L: y = 3 - x$ ,

p + p + 1 - 3 > 0

2p > 2

which gives

p > 1.

For $A$ to lie below the semicircle $S: y = \sqrt{9 - x^2}$ ,

p^2 + (p+1)^2 - 9 < 0.

Now simplify:

\begin{aligned} p^2 + p^2 + 2p + 1 - 9 &< 0 \\ 2p^2 + 2p - 8 &< 0 \\ p^2 + p - 4 &< 0 \end{aligned}

Root Interval Evaluation

The roots of

p^2 + p - 4 = 0

are

p = \frac{-1 \pm \sqrt{17}}{2}.

Hence,

p^2 + p - 4 < 0

for

\frac{-1-\sqrt{17}}{2} < p < \frac{-1+\sqrt{17}}{2}.

From the region, we also need $0 < p < 3$ , and from the line condition we already have $p > 1$ . Therefore, combining all conditions,

1 < p < \frac{\sqrt{17}-1}{2}.

So,

a = 1, \quad b = \frac{\sqrt{17}-1}{2}.

Now evaluate the required expression:

b^2 + b - a^2 = 3.

Therefore, the final answer is $3$ .

Common mistakes

Using non-strict inequalities for a point lying inside the region. This is wrong because interior points must satisfy strict conditions relative to the boundaries used here. Use $p > 1$ and the corresponding strict upper bound on $p$ .
Checking only the inequality $3-x \le y \le \sqrt{9-x^2}$ but forgetting that the point is $(p,p+1)$ , so both $x$ and $y$ must be substituted correctly. Replace $x$ by $p$ and $y$ by $p+1$ before solving.
Solving $p^2 + p - 4 < 0$ incorrectly outside its roots. A quadratic with positive leading coefficient is negative between its roots, not outside them. First find both roots, then take the interval between them.

Practice more Basics: Distance, Section Formula, Locus questions

Get unlimited AI-adaptive practice, mastery tracking, and an AI tutor that explains every step — free to start.

Start free trial More on Basics: Distance, Section Formula, Locus

Answer

Step-by-step solution

Common mistakes

Practice more Basics: Distance, Section Formula, Locus questions

Related questions