Question

Let the point (p, p+1) lie inside the region \(E = \{(x,y):3-x≤y≤\sqrt{9-x^2}, 0 ≤ x ≤ 3\} \)If the set of all values of p is the interval (a,b). then b2 + b - a2 is equal to_____

Answer 1

Correct Answer: 3

The region \(E\) is defined by the inequalities: \[ 3 - x \leq y \leq \sqrt{9 - x^2}, \quad 0 \leq x \leq 3. \] The point \(A = (p, p+1)\) lies inside the region. For \(A\) to lie above the line \(L: y = 3 - x\): \[ p + p + 1 - 3 > 0 \implies 2p > 2 \implies p > 1. \] For \(A\) to lie below the semicircle \(S: y = \sqrt{9 - x^2}\): \[ p^2 + (p+1)^2 - 9 < 0. \] Solving: \[ p^2 + p^2 + 2p + 1 - 9 < 0 \implies 2p^2 + 2p - 8 < 0 \implies p^2 + p - 4 < 0. \] The roots of \(p^2 + p - 4 = 0\) are: \[ p = \frac{-1 \pm \sqrt{17}}{2}. \] So, the valid interval is: \[ 0 < p < \frac{\sqrt{17} - 1}{2}. \] Combining with \(p > 1\): \[ 1 < p < \frac{\sqrt{17} - 1}{2}. \] Calculating: \[ b^2 + b - a^2 = 3. \] ✅ Therefore, the final answer is: \(3\).