ExamBro
ExamBro
JEE Mains · Maths · STD 11 - 9. straight line

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 of \(p\) is the interval \((a, b)\). then \(b^2+b-a^2\) is equal to \(.................\).

  1. A \(2\)
  2. B \(1\)
  3. C \(4\)
  4. D \(3\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(3\)

Step-by-step Solution

Detailed explanation

\(3-x \leq y \leq \sqrt{9-x^2}\) Points \(( p , p +1)\) lies on \(y = x +1\) So point of intersection between \(y = x +1 y =3- x \text { is } x =1, y =2\) and point of intersection between \(x+1=\sqrt{9-x^2}\) is \(x=\frac{-1+\sqrt{17}}{2}\) Hence…
Same subject
Explore more questions on app
From JEE Mains
Explore more questions on app