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

माना बिन्दु \((\mathrm{p}, \mathrm{p}+1)\) क्षेत्र \(\mathrm{E}=\left\{(\mathrm{x}, \mathrm{y}): 3-\mathrm{x} \leq \mathrm{y} \leq \sqrt{9-\mathrm{x}^2}, 0 \leq \mathrm{x} \leq 3\right\}\) के अन्दर स्थित है। यदि \(\mathrm{p}\) के सभी मानों का समुच्चय अन्तराल \((\mathrm{a}, \mathrm{b})\) है, तब \(\mathrm{b}^2+\mathrm{b}-\mathrm{a}^2\) बराबर है

  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…
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