JEE Mains · Maths · STD 12 - 8. Application and integration
The area (in sq. units) of the smaller portion enclosed between the curves, \(x^2 + y^2 = 4\) and \(y^2 =3x\), is
- A \(\frac{1}{{2\sqrt 3 }} + \frac{\pi }{3}\)
- B \(\frac{1}{{\sqrt 3 }} + \frac{{2\pi }}{3}\)
- C \(\frac{1}{{2\sqrt 3 }} + \frac{{2\pi }}{3}\)
- D \(\frac{1}{{\sqrt 3 }} + \frac{{4\pi }}{3}\)
Answer & Solution
Correct Answer
(D) \(\frac{1}{{\sqrt 3 }} + \frac{{4\pi }}{3}\)
Step-by-step Solution
Detailed explanation
From the given equations, we get; \(x^{2}+3 x-4=0\) \( \Rightarrow \) \((x+4)(x-1)=0\) \(x=-4, x=1\) when \(x=1, y=\sqrt{3}\) \(\therefore \) req. area \(=\)…
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