JEE Mains · Maths · STD 12 - 8. Application and integration
The area of the region bounded by the curves \(x+3y^2=0\) and \(x+4y^2=1\) is equal to:
- A \(\dfrac{1}{3}\)
- B \(\dfrac{2}{3}\)
- C \(\dfrac{4}{3}\)
- D \(\dfrac{5}{3}\)
Answer & Solution
Correct Answer
(C) \(\dfrac{4}{3}\)
Step-by-step Solution
Detailed explanation
The equations of the given curves are \(x = -3y^2\) and \(x = 1 - 4y^2\). To find the points of intersection, equate the values of \(x\): \(-3y^2 = 1 - 4y^2\) \(y^2 = 1 \Rightarrow y = \pm 1\) The region is bounded between \(y = -1\) and \(y = 1\). In this interval,…
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