JEE Mains · Maths · STD 12 - 8. Application and integration
Let \(\Delta\) be the area of the region \(\left\{( x , y ) \in R ^2: x ^2+ y ^2 \leq 21, y ^2 \leq 4 x , x \geq 1\right\}\). Then \(\frac{1}{2}\left(\Delta-21 \sin ^{-1} \frac{2}{\sqrt{7}}\right)\) is equal to
- A \(2 \sqrt{3}-\frac{1}{3}\)
- B \(\sqrt{3}-\frac{2}{3}\)
- C \(2 \sqrt{3}-\frac{2}{3}\)
- D \(\sqrt{3}-\frac{4}{3}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{3}-\frac{4}{3}\)
Step-by-step Solution
Detailed explanation
\(\text { Area } 2 \int \limits_1^3 2 \sqrt{x} d x+2 \int_3^{\sqrt{21}} \sqrt{21-x^2 d x}\) \(\Delta=\frac{8}{3}(3 \sqrt{3}-1)+21 \sin ^{-1}\left(\frac{2}{\sqrt{7}}\right)-6 \sqrt{3}\)…
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