JEE Mains · Maths · STD 12 - 8. Application and integration
Let \(A _{1}\) be the area of the region bounded by the curves \(y =\sin x , y =\cos x\) and \(y\) -axis in the first quadrant. Also, let \(A _{2}\) be the area of the region bounded by the curves \(y=\sin x\) \(y =\cos x , x\) -axis and \(x =\frac{\pi}{2}\) in the first quadrant. Then ..... .
- A \(A _{1}: A _{2}=1: \sqrt{2}\) and \(A _{1}+ A _{2}=1\)
- B \(A _{1}= A _{2}\) and \(A _{1}+ A _{2}=\sqrt{2}\)
- C \(2 A _{1}= A _{2}\) and \(A _{1}+ A _{2}=1+\sqrt{2}\)
- D \(A _{1}: A _{2}=1: 2\) and \(A _{1}+ A _{2}=1\)
Answer & Solution
Correct Answer
(A) \(A _{1}: A _{2}=1: \sqrt{2}\) and \(A _{1}+ A _{2}=1\)
Step-by-step Solution
Detailed explanation
\(A _{1}=\int_{0}^{\pi / 4}(\cos x -\sin x ) d x\) \(A _{1}=(\sin x +\cos x )_{0}^{\pi / 4}=\sqrt{2}-1\) \(A _{2}=\int_{0}^{\pi / 4} \sin x d x +\int_{\pi / 4}^{\pi / 2} \cos x dx\) \(=(-\cos x)_{0}^{\pi / 4}+(\sin x)_{\pi / 4}^{\pi / 2}\) \(A_{2}=\sqrt{2}(\sqrt{2}-1)\)…
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