JEE Mains · Maths · STD 12 - 9. differential equations
Let \(S =(0,2 \pi)-\left\{\frac{\pi}{2}, \frac{3 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}\right\}\). Let \(y =\) \(y ( x ), x \in S\), be the solution curve of the differential equation \(\frac{ dy }{ dx }=\frac{1}{1+\sin 2 x }, y \left(\frac{\pi}{4}\right)=\frac{1}{2}\). if the sum of abscissas of all the points of intersection of the curve \(y=y(x)\) with the curve \(y=\sqrt{2} \sin x\) is \(\frac{k \pi}{12}\), then \(k\) is equal to
- A \(50\)
- B \(40\)
- C \(41\)
- D \(42\)
Answer & Solution
Correct Answer
(D) \(42\)
Step-by-step Solution
Detailed explanation
\(\frac{ dy }{ dx }=\frac{1}{1+\sin 2 x }\) \(\int d y=\int \frac{d x}{(\sin x+\cos x)^{2}}\) \(\int d y=\int \frac{\sec ^{2} x}{(1+\tan x)^{2}}\) \(y(x)=-\frac{1}{1+\tan x}+C\) \(y\left(\frac{\pi}{4}\right)=\frac{1}{2}=-\frac{1}{2}+C\) \(C =1\) \(y(x)=\frac{-1}{1+\tan x}+1\)…
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