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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

  1. A \(50\)
  2. B \(40\)
  3. C \(41\)
  4. D \(42\)
Verified Solution

Answer & Solution

Correct Answer

(D) \(42\)

Step-by-step Solution

Detailed explanation

\(\frac{ dy }{ dx }=\frac{1}{1+\sin 2 x }\) \(\int d y=\int \frac{d x}{(\sin x+\cos x)^{2}}\) \(\int d y=\int \frac{\sec ^{2} x}{(1+\tan x)^{2}}\) \(y(x)=-\frac{1}{1+\tan x}+C\) \(y\left(\frac{\pi}{4}\right)=\frac{1}{2}=-\frac{1}{2}+C\) \(C =1\) \(y(x)=\frac{-1}{1+\tan x}+1\)…