Let $S=(0,2 \pi)-\left\{\frac{\pi}{2}, \frac{3 \pi}{4}, \frac{3 \pi}{2}, \frac{7 \pi}{4}\right\}$. L

Question

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 $\frac{d y}{d x}=\frac{1}{1+\sin 2 x}, y\left(\frac{\pi}{4}\right)=\frac{1}{2}$. 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

Professor PC Thomas Classes · Answer

(42) : We have\ $\frac{d y}{d x}=\frac{1}{1+\sin 2 x}, y\left(\frac{\pi}{4}ight)=\frac{1}{2}$\ $d y=\frac{\sec ^{2} x}{(1+	an x)^{2}} d x \Rightarrow y=-\frac{1}{1+	an x}+c$\ when $x=\frac{\pi}{4}, y=\frac{1}{2} \Rightarrow c=1$\ $	herefore y=\frac{	an x}{1+	an x}=\frac{\sin x}{\sin x+\cos x}$\ We have $y=\sqrt{2} \sin x$\ For intersection point\ $\sqrt{2} \sin x=\frac{\sin x}{\sin x+\cos x}$\ $\Rightarrow \sin x\left(\sqrt{2}-\frac{1}{\sin x+\cos x}ight)=0$\ $\sin x=0$ or $\sin x+\cos x=\frac{1}{\sqrt{2}}$\ $x=\pi$ or $\sin \left(x+\frac{\pi}{4}ight)=\frac{1}{2}$\ $\Rightarrow x+\frac{\pi}{4}=\frac{5 \pi}{6}$ or $\frac{13 \pi}{6} 	herefore x=\pi, \frac{7 \pi}{12}, \frac{23 \pi}{12}$\ Sum of abscissas $=\pi+\frac{7 \pi}{12}+\frac{23 \pi}{12}=\frac{42 \pi}{12}$\ $	herefore k=42$\