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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 $\frac{\mathrm{k} \pi}{12}$, then $\mathrm{k}$ is equal to ___________.
Option: 1
42

Option: 2
-

Option: 3
-

Option: 4
-

Answers (1)

best_answer

$\mathrm{\frac{d y}{d x} =\frac{1}{1+\sin 2 x}} \\$

$\mathrm{\int d y =\int \frac{d x}{ (\sin x+\cos x)^{2}} }\\$

$\mathrm{\int d y =\int \frac{\sec ^{2} x}{(1+\tan x)^{2}} }$

$\mathrm{y(x) =-\frac{1}{1+\tan x}+c} \\$

$\mathrm{y\left(\frac{\pi}{4}\right) =\frac{1}{2}=\frac{-1}{2}+c } \\$

$\mathrm{c=1 } \\$

$\mathrm{y(x) =\frac{-1}{1+\tan x}+1 }$

$\mathrm{y(x)=\frac{-1+1+\tan x}{1+\tan x}} \\$

$\mathrm{y(x)=\frac{\tan x}{1+\tan x} }$

Solving with $\mathrm{y=\sqrt{2}\sin x }$

$\mathrm{\frac{\tan x}{1+\tan x}=\sqrt{2} \sin x} \\$

$\mathrm{\sin x=0 \quad, \frac{1}{\sqrt{2}}=\sin x+\cos x}$

$\mathrm{x=\pi \qquad \frac{1}{2}=\sin \left(x+\frac{\pi}{4}\right) }\\$

$\mathrm{\sin \frac{\pi}{6}=\sin \left(x+\frac{\pi}{4}\right)} \\$

$\mathrm{x+\frac{\pi}{4}=\pi-\frac{\pi}{6}, 2 \pi+\frac{\pi}{6}}$

$\mathrm{x=\frac{5 \pi}{6}-\frac{\pi}{4}} \\$

$\mathrm{x=\frac{7 \pi}{12}} \\$

$\mathrm{x=\frac{13 \pi}{6}-\frac{\pi}{4}}$

$\mathrm{x=\frac{23 \pi}{12}}$

Sum of solution

$\mathrm{=\pi+\frac{7 \pi}{12}+\frac{23 \pi}{12}} \\$

$\mathrm{=\frac{42 \pi}{12}=\frac{k \pi}{12} }\\$

$\mathrm{\Rightarrow k=42}$

Hence answer is $\mathrm{42}$

Posted by

Shailly goel

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