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∈S, be the solution curve of the differential equation
$\frac{dy}{dx}=\frac{1}{1+sin⁡2x},y(\frac{π}{4})=\frac{1}{2}$
.If the sum of abscissas of all the points of intersection of the curve y = y(x) with the curve
$y=\sqrt2sin⁡x$ is $\frac{kπ}{12}$,
then k is equal to _________.

Answer 1

The correct answer is 42
$\frac{dy}{dx}=\frac{1}{1+sin⁡2x}$
$⇒dy=\frac{sec^2⁡xdx}{(1+tan⁡ x)^2}$
$⇒y=−\frac{1}{1+tan⁡x}+c$
When
$x=\frac{π}{4}, y=\frac{1}{2}$ gives c=1
So $y=\frac{tan⁡x}{1+tan⁡x}⇒y=\frac{sin⁡x}{sin⁡x+cos⁡x}$
Now, $y=\sqrt2sin⁡x ⇒sin⁡x=0$
or
$sin⁡x+cos⁡x=\frac{1}{\sqrt2}$
sinx = 0 gives x = π only
and
$sin⁡x+cos⁡x=\frac{1}{\sqrt2}⇒sin⁡(x+\frac{π}{4})=\frac{1}{2}$
So
$x+\frac{π}{4}=\frac{5π}{6}$ or $\frac{13π}{6}⇒x=\frac{7π}{12}$ or $\frac{23π}{12}$

Sum of all solutions $= π+\frac{7π}{12}+\frac{23π}{12}=\frac{42π}{12}$
Therefore, k = 42.