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

Correct Answer: 42

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.