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

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Differential Equations

Let

S = (0, 2 π) - {\frac{π}{2}, \frac{3 π}{4}, \frac{3 π}{2}, \frac{7 π}{4}}

. Let

y = y (x)

x \in S

, be the solution curve of the differential equation

\frac{d y}{d x} = \frac{1}{1 + s i n 2 x}

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 = 2 sin x

\frac{k π}{12}

, then

k

is equal to ____.

Answer: 42

Solving the differential equation:

\frac{d y}{d x} = \frac{1}{1 + sin 2 x} = \frac{1}{( sin x + cos x ) ^{2}} = \frac{sec ^{2} x}{( 1 + tan x ) ^{2}}

Integrating:

y = - \frac{1}{1 + tan x} + C

Using

y (\frac{π}{4}) = \frac{1}{2}

\frac{1}{2} = - \frac{1}{2} + C \Rightarrow C = 1

Thus:

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

Solving

\frac{t a n x}{1 + t a n x} = 2 sin x

:

This simplifies to

sin x = 0

sin x + cos x = \frac{1}{2}

.

For

sin x = 0

x = π

(in the domain).

For

sin (x + \frac{π}{4}) = \frac{1}{2}

x + \frac{π}{4} = \frac{5 π}{6}

\frac{13 π}{6}

, giving

x = \frac{7 π}{12}

\frac{23 π}{12}

.

Sum:

π + \frac{7 π}{12} + \frac{23 π}{12} = \frac{42 π}{12}

Therefore,

k = 42

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Integer answer