QJEEM22Ma361a: JEE Main 2022 Mathematics - Class 12 JEEM22Ma361a Solution: Let . Let , x S, be the solution curve of the differential…

${{dy} \over {dx}} = {1 \over {1 + \sin 2x}}$

$\Rightarrow dy = {{{{\sec }^2}xdx} \over {{{(1 + \tan x)}^2}}}$

$\Rightarrow y = - {1 \over {1 + \tan x}} + c$

When $x = {\pi \over 4}$ , $y = {1 \over 2}$ gives c = 1

So $y = {{\tan x} \over {1 + \tan x}} \Rightarrow y = {{\sin x} \over {\sin x + \cos x}}$

Now, $y = \sqrt 2 \sin x \Rightarrow \sin x = 0$

or $\sin x + \cos x = {1 \over {\sqrt 2 }}$

$\sin x = 0$ gives $x = \pi$ only.

and $\sin x + \cos x = {1 \over {\sqrt 2 }} \Rightarrow \sin \left( {x + {\pi \over 4}} \right) = {1 \over 2}$

So $x + {\pi \over 4} = {{5\pi } \over 6}$ or ${{13\pi } \over 6} \Rightarrow x = {{7\pi } \over {12}}$ or ${{23\pi } \over {12}}$

Sum of all solutions $= \pi + {{7\pi } \over {12}} + {{23\pi } \over {12}} = {{42\pi } \over {12}}$

Hence, $k = 42$ .