Question

Let $S=(0,2\pi )-\{\frac{\pi }{2},\frac{3\pi }{4},\frac{3\pi }{2},\frac{7\pi }{4}\}$. Let $y=y\left(x\right),x\in S$ 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 $y=y\left(x\right)$ with the curve $y=\sqrt{2}\sin x\text{is}\frac{k\pi }{12}$, then $k$ is equal to

• A. 42.0
• B. 42.00
• C. 42

Answer 1

Answer: (A), (B), (C)

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

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

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

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

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

Now, $y=\sqrt{2}\sin x$
$\Rightarrow \sin x=0$ or $\sin x+\cos x=\frac{1}{\sqrt{2}}$
$\sin x=0\text{gives}x=\pi$ only and $\sin x+\cos x=\frac{1}{\sqrt{2}}\Rightarrow \sin (x+\frac{\pi }{4})=\frac{1}{2}$
So, $x+\frac{\pi }{4}=\frac{5\pi }{6}\text{or}\frac{13\pi }{6}\Rightarrow x=\frac{7\pi }{12}\text{or}\frac{23\pi }{12}$
Sum of all solutions $=\pi +\frac{7\pi }{12}+\frac{23\pi }{12}=\frac{42\pi }{12}$
Hence $k=42$.