Question

Let $A=\{{\lambda }_1,{\lambda }_2,\dots ,{\lambda }_m\},B=\{{\mu }_1,{\mu }_2,\dots ,{\mu }_n\}$ be two sets of values of $\lambda$ and $\mu$, where $\lambda ,\mu \in (0,100\pi ]$. The equation $x^2(\sin \lambda x+\cos \mu x)-2x\left[\sin \right(\lambda +\mu )x+\sin (\lambda -\mu \left)x\right]+(\sin \lambda x+\cos \mu x)=0$ has a positive solution for the ordered pair ${\lambda }_i,{\mu }_j$. If $\sum _{i=1}^m{\lambda }_i+\sum _{j=1}^n{\mu }_j=25k\pi$, then the value of $k$ is

Answer 1

Given equation is $x^2(\sin \lambda x+\cos \mu x)-2x\left[\sin \right(\lambda +\mu )x+\sin (\lambda -\mu \left)x\right]+(\sin \lambda x+\cos \mu x)=0$
$\Rightarrow (x^2+1)(\sin \lambda x+\cos \mu x)-2x\left(2\sin \lambda x\cos \mu x\right)=0\Rightarrow (x^2+1)(\sin \lambda x+\cos \mu x)=4x\sin \lambda x\cos \mu x\Rightarrow \frac{\sin \lambda x+\cos \mu x}{\sin \lambda x\cos \mu x}=\frac{4x}{x^2+1}\Rightarrow \text{cosec}\lambda x+\sec \mu x=\frac{4x}{x^2+1}$
When $x>0,$
$\text{cosec}\lambda x+\sec \mu x=\frac{4}{x+\frac{1}{x}}$
We know that
$x+\frac{1}{x}\ge 2\Rightarrow \frac{4}{x+\frac{1}{x}}\le 2\text{cosec}\lambda x+\sec \mu x\ge 2$
So, the equation holds true only when $x+\frac{1}{x}=2\Rightarrow x=1$
$\text{cosec}\lambda +\sec \mu =2\Rightarrow \text{cosec}\lambda =1\text{and}\sec \mu =1\Rightarrow \lambda =\frac{(4p+1)\pi }{2}\text{and}\mu =2r\pi ,p,r\in I\Rightarrow \lambda =\frac{\pi }{2},\frac{5\pi }{2},\dots ,\frac{197\pi }{2}\Rightarrow \mu =2\pi ,4\pi ,\dots ,100\pi$

Now, $\sum _{i=1}^m{\lambda }_i+\sum _{j=1}^n{\mu }_j=25k\pi$
$\Rightarrow \frac{50}{2}[\frac{\pi }{2}+\frac{197\pi }{2}]+\frac{50}{2}[2\pi +100\pi ]=25k\pi \Rightarrow 99+102=k\therefore k=201$