For $x\in R,$ let the function $y\left(x\right)$ be the solution of the differential equation $\frac{dy}{dx}+12y=\cos \left(\frac{\pi }{12}x\right);y\left(0\right)=0$

Then, which of the following statements is/are TRUE?

$\frac{dy}{dx}+12y=\cos \left(\frac{\pi }{12}x\right)$

Linear D.E. I.F $=e^{\int 12.dx}=e^{12x}.$

Solution of DE $y.e^{12x}=\int e^{12x}.\cos \left(\frac{\pi }{12}x\right)dx$

$y.e^{12x}=\frac{e^{12x}}{{\left(12\right)}^2+{\left(\frac{\pi }{12}\right)}^2}(12\cos \frac{\pi }{12}x+\frac{\pi }{12}\sin \frac{\pi }{12}x)+C$

$\Rightarrow y=\frac{\left(12\right)}{{\left(12\right)}^4+{\pi }^2}({\left(12\right)}^2\cos \left(\frac{\pi x}{12}\right)+\pi \sin \left(\frac{\pi x}{12}\right))+\frac{C}{e^{12x}}$

Given y(0) = 0 $\Rightarrow 0=\frac{12}{{12}^4+{\pi }^2}({12}^2+0)+C\Rightarrow C=\frac{-{12}^3}{{12}^4+{\pi }^2}$

$\therefore y=\frac{12}{{12}^4+{\pi }^2}[{\left(12\right)}^2\cos \left(\frac{\pi x}{12}\right)+\pi \sin \left(\frac{\pi x}{12}\right)-{12}^2.e^{-12x}]$

Now $\frac{dy}{dx}=\frac{12}{{12}^4+{\pi }^2}[\underset{\min .value}{\underbrace{-12\pi sin\left(\frac{\pi x}{12}\right)+\frac{{\pi }^2}{12}\cos \left(\frac{\pi x}{12}\right)}}+{12}^3{e}^{-12x}]$

$(-\sqrt{144{\pi }^2+\frac{{\pi }^4}{144}}=-12\pi \sqrt{1+\frac{{\pi }^2}{{12}^4}})\Rightarrow \frac{dy}{dx}>0\forall x\le 0$ & may be negative/positive for x > 0
So, f (x) is neither increasing nor decreasing For some $\beta \in R,y=\beta$ intersects $y=f\left(x\right)$ at infinitely many points So option C is correct