Given, $$ \frac{d y}{d x}+12 y=\cos \left(\frac{\pi}{12} x\right) $$

This is a linear differential equation.

Where P = 12 and Q = $\cos \left(\frac{\pi}{12} x\right)$

$$ \text { I.F. }=e^{\int P d x}=e^{\int 12 d x}=e^{12 x} $$

Solution of differencial equation is

$$y \times I.F = \int {Q \times \left( {I.F} \right)} dx$$

$$y \times \text { I.F. }=\int e^{12 x} \cdot \cos \left(\frac{\pi}{12} x\right) d x $$

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

[As $$\int {{e^{ax}}.\cos \left( {bx} \right)dx} = {{{e^{ax}}} \over {{a^2} + {b^2}}}\left\{ {a\cos \left( {bx} \right) + b\sin \left( {bx} \right)} \right\} + C$$]

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

$$ \begin{aligned} & \text { Given }, y(0)=0 \\\\ & \Rightarrow 0=\frac{12 \cdot\left(12^2\right)}{12^4+\pi^2}+C \\\\ & \Rightarrow C=\frac{-12^3}{12^4+\pi^2} \\\\ & \therefore y=\frac{12}{12^4+\pi^2}\left[12^2 \cos \left(\frac{\pi x}{12}\right)+\pi \sin \left(\frac{\pi x}{12}\right)-12^2 \cdot e^{-12 x}\right] \end{aligned} $$

$y(x)$ is not a periodic function as $e^{-12 x}$ is a non-periodic function.

$$ \therefore $$ Option (D) is a wrong statement.

$$ \begin{aligned} & \text { Now } \frac{d y}{d x}=\frac{12}{12^4+\pi^2}[{-12 \pi \sin \left(\frac{\pi x}{12}\right)+\frac{\pi^2}{12} \cos \left(\frac{\pi x}{12}\right)}+12^3 \mathrm{e}^{-12 x}] \end{aligned} $$

Values of ${-12 \pi \sin \left(\frac{\pi x}{12}\right)+\frac{\pi^2}{12} \cos \left(\frac{\pi x}{12}\right)}$ varies between $\left(-\sqrt{144 \pi^2+\frac{\pi^4}{144}},-12 \pi \sqrt{1+\frac{\pi^2}{12^4}}\right)$

So, $\mathrm{f}(\mathrm{x})$ is neither increasing nor decreasing.

$$ \therefore $$ Option (A) and (B) are wrong statements.

For some $\beta \in R, y=\beta$ intersects $y=f(x)$ at infinitely many points.

So option $\mathrm{C}$ is correct.