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

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

The given differential equation is linear of first order:
$$\frac{dy}{dx}+12y=\cos\!\left(\frac{\pi}{12}\,x\right)$$

Step 1 Integrating factor (IF).
For $$\frac{dy}{dx}+P\,y=Q(x)$$ the integrating factor is $$IF=\exp\!\left(\int P\,dx\right)=\exp\!\left(\int 12\,dx\right)=e^{12x}.$$

Step 2 Multiply the equation by the IF and integrate.
$$e^{12x}\frac{dy}{dx}+12e^{12x}y=e^{12x}\cos\!\left(\tfrac{\pi}{12}x\right)$$
Recognising the left side as $$\dfrac{d}{dx}\!\left(y\,e^{12x}\right)$$, integrate:
$$y\,e^{12x}= \int e^{12x}\cos\!\left(\tfrac{\pi}{12}x\right)\,dx + C.$$

Step 3 Evaluate the integral.
Use the standard formula $$\int e^{ax}\cos(bx)\,dx=\frac{e^{ax}\!\left(a\cos bx + b\sin bx\right)}{a^{2}+b^{2}}.$$
Here $$a=12,\; b=\frac{\pi}{12} \;(\text{denote }k=\tfrac{\pi}{12}).$$
Hence $$\int e^{12x}\cos(kx)\,dx=\frac{e^{12x}\!\left(12\cos kx+k\sin kx\right)}{12^{2}+k^{2}}.$$ Put $$D=12^{2}+k^{2}=144+\frac{\pi^{2}}{144}.$$ Therefore $$y\,e^{12x}=e^{12x}\frac{12\cos kx+k\sin kx}{D}+C,$$ and dividing by $$e^{12x}$$ gives the general solution $$y(x)=\frac{12\cos kx+k\sin kx}{D}+C\,e^{-12x}.$$

Step 4 Apply the initial condition $$y(0)=0.$$
At $$x=0:\; \cos0=1,\; \sin0=0.$$
$$0=\frac{12}{D}+C \;\Longrightarrow\; C=-\frac{12}{D}.$$

Step 5 Final explicit solution.
$$y(x)=\frac{12\cos kx+k\sin kx}{D}-\frac{12}{D}\,e^{-12x},\qquad k=\frac{\pi}{12},\; D=144+\frac{\pi^{2}}{144}.$$

Analysis of the four statements.

Statement A “$$y(x)$$ is an increasing function.”
For monotonic increase we would need $$\dfrac{dy}{dx}\gt0$$ for every $$x$$.
Using the original equation $$\dfrac{dy}{dx}=\cos kx-12y$$, the right side changes sign because $$\cos kx$$ oscillates between $$1$$ and $$-1$$ while $$y(x)$$ stays bounded. Consequently $$\dfrac{dy}{dx}$$ is sometimes positive and sometimes negative. Hence $$y(x)$$ is not always increasing. Statement A is false.

Statement B “$$y(x)$$ is a decreasing function.”
The same reasoning shows $$\dfrac{dy}{dx}$$ is not always negative, so $$y(x)$$ is not always decreasing. Statement B is false.

Statement D “$$y(x)$$ is a periodic function.”
The term $$-\dfrac{12}{D}\,e^{-12x}$$ decays exponentially and destroys exact periodicity; no single positive period $$T$$ satisfies $$y(x+T)=y(x)$$ for all $$x$$. Thus $$y(x)$$ is not periodic. Statement D is false.

Statement C “ There exists $$\beta\in\mathbb{R}$$ such that the horizontal line $$y=\beta$$ meets the curve $$y=y(x)$$ at infinitely many points.”
As $$x\to\infty$$, the exponential term vanishes and $$y(x)$$ approaches the purely sinusoidal part $$y_{p}(x)=\frac{12\cos kx+k\sin kx}{D},$$ which has period $$\frac{2\pi}{k}=24$$ and oscillates between fixed maximum and minimum values. Because this limiting waveform is bounded above and below and is attained arbitrarily closely for large $$x$$, choose any $$\beta$$ strictly between its maximum and minimum, for example $$\beta=0$$ (its average value). For sufficiently large $$x$$, $$y(x)$$ will cross this horizontal line once every half-period, giving infinitely many intersection points. Hence such a $$\beta$$ exists. Statement C is true.

Thus, the only correct choice is:
Option C which is: There exists a real number $$\beta$$ such that the line $$y=\beta$$ intersects the curve $$y=y(x)$$ at infinitely many points.