For $x\in ℝ$, let the function $y\left(x\right)$ be the solution of the differential equation

$\frac{dy}{dx}+12y=\cos \left(\frac{\pi }{12}x\right)$, with $y\left(0\right)=0$.

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

The correct answer is: There exists a real number β such that the line y = β intersects the curve y = y(x) at infinitely many points.

We are given the first-order linear ODE:

$\frac{dy}{dx}+12y=\cos \left(\frac{\pi x}{12}\right)$, with $y\left(0\right)=0$.

Step 1: Find the Integrating Factor.

This is of the standard form $\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$ where $P\left(x\right)=12$.

The integrating factor is $\mu \left(x\right)=e^{\int 12dx}=e^{12x}$.

Step 2: Multiply both sides by the integrating factor and integrate.

Multiplying gives:

$\frac{d}{dx}\left(y·e^{12x}\right)=e^{12x}·\cos \left(\frac{\pi x}{12}\right)$

Integrating the right side using the standard formula $\int e^{ax}\cos \left(bx\right)dx=\frac{e^{ax}\left(a\cos \left(bx\right)+b\sin \left(bx\right)\right)}{a^2+b^2}$

with $a=12$ and $b=\frac{\pi }{12}$, we get $a^2+b^2=144+\frac{{\pi }^2}{144}=\frac{144·144+{\pi }^2}{144}=\frac{20736+{\pi }^2}{144}$.

Therefore the general solution is:

$y\left(x\right)=\frac{1728\cos \left(\frac{\pi x}{12}\right)+12\pi \sin \left(\frac{\pi x}{12}\right)}{20736+{\pi }^2}+C{e}^{-12x}$

Step 3: Apply the initial condition y(0) = 0.

$0=\frac{1728}{20736+{\pi }^2}+C⟹C=\frac{-1728}{20736+{\pi }^2}$

So the particular solution is:

$y\left(x\right)=\frac{1728\cos \left(\frac{\pi x}{12}\right)+12\pi \sin \left(\frac{\pi x}{12}\right)-1728e^{-12x}}{20736+{\pi }^2}$

Step 4: Analyze the long-term behavior (as x → +∞).

As $x\to +\mathrm{\infty }$, the transient term $1728e^{-12x}\to 0$ exponentially. The solution settles into a purely oscillatory (steady-state) behaviour:

$y\left(x\right)\to \frac{1728\cos \left(\frac{\pi x}{12}\right)+12\pi \sin \left(\frac{\pi x}{12}\right)}{20736+{\pi }^2}$ as $x\to +\mathrm{\infty }$

This is a sinusoidal oscillation. The amplitude of the numerator is:

$\sqrt{{1728}^2+{\left(12\pi \right)}^2}=12\sqrt{{144}^2+{\pi }^2}=12\sqrt{20736+{\pi }^2}$

So the amplitude of the oscillation of y(x) for large x is:

$\frac{12\sqrt{20736+{\pi }^2}}{20736+{\pi }^2}=\frac{12}{\sqrt{20736+{\pi }^2}}$

This is a nonzero amplitude, meaning the curve y = y(x) oscillates indefinitely between a positive peak and a negative trough for all large x.

Step 5: Verify each option.

Option A (Increasing function): FALSE. Since y(x) oscillates for large x, it cannot be globally increasing.

Option B (Decreasing function): FALSE. Similarly, y(x) cannot be globally decreasing since it has positive values for large x.

Option C (Line y = β intersects the curve at infinitely many points): TRUE.
Since the solution oscillates about 0 with a nonzero amplitude as x → +∞, the value y = 0 lies strictly inside the oscillation band. Therefore the horizontal line $y=\beta$ with $\beta =0$ (or any β with $\left|\beta \right|<\frac{12}{\sqrt{20736+{\pi }^2}}$) crosses the curve infinitely many times, once in every oscillation cycle.

Option D (Periodic function): FALSE. The full solution has the extra decaying term $\frac{-1728e^{-12x}}{20736+{\pi }^2}$. This transient term breaks periodicity; a periodic function must repeat its values exactly, but e^-12x is strictly decreasing and never repeats. So y(x) is NOT periodic.

Conclusion: Only Option C is TRUE. For example, the line $y=0$ (i.e., β = 0) intersects the curve y = y(x) at infinitely many points, since the solution oscillates through zero infinitely often as x → +∞.