For $x \in R,$ let the function $y (x)$ be the solution of the differential equation $\frac{d y}{d x} + 12 y = \cos (\frac{π}{12} x); y (0) = 0$
Then, which of the following statements is/are TRUE?

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$y (x)$ is an increasing function

$y (x)$ is a decreasing function

There exists a real number $β$ such that the line $y = β$ intersects the curve $y = y (x)$ at infinitely many points

$y (x)$ is a periodic function

answer is C.

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Detailed Solution

$\frac{d y}{d x} + 12 y = \cos (\frac{π}{12} x)$

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

Solution of DE $y . e^{12 x} = \int e^{12 x} . \cos (\frac{π}{12} x) d x$

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

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

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

$∴ y = \frac{12}{12^{4} + π^{2}} [{(12)}^{2} \cos (\frac{π x}{12}) + π \sin (\frac{π x}{12}) - 12^{2} . e^{- 12 x}]$

Now $\frac{d y}{d x} = \frac{12}{12^{4} + π^{2}} [\underset{\min . v a l u e}{\underset{⏟}{- 12 π s i n (\frac{π x}{12}) + \frac{π^{2}}{12} \cos (\frac{π x}{12})}} + 12^{3} e^{- 12 x}]$

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