How many times does the line $y=\frac{1}{4} x$ cut the curve $y=\sin(x)+\cos(2x)$?

Question

This is a multiple choice question, where using calculators is not allowed. Candidates have, on average, two minutes to solve the problem.

PROBLEM:

How many times does the line $y=\frac{1}{4} x$ cut the curve $y=\sin(x)+\cos(2x)$?

CHOICES:

(A) $4$ (B) $5$ (C) $6$ (D) $7$ (E) $8$

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What I have tried was not feasible:

• Using differentiation, I found that the curve is bounded between $y=\frac{9}{8}$ and $y=-2$.

• When $x<-8$, $\frac{1}{4}x<-2$ , and when $x>\frac{9}{2}$, $\frac{1}{4}x>\frac{9}{8}$. So, any solution to the equation $y=\sin(x)+\cos(2x)=\frac{1}{4}x$ must lie in the interval $(-8,\frac{9}{2})$.

I do not know how to proceed from these two points. Even I know, that takes a long time.

Also, I do not know if using the fact that the curve is periodic will really help.

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By Desmos, I took a look to the graph, the line cuts the curve $7$ times. So the correct choice is $D$.

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Any help on how to solve this would be appreciated. THANKS!

Answer 1

The function is periodic with period $2\pi$ and achieves local extrema

$$y' = \cos x - 2\sin 2x = 0 \implies \cos x (1-4\sin x) = 0$$

at

$$\pm\sqrt{1-\cos^2x}+2\cos^2x-1\Bigr|_{\cos x = 0} = 0,-2$$

and

$$\sin x + 1-2\sin^2x\Bigr|_{\sin x = \frac{1}{4}} = \frac{9}{8}$$

which means the only interval we can expect intersections is on $\left[-8,\frac{9}{2}\right]$. The function has four extrema per period and the interval contains

$$\frac{25}{4\pi} = \frac{8\pi+(25-8\pi)}{4\pi} = 2 + \epsilon$$

two whole periods by estimating since $\epsilon < \frac{1}{12}$. $2\cdot 4 = 8$ turn arounds means $7$ intersections between them.

Answer 2

This answer does not pretend to be rigorous but just suggests an approach to solve the problem in two minutes (you need a peace of paper and a pencil).

You have already almost done the job. The function is periodic with period $2\pi$ and has on this period two minima $(\frac\pi2,0),(\frac{3\pi}2,-2)$ and two maxima: $(\arcsin(\frac14),\frac98),(\pi-\arcsin(\frac14),\frac98)$.

Now just draw a picture, putting on the plot approximate positons of maxima and minima and connecting them with straight lines. We know that $\arcsin(\frac14)\approx\frac\pi{12}$, but in fact the value does not matter as the figure tells you!

The only doubt is about the leftmost intersection and it is resolved by observing $$ -8<-\frac{5\pi}2\iff \pi<\frac{16}5. $$