How do you make a Jello

Let the function of the curve be $f(x)$ . We note that $f(x)$ has a period of $\pi$ and that $f(n\pi) = 0$ , where $n$ is an integer. All the answer options pass the test of $f(n\pi)=0$ . Now let us test $f\left(n\pi + \frac \pi 2\right) = 2$ .

$\begin{cases} f_1 (x) = |\sin x| + |\sin 2x| +|\sin 3x| & \implies f_1\left(\frac \pi 2\right) = 1+0+1 = \color{#3D99F6} 2 & \color{#3D99F6} \text{Pass} \\ f_2 (x) = |\sin x| + |\sin 2x| +|\sin 4x| & \implies f_2 \left(\frac \pi 2\right) = 1+0+0 = \color{#D61F06} 1 & \color{#D61F06} \text{Fail} \\ f_3 (x) = |\sin 3x| - |\sin 6x| + \sin 9x & \implies f_3 \left(\frac \pi 2\right) = 1-0+1 = \color{#3D99F6} 2 & \color{#3D99F6} \text{Pass} \\ f_4 (x) = 2.5|\sin x| + \sin 2x -|\sin 3x| & \implies f_4 \left(\frac \pi 2\right) = 2.5+0-1 = \color{#D61F06} 1.5 & \color{#D61F06} \text{Fail} \\ f_5 (x) = 2.5|\sin 2x| -|\sin 4x|+|\sin 6x| & \implies f_5 \left(\frac \pi 2\right) = 0-0+0 = \color{#D61F06} 0 & \color{#D61F06} \text{Fail} \end{cases}$

We note that the maximum of $f(x)$ is about 2.5 occurring at about $x = n\pi \pm \frac \pi 6$ . Let us test that on $f_1(x)$ and $f_3(x)$ .

$\begin{cases} f_1 (x) = |\sin x| + |\sin 2x| +|\sin 3x| & \implies f_1\left(\frac \pi 6\right) = \frac 12 + \frac {\sqrt 3}2 +1 = \color{#3D99F6} 2.366 & \color{#3D99F6} \text{Pass} \\ f_3 (x) = |\sin 3x| - |\sin 6x| + \sin 9x & \implies f_3 \left(\frac \pi 6\right) = \frac 12 -0-\frac 12 = \color{#D61F06} 0 & \color{#D61F06} \text{Fail} \end{cases}$

Therefore, $\boxed{y=|\sin x| + |\sin 2x| +|\sin 3x|}$ . A plot of the curve confirms it. I used an Excel spreadsheet.