SAT1000 - P333

Since $f(x) = \cos 2x$ , it has a period of $\pi$ . The $a$ in $ag(x) = a\sin x$ changes the amplitude of the $\sin x$ curve. When $a=1$ , $f(x) = g(x) = -1$ when $x = 2k\pi + \frac 32\pi$ , where $k$ is a positive integer. Therefore, when $a=1$ , $f(x) + ag(x) = 0$ has $3$ roots every $2\pi$ cycle. Similarly, when $a=-1$ , $f(x) + ag(x) = 0$ also has $3$ roots every $2\pi$ cycle. Therefore the $n \pi$ cycles that give $2013$ roots is $n \pi = \frac {2013}3 \times 2 \pi \implies n=1342$ .

When $|a| < 1$ , we note that there are $2$ roots for every $\pi$ cycle. Since $2013$ is not divisible by $2$ , there is no solution for $n$ for $|a| < 1$ . When $|a| > 1$ , there are $2$ roots in the $n$ th $\pi$ cycle when $n$ is odd and $0$ root when $n$ is even. Therefore the number of roots for $n\pi$ cycles is always even so there is no solution.

This means there are solutions when $a = \pm 1$ and $n=1342$ . Therefore $N=2$ , $M = 1(-1+1342) + 2(1+1342) = 4027$ , and $\lfloor M+N \rfloor = \boxed{4029}$ .