2013 Mock AIME I Problems/Problem 6

Problem 6

Find the number of integer values $k$ can have such that the equation $7\cos x+5\sin x=2k+1$ has a solution.

Solution

$f(x)=7\cos x+5\sin x$ is a continuous function, so every value between its minimum and maximum is attainable. By Cauchy-Schwarz, $(7\cos x+5\sin x)^2 \le (7^2+5^2)(\cos^2 x+\sin^2 x)=74$ Giving a maximum of $\sqrt{74}$ , which is achievable when $\frac{\cos x}{7}=\frac{\sin x}{5}$ . Note that a minimum of $-\sqrt{74}$ can be attained at $f(x+\pi)$ . Thus the values of $k$ that work are the integers from $-4$ to $3$ , inclusive, giving a total of $\boxed{8}$ .

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2013 Mock AIME I Problems/Problem 6

Problem 6

Solution