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 1

$f(x)=7\cos x+5\sin x$ is a continuous function, so every value between its minimum and maximum is attainable by the Intermediate Value Theorem. 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 $\boxed{008}$ .

Solution 2 (calculus)

As in the first solution, let $f(x)=7\cos x+5\sin x$ . Then, $f'(x)=-7\sin x+5\cos x$ . Thus, $f(x)$ has maxima and minima when $7\sin x = 5\cos x$ . After squaring both sides and applying the Pythagorean Identity, we solve for $\sin x$ : \begin{align*} 49\sin^2x &= 25\cos^2x \\ 49\sin^2x &= 25(1-\sin^2x) \\ 74\sin^2x &= 25 \\ \sin^2x &= \frac{25}{74} \\ \sin x &= \pm \frac5{\sqrt{74}} \end{align*} Thus, $\cos^2x=1-\tfrac{25}{74}=\tfrac{49}{74}$ , so $\cos x = \pm \tfrac7{\sqrt{74}}$ . The maximum of the function occurs when $\sin x,\cos x > 0$ , so the maximum is $\tfrac{25}{\sqrt{74}}+\tfrac{49}{\sqrt{74}}=\tfrac{74}{\sqrt{74}}=\sqrt{74}$ . Likewise, when both $\sin x$ and $\cos x$ are negative, the minimum $-\sqrt{74}$ occurs. We can now proceed as in the first solution to solve the problem to get our answer $\boxed{008}$ .

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2013 Mock AIME I Problems/Problem 6

Contents

Problem 6

Solution 1

Solution 2 (calculus)

See also