2013 Mock AIME I Problems/Problem 6
Problem 6
Find the number of integer values can have such that the equation has a solution.
Solution 1
is a continuous function, so every value between its minimum and maximum is attainable by the Intermediate Value Theorem. By Cauchy-Schwarz, Giving a maximum of , which is achievable when . Note that a minimum of can be attained at . Thus the values of that work are the integers from to , inclusive, giving a total of .
Solution 2 (calculus)
As in the first solution, let . Then, . Thus, has maxima and minima when . After squaring both sides and applying the Pythagorean Identity, we solve for : \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, , so . The maximum of the function occurs when , so the maximum is . Likewise, when both and are negative, the minimum occurs. We can now proceed as in the first solution to solve the problem to get our answer .