Difference between revisions of "2013 Mock AIME I Problems/Problem 6"

(Created page with "==Problem 6== Find the number of integer values <math>k</math> can have such that the equation <cmath>7\cos x+5\sin x=2k+1</cmath> has a solution. ==Solution== <math>f(x)=7\c...")
 
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==Solution==
 
==Solution==
<math>f(x)=7\cos x+5\sin x</math> is a continuous function, so every value between its minimum and maximum is attainable. By Cauchy-Schwarz, <cmath>(7\cos x+5\sin x)^2 \ge (7^2+5^2)(\cos^2 x+\sin^2 x)=74</cmath> Giving a maximum of <math>\sqrt{74}</math>, which is achievable when <math>\frac{\cos x}{7}=\frac{\sin x}{5}</math>. Note that a minimum of <math>-\sqrt{74}</math> can be attained at <math>f(x+\pi)</math>. Thus the values of <math>k</math> that work are the integers from <math>-4</math> to <math>3</math>, inclusive, giving a total of <math>\boxed{8}</math>.
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<math>f(x)=7\cos x+5\sin x</math> is a continuous function, so every value between its minimum and maximum is attainable. By Cauchy-Schwarz, <cmath>(7\cos x+5\sin x)^2 \le (7^2+5^2)(\cos^2 x+\sin^2 x)=74</cmath> Giving a maximum of <math>\sqrt{74}</math>, which is achievable when <math>\frac{\cos x}{7}=\frac{\sin x}{5}</math>. Note that a minimum of <math>-\sqrt{74}</math> can be attained at <math>f(x+\pi)</math>. Thus the values of <math>k</math> that work are the integers from <math>-4</math> to <math>3</math>, inclusive, giving a total of <math>\boxed{8}</math>.

Revision as of 08:22, 23 October 2023

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 $\boxed{8}$.