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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− | <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 \ | + | <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 09:22, 23 October 2023
Problem 6
Find the number of integer values can have such that the equation
has a solution.
Solution
is a continuous function, so every value between its minimum and maximum is attainable. 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
.