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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...")
 
(added links to first solution, added second solution)
 
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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.
 
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.
  
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==Solution==
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== Solution 1 ==
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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 \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 the [[Intermediate Value Theorem]]. 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{008}</math>.
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== Solution 2 (calculus) ==
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As in the first solution, let <math>f(x)=7\cos x+5\sin x</math>. Then, <math>f'(x)=-7\sin x+5\cos x</math>. Thus, <math>f(x)</math> has maxima and minima when <math>7\sin x = 5\cos x</math>. After squaring both sides and applying the [[Trigonometric identities#Pythagorean identities|Pythagorean Identity]], we solve for <math>\sin x</math>:
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\begin{align*}
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49\sin^2x &= 25\cos^2x \\
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49\sin^2x &= 25(1-\sin^2x) \\
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74\sin^2x &= 25 \\
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\sin^2x &= \frac{25}{74} \\
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\sin x &= \pm \frac5{\sqrt{74}}
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\end{align*}
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Thus, <math>\cos^2x=1-\tfrac{25}{74}=\tfrac{49}{74}</math>, so <math>\cos x = \pm \tfrac7{\sqrt{74}}</math>. The maximum of the function occurs when <math>\sin x,\cos x > 0</math>, so the maximum is <math>\tfrac{25}{\sqrt{74}}+\tfrac{49}{\sqrt{74}}=\tfrac{74}{\sqrt{74}}=\sqrt{74}</math>. Likewise, when both <math>\sin x</math> and <math>\cos x</math> are negative, the minimum <math>-\sqrt{74}</math> occurs. We can now proceed as in the first solution to solve the problem to get our answer <math>\boxed{008}</math>.
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== See also ==
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* [[2013 Mock AIME I Problems]]
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* [[2013 Mock AIME I Problems/Problem 5|Preceded by Problem 5]]
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* [[2013 Mock AIME I Problems/Problem 7|Followed by Problem 7]]
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[[Category:Intermediate Algebra Problems]]

Latest revision as of 12:31, 30 July 2024

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 $\boxed{008}$.

See also

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