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) |
||
(2 intermediate revisions by 2 users not shown) | |||
Line 2: | Line 2: | ||
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. | ||
− | ==Solution== | + | == Solution 1 == |
− | <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 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>. |
+ | |||
+ | == Solution 2 (calculus) == | ||
+ | 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>: | ||
+ | \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, <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>. | ||
+ | |||
+ | == See also == | ||
+ | * [[2013 Mock AIME I Problems]] | ||
+ | * [[2013 Mock AIME I Problems/Problem 5|Preceded by Problem 5]] | ||
+ | * [[2013 Mock AIME I Problems/Problem 7|Followed by Problem 7]] | ||
+ | |||
+ | [[Category:Intermediate Algebra Problems]] |
Latest revision as of 13:31, 30 July 2024
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
.