Korean CSAT Mock Exam September #21

Calculus Level 4

Two functions f ( x ) = sin ( k x ) + 2 f(x) = \sin(kx)+2 and g ( x ) = 3 cos 12 x g(x) = 3\cos{12x} are defined at a closed interval [ 2 π , 2 π ] [-2\pi,2\pi] . Let a a be the y y value(s) of the point(s) where f ( x ) f(x) and g ( x ) g(x) intersect.

How many natural numbers k k satisfies { x f ( x ) = a } { x g ( x ) = a } \{ x|f(x)=a \} \subset \{ x|g(x)=a \} ?

BONUS: How many natural numbers k k that satisfies { x g ( x ) = a } { x f ( x ) = a } \{ x|g(x)=a \} \subset \{ x|f(x)=a \} instead?

This is a problem from 2020.
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1 solution

Inquisitor Math
Nov 10, 2020

The following is my approach for this problem: Let’s focus on one period of each function. (Denote those as f 1 , g 1 f_1, g_1 .) It is easy to observe that an even number of g 1 g_1 should ‘fit into’ f 1 f_1 . (This is because both functions f,g are periodic and symmetric.) Thus T g T f = 12 k \dfrac{T_g}{T_f} = \dfrac{12}k is an even number. This is equivalent to saying that k k divides 6. Therefore, k = 1 , 2 , 3 , 6 k=1,2,3,6 .

(If you solved this differently, plz share!)

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