Track / Stage 5 / 37 of 400 #637 of 1964
Problem 637 AIME late Difficulty 5.0 Find the answer
Let the function y = tan ω x ( ω > 0 ) y=\tan \omega x(\omega>0) y = tan ω x ( ω > 0 ) y = a y=a y = a A A A B B B ∣ A B ∣ |A B| ∣ A B ∣ π \pi π f ( x ) = 3 sin ω x − cos ω x
f(x)=\sqrt{3} \sin \omega x-\cos \omega x
f ( x ) = 3 sin ω x − cos ω x [ 2 k π − π 6 , 2 k π + π 6 ] ( k ∈ Z ) \left[2 k \pi-\frac{\pi}{6}, 2 k \pi+\frac{\pi}{6}\right](k \in \mathbf{Z}) [ 2 k π − 6 π , 2 k π + 6 π ] ( k ∈ Z ) [ 2 k π − π 3 , 2 k π + 2 π 3 ] ( k ∈ Z ) \left[2 k \pi-\frac{\pi}{3}, 2 k \pi+\frac{2 \pi}{3}\right](k \in \mathbf{Z}) [ 2 k π − 3 π , 2 k π + 3 2 π ] ( k ∈ Z ) [ 2 k π − 2 π 3 , 2 k π + π 3 ] ( k ∈ Z ) \left[2 k \pi-\frac{2 \pi}{3}, 2 k \pi+\frac{\pi}{3}\right](k \in \mathbf{Z}) [ 2 k π − 3 2 π , 2 k π + 3 π ] ( k ∈ Z ) [ 2 k π − π 6 , 2 k π + 5 π 6 ] ( k ∈ Z ) \left[2 k \pi-\frac{\pi}{6}, 2 k \pi+\frac{5 \pi}{6}\right](k \in \mathbf{Z}) [ 2 k π − 6 π , 2 k π + 6 5 π ] ( k ∈ Z )
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Official solution 7. B.
It is known that, π ω = π ⇒ ω = 1 \frac{\pi}{\omega}=\pi \Rightarrow \omega=1 ω π = π ⇒ ω = 1 f ( x ) = 2 sin ( x − π 6 ) f(x)=2 \sin \left(x-\frac{\pi}{6}\right) f ( x ) = 2 sin ( x − 6 π ) ⇒ − π 2 + 2 k π ⩽ x − π 6 ⩽ π 2 + 2 k π .
\Rightarrow-\frac{\pi}{2}+2 k \pi \leqslant x-\frac{\pi}{6} \leqslant \frac{\pi}{2}+2 k \pi \text {. }
⇒ − 2 π + 2 k π ⩽ x − 6 π ⩽ 2 π + 2 k π .
Therefore, the monotonic increasing interval of f ( x ) f(x) f ( x ) [ 2 k π − π 3 , 2 k π + 2 π 3 ] ( k ∈ Z ) .
\left[2 k \pi-\frac{\pi}{3}, 2 k \pi+\frac{2 \pi}{3}\right](k \in \mathbf{Z}) .
[ 2 k π − 3 π , 2 k π + 3 2 π ] ( k ∈ Z ) .
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Source: NuminaMath-1.5 ,
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