Questions: Graf ktorej z nasledujúcich funkcií má na intervale ⟨0,2 π⟩ najviac priesečníkov s osou x ?
(A) f1: y=2+sin x
(B) f2: y=2 sin x
(C) f3: y=sin (x/2)
(D) f4: y=sin 2x
(E) f5: y=sin x
Transcript text: Graf ktorej $z$ nasledujúcich funkcií má na intervale $\langle 0,2 \pi\rangle$ najviac priesečníkov s osou $x$ ?
(A) $f_{1}: y=2+\sin x$
(B) $f_{2}: y=2 \sin x$
(C) $f_{3}: y=\sin \frac{x}{2}$
(D) $f_{4}: y=\sin 2 x$
(E) $f_{5}: y=\sin x$
Solution
Solution Steps
To determine which function has the fewest intersections with the x-axis on the interval \(\langle 0, 2\pi \rangle\), we need to find the number of roots (or zeros) of each function within this interval. A function intersects the x-axis where its value is zero. We will evaluate each function to find the number of such points.
Step 1: Identify the Functions
We are given five functions to analyze for their intersections with the x-axis on the interval \( \langle 0, 2\pi \rangle \):
\( f_1: y = 2 + \sin x \)
\( f_2: y = 2 \sin x \)
\( f_3: y = \sin \frac{x}{2} \)
\( f_4: y = \sin 2x \)
\( f_5: y = \sin x \)
Step 2: Count the Intersections
We evaluate each function to determine the number of roots (intersections with the x-axis) within the specified interval:
For \( f_1 \), \( 2 + \sin x = 0 \) has 0 intersections.
For \( f_2 \), \( 2 \sin x = 0 \) has 2 intersections.
For \( f_3 \), \( \sin \frac{x}{2} = 0 \) has 2 intersections.
For \( f_4 \), \( \sin 2x = 0 \) has 2 intersections.
For \( f_5 \), \( \sin x = 0 \) has 2 intersections.
Step 3: Determine the Function with the Fewest Intersections
From the counts:
\( f_1 \): 0 intersections
\( f_2 \): 2 intersections
\( f_3 \): 2 intersections
\( f_4 \): 2 intersections
\( f_5 \): 2 intersections
The function \( f_1 \) has the fewest intersections with the x-axis.
Final Answer
The function with the fewest intersections with the x-axis on the interval \( \langle 0, 2\pi \rangle \) is \( f_1 \).