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Choose the function matching the displayed graph.

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TOPIC 01

2021 JM01

Frequency is obtained from a full cycle, not the peak-to-trough half-cycle.

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Use one hint at a time. A correction explains what changed, not just the final answer.

01 / Standard#Your turn

Choose the function matching the displayed graph.

2021 JM01 I.11 graph / 第 11 題圖像0135π/2πxy

Official paper · jm01-2021 · I.11 · PDF 3

Official original and suggested answers ↗ · Suggested answer PDF page 5

Skills and prerequisite lessons
  1. Option Ay=3+2cos⁡(x/2)y=3+2\cos(x/2)
  2. Option By=3+2cos⁡2xy=3+2\cos2x
  3. Option Cy=3+2cos⁡xy=3+2\cos x
  4. Option Dy=1+2cos⁡(x/2)y=1+2\cos(x/2)
  5. Option Ey=1+2cos⁡2xy=1+2\cos2x

Working and explanation

BUILD THE REASONING

Hint 1
Read the maximum and minimum to find the midline and amplitude.
Hint 2
Measure the distance between consecutive peaks.
Worked solution
  1. The graph ranges from 1 to 5, so its midline is 3 and amplitude is 2.

    c=(5+1)/2=3,A=(5−1)/2=2c=(5+1)/2=3,\quad A=(5-1)/2=2
  2. Peaks at zero and π give period π and frequency 2.

    T=π⇒ω=2π/T=2T=\pi\Rightarrow\omega=2\pi/T=2
  3. The graph begins at a maximum, agreeing with positive cosine.

    y=3+2cos⁡2xy=3+2\cos2x

B: y=3+2cos 2x.

Checks and common pitfalls: Frequency is obtained from a full cycle, not the peak-to-trough half-cycle.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • The graph ranges from 1 to 5, so its midline is 3 and amplitude is 2.
    c=(5+1)/2=3,A=(5−1)/2=2c=(5+1)/2=3,\quad A=(5-1)/2=2
  • Peaks at zero and π give period π and frequency 2.
    T=π⇒ω=2π/T=2T=\pi\Rightarrow\omega=2\pi/T=2
  • The graph begins at a maximum, agreeing with positive cosine.
    y=3+2cos⁡2xy=3+2\cos2x

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Curriculum and source notes ↗