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An even function decreases strictly on (−∞,0). Choose the correct ordering; write a=2^(−7/3), b=3^(−2/7), c=log₃(2/7).

Read the idea, work independently, then explain what changed.

TOPIC 01

2024 JM01

A negative argument can have the largest value because f is even.

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

01 / Standard#Your turn

An even function decreases strictly on (−∞,0). Choose the correct ordering; write a=2^(−7/3), b=3^(−2/7), c=log₃(2/7).

Official paper · jm01-2024 · I.15 · PDF 3

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

Skills and prerequisite lessons
  1. Option Af(a)>f(b)>f(c)f(a)>f(b)>f(c)
  2. Option Bf(b)>f(c)>f(a)f(b)>f(c)>f(a)
  3. Option Cf(c)>f(a)>f(b)f(c)>f(a)>f(b)
  4. Option Df(b)>f(a)>f(c)f(b)>f(a)>f(c)
  5. Option Ef(c)>f(b)>f(a)f(c)>f(b)>f(a)

Working and explanation

BUILD THE REASONING

Hint 1
Evenness makes f increase with positive |x|.
Hint 2
Compare a,b with 1/2 and |c| with 1.
Worked solution
  1. Obtain strict size bounds.

    0<a<1/4<1/2<b<1,2/7<1/3  ⟹  c<−10<a<1/4<1/2<b<1,\quad2/7<1/3\implies c<-1
  2. Order by absolute arguments.

    ∣c∣>b>a>0  ⟹  f(c)=f(∣c∣)>f(b)>f(a)|c|>b>a>0\implies f(c)=f(|c|)>f(b)>f(a)

E.

Checks and common pitfalls: A negative argument can have the largest value because f is even.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Obtain strict size bounds.
    0<a<1/4<1/2<b<1,2/7<1/3  ⟹  c<−10<a<1/4<1/2<b<1,\quad2/7<1/3\implies c<-1
  • Order by absolute arguments.
    ∣c∣>b>a>0  ⟹  f(c)=f(∣c∣)>f(b)>f(a)|c|>b>a>0\implies f(c)=f(|c|)>f(b)>f(a)

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Curriculum and source notes ↗