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Find all a for which f is increasing throughout (2,4).

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

TOPIC 01

2023 JM01

The endpoint case a=−1/2 is still strictly increasing on the open interval.

Try it. Leave your reasoning visible.

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

01 / Standard#Your turn

Find all a for which f is increasing throughout (2,4).

f(x)=ax2+4x+1f(x)=ax^2+4x+1

Official paper · jm01-2023 · I.7 · PDF 2

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

Skills and prerequisite lessons
  1. Option A[−1/2,0)[-1/2,0)
  2. Option B(0,1/2](0,1/2]
  3. Option C[−1/2,1/2][-1/2,1/2]
  4. Option D[−1/2,∞)[-1/2,\infty)
  5. Option E[1/2,∞)[1/2,\infty)

Working and explanation

BUILD THE REASONING

Hint 1
For a≥0 the derivative is positive on the interval.
Hint 2
For a<0 its smallest limiting derivative occurs at x=4.
Worked solution
  1. Differentiate.

    f′(x)=2ax+4f'(x)=2ax+4
  2. In the negative-a case require 8a+4≥0; equality is allowed because 4 is excluded.

    a≥−1/2a\ge-1/2

D: a≥−1/2.

Checks and common pitfalls: The endpoint case a=−1/2 is still strictly increasing on the open interval.

Reasoning checklist · self / teacher assessment
  • Teaching assessment checklist, independently authored. Use the original paper for official marks.
  • Differentiate.
    f′(x)=2ax+4f'(x)=2ax+4
  • In the negative-a case require 8a+4≥0; equality is allowed because 4 is excluded.
    a≥−1/2a\ge-1/2

Think first. Reveal a hint when the class is ready.

Focus on one question

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Curriculum and source notes ↗