M1 awarded
missed A1
A*
18/20
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MATHEMATICS FURTHER MATHS PHYSICS CHEMISTRY BIOLOGY COMPUTER SCIENCE ENGLISH LIT ENGLISH LANG HISTORY MATHEMATICS FURTHER MATHS PHYSICS CHEMISTRY BIOLOGY COMPUTER SCIENCE ENGLISH LIT ENGLISH LANG HISTORY
QUESTION 7 — A-LEVEL MATHEMATICS (9709) 4 MARKS
Solve: 2|2x − a| < |x + 3a|
Squared both sides: 4(2x−a)² < (x+3a)²M1 — valid method to remove the modulus
Simplified to 15x² − 22ax − 5a² < 0A1 — correct expansion
Final answer: x < 5/3 aMissed −1/5a < x and used ≤ instead of <
Mark Analysis
2 / 4 MARKS
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EXPLAIN ANSWER CHECK MY ANSWER ASK ANYTHING
TUTOR CHATQ7 · MATHS
"Missed −1/5a < x and used ≤ instead of <"
A
why do I need the other inequality as well?
M
Because |x + 3a| splits into two cases once you unsquare — you solved the positive case but the negative case gives the second bound. Modulus inequalities almost always need both.
A
can you show me in code how to solve this?
M
Python
def solve_modulus(a):
    # Solve 2|2x - a| < |x + 3a|
    # Case 1: positive values
    return (-a/5, 5*a/3)
✦ Explain Answer ✓ Check My Answer
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AUTO-GENERATED WEAK-AREA TARGETED MCQ BOSS FIGHTS
FLASHCARDCARD 3 OF 5
EASY
What condition makes |2x−a| < |x+3a| flip to ≥ instead?
Squaring both sides is only valid because both sides are non-negative by definition — no sign flip needed here.
tap the card to flip
EASY +8 XP
MEDIUM +5 XP
HARD +2 XP
⚠️ BOSS FIGHT: PROVE YOU UNDERSTAND
Which step is the M1 mark actually awarded for?
A. Writing the final inequality
B. Squaring both sides validly ✓
C. Stating x is real
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STREAK MULTIPLIERS TOPIC PROFICIENCY MEDAL COLLECTION
PROGRESSMATHEMATICS 9709
LV 6
440 XP700 XP
🔥 7-day streak · 1.5× XP
Modulus inequalities82%
Integration by parts46%
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