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Practise

Question bank

Original exam-style questions, mapped to your syllabus. Answers are checked for mathematical equivalence and every attempt updates your mastery.

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14 questions

SA-9709-P2-AL-0022 marksL1 FoundationCalculatorSimple
The solution of ∣2x−1∣<5|2x-1|<5 is a<x<ba<x<b. Find aa and bb.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P2-AL-Q021 markL1 FoundationCalculatorSimple
The inequality ∣x−1∣<4|x-1|<4 is equivalent to
Choose one option

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-AL-0022 marksL1 FoundationCalculatorSimple
The solution of ∣x−3∣<2|x-3|<2 is a<x<ba<x<b. Find aa and bb.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-AL2-G012 marksL1 FoundationCalculatorSimple
The solution of ∣x−1∣<3|x-1|<3 is a<x<ba<x<b. Find aa and bb.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-AL2-Q011 markL1 FoundationCalculatorSimple
The inequality ∣x∣>5|x|>5 is equivalent to
Choose one option

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-AL-Q011 markL1 FoundationCalculatorSimple
For a>0a>0, the inequality ∣x∣<a|x|<a means
Choose one option

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P2-AL-0073 marksL2 StandardCalculatorSimple
The solution of ∣x−2∣≥3|x-2|\ge3 is x≤ax\le a or x≥bx\ge b. Find aa and bb.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-AL2-0013 marksL2 StandardCalculatorSimple
The solution of ∣3x+2∣≥4|3x+2|\ge4 is x≤ax\le a or x≥bx\ge b. Find aa and bb.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P2-AL-0034 marksL3 ExamCalculatorTwo moduli
The solution of the inequality ∣x−3∣>∣2x∣|x-3|>|2x| is a<x<ba<x<b. Find aa and bb.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-AL-0034 marksL3 ExamCalculatorTwo moduli
The solution of the inequality ∣2x+1∣>∣x−4∣|2x+1|>|x-4| is x<ax<a or x>bx>b. Find aa and bb.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-AL2-0023 marksL3 ExamCalculatorTwo moduli
The solution of ∣x+2∣<∣x−4∣|x+2|<|x-4| is x<ax<a. Find aa.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-AL2-0034 marksL3 ExamCalculatorModulus and a line
The solution of ∣2x−5∣<x+1|2x-5|<x+1 is a<x<ba<x<b. Find aa and bb.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-AL2-0044 marksL4 ChallengingCalculatorTwo moduli
The solution of 2∣x−1∣>∣x+2∣2|x-1|>|x+2| is x<ax<a or x>bx>b. Find aa and bb.

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Salam Academy Original Practice · Proposed mark scheme
SA-9709-P3-AL2-0054 marksL5 ExtensionCalculatorModulus and a line
The solution of ∣x−3∣>2x|x-3|>2x is x<ax<a. Find aa.

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Salam Academy Original Practice · Proposed mark scheme