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Edexcel IAL P2 January 2021 question paper and mark scheme (WMA12/01)

Pure Mathematics 2 (P2), sat in January 2021: 1 h 30 min, 75 marks, calculator allowed. The buttons open Pearson's own files.

The official paper and mark scheme

Sit it under exam conditions: open the paper, press Start timer, press Finish when time is up, then mark it against the mark scheme.

Try 3 P2 questions like these

Written and checked by us, not taken from the Pearson paper. Tap "Worked answer" to see every step.

  1. Circles · 6 marks

    The circle $C$ has centre $(a, b)$ and passes through the points $P(0, 0)$, $Q(6, 0)$, and $R(0, 8)$. Find the values of $a$, $b$, and the radius of $C$.

    Worked answer

    $P$ and $Q$ both on circle ⇒ centre lies on the perpendicular bisector of $PQ$, so $a = 3$   M1 A1. Similarly $P$ and $R$ give $b = 4$   M1 A1. Radius $= |CP| = \sqrt{9 + 16} = 5$   M1 A1.

  2. Proof · 5 marks

    (a) Prove by exhaustion that the square of any integer leaves a remainder of 0, 1 or 4 when it is divided by 5. [4]

    (b) Hence show that the last digit of a square number is never 2, 3, 7 or 8. [1]

    Worked answer

    (a) Considers every case: \(n=5k,\ 5k\pm1,\ 5k\pm2\) (or \(5k,\ 5k+1,\ 5k+2,\ 5k+3,\ 5k+4\)), \(k\) an integer M1

    Squares at least two cases correctly, e.g. \((5k\pm1)^2=25k^2\pm10k+1\) M1

    All cases correct: \((5k)^2=5(5k^2)\); \((5k\pm1)^2=5(5k^2\pm2k)+1\); \((5k\pm2)^2=5(5k^2\pm4k)+4\) A1

    Conclusion: every integer is in one of the cases, so the remainder is always 0, 1 or 4 A1 (cso)

    (b) A whole number ending in 2 or 7 leaves remainder 2, and one ending in 3 or 8 leaves remainder 3, on division by 5; by (a) a square leaves remainder 0, 1 or 4, so it cannot end in 2, 3, 7 or 8 B1

  3. The Binomial Expansion · 4 marks

    The first three terms of the expansion $(a+bx)^6$ are $64 - 192x + cx^2$. Find the values of the constants $a$, $b$, and $c$.

    Worked answer
    1. M1 for equating the first term: $a^6 = 64 \implies a=2$.
    2. M1 for equating the $x$ coefficient to find $b$: $\binom{6}{1}a^5 b = -192 \implies 6(32)b = -192 \implies 192b = -192$.
    3. A1 for obtaining $b = -1$.
    4. A1 for finding the $x^2$ coefficient: $c = \binom{6}{2}a^4 b^2 = 15(16)(-1)^2 = 240$.

More on these topics: Circles · Proof · The Binomial Expansion

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Formula booklet

Mathematical Formulae and Statistical Tables (International AS/A Level Mathematics) · Pearson Edexcel · For the 2018 IAL specification (current)

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Questions students ask

Where is the P2 January 2021 mark scheme?

The mark scheme button above opens Pearson's own copy.

What are the grade boundaries for IAL P2 January 2021?

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Is a calculator allowed in P2?

1h 30min written paper (75 marks), calculator permitted.