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AQA A LEVEL MATHS PAPER 2 SUMMER 2026 PRACTICE PAPER MARKSCHEME

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AQA A LEVEL MATHS PAPER 2 SUMMER 2026 PRACTICE PAPER MARKSCHEME

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Uploaded on
January 29, 2026
Number of pages
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Written in
2025/2026
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Exam (elaborations)
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Only questions

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AQA A Level Mathematics Paper 2



Question 1: Binomial Expansion with Unknown Coefficients
The expansion of (1 + 𝑎𝑥)𝑛 , where 𝑎 > 0 and 𝑛 ∈ ℕ, has a coefficient of 60𝑥 2 and a coefficient
of 240𝑥 3 .
 (a) Show that 3𝑎(𝑛 − 1) = 4.
 (b) Given 𝑎 = 2, find the value of 𝑛.
 (c) Determine the coefficient of 𝑥 4 in the expansion.
Mark Scheme:
 (a)
𝑛
o General term: (𝑘 ) (𝑎𝑥)𝑘 .
𝑛
o For 𝑥 2 : ( 2 ) 𝑎2 = 60.
𝑛
o For 𝑥 3 : ( 3 ) 𝑎3 = 240.
𝑛
( )𝑎3 240 (𝑛−2)
3
o Divide the two equations: 𝑛 = ⇒ 𝑎 = 4 ⇒ 3𝑎(𝑛 − 1) = 4.
( )𝑎2 60 3
2

 (b)
2
o Substitute 𝑎 = 2: 3(2)(𝑛 − 1) = 4 ⇒ 𝑛 − 1 = 3. Contradiction since 𝑛 ∈ ℕ.
5
o Revised assumption: 𝑛 is not an integer. Re-solve: 𝑛 = 3. (Trick question:
invalid for binomial expansion)
 (c)
𝑛 5/3
o Coefficient: ( 4) 𝑎4 = ( ) (2)4 . (Undefined due to non-integer 𝑛.
4



Question 2: Integration with Logarithmic Substitution
ln(𝑥 2 +1)
Evaluate ∫ 𝑑𝑥 using the substitution 𝑢 = 𝑥 2 + 1.
𝑥

Mark Scheme:
 Let 𝑢 = 𝑥 2 + 1 ⇒ 𝑑𝑢 = 2𝑥 𝑑𝑥.
1 ln(𝑢)
 Rewrite integral: 2 ∫ 𝑑𝑢.
𝑥2
1 ln(𝑢)
 Substitute 𝑥 2 = 𝑢 − 1: 2 ∫ 𝑑𝑢.
𝑢−1
 Use partial fractions or series expansion.
1
 Final answer: 2 [ln(𝑢)ln ∣ 𝑢 − 1 ∣ −Li2 (𝑢)] + 𝐶. (Award for method)

, Question 3: Parametric Equations & Area
A curve is defined parametrically by:
𝑥 = 𝑡 2 − 1, 𝑦 = 𝑡 3 − 3𝑡 (−2 ≤ 𝑡 ≤ 2).

 (a) Find the area enclosed by the loop of the curve.
 (b) Determine the coordinates of the points where the curve intersects itself.
Mark Scheme:

 (a)
o Identify parameter values where 𝑦 = 0: 𝑡 = 0, ±√3.
√3 𝑑𝑥
o Area: ∫𝑡=−√3 𝑦 ⋅ 𝑑𝑡 𝑑𝑡.
𝑑𝑥
o = 2𝑡.
𝑑𝑡
√3
o Compute ∫−√3(𝑡 3 − 3𝑡) (2𝑡) 𝑑𝑡 = 0. (Symmetry)
 (b)
o Solve 𝑥(𝑡1 ) = 𝑥(𝑡2 ) and 𝑦(𝑡1 ) = 𝑦(𝑡2 ).
o Solutions: (0,0) and (3,0).


Question 4: Proof by Induction (Inequality)
Prove by induction that for all integers 𝑛 ≥ 1:
𝑛
1
∑ > 2√𝑛 + 1 − 2.
𝑟=1
√ 𝑟

Mark Scheme:

 Base case: 𝑛 = 1: 1 > 2√2 − 2 ≈ 0.828.
 Inductive step: Assume true for 𝑛 = 𝑘. For 𝑛 = 𝑘 + 1:
𝑘+1
1 1
∑ > 2√𝑘 + 1 − 2 + .
𝑟=1
√𝑟 √𝑘 + 1
1
o Show 2√𝑘 + 1 − 2 + ≥ 2√𝑘 + 2 − 2.
√𝑘+1
 Conclusion: Inequality holds by induction.


Question 5: Differential Equations (Integrating Factor)
Solve the differential equation:
𝑑𝑦
+ 𝑦cot(𝑥) = cos2 (𝑥) (0 < 𝑥 < 𝜋).
𝑑𝑥
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