AQA A-Level Mathematics Complete Practice Papers | QUESTIONs, Fully Worked Solutions & Mark Scheme (2026/2027)

AQA A-Level Mathematics Complete Practice
Papers | QUESTIONs, Fully Worked Solutions &
Mark Scheme (2026/2027)

QUESTION 1
2
Find the derivative of 𝑓(𝑥 ) = 4𝑥 3 − + 5 with respect to 𝑥.
𝑥2
4
• A. 12𝑥 2 +
𝑥3
4
• B. 12𝑥 2 −
𝑥3

• C. 12𝑥 2 + 4𝑥 −3
• D. 12𝑥 2 − 4𝑥 −1
4
Correct Answer: A. 12𝑥 2 +
𝑥3

Detailed Rationale: Differentiating term by term using the power rule,
2
the derivative of 4𝑥 3 is 12𝑥 2 , and rewriting − as −2𝑥 −2 gives a
𝑥2
4
derivative of −2(−2𝑥 −3 ) = 4𝑥 −3 = . The constant 5 differentiates to
𝑥3
0.
QUESTION 2
√50+√18
Simplify the expression completely.
√2

• A. 4
• B. 8
• C. 16

, • D. 34
Correct Answer: B. 8

Detailed Rationale: Simplifying the surds in the numerator gives √50 =
√25 × 2 = 5√2 and √18 = √9 × 2 = 3√2. Their sum is 8√2. Dividing
by √2 yields 8.
QUESTION 3
Determine the range of values of 𝑘 for which the quadratic equation
𝑘𝑥 2 − 4𝑥 + 𝑘 = 0 has two distinct real roots.
• A. −2 < 𝑘 < 2
• B. 𝑘 < −2 or 𝑘 > 2
• C. −2 < 𝑘 < 2, 𝑘 ≠ 0
• D. 𝑘 ≤ −2 or 𝑘 ≥ 2
Correct Answer: C. −2 < 𝑘 < 2, 𝑘 ≠ 0
Detailed Rationale: For two distinct real roots, the discriminant must be
strictly greater than zero (𝑏 2 − 4𝑎𝑐 > 0). Here, (−4)2 − 4(𝑘)(𝑘) >
0 ⟹ 16 − 4𝑘 2 > 0 ⟹ 𝑘 2 < 4, which means −2 < 𝑘 < 2. Since 𝑘 is
the coefficient of 𝑥 2 , 𝑘 cannot equal 0 for it to remain a quadratic
equation.
QUESTION 4
Find the center and radius of the circle given by the equation 𝑥 2 + 𝑦 2 −
6𝑥 + 8𝑦 − 11 = 0.
• A. Center (3, −4), Radius 6
• B. Center (−3,4), Radius 6

, • C. Center (3, −4), Radius 36
• D. Center (−3,4), Radius 36
Correct Answer: A. Center (3, −4), Radius 6
Detailed Rationale: Completing the square for both 𝑥 and 𝑦 gives
(𝑥 − 3)2 − 9 + (𝑦 + 4)2 − 16 − 11 = 0, which simplifies to
(𝑥 − 3)2 + (𝑦 + 4)2 = 36. Comparing this to the standard circle
equation (𝑥 − 𝑎)2 + (𝑦 − 𝑏)2 = 𝑟 2 gives a center of (3, −4) and a
radius of √36 = 6.
QUESTION 5
Solve the equation log 2 (𝑥 + 3) + log 2 (𝑥 − 1) = 5.
• A. 𝑥 = 5
• B. 𝑥 = −5 and 𝑥 = 5
• C. 𝑥 = 7
• D. 𝑥 = −7 and 𝑥 = 5
Correct Answer: A. 𝑥 = 5

Detailed Rationale: Using the logarithm addition law, log 2 ((𝑥 + 3)(𝑥 −
1)) = 5). Converting to exponential form gives 𝑥 2 + 2𝑥 − 3 = 25 =
32, leading to the quadratic equation 𝑥 2 + 2𝑥 − 35 = 0. Factoring
yields (𝑥 + 7)(𝑥 − 5) = 0, giving potential solutions 𝑥 = −7 and 𝑥 =
5. Since log 2 (𝑥 − 1) requires 𝑥 > 1, only 𝑥 = 5 is valid.
QUESTION 6
Find the coefficient of 𝑥 3 in the binomial expansion of (2 − 3𝑥 )5 .
• A. −720

, • B. −1080
• C. 720
• D. 1080
Correct Answer: B. −1080
Detailed Rationale: The general term in the expansion is given by
(5𝑟)(2)5−𝑟 (−3𝑥 )𝑟 . For the term containing 𝑥 3 , set 𝑟 = 3:
(53)(2)2 (−3𝑥 )3 = 10 × 4 × (−27𝑥 3 ) = −1080𝑥 3 .

QUESTION 7
Solve the trigonometric equation 2 cos 2 𝜃 + sin 𝜃 − 1 = 0 for 0∘ ≤
𝜃 < 360∘ .
• A. 𝜃 = 30∘ , 150∘ , 270∘
• B. 𝜃 = 90∘ , 210∘ , 330∘
• C. 𝜃 = 60∘ , 120∘ , 180∘
• D. 𝜃 = 45∘ , 135∘ , 225∘
Correct Answer: B. 𝜃 = 90∘ , 210∘ , 330∘
Detailed Rationale: Using the identity cos2 𝜃 = 1 − sin2 𝜃, substitute
into the equation: 2(1 − sin2 𝜃 ) + sin 𝜃 − 1 = 0 ⟹ 2 sin2 𝜃 − sin 𝜃 −
1 = 0. Factoring yields (2 sin 𝜃 + 1)(sin 𝜃 − 1) = 0, giving sin 𝜃 = 1
(𝜃 = 90∘ ) and sin 𝜃 = −0.5 (𝜃 = 210∘ , 330∘ ).
QUESTION 8
Given the vectors 𝐚 = 3𝐢 − 2𝐣 + 𝐤 and 𝐛 = 𝐢 + 4𝐣 − 2𝐤, find the scalar
product 𝐚 ⋅ 𝐛.
• A. −7