A-Level Mathematics — Practice Paper 1

Differentiation, integration, sequences, binomial expansion, trigonometry and proof — with the method marks separated from the accuracy marks, as they are in the real mark scheme.

How to use this: at A-Level, a question worth 5 marks typically awards 3–4 for method. Structure your answer so each stage is visible and attributable, and never erase working.

Section A

1. Differentiate $y = (3x^2 + 1)^5$ with respect to $x$ (2 marks)

Chain rule:  dy/dx = n[f(x)]ⁿ⁻¹ · f'(x)

f(x)  = 3x² + 1        f'(x) = 6x

dy/dx = 5(3x² + 1)⁴ · 6x
      = 30x(3x² + 1)⁴

Marks: M1 for the chain rule structure, A1 for the simplified result.

2. Find $\displaystyle\int \frac{6x}{3x^2+1}\,dx$ (3 marks)

Notice the numerator is exactly the derivative of the denominator.

⌠  f'(x)
│ ------- dx = ln|f(x)| + C
⌡  f(x)

∫ 6x/(3x² + 1) dx = ln|3x² + 1| + C

Answer: $\ln|3x^2+1| + C$

Marks: M1 for recognising the $f'/f$ form, A1 for the logarithm, B1 for $+C$ and the modulus.

The recognition worth drilling: whenever the top is a constant multiple of the derivative of the bottom, the answer is a logarithm. Attempting substitution works but takes three times as long.

3. The first three terms of a geometric series are $8$, $12$, $18$. Find the sum to infinity, or explain why it does not exist (3 marks)

Common ratio:  r = 12/8 = 1.5

For a sum to infinity we require |r| < 1.
Here |r| = 1.5 > 1.

∴ the sum to infinity does not exist — the series diverges.

Marks: M1 for finding $r$, M1 for stating the convergence condition, A1 for the conclusion.

This is a trap question. Candidates who apply $S_\infty = \frac{a}{1-r}$ mechanically get $\frac{8}{-0.5} = -16$ — a negative sum for a series of positive increasing terms, which is plainly impossible. Always check $|r|<1$ before using the formula.

4. Find the first three terms in the expansion of $(1+2x)^8$ in ascending powers of $x$ (3 marks)

(1 + y)ⁿ = 1 + ny + [n(n-1)/2!] y² + ...

With y = 2x, n = 8:

Term 1: 1

Term 2: 8(2x) = 16x

Term 3: [8 × 7 / 2] (2x)²
      = 28 × 4x²
      = 112x²

Answer: $1 + 16x + 112x^2$

Marks: M1 for the binomial structure, A1 for the $x$ term, A1 for the $x^2$ term.

Where marks are lost: squaring only the $x$ and not the 2 — writing $28 \times 2x^2 = 56x^2$. The whole term $(2x)$ is squared.

5. Solve $2\sin^2\theta + \sin\theta - 1 = 0$ for $0° \le \theta \le 360°$ (4 marks)

Treat as a quadratic in sin θ:

(2 sin θ - 1)(sin θ + 1) = 0

sin θ = 1/2      or      sin θ = -1

sin θ = 1/2   ⟹   θ = 30°, 150°
sin θ = -1    ⟹   θ = 270°

θ = 30°, 150°, 270°

Marks: M1 for factorising, A1 for both values of $\sin\theta$, A1 for the first pair of angles, A1 for all three.

The most common loss is finding only $30°$. Sine is positive in the first and second quadrants, so $180° - 30° = 150°$ is also a solution. Sketch the sine curve over the given range before writing your answers — it takes ten seconds and catches every missing root.

Section B

6. A curve has equation $y = x^3 - 6x^2 + 9x + 2$. Find the coordinates of the stationary points and determine their nature (6 marks)

dy/dx = 3x² - 12x + 9 = 3(x² - 4x + 3) = 3(x - 1)(x - 3)

Stationary where dy/dx = 0:   x = 1  or  x = 3

y(1) = 1 - 6 + 9 + 2 = 6      ⟹  (1, 6)
y(3) = 27 - 54 + 27 + 2 = 2   ⟹  (3, 2)

Second derivative:  d²y/dx² = 6x - 12

At x = 1:  6(1) - 12 = -6  < 0  ⟹  MAXIMUM
At x = 3:  6(3) - 12 = +6  > 0  ⟹  MINIMUM

Answer: $(1,6)$ is a maximum; $(3,2)$ is a minimum.

Marks: M1 differentiate, A1 both $x$ values, A1 both $y$ values, M1 second derivative, A1 both signs evaluated, A1 both natures stated.

Note the sign convention: second derivative negative means maximum. Many candidates invert this. A quick sanity check: at a maximum the curve is concave down, so the gradient is decreasing, so $\frac{d^2y}{dx^2} < 0$.

7. Prove by contradiction that there is no greatest even integer (4 marks)

Assume the opposite: there exists a greatest even integer, call it N.

Since N is even, N = 2k for some integer k.

Consider M = N + 2 = 2k + 2 = 2(k + 1).

Since (k + 1) is an integer, M is even.
And M = N + 2 > N.

So M is an even integer greater than N, contradicting the
assumption that N is the greatest even integer.

∴ the assumption is false, and there is no greatest even integer.  ∎

Marks: M1 for stating the negation correctly, M1 for constructing a larger even integer, A1 for showing it is even, A1 for stating the contradiction and concluding.

Proof by contradiction is marked on structure. The final line — explicitly naming the contradiction and rejecting the assumption — carries its own mark and is routinely omitted.

8. Find $\displaystyle\int_1^4 \frac{1}{\sqrt{x}}\,dx$ (4 marks)

Rewrite with an index:   1/√x = x^(-1/2)

⌠            x^(1/2)
│ x^(-1/2) dx = --------- = 2√x
⌡              1/2

Evaluate between 1 and 4:

[2√x]₁⁴ = 2√4 - 2√1
        = 4 - 2
        = 2

Answer: 2

Marks: M1 for converting to index form, A1 for the antiderivative, M1 for substituting limits, A1 for the value.

Always convert roots and fractions to indices before integrating. Attempting to integrate $1/\sqrt{x}$ in its original form is where this question goes wrong.

How A-Level Maths is marked

The practical consequences:

1. Never erase. Crossed-out work that is correct can still be marked; erased work cannot.
2. One error rarely costs a whole question. Follow-through means a slip early on usually costs one A mark, not all of them — provided the subsequent method is visible.
3. Attempt every part. Later parts frequently do not depend on earlier answers, and even where they do, follow-through applies.
4. Check the domain. Trigonometric equations, logarithms and square roots all have range restrictions that carry marks of their own.

Where to go next

• A-Level Chemistry Practice Paper 1 — if you are taking both
• AP Calculus AB Practice Set 1 — the same calculus in the US system
• GCSE Mathematics Practice Paper 1 — for algebra fluency underneath this

Ask StudyAI to mark your working line by line and tell you which M mark your method stopped earning.