EXAM FOR AQA AS LEVEL MATHEMATICS PAPER 1 AND 2
,Table of Contents
AQA AS-Level Mathematics (7356)
Premium Original Practice Examination — 2026 -2027 Edition
No. Section
1 Cover Page
2 Examination Overview
3 Paper 1 — Pure Mathematics
4 Paper 2 — Statistics and Mechanics
AQA AS-Level Mathematics (7356)
Premium Original Practice Examination — 2026 Edition
Paper 1: Pure Mathematics
Difficult Questions with Answers, Explanations & Exam Strategy
Time allowed: 1 hour 30 minutes | Total: 80 marks
1. Algebraic proof
Prove that 𝑛5 − 𝑛is divisible by 30 for every positive integer 𝑛.
[5 marks]
Answer and rationale
𝑛 5 − 𝑛 = 𝑛 ( 𝑛 4 − 1)
= 𝑛(𝑛2 − 1)(𝑛2 + 1)
= 𝑛(𝑛 − 1)(𝑛 + 1)(𝑛2 + 1)
Among the three consecutive integers 𝑛 − 1, 𝑛, 𝑛 + 1:
• one is divisible by 3;
• at least one is even, so the product is divisible by 2.
Therefore, 𝑛(𝑛 − 1)(𝑛 + 1)is divisible by 6.
Also, by considering the possible remainders when 𝑛is divided by 5, or by using Fermat’s theorem,
𝑛5 ≡ 𝑛 (mod 5)
so
𝑛5 − 𝑛 ≡ 0 (mod 5).
The expression is divisible by 2, 3 and 5. Since these factors are coprime,
𝑛5 − 𝑛 is divisible by 30.
Exam strategy
Factor first. When a proof asks about divisibility, look for consecutive integers because they automatically create
factors of 2 and 3.
2. Factor theorem and algebraic roots
,The polynomial
𝑃(𝑥 ) = 2𝑥 3 − 3𝑥 2 − 8𝑥 + 12
has three real roots. Solve 𝑃(𝑥 ) = 0.
[5 marks]
Answer and rationale
Group the terms:
2𝑥 3 − 3𝑥 2 − 8𝑥 + 12
= 𝑥 2 (2𝑥 − 3) − 4(2𝑥 − 3)
= (2𝑥 − 3)(𝑥 2 − 4)
= (2𝑥 − 3)(𝑥 − 2)(𝑥 + 2).
Therefore,
2𝑥 − 3 = 0, 𝑥 − 2 = 0, 𝑥 + 2 = 0.
3
𝑥 = , 2, − 2
2
Exam strategy
For a cubic with four terms, try grouping before using a calculator. Once a common bracket appears, factor it out
immediately.
3. Modulus equation
Solve
∣ 2𝑥 − 3 ∣= 𝑥 2 − 5𝑥 + 6.
[5 marks]
Answer and rationale
Since the left-hand side is non-negative,
𝑥 2 − 5𝑥 + 6 ≥ 0.
(𝑥 − 2)(𝑥 − 3) ≥ 0
so
𝑥 ≤ 2 or 𝑥 ≥ 3.
Case 1: 𝟐𝒙 − 𝟑 ≥ 𝟎
∣ 2𝑥 − 3 ∣= 2𝑥 − 3
2𝑥 − 3 = 𝑥 2 − 5𝑥 + 6
𝑥 2 − 7𝑥 + 9 = 0.
7 ± √13
𝑥= .
2
Both values satisfy the required conditions.
Case 2: 𝟐𝒙 − 𝟑 < 𝟎
∣ 2𝑥 − 3 ∣= 3 − 2𝑥
3 − 2𝑥 = 𝑥 2 − 5𝑥 + 6
𝑥 2 − 3𝑥 + 3 = 0.
𝑏 2 − 4𝑎𝑐 = 9 − 12 = −3,
so there are no real roots from this case.
, 7 − √13 7 + √13
𝑥= ,
2 2
Exam strategy
Always split a modulus equation into cases. Check every final answer against the condition for its case.
4. Tangents to a circle
The circle 𝐶has equation
𝑥 2 + 𝑦 2 − 6𝑥 + 4𝑦 − 12 = 0.
Tangents are drawn to 𝐶from the point 𝑃(0, 4).
Find the equations of the two tangents.
[5 marks]
Answer and rationale
Complete the square:
𝑥 2 − 6𝑥 + 𝑦 2 + 4𝑦 = 12
(𝑥 − 3)2 + (𝑦 + 2)2 = 25.
Therefore, the centre is (3, −2)and the radius is 5.
A line through 𝑃(0, 4)has equation
𝑦 = 𝑚𝑥 + 4.
Rearrange:
𝑚𝑥 − 𝑦 + 4 = 0.
For a tangent, the perpendicular distance from the centre to the line equals the radius:
∣ 3𝑚 + 6 ∣
= 5.
√𝑚2 +1
Squaring:
(3𝑚 + 6)2 = 25(𝑚2 + 1)
9𝑚2 + 36𝑚 + 36 = 25𝑚2 + 25
16𝑚2 − 36𝑚 − 11 = 0.
36 ± √2000
𝑚=
32
9 ± 5√5
𝑚= .
8
Hence the tangent equations are
9 + 5√5
𝑦=( )𝑥 + 4
8
and
9 − 5√5
𝑦=( ) 𝑥 + 4.
8
Exam strategy
For a tangent, use the key fact:
distance from centre to line = radius.
,Do not try to find the tangent points unless the question specifically requires them.
5. Recurrence relations
A sequence is defined by
3
𝑢1 = 2, 𝑢𝑛+1 = 𝑢 + 4.
2 𝑛
(a) Find 𝑢2 and 𝑢3 .
(b) Show that**
3 𝑛−1
𝑢𝑛 = 10 ( ) − 8.
2
(c) Find the first value of 𝑛for which 𝑢𝑛 > 100.
[5 marks]
Answer and rationale
(a)
3
𝑢2 = (2) + 4 = 7
2
3 29
𝑢3 = (7) + 4 = .
2 2
29
𝑢2 = 7, 𝑢3 =
2
(b) Let
𝑣𝑛 = 𝑢𝑛 + 8.
Then
𝑣𝑛+1 = 𝑢𝑛+1 + 8
3
= 𝑢𝑛 + 12
2
3
= (𝑢𝑛 + 8)
2
3
= 𝑣𝑛 .
2
So 𝑣𝑛 is geometric. Since
𝑣1 = 2 + 8 = 10,
3 𝑛−1
𝑣𝑛 = 10 ( ) .
2
Therefore,
3 𝑛−1
𝑢𝑛 = 10 ( ) − 8.
2
(c)
3 𝑛−1
10 ( ) − 8 > 100
2
3 𝑛−1
( ) > 10.8.
2
Using logarithms,
ln(10.8)
𝑛−1>
ln(1.5)
, 𝑛 > 6.86.
𝑛=7
Exam strategy
For a recurrence of the form 𝑢𝑛+1 = 𝑎𝑢𝑛 + 𝑏, shift the sequence by the equilibrium value to turn it into a
geometric sequence.
6. Binomial expansion
Use the binomial expansion of (1 − 2𝑥 )7 to find an approximation for 0.987 .
[5 marks]
Answer and rationale
(1 − 2𝑥 )7
= 1 + 7(−2𝑥 ) + 21(−2𝑥 )2 + 35(−2𝑥 )3 + ⋯
= 1 − 14𝑥 + 84𝑥 2 − 280𝑥 3 + ⋯
Since
0.98 = 1 − 0.02 = 1 − 2(0.01),
use 𝑥 = 0.01:
0.987 ≈ 1 − 14(0.01) + 84(0.01)2 − 280(0.01)3 .
= 1 − 0.14 + 0.0084 − 0.00028
0.987 ≈ 0.86812
Exam strategy
First rewrite the decimal in the form 1 + small number. Keep enough terms for the accuracy requested.
7. Trigonometric equation
Solve, for 0∘ ≤ 𝑥 < 360∘ ,
2cos2 𝑥 − 3 sin 𝑥 = 0.
[5 marks]
Answer and rationale
Use
cos2 𝑥 = 1 − sin2 𝑥.
2(1 − sin2 𝑥 ) − 3 sin 𝑥 = 0
2 − 2sin2 𝑥 − 3 sin 𝑥 = 0.
Rearrange:
2sin2 𝑥 + 3 sin 𝑥 − 2 = 0.
(2 sin 𝑥 − 1)(sin 𝑥 + 2) = 0.
1
sin 𝑥 =
2
or
sin 𝑥 = −2.
Since sin 𝑥 = −2is impossible,
1
sin 𝑥 = .
2
𝑥 = 30∘ , 150∘
,Exam strategy
When both cos2 𝑥and sin 𝑥appear, convert cos2 𝑥into 1 − sin2 𝑥, then solve a quadratic.
8. Exponentials and logarithms
Solve
𝑒 2𝑥 − 5𝑒 𝑥 + 6 = 0.
[5 marks]
Answer and rationale
Let
𝑢 = 𝑒𝑥.
Then
𝑒 2𝑥 = (𝑒 𝑥 )2 = 𝑢2 .
The equation becomes
𝑢2 − 5𝑢 + 6 = 0.
(𝑢 − 2)(𝑢 − 3) = 0.
𝑢 = 2 or 𝑢 = 3.
Therefore,
𝑒 𝑥 = 2 or 𝑒 𝑥 = 3.
𝑥 = ln 2 , ln 3
Exam strategy
For an equation containing 𝑒 2𝑥 and 𝑒 𝑥 , substitute 𝑢 = 𝑒 𝑥 . It converts the problem into an ordinary quadratic.
9. Differentiation and stationary points
The curve 𝐶has equation
𝑦 = 𝑥 2 𝑒 −𝑥 .
Find and classify all stationary points of 𝐶.
[5 marks]
Answer and rationale
Differentiate using the product rule:
𝑑𝑦
= 2𝑥𝑒 −𝑥 + 𝑥 2 (−𝑒 −𝑥 )
𝑑𝑥
= 𝑒 −𝑥 (2𝑥 − 𝑥 2 )
= 𝑥𝑒 −𝑥 (2 − 𝑥 ).
For stationary points:
𝑥𝑒 −𝑥 (2 − 𝑥 ) = 0.
Since 𝑒 −𝑥 ≠ 0,
𝑥 = 0 or 𝑥 = 2.
At 𝑥 = 0,
𝑦 = 0.
At 𝑥 = 2,
, 4
𝑦 = 4𝑒 −2 = .
𝑒2
To classify, inspect the sign of
𝑑𝑦
= 𝑥𝑒 −𝑥 (2 − 𝑥 ).
𝑑𝑥
𝑑𝑦
Interval Sign of Behaviour
𝑑𝑥
𝑥<0 Negative Decreasing
0 < 𝑥 < 2 Positive Increasing
𝑥>2 Negative Decreasing
Therefore,
(0, 0) is a minimum
and
4
(2, ) is a maximum.
𝑒2
Exam strategy
After finding stationary values, always classify them. A sign table is often clearer and safer than a second
derivative.
10. Tangent and normal
The curve 𝐶has equation
𝑥+1
𝑦= .
𝑥−1
Find the equation of the normal to 𝐶at the point where 𝑥 = 2.
[5 marks]
Answer and rationale
First find the point:
2+1
𝑦= = 3.
2−1
So the point is (2, 3).
Differentiate using the quotient rule:
𝑑𝑦 (𝑥 − 1)(1) − (𝑥 + 1)(1)
=
𝑑𝑥 (𝑥 − 1)2
𝑥−1−𝑥−1
=
(𝑥 − 1)2
−2
= .
(𝑥 − 1)2
At 𝑥 = 2,
𝑑𝑦
= −2.
𝑑𝑥
The gradient of the normal is
1
.
2
,Using point–gradient form:
1
𝑦−3= (𝑥 − 2).
2
1
𝑦 = 𝑥+2
2
Exam strategy
A normal gradient is the negative reciprocal of the tangent gradient. Write the point down before you begin the
equation.
11. Parametric differentiation
A curve is defined parametrically by
𝑥 = 𝑡 2 + 1, 𝑦 = 𝑡 3 − 3𝑡.
𝑑𝑦
(a) Find in terms of 𝑡.
𝑑𝑥
(b) Find the points where the tangent is horizontal.
[5 marks]
Answer and rationale
𝑑𝑥
= 2𝑡
𝑑𝑡
𝑑𝑦
= 3𝑡 2 − 3.
𝑑𝑡
Therefore,
𝑑𝑦
𝑑𝑦
= 𝑑𝑡
𝑑𝑥 𝑑𝑥
𝑑𝑡
2
3𝑡 − 3
= .
2𝑡
𝑑𝑦 3(𝑡 2 − 1)
=
𝑑𝑥 2𝑡
For a horizontal tangent,
𝑑𝑦
= 0.
𝑑𝑥
So,
3(𝑡 2 − 1) = 0
𝑡 = ±1.
When 𝑡 = 1,
𝑥 = 2, 𝑦 = −2.
When 𝑡 = −1,
𝑥 = 2, 𝑦 = 2.
(2, −2) and (2, 2)
Exam strategy
𝑑𝑦 𝑑𝑥
For parametric curves, a horizontal tangent occurs when = 0, provided ≠ 0.
𝑑𝑡 𝑑𝑡
, 12. Areas under a curve
The curve 𝐶has equation
𝑦 = 𝑥 2 − 4𝑥 + 3.
Find the exact total area between 𝐶and the 𝑥-axis for 0 ≤ 𝑥 ≤ 4.
[5 marks]
Answer and rationale
First find where the curve crosses the 𝑥-axis:
𝑥 2 − 4𝑥 + 3 = 0
(𝑥 − 1)(𝑥 − 3) = 0.
The roots are
𝑥 = 1, 𝑥 = 3.
The curve is above the axis from 0to 1, below from 1to 3, and above from 3to 4.
2
𝑥3
∫( 𝑥 − 4𝑥 + 3) 𝑑𝑥 = − 2𝑥 2 + 3𝑥.
3
The three positive areas are:
4
𝐴1 = ,
3
4
𝐴2 = ,
3
4
𝐴3 = .
3
Therefore,
4 4 4
Total area = + + .
3 3 3
4 square units
Exam strategy
Area is always positive. Split an integral at every 𝑥-intercept and reverse the sign of any region below the axis.
13. Numerical methods
The equation
𝑥 3 − 2𝑥 − 5 = 0
has a root 𝛼.
Use the iteration
3
𝑥𝑛+1 = √2𝑥𝑛 + 5,
starting with 𝑥1 = 2, to find 𝛼correct to 3 decimal places.
[5 marks]
Answer and rationale
𝑛 𝑥𝑛
1 2.000000
2 2.080084
,Table of Contents
AQA AS-Level Mathematics (7356)
Premium Original Practice Examination — 2026 -2027 Edition
No. Section
1 Cover Page
2 Examination Overview
3 Paper 1 — Pure Mathematics
4 Paper 2 — Statistics and Mechanics
AQA AS-Level Mathematics (7356)
Premium Original Practice Examination — 2026 Edition
Paper 1: Pure Mathematics
Difficult Questions with Answers, Explanations & Exam Strategy
Time allowed: 1 hour 30 minutes | Total: 80 marks
1. Algebraic proof
Prove that 𝑛5 − 𝑛is divisible by 30 for every positive integer 𝑛.
[5 marks]
Answer and rationale
𝑛 5 − 𝑛 = 𝑛 ( 𝑛 4 − 1)
= 𝑛(𝑛2 − 1)(𝑛2 + 1)
= 𝑛(𝑛 − 1)(𝑛 + 1)(𝑛2 + 1)
Among the three consecutive integers 𝑛 − 1, 𝑛, 𝑛 + 1:
• one is divisible by 3;
• at least one is even, so the product is divisible by 2.
Therefore, 𝑛(𝑛 − 1)(𝑛 + 1)is divisible by 6.
Also, by considering the possible remainders when 𝑛is divided by 5, or by using Fermat’s theorem,
𝑛5 ≡ 𝑛 (mod 5)
so
𝑛5 − 𝑛 ≡ 0 (mod 5).
The expression is divisible by 2, 3 and 5. Since these factors are coprime,
𝑛5 − 𝑛 is divisible by 30.
Exam strategy
Factor first. When a proof asks about divisibility, look for consecutive integers because they automatically create
factors of 2 and 3.
2. Factor theorem and algebraic roots
,The polynomial
𝑃(𝑥 ) = 2𝑥 3 − 3𝑥 2 − 8𝑥 + 12
has three real roots. Solve 𝑃(𝑥 ) = 0.
[5 marks]
Answer and rationale
Group the terms:
2𝑥 3 − 3𝑥 2 − 8𝑥 + 12
= 𝑥 2 (2𝑥 − 3) − 4(2𝑥 − 3)
= (2𝑥 − 3)(𝑥 2 − 4)
= (2𝑥 − 3)(𝑥 − 2)(𝑥 + 2).
Therefore,
2𝑥 − 3 = 0, 𝑥 − 2 = 0, 𝑥 + 2 = 0.
3
𝑥 = , 2, − 2
2
Exam strategy
For a cubic with four terms, try grouping before using a calculator. Once a common bracket appears, factor it out
immediately.
3. Modulus equation
Solve
∣ 2𝑥 − 3 ∣= 𝑥 2 − 5𝑥 + 6.
[5 marks]
Answer and rationale
Since the left-hand side is non-negative,
𝑥 2 − 5𝑥 + 6 ≥ 0.
(𝑥 − 2)(𝑥 − 3) ≥ 0
so
𝑥 ≤ 2 or 𝑥 ≥ 3.
Case 1: 𝟐𝒙 − 𝟑 ≥ 𝟎
∣ 2𝑥 − 3 ∣= 2𝑥 − 3
2𝑥 − 3 = 𝑥 2 − 5𝑥 + 6
𝑥 2 − 7𝑥 + 9 = 0.
7 ± √13
𝑥= .
2
Both values satisfy the required conditions.
Case 2: 𝟐𝒙 − 𝟑 < 𝟎
∣ 2𝑥 − 3 ∣= 3 − 2𝑥
3 − 2𝑥 = 𝑥 2 − 5𝑥 + 6
𝑥 2 − 3𝑥 + 3 = 0.
𝑏 2 − 4𝑎𝑐 = 9 − 12 = −3,
so there are no real roots from this case.
, 7 − √13 7 + √13
𝑥= ,
2 2
Exam strategy
Always split a modulus equation into cases. Check every final answer against the condition for its case.
4. Tangents to a circle
The circle 𝐶has equation
𝑥 2 + 𝑦 2 − 6𝑥 + 4𝑦 − 12 = 0.
Tangents are drawn to 𝐶from the point 𝑃(0, 4).
Find the equations of the two tangents.
[5 marks]
Answer and rationale
Complete the square:
𝑥 2 − 6𝑥 + 𝑦 2 + 4𝑦 = 12
(𝑥 − 3)2 + (𝑦 + 2)2 = 25.
Therefore, the centre is (3, −2)and the radius is 5.
A line through 𝑃(0, 4)has equation
𝑦 = 𝑚𝑥 + 4.
Rearrange:
𝑚𝑥 − 𝑦 + 4 = 0.
For a tangent, the perpendicular distance from the centre to the line equals the radius:
∣ 3𝑚 + 6 ∣
= 5.
√𝑚2 +1
Squaring:
(3𝑚 + 6)2 = 25(𝑚2 + 1)
9𝑚2 + 36𝑚 + 36 = 25𝑚2 + 25
16𝑚2 − 36𝑚 − 11 = 0.
36 ± √2000
𝑚=
32
9 ± 5√5
𝑚= .
8
Hence the tangent equations are
9 + 5√5
𝑦=( )𝑥 + 4
8
and
9 − 5√5
𝑦=( ) 𝑥 + 4.
8
Exam strategy
For a tangent, use the key fact:
distance from centre to line = radius.
,Do not try to find the tangent points unless the question specifically requires them.
5. Recurrence relations
A sequence is defined by
3
𝑢1 = 2, 𝑢𝑛+1 = 𝑢 + 4.
2 𝑛
(a) Find 𝑢2 and 𝑢3 .
(b) Show that**
3 𝑛−1
𝑢𝑛 = 10 ( ) − 8.
2
(c) Find the first value of 𝑛for which 𝑢𝑛 > 100.
[5 marks]
Answer and rationale
(a)
3
𝑢2 = (2) + 4 = 7
2
3 29
𝑢3 = (7) + 4 = .
2 2
29
𝑢2 = 7, 𝑢3 =
2
(b) Let
𝑣𝑛 = 𝑢𝑛 + 8.
Then
𝑣𝑛+1 = 𝑢𝑛+1 + 8
3
= 𝑢𝑛 + 12
2
3
= (𝑢𝑛 + 8)
2
3
= 𝑣𝑛 .
2
So 𝑣𝑛 is geometric. Since
𝑣1 = 2 + 8 = 10,
3 𝑛−1
𝑣𝑛 = 10 ( ) .
2
Therefore,
3 𝑛−1
𝑢𝑛 = 10 ( ) − 8.
2
(c)
3 𝑛−1
10 ( ) − 8 > 100
2
3 𝑛−1
( ) > 10.8.
2
Using logarithms,
ln(10.8)
𝑛−1>
ln(1.5)
, 𝑛 > 6.86.
𝑛=7
Exam strategy
For a recurrence of the form 𝑢𝑛+1 = 𝑎𝑢𝑛 + 𝑏, shift the sequence by the equilibrium value to turn it into a
geometric sequence.
6. Binomial expansion
Use the binomial expansion of (1 − 2𝑥 )7 to find an approximation for 0.987 .
[5 marks]
Answer and rationale
(1 − 2𝑥 )7
= 1 + 7(−2𝑥 ) + 21(−2𝑥 )2 + 35(−2𝑥 )3 + ⋯
= 1 − 14𝑥 + 84𝑥 2 − 280𝑥 3 + ⋯
Since
0.98 = 1 − 0.02 = 1 − 2(0.01),
use 𝑥 = 0.01:
0.987 ≈ 1 − 14(0.01) + 84(0.01)2 − 280(0.01)3 .
= 1 − 0.14 + 0.0084 − 0.00028
0.987 ≈ 0.86812
Exam strategy
First rewrite the decimal in the form 1 + small number. Keep enough terms for the accuracy requested.
7. Trigonometric equation
Solve, for 0∘ ≤ 𝑥 < 360∘ ,
2cos2 𝑥 − 3 sin 𝑥 = 0.
[5 marks]
Answer and rationale
Use
cos2 𝑥 = 1 − sin2 𝑥.
2(1 − sin2 𝑥 ) − 3 sin 𝑥 = 0
2 − 2sin2 𝑥 − 3 sin 𝑥 = 0.
Rearrange:
2sin2 𝑥 + 3 sin 𝑥 − 2 = 0.
(2 sin 𝑥 − 1)(sin 𝑥 + 2) = 0.
1
sin 𝑥 =
2
or
sin 𝑥 = −2.
Since sin 𝑥 = −2is impossible,
1
sin 𝑥 = .
2
𝑥 = 30∘ , 150∘
,Exam strategy
When both cos2 𝑥and sin 𝑥appear, convert cos2 𝑥into 1 − sin2 𝑥, then solve a quadratic.
8. Exponentials and logarithms
Solve
𝑒 2𝑥 − 5𝑒 𝑥 + 6 = 0.
[5 marks]
Answer and rationale
Let
𝑢 = 𝑒𝑥.
Then
𝑒 2𝑥 = (𝑒 𝑥 )2 = 𝑢2 .
The equation becomes
𝑢2 − 5𝑢 + 6 = 0.
(𝑢 − 2)(𝑢 − 3) = 0.
𝑢 = 2 or 𝑢 = 3.
Therefore,
𝑒 𝑥 = 2 or 𝑒 𝑥 = 3.
𝑥 = ln 2 , ln 3
Exam strategy
For an equation containing 𝑒 2𝑥 and 𝑒 𝑥 , substitute 𝑢 = 𝑒 𝑥 . It converts the problem into an ordinary quadratic.
9. Differentiation and stationary points
The curve 𝐶has equation
𝑦 = 𝑥 2 𝑒 −𝑥 .
Find and classify all stationary points of 𝐶.
[5 marks]
Answer and rationale
Differentiate using the product rule:
𝑑𝑦
= 2𝑥𝑒 −𝑥 + 𝑥 2 (−𝑒 −𝑥 )
𝑑𝑥
= 𝑒 −𝑥 (2𝑥 − 𝑥 2 )
= 𝑥𝑒 −𝑥 (2 − 𝑥 ).
For stationary points:
𝑥𝑒 −𝑥 (2 − 𝑥 ) = 0.
Since 𝑒 −𝑥 ≠ 0,
𝑥 = 0 or 𝑥 = 2.
At 𝑥 = 0,
𝑦 = 0.
At 𝑥 = 2,
, 4
𝑦 = 4𝑒 −2 = .
𝑒2
To classify, inspect the sign of
𝑑𝑦
= 𝑥𝑒 −𝑥 (2 − 𝑥 ).
𝑑𝑥
𝑑𝑦
Interval Sign of Behaviour
𝑑𝑥
𝑥<0 Negative Decreasing
0 < 𝑥 < 2 Positive Increasing
𝑥>2 Negative Decreasing
Therefore,
(0, 0) is a minimum
and
4
(2, ) is a maximum.
𝑒2
Exam strategy
After finding stationary values, always classify them. A sign table is often clearer and safer than a second
derivative.
10. Tangent and normal
The curve 𝐶has equation
𝑥+1
𝑦= .
𝑥−1
Find the equation of the normal to 𝐶at the point where 𝑥 = 2.
[5 marks]
Answer and rationale
First find the point:
2+1
𝑦= = 3.
2−1
So the point is (2, 3).
Differentiate using the quotient rule:
𝑑𝑦 (𝑥 − 1)(1) − (𝑥 + 1)(1)
=
𝑑𝑥 (𝑥 − 1)2
𝑥−1−𝑥−1
=
(𝑥 − 1)2
−2
= .
(𝑥 − 1)2
At 𝑥 = 2,
𝑑𝑦
= −2.
𝑑𝑥
The gradient of the normal is
1
.
2
,Using point–gradient form:
1
𝑦−3= (𝑥 − 2).
2
1
𝑦 = 𝑥+2
2
Exam strategy
A normal gradient is the negative reciprocal of the tangent gradient. Write the point down before you begin the
equation.
11. Parametric differentiation
A curve is defined parametrically by
𝑥 = 𝑡 2 + 1, 𝑦 = 𝑡 3 − 3𝑡.
𝑑𝑦
(a) Find in terms of 𝑡.
𝑑𝑥
(b) Find the points where the tangent is horizontal.
[5 marks]
Answer and rationale
𝑑𝑥
= 2𝑡
𝑑𝑡
𝑑𝑦
= 3𝑡 2 − 3.
𝑑𝑡
Therefore,
𝑑𝑦
𝑑𝑦
= 𝑑𝑡
𝑑𝑥 𝑑𝑥
𝑑𝑡
2
3𝑡 − 3
= .
2𝑡
𝑑𝑦 3(𝑡 2 − 1)
=
𝑑𝑥 2𝑡
For a horizontal tangent,
𝑑𝑦
= 0.
𝑑𝑥
So,
3(𝑡 2 − 1) = 0
𝑡 = ±1.
When 𝑡 = 1,
𝑥 = 2, 𝑦 = −2.
When 𝑡 = −1,
𝑥 = 2, 𝑦 = 2.
(2, −2) and (2, 2)
Exam strategy
𝑑𝑦 𝑑𝑥
For parametric curves, a horizontal tangent occurs when = 0, provided ≠ 0.
𝑑𝑡 𝑑𝑡
, 12. Areas under a curve
The curve 𝐶has equation
𝑦 = 𝑥 2 − 4𝑥 + 3.
Find the exact total area between 𝐶and the 𝑥-axis for 0 ≤ 𝑥 ≤ 4.
[5 marks]
Answer and rationale
First find where the curve crosses the 𝑥-axis:
𝑥 2 − 4𝑥 + 3 = 0
(𝑥 − 1)(𝑥 − 3) = 0.
The roots are
𝑥 = 1, 𝑥 = 3.
The curve is above the axis from 0to 1, below from 1to 3, and above from 3to 4.
2
𝑥3
∫( 𝑥 − 4𝑥 + 3) 𝑑𝑥 = − 2𝑥 2 + 3𝑥.
3
The three positive areas are:
4
𝐴1 = ,
3
4
𝐴2 = ,
3
4
𝐴3 = .
3
Therefore,
4 4 4
Total area = + + .
3 3 3
4 square units
Exam strategy
Area is always positive. Split an integral at every 𝑥-intercept and reverse the sign of any region below the axis.
13. Numerical methods
The equation
𝑥 3 − 2𝑥 − 5 = 0
has a root 𝛼.
Use the iteration
3
𝑥𝑛+1 = √2𝑥𝑛 + 5,
starting with 𝑥1 = 2, to find 𝛼correct to 3 decimal places.
[5 marks]
Answer and rationale
𝑛 𝑥𝑛
1 2.000000
2 2.080084
