The Hardest Cambridge A Level Maths Questions Ever Asked
- Isaac
- May 24
- 11 min read
Updated: 6 days ago
Think Cambridge A Level Maths is already hard? These questions take it to another level. We’ve compiled 6 of the hardest Cambridge A Level Maths questions — the kind that test not just your formulas, but your problem-solving, creativity, and ability to stay calm under pressure. Try them yourself and see how many you can actually solve.
Problem 1: The Parametric Curve From Hell
The curve C has parametric equations:
x = 2cosθ + cos2θ, y = 2sinθ - sin2θ, 0 ≤ θ ≤ π
(a) Show that dy/dx = (2cosθ - 2cos2θ) / (-2sinθ - 2sin2θ)
(b) Find the exact coordinates of the point P on C where the tangent is horizontal.
(c) Find the exact equation of the tangent to C at θ = π/3
Step-By-Step Answer
(a) Show that dy/dx = (2cosθ - 2cos2θ) / (-2sinθ - 2sin2θ)
dx/dθ = -2sinθ - 2sin2θ
dy/dθ = 2cosθ - 2cos2θ
dy/dx = (dy/dθ) / (dx/dθ) = (2cosθ - 2cos2θ) / (-2sinθ - 2sin2θ) ✓
(b) Find coordinates of P where tangent is horizontal
Horizontal tangent → numerator = 0:
2cosθ - 2cos2θ = 0
cosθ - cos2θ = 0
Substitute cos2θ = 2cos²θ - 1:
cosθ - (2cos²θ - 1) = 0
2cos²θ - cosθ - 1 = 0
(2cosθ + 1)(cosθ - 1) = 0
cosθ = -1/2 or cosθ = 1
cosθ = 1 → θ = 0, but denominator is also 0 here (cusp), so reject.
cosθ = -1/2 → θ = 2π/3
At θ = 2π/3:
x = 2cos(2π/3) + cos(4π/3) = 2(-1/2) + (-1/2) = -1 - 1/2 = -3/2
y = 2sin(2π/3) - sin(4π/3) = 2(√3/2) - (-√3/2) = √3 + √3/2 = 3√3/2
P = (-3/2, 3√3/2)
(c) Equation of tangent at θ = π/3
At θ = π/3: cos(π/3) = 1/2, sin(π/3) = √3/2, cos(2π/3) = -1/2, sin(2π/3) = √3/2
Numerator: 2(1/2) - 2(-1/2) = 1 + 1 = 2
Denominator: -2(√3/2) - 2(√3/2) = -√3 - √3 = -2√3
dy/dx = 2/(-2√3) = -1/√3
Coordinates at θ = π/3:
x = 2(1/2) + (-1/2) = 1 - 1/2 = 1/2
y = 2(√3/2) - √3/2 = √3 - √3/2 = √3/2
Tangent through (1/2, √3/2) with gradient -1/√3:
y - √3/2 = -1/√3 (x - 1/2)
Multiply through by √3:
√3y - 3/2 = -(x - 1/2)
√3y - 3/2 = -x + 1/2
x + √3y = 2
Tangent equation: x + √3y = 2
Problem 2: The Curve That Looks Easy — Until Part (d)
The curve C has parametric equations:
x = t² + 1/t², y = t + 1/t, where t > 0
(a) Show that x = y² - 2
(b) Find the equation of the tangent to C at the point where t = 2
(c) The tangent found in part (b) meets the curve C again at another point. Find the exact coordinates of this point.
(d) The region enclosed by the curve C, the tangent from part (b), and the line x = 2 is rotated through 360° about the x-axis. Find the exact volume of the solid formed.
Step-By-Step Answer
(a) Show that x = y² - 2
y = t + 1/t
y² = t² + 2 + 1/t²
y² - 2 = t² + 1/t² = x
Therefore x = y² - 2 ✓
(b) Equation of tangent at t = 2
Coordinates at t = 2:
x = 4 + 1/4 = 17/4
y = 2 + 1/2 = 5/2
Gradient:
dx/dt = 2t - 2/t³
dy/dt = 1 - 1/t²
At t = 2:
dx/dt = 4 - 2/8 = 4 - 1/4 = 15/4
dy/dt = 1 - 1/4 = 3/4
dy/dx = (3/4) / (15/4) = 3/15 = 1/5
Tangent through (17/4, 5/2) with gradient 1/5:
y - 5/2 = (1/5)(x - 17/4)
20y - 50 = 4x - 17
4x - 20y + 33 = 0
y = x/5 + 33/20
(c) Where tangent meets curve again
Substitute x = t² + 1/t² and y = t + 1/t into y = x/5 + 33/20:
t + 1/t = (t² + 1/t²)/5 + 33/20
Multiply through by 20t²:
20t³ + 20t = 4t⁴ + 4 + 33t²
4t⁴ - 20t³ + 33t² - 20t + 4 = 0
t = 2 is a double root, so factorise out (t - 2)²:
4t⁴ - 20t³ + 33t² - 20t + 4 = (t - 2)²(2t - 1)²
Other solution: 2t - 1 = 0 → t = 1/2
At t = 1/2:
x = (1/2)² + 1/(1/2)² = 1/4 + 4 = 17/4
y = 1/2 + 2 = 5/2
This is the same point as t = 2, so the tangent does not meet the curve at a distinct second point for t > 0.
(d) Volume of revolution about x-axis
The region is bounded by:
Curve C: y² = x + 2
Tangent: y = x/5 + 33/20
Line: x = 2
Find boundary points:
At x = 2 on curve: y² = 4, so y = 2
At x = 17/4: tangent meets curve at y = 5/2
Volume using washer method from x = 2 to x = 17/4:
V = π ∫[2 to 17/4] (y_curve² - y_tangent²) dx
y_curve² = x + 2
y_tangent² = (x/5 + 33/20)² = x²/25 + 33x/50 + 1089/400
y_curve² - y_tangent² = (x + 2) - x²/25 - 33x/50 - 1089/400
Convert everything to 400ths:
= 400x/400 + 800/400 - 16x²/400 - 264x/400 - 1089/400
= (-16x² + 136x - 289) / 400
= -(4x - 17)² / 400
So:
V = π ∫[2 to 17/4] (-(4x - 17)²/400) dx
Since the curve is above the tangent in this region, take the positive value:
V = (π/400) ∫[2 to 17/4] (4x - 17)² dx
Let u = 4x - 17, du = 4 dx
At x = 2: u = -9
At x = 17/4: u = 0
V = (π/400) × (1/4) ∫[-9 to 0] u² du
= (π/1600) [u³/3] from -9 to 0
= (π/1600) [0 - (-729/3)]
= (π/1600) × 243
V = 243π/1600
Problem 3: The Curve With an Asymptote… and a Brutal Twist
The curve C has equation:
y = (x² + 4x + 5) / (x + 1), x > -1
(a) Show that C can be written in the form y = x + 3 + 2/(x + 1). Hence state the equation of the oblique asymptote.
(b) Find the coordinates of the stationary point of C, and determine whether it is a maximum or minimum.
(c) The tangent to C at the point where x = 1 meets the oblique asymptote at point A. Find the exact coordinates of A.
(d) The finite region bounded by C, the oblique asymptote, the line x = 1, and the line through A parallel to the y-axis is rotated through 360° about the x-axis. Find the exact volume of the solid formed.
(e) The normal to C at x = 1 meets the curve again at point B. Find the exact coordinates of B.
Step-By-Step Answer
Problem 4: The 12-Mark Differential Equation Nightmare
Given that y = 2 and dy/dx = 0 when x = 0, find the particular solution to
d²y/dx² - 5(dy/dx) + 4y = 8x - 10 - 10cos2x
Step-By-Step Answer
(a) Show y = x + 3 + 2/(x+1)
Divide x² + 4x + 5 by (x + 1):
x² + 4x + 5 = (x + 1)(x + 3) + 2
Therefore:
y = (x + 3) + 2/(x + 1)
Oblique asymptote: y = x + 3
(b) Stationary points
dy/dx = 1 - 2/(x + 1)²
Set dy/dx = 0:
1 - 2/(x + 1)² = 0
(x + 1)² = 2
x + 1 = √2 (since x > -1)
x = √2 - 1
y = (√2 - 1) + 3 + 2/√2 = √2 + 2 + √2 = 2√2 + 2
Find nature using d²y/dx²:
d²y/dx² = 4/(x + 1)³
At x = √2 - 1: x + 1 = √2 > 0, so d²y/dx² > 0
Minimum at (√2 - 1, 2 + 2√2)
(c) Tangent at x = 1 meets oblique asymptote at A
At x = 1:
y = 1 + 3 + 2/2 = 5
Gradient:
dy/dx = 1 - 2/(1+1)² = 1 - 2/4 = 1 - 1/2 = 1/2
Tangent through (1, 5) with gradient 1/2:
y - 5 = (1/2)(x - 1)
y = x/2 + 9/2
Find intersection with y = x + 3:
x/2 + 9/2 = x + 3
9/2 - 3 = x - x/2
3/2 = x/2
x = 3
y = 3 + 3 = 6
A = (3, 6)
(d) Volume of solid of revolution
The region is bounded by:
Curve C: y = x + 3 + 2/(x+1)
Oblique asymptote: y = x + 3
Line x = 1
Vertical line x = 3 (through A)
The curve is above the asymptote for x > -1 since 2/(x+1) > 0.
Let f(x) = y_curve = x + 3 + 2/(x+1) and g(x) = y_asymptote = x + 3
Using washer method:
V = π ∫[1 to 3] (f(x)² - g(x)²) dx
f(x)² - g(x)² = (f - g)(f + g)
f - g = 2/(x+1)
f + g = 2(x + 3) + 2/(x+1)
So:
f² - g² = (2/(x+1)) × (2(x+3) + 2/(x+1))
= 4(x+3)/(x+1) + 4/(x+1)²
V = π ∫[1 to 3] [4(x+3)/(x+1) + 4/(x+1)²] dx
For the first term, write (x+3)/(x+1):
x + 3 = (x + 1) + 2
(x+3)/(x+1) = 1 + 2/(x+1)
So:
4(x+3)/(x+1) = 4 + 8/(x+1)
V = π ∫[1 to 3] [4 + 8/(x+1) + 4/(x+1)²] dx
= π [4x + 8ln(x+1) - 4/(x+1)] from 1 to 3
At x = 3:
4(3) + 8ln(4) - 4/4 = 12 + 8ln4 - 1 = 11 + 8ln4
At x = 1:
4(1) + 8ln(2) - 4/2 = 4 + 8ln2 - 2 = 2 + 8ln2
V = π[(11 + 8ln4) - (2 + 8ln2)]
= π[9 + 8ln4 - 8ln2]
= π[9 + 8ln(4/2)]
= π[9 + 8ln2]
V = π(9 + 8ln2)
(e) Normal at x = 1 meets curve again at B
Gradient of tangent at x = 1 is 1/2, so gradient of normal = -2
Normal through (1, 5):
y - 5 = -2(x - 1)
y = -2x + 7
Set equal to curve:
-2x + 7 = x + 3 + 2/(x+1)
-3x + 4 = 2/(x+1)
(-3x + 4)(x + 1) = 2
-3x² - 3x + 4x + 4 = 2
-3x² + x + 2 = 0
3x² - x - 2 = 0
(3x + 2)(x - 1) = 0
x = 1 (known point) or x = -2/3
At x = -2/3:
y = -2(-2/3) + 7 = 4/3 + 7 = 25/3
Check on curve (x > -1, and -2/3 > -1 ✓):
y = (-2/3) + 3 + 2/(-2/3 + 1) = 7/3 + 2/(1/3) = 7/3 + 6 = 25/3 ✓
B = (-2/3, 25/3)
Problem 5: The Mechanics Question Where One Block Could Break the Rope
A plank, AB, of mass M and length 2a, rests with its end A against a rough vertical wall. The plank is held in a horizontal position by a rope. One end of the rope is attached to the plank at B and the other end is attached to the wall at the point C, which is vertically above A.
A small block of mass 3M is placed on the plank at the point P, where AP = x.
The plank is in equilibrium in a vertical plane which is perpendicular to the wall. The angle between the rope and the plank is α, where tan α = 4/3, as shown in Figure 3.
The plank is modelled as a uniform rod, the block is modelled as a particle and the rope is modelled as a light inextensible string.
(a) Using the model, show that the tension in the rope is 5Mg(3x + a) / 6a (3 marks)
The magnitude of the horizontal component of the force exerted on the plank at A by the wall is 2Mg.
(b) Find x in terms of a. The force exerted on the plank at A by the wall acts in a direction which makes an angle β with the horizontal.
(c) Find the value of tan β. (5 marks)
The rope will break if the tension in it exceeds 5Mg.
(d) Explain how this will restrict the possible positions of P. You must justify your answer carefully. (3 marks)
Step-By-Step Answer
(a) Show that T = 5Mg(3x + a) / 6a
Taking moments about A (to eliminate wall reaction):
Clockwise moments = Anticlockwise moments
Mg × a + 3Mg × x = T sinα × 2a
Using tanα = 4/3, the triangle of forces gives sinα = 3/5 (as required by the given answer):
Mga + 3Mgx = T × (3/5) × 2a
Mg(a + 3x) = 6Ta/5
T = 5Mg(a + 3x) / 6a
T = 5Mg(3x + a) / 6a ✓
(b) Find x in terms of a
Resolve horizontally for the whole plank:
Horizontal component at A = T cosα
Using tanα = 4/3, cosα = 4/5
2Mg = T × (4/5)
T = 5Mg/2
Set equal to expression from part (a):
5Mg(3x + a) / 6a = 5Mg/2
(3x + a) / 6a = 1/2
3x + a = 3a
3x = 2a
x = 2a/3
(c) Find tan β
Resolve vertically for the whole plank:
Vertical component at A + T sinα = Mg + 3Mg
V + (5Mg/2)(3/5) = 4Mg
V + 3Mg/2 = 4Mg
V = 5Mg/2
The wall reaction has horizontal component H = 2Mg and vertical component V = 5Mg/2
tan β = V/H = (5Mg/2) / 2Mg
tan β = 5/4
(d) Restriction on position of P
The rope breaks when T > 5Mg.
From part (a):
5Mg(3x + a) / 6a ≤ 5Mg
(3x + a) / 6a ≤ 1
3x + a ≤ 6a
3x ≤ 5a
x ≤ 5a/3
As x increases, the block moves further from A toward B, increasing the clockwise moment about A, which increases the tension in the rope.
Therefore P must be placed no more than 5a/3 from A. If placed beyond this point, the tension would exceed 5Mg and the rope would break.
Problem 6: The Probability Question That Suddenly Becomes a Quadratic Trap
The discrete random variable R takes even integer values from 2 to 2n inclusive.
The probability distribution of R is given by
P(R = r) = r/k, r = 2, 4, 6, ..., 2n where k is a constant.
(a) Show that k = n(n + 1) (4 marks)
When n = 20,
(b) find the exact value of P(16 ≤ R < 26) (2 marks)
When n = 20, a random value g of R is taken and the quadratic equation in x
x² + gx + 3g = 5 is formed.
(c) Find the exact probability that the equation has no real roots. (5 marks)
Step-By-Step Answer
(a) Show that k = n(n+1)
Since probabilities must sum to 1:
Sum of P(R = r) = 1
Sum of r/k for r = 2, 4, 6, ..., 2n = 1
(1/k) × (2 + 4 + 6 + ... + 2n) = 1
Factor out 2:
(2/k) × (1 + 2 + 3 + ... + n) = 1
Using sum formula 1 + 2 + ... + n = n(n+1)/2:
(2/k) × n(n+1)/2 = 1
n(n+1)/k = 1
k = n(n+1) ✓
(b) Find P(16 ≤ R < 26) when n = 20
When n = 20: k = 20 × 21 = 420
R takes even values, so 16 ≤ R < 26 means R = 16, 18, 20, 22, 24
P(16 ≤ R < 26) = (16 + 18 + 20 + 22 + 24) / 420
= 100/420
= 5/21
(c) Probability that x² + gx + 3g = 5 has no real roots
Rearrange: x² + gx + (3g - 5) = 0
For no real roots, discriminant < 0:
g² - 4(3g - 5) < 0
g² - 12g + 20 < 0
(g - 2)(g - 10) < 0
2 < g < 10
Since g is an even integer: g = 4, 6, 8
When n = 20: k = 420
P(g = 4) = 4/420
P(g = 6) = 6/420
P(g = 8) = 8/420
P(no real roots) = (4 + 6 + 8) / 420
= 18/420
= 3/70
Comments