These are real past-paper questions. Think of your answer first, then open the explanation.
May/June 2021 · Paper 1 · Series
The sum of the first 20 terms of an arithmetic progression is 405 and the sum of the first 40 terms is 1410.
Find the 60th term of the progression. [5]
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Using Sₙ = n/2[2a + (n-1)d]:
S₂₀ = 10(2a + 19d) = 405 → 2a + 19d = 40.5
S₄₀ = 20(2a + 39d) = 1410 → 2a + 39d = 70.5
Subtracting: 20d = 30, so d = 1.5
Then 2a + 19(1.5) = 40.5 → 2a = 12 → a = 6
60th term = a + 59d = 6 + 59(1.5) = 6 + 88.5 = 94.5
Set up two equations from the sum formula for n=20 and n=40, solve simultaneously for a and d, then use a+59d for the 60th term.
May/June 2021 · Paper 1 · Trigonometry
(a) Prove the identity (1 - 2sin²θ)/(1 - sin²θ) ≡ 1 - tan²θ. [2]
(b) Hence solve the equation (1 - 2sin²θ)/(1 - sin²θ) = 2tan⁴θ for 0° ≤ θ ≤ 180°. [3]
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(a) Since 1 - sin²θ = cos²θ:
(1 - 2sin²θ)/(1 - sin²θ) = (1 - 2sin²θ)/cos²θ = (cos²θ - sin²θ)/cos²θ [since cos²θ = 1 - sin²θ, so cos²θ - sin²θ = 1 - 2sin²θ]
= cos²θ/cos²θ - sin²θ/cos²θ
= 1 - tan²θ ✓
(b) Using the identity from (a):
1 - tan²θ = 2tan⁴θ
2tan⁴θ + tan²θ - 1 = 0
Let u = tan²θ: 2u² + u - 1 = 0 → (2u - 1)(u + 1) = 0
u = 0.5 or u = -1 (rejected, since tan²θ ≥ 0)
tan²θ = 0.5 → tanθ = ±0.7071
θ = 35.3° or θ = 180° - 35.3° = 144.7°
Rewrite 1-sin²θ as cos²θ to prove the identity, then substitute it into the equation to form and solve a quadratic in tan²θ.
May/June 2021 · Paper 1 · Functions
Functions f and g are defined as follows:
f(x) = (x - 2)² - 4 for x ≥ 2,
g(x) = ax + 2 for x ∈ ℝ,
where a is a constant.
(a) State the range of f. [1]
(b) Find f⁻¹(x). [2]
(c) Given that a = -5/3, solve the equation f(x) = g(x). [3]
(d) Given instead that gg(f⁻¹(12)) = 62, find the possible values of a. [5]
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(a) Since x ≥ 2, (x-2)² ≥ 0, so the range of f is f(x) ≥ -4.
(b) y = (x - 2)² - 4
y + 4 = (x - 2)²
x - 2 = √(y + 4) (taking the positive root since x ≥ 2)
x = 2 + √(y + 4)
f⁻¹(x) = 2 + √(x + 4)
(c) With a = -5/3:
(x - 2)² - 4 = -5/3 x + 2
x² - 4x = -5/3 x + 2
x² - 4x + 5/3 x - 2 = 0
x² - 7/3 x - 2 = 0
Multiply by 3: 3x² - 7x - 6 = 0
(3x + 2)(x - 3) = 0
x = -2/3 or x = 3
Since f requires x ≥ 2, only x = 3 is valid.
(d) f⁻¹(12) = 2 + √(12 + 4) = 2 + 4 = 6
g(f⁻¹(12)) = g(6) = 6a + 2
gg(f⁻¹(12)) = g(6a + 2) = a(6a + 2) + 2 = 6a² + 2a + 2
Set equal to 62:
6a² + 2a + 2 = 62
6a² + 2a - 60 = 0
3a² + a - 30 = 0
a = [-1 ± √(1 + 360)]/6 = [-1 ± 19]/6
a = 3 or a = -10/3
Rearrange y=(x-2)²-4 to isolate x for part (b)'s inverse; for later parts substitute into f(x)=g(x), or chain g(g(f⁻¹(x))) step by step and solve the resulting equation in a.
May/June 2021 · Paper 1 · Coordinate geometry
The equation of a circle is x² + y² - 4x + 6y - 77 = 0.
(a) Find the x-coordinates of the points A and B where the circle intersects the x-axis. [2]
(b) Find the point of intersection of the tangents to the circle at A and B. [6]
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(a) Set y = 0:
x² - 4x - 77 = 0
(x + 7)(x - 11) = 0
x = -7 or x = 11
So A = (-7, 0) and B = (11, 0).
(b) Completing the square: (x-2)² + (y+3)² = 90, so the centre is C = (2, -3).
Gradient of CA = (0-(-3))/(-7-2) = 3/(-9) = -1/3
Gradient of tangent at A (perpendicular to CA) = 3
Tangent at A: y - 0 = 3(x + 7) → y = 3x + 21
Gradient of CB = (0-(-3))/(11-2) = 3/9 = 1/3
Gradient of tangent at B (perpendicular to CB) = -3
Tangent at B: y - 0 = -3(x - 11) → y = -3x + 33
Setting the tangents equal:
3x + 21 = -3x + 33
6x = 12
x = 2
y = 3(2) + 21 = 27
Point of intersection: (2, 27)
Find A and B by setting y=0, locate the circle's centre by completing the square, use perpendicular gradients to write both tangent equations, then solve them simultaneously.
