AQA C2 2012 Unofficial Worked Answers

A lot of people found this paper very difficult as some of the questions required more thought than usual. I didn't actually sit this paper, so I couldn't wait to get the paper to see why people found it so difficult. I managed to get hold of the paper so I thought I would share my answers.

You may wish to first download the June C2 2012 paper.

a.) d = 9

b.) U₁₀₀ = 23 + 99(9)

U₁₀₀ = 914

c.) S_n = 0.5n[2a + (n-1)d]

S₂₈₀ = 140[2(23) + 279(9)]

S₂₈₀ = 357,980

a.) Area = 0.5 × 26 × 31.5 × 5/13

Area = 157.5cm²

b.) cos(sin^-1(5/13))

12/13

c.) AC² = 26² + 31.5² - (2 × 26 × 31.5 × 12/13)

AC² = 156.25

AC = 12.5cm

a.) x³ - 2x^3/2 + 1

b.) 0.25x⁴ - 0.8x^5/2 + x + c

c.) (0.25(4)⁴ - 0.8(4)^5/2 + 4) - (0.25(1)⁴ - 0.8(1)^5/2 + (1)) 

42.4 - 0.45

41.95

a.) U₁ = 12

U₂ = 3

b.) r = 0.25

c.) S∞ = a/(1-r)

S∞ = 12/(1-0.25)

S∞ = 16

d.) U₄ = 48(0.25)⁴ = 0.1875

S∞ = 0.1875/(1-0.25)

S∞ = 0.25

a.) rθ = 18(2π/3)
rθ = 12π

b.) i.) α = 2(π - π/3 - π/2)

α = π/3

ii.) PT = QT = 18tan(π/3) = 18√3

Area of two triangles = 2 × 0.5 × 18 × 18√3 = 324√3

Area of sector = 0.5 × 18² × 2π/3 = 108π

Shaded area = 324√3 - 108π = 221.8924551m²

Shaded area = 222m² to 3 significant figures

a.) i.) f'(2) = 3(2)² - 4/2² - 11

12 - 1 - 11 = 0

ii.) f''(x) = 6x + 8x^-3

f''(2) = 6(2) + 8(2)^-3

f''(2) = 13

iii.) f''(x) > 0, so it is a minimum point

b.) y = x³ + 4x^-1 -11x + c

1 = 2³ + 4(2)^-1 -11(2) + c

1 = 8 + 2 - 22 + c

c = 13, so the equation is:

y = x³ + 4x^-1 -11x + 13

a.) tanθ + 1 = 0 or sin²θ - 3cos²θ = 0

 tanθ = -1

sin²θ = 3cos²θ, divide through by cos²θ to get:

tan²θ = 3

tanθ = ±√3

tanθ = -√3, -1, √3

b.) θ = -45°, 135°, -60°, 120°, 60°

θ = 60°, 120°, 135°

a.) It must go through (0,1)

b.) i.) 7^x = 7^2x - 12

Let a = 7^x

a = a² - 12

a² - a - 12 = 0

(a - 4)(a + 3) = 0

7^x = 4 or 7^x = -3; this is illogical so they meet at y = 4

ii.) 7^x = 4

log7^x = log4

xlog7 = log4

x = log4/log7

x = 0.712 to three significant figures

a.) h = 0.25

x = 0, y = 0; x = 0.25, y = 0.0263; x = 0.5, y = 0.0969; x = 0.75, y = 0.1938; x = 1, y = 0.3010

0.25 × 0.5[(0 + 0.3010) + 2(0.0263 + 0.0969 + 0.1938)]

= 0.117 to 3 significant figures

b.) Translated by: 

c.) i.) log₁₀10 + log₁₀x²

log₁₀10 + 2log₁₀x

1 + 2log₁₀x

ii.) 1 + 2log₁₀x = log₁₀(10x²) = log₁₀(√10x)² = 2log₁₀(√10x)

Stretch parallel to the x-axis scale factor of 1/√10

iii.) 1 + 2log₁₀x = log₁₀(x² + 1)

log₁₀(10x²) = log₁₀(x² + 1)

10x² = x² + 1

x² = 1/9

x = 1/3 as x > 0

m_OP = log₁₀(10(1/3)²) ÷ 1/3

m_OP = 3log₁₀(10/9)

m_OP = log₁₀(1000/729)