## Thursday, 7 June 2012

### 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.

a.) d = 9
b.) U100 = 23 + 99(9)
U100 = 914
c.) Sn = 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 - 2x3/2 + 1
b.) 0.25x4 - 0.8x5/2 + 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) = 183
Area of two triangles = 2 × 0.5 × 18 × 183 = 3243
Area of sector = 0.5 × 182 × 2π/3 = 108π
Shaded area = 3243 - 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.) 7x = 72x - 12
Let a = 7x
a = a2 - 12
a2 - a - 12 = 0
(a - 4)(a + 3) = 0
7x = 4 or 7x = -3; this is illogical so they meet at y = 4
ii.) 7x = 4
log7x = 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
mOP = log10(10(1/3)2÷ 1/3
mOP = 3log10(10/9)
mOP = log10(1000/729)