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

a.) d = 9

b.) U

_{100}= 23 + 99(9)
U

_{100}= 914
c.) S

_{n}= 0.5n[2a + (n-1)d]
S

_{280}= 140[2(23) + 279(9)]
S

_{280}= 357,980
a.) Area = 0.5 × 26 × 31.5 × 5/13

Area = 157.5cm

^{2}
b.) cos(sin

^{-1}(5/13))
12/13

c.) AC

^{2}= 26^{2}+ 31.5^{2}- (2 × 26 × 31.5 × 12/13)
AC

^{2}= 156.25
AC = 12.5cm

a.) x

^{3}- 2x^{3/2}+ 1
b.) 0.25x

^{4}- 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.) U

_{1}= 12
U

_{2}= 3
b.) r = 0.25

c.) S

_{}∞ = a/(1-r)
S

_{}∞ = 12/(1-0.25)
S

_{}∞ = 16
d.) U

_{4}= 48(0.25)^{4}= 0.1875
S

_{}∞ = 0.1875/(1-0.25)
S

_{}∞ = 0.25
a.) rθ = 18(2π/3)

rθ = 12π

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 × 18

^{2}× 2π/3 = 108π
Shaded area = 324√3 - 108π = 221.8924551m

^{2}
Shaded area = 222m

^{2}to 3 significant figures
a.) i.) f'(2) = 3(2)

^{2}- 4/2^{2}- 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 = x

^{3}+ 4x^{-1}-11x + c
1 = 2

^{3}+ 4(2)^{-1}-11(2) + c
1 = 8 + 2 - 22 + c

c = 13, so the equation is:

y = x

^{3}+ 4x^{-1}-11x + 13
a.) tanθ + 1 = 0 or sin

^{2}θ - 3cos^{2}θ = 0
tanθ = -1

sin

^{2}θ = 3cos^{2}θ, divide through by cos^{2}θ to get:
tan

^{2}θ = 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 = a

^{2}- 12
a

^{2}- 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

c.) i.) log

_{10}10 + log_{10}x^{2}
log

_{10}10 + 2log_{10}x
1 + 2log

_{10}x
ii.) 1 + 2log

_{10}x = log_{10}(10x^{2}) = log_{10}(√10x)^{2}= 2log_{10}(√10x)
Stretch parallel to the x-axis scale factor of 1/√10

iii.) 1 + 2log

_{10}x = log_{10}(x^{2}+ 1)
log

_{10}(10x^{2}) = log_{10}(x^{2}+ 1)
10x

^{2}= x^{2}+ 1
x

^{2}= 1/9
x = 1/3 as x > 0

m

_{OP}= log_{10}(10(1/3)^{2}) ÷ 1/3
m

_{OP}= 3log_{10}(10/9)
m

_{OP}= log_{10}(1000/729)
Thanks for posting the link to the paper. I was wondering, do you have any of the other AQA papers from May/June 2012? They are not easy to find!!!

ReplyDeleteYeah they try to keep them for themselves for six months or so and then sell them to schools!

Deletei have all paper please email me at g.milton@live.co.uk to get them

ReplyDeletekind regardssssssssssssssssssssssssssssssssss