Paper 1, Section A 1. A 2. C 3. D 4. A 5. B 6. D 7. C 8. B 9. C 10. B 11. D 12. A 13. B 14. C 15. C 16. A 17. B 18. B 19. C 20. A
For Q20: In the graph, the x coordinate has been halved because of the 2x (it has not been doubled) and the y co-ordinate has had 3 subtracted from it. We want the original equation (f(x)) before the x coordinate was halved and the y coordinate had 3 subtracted from it. So we have to do the opposite to get back to (f(x)) which is to double the x coordinate (6x2=12) and add 3 onto the y coordinate (4+3=7) therefore the answer must be A (12, 7) as stated above.
Paper 1, Section B 21)a) 5y=-2x+72 b)5(12)=-2(6)+72 60=-12+72 60=60 (Proof) c)2:1 22)a)i)f(1)=0 ii)(x-1)(x-1)(2x+5) c) (1,-1) d)(-5/2,-8) 23)a)i)m=v/h (use triangle) therefore tan a=3/2 ii) sin a=3/√13 b)sin b=3/5 cos b=4/5 c)i)6/5√13 ii)-6/5√13
Ukdragon37
Higher Maths Paper 2
2010
3ai)
1a)
M 0,1, 0 N 4, 2, 2
y 3 x
x 2 y 2 14 x 4 y 19 0
b)
x 2 3 x 14 x 4 3 x 19 0
0 0 0 VM m v 1 2 1 0 3 3 4 0 4 VN n v 2 2 0 2 3 1
x 2 9 6 x x 2 14 x 12 4 x 19
c)
ii)
2
cos MVN
2 x2 4x 2 2 x 1
2
Exactly one root so tangent (alternatively show determinant 0)
x 1 y 3 x 3 1 4
3 VM VN VM VN 10 17
MVN 76.7
P 1, 4
2a)
b)
12 cos x 5sin x k cos x a
Let centre of large circle A
k cos x cos a k sin x sin a
A 7, 2
k 122 52 13
6 AP 6
12 k cos a 13cos a 5 k sin a 13sin a tan a
PC
5 a 22.62 12
2 1 AP c p PC C 1, 6 3 2
bi) Max = 13, Min = ‐13
Radius of large circle AP 62 62 6 2
ii) 12 cos x 5sin x 13cos x 22.62
Radius of small circle 2 2
Max when cos x 22.62 1, 22.62 x 22.62 382.62
x 22.62 360
x 1 y 6 2
x 337.38
Min when cos x 22.62 1, 22.62 x 22.62 382.62
x 22.62 180 x 157.38
1
2
8
{Date In}
Ukdragon37 4)
b)
2 cos 2 x 5cos x 4 0, 0 x 2
For stationary points
2 2 cos x 1 5cos x 4 0
A ' x 12 6 x 2 0
4 cos 2 x 5cos x 6 0
x2 2
4 cos x 3 cos x 2 0
x 2 but area cannot be negative so x 2
2
A
cos x 2 0
2 12
cos x 2 No solutions 2 A' x
4 cos x 3 0 3 4 x 2.42, 3.86
cos x
2 2
2
2
2
0
/ So maximum
\
3
6a)
5ai)
1
y 2x 9 2
Qy 10 x 2
1 dy 1 1 2 x 9 2 2 dx 2 2x 9 dy 1 1 m and y 3 at x 9, dx 9 3
2 10 0 4 5 PQ Qy Py 6 x 2 Py
ii)
A x PQRS 2TP PQ
2 x 6 x 2 12 x 2 x3
y b m x a
y 3
1 1 x 9 x 3 3 3
1 y x 3
b) 1
y 2x 9 2 0
2x 9 0
2x 9
x 4.5
4.5, 0
2
12 2 4 2 8 2
Ukdragon37 6c)
From O to A the area under the straight line 1 1 27 is a triangle with area 4.5 4.5 2 3 8 For the rest of the area: 9
1 3 1 1 2 1 4.5 3 x 2 x 9 2 dx 6 x 3 2 x 9 2 4.5 9
81 27 81 9 0 6 3 24 8
Total area is
27 9 36 18 9 8 8 8 4 2
7a)
log 4 x P 4P x P
12 16 x P
16 2 x log16 x
P 2
b)
log 3 x log 9 x 12 2 log 9 x log 9 x 12 3log 9 x 12
log 9 x 4 x 94
3