2005 paper 2 solution

Higher Maths 2005
Paper 2 Solutions
1.
3
4x
−1
x
2
x
=
=
4x
=
2
= 2x
4x
2
x
1
−
x
−2
( 4x
− x )
2
2
3
−1
2
x
− + c
−1
1
+ + c
x
dx
dx
2. (a) sin(p + q) = sinpcosq + cospsinq (b) cos(p + q) = cospcosq – sinpsinq
=
15
17
4 8 3
⋅ + ⋅
=
5 17 5
8
17
4 15 3
⋅ − ⋅
5 17 5
=
60 24
32 45
+ = −
85 85
85 85
84
= 85
=
13
−
85
(c) tan(p + q) =
=
=
sin( p + q)
cos( p + q)
84
85
13
−
85
84 85 × −
85 13
=
84
−
13
Solutions by MAitchison www.mathsroom.co.uk

3. (a) m AB =
( y
( x
2
2
− y1)
− x )
1
midpoint = (3,2)
(4 − 0)
=
(5 −1)
4
= 4
y – b = m(x – a)
y – 2 = –1(x – 3)
= 1 y – 2 = –x + 3
m chord = –1 as m 1 .m 2 = –1 for lines x + y = 5
(b) m T =
1
− (c) Perpendicular bisector of AB (answer (a))
3
m r = 3 and radius CA will intersect at centre, C.
y – b = m(x – a) x + (3x – 3) = 5
y – 0 = 3(x – 1) 4x = 8
y = 3x – 3 x = 2
y = 3
C(2,3)
r =
r =
2
( xC
− xA
) + ( yC
− y A
2
( 2 − 1) + (3 − 0)
2
)
2
r =
2
( 1) + (3)
2
r = 10
Equation of circle: (x – 2) 2 + (y – 3) 2 = 10
Solutions by MAitchison www.mathsroom.co.uk

4. (a)
TA =
− 5
15
1
− 40
TB = 15
(b) TA =
2
2
( − 5) + (15) + (1)
= 251
2
2
TB =
2
( − 40) + (15) + (2)
2
2
= 1829
cosA =
cosA =
a ⋅b
a b
cosA = 0.631
A = 50.9°
( −5)(
−40)
+ (15)(15) + (1)(2)
251. 1829
5. For points of intersection y = y
2x 2 – 9 = x 2 2 2
A = ( x − (2x
− 9)) dx
−
x 2 – 9 = 0 = ( 9 − x dx
−
3
3
3
3
2 )
(x + 3)(x – 3) = 0 =
3
9x
− x
3
3
− 3
x = –3, 3 =
(3)
9(3) −
3
= (27 – 9) – (–27 + 9)
= 18 – (–18)
= 36 units 2
3
( −3)
− 9( −3)
−
3
3
Solutions by MAitchison www.mathsroom.co.uk

6. y =
24
x
2
y = 24x −1
y =
when x = 4
dy =
2
−12x −3
y = 12
dx
dy −12
=
dx
3
( x)
24
(4)
dy −12 =
dx
3
(4)
dy 12 = −
dx 8
dy 3 = −
dx 2
P(4,12) m =
y – b = m(x – a)
y – 12 =
3
− (x – 4)
2
2y – 24 = –3x + 12
3x + 2y = 36
3
−
2
7. log 4 (5 – x) – log 4 (3 – x) = 2
(5 − x)
log 4
(3 − x)
4 2 =
16 =
= 2
(5 − x)
(3 − x)
(5 − x)
(3 − x)
16(3 – x) = 5 – x
48 – 16x = 5 – x
43 = 15x
43 = x
15
Solutions by MAitchison www.mathsroom.co.uk

8. For points of intersection y = y
ksin2x = sinx
k × 2sinxcosx = sinx
2ksinxcosx – sinx = 0
sinx(2kcosx – 1) = 0
sinx = 0 or 2kcosx – 1 = 0
x = 0, , 2 2kcosx = 1
cosx =
1
2k
From diagram, x = 0, , 2 correspond to points O, B and D respectively.
Hence, at A and C, cosx =
1 .
2k
9. (a) At launch, t = 0 (b) 20 = 252e –0.06335t
V = £252 million
20
252
= e –0.06335t
20
.06335
log e = loge ( )
252
e−0 t
20
log e = –0.06335t
252
20
loge
252
t =
− 0.06335
t = 40 years
Solutions by MAitchison www.mathsroom.co.uk

10. a(a + b + c)
= aa + ab + ac
= a 2 + abcos90° + accos60°
= (3) 2 + 0 + (3)(3)cos60°
= 9 + 9. 2
1
= 13.5
11. (a)
–1 1 p p 1
–1 1 – p –1
1 p – 1 1 0
x = –1 is a solution
(b) Quotient: x 2 – (p – 1)x + 1
For real roots b 2 – 4ac 0
y
(p – 1) 2 – 4(1)(1) 0
p 2 – 2p + 1 – 4 0
p 2 – 2p – 3 0
(p + 1)(p – 3) 0
O
–1 3
x
p –1 and p 3 for all roots of the cubic function to be real.
Solutions by MAitchison www.mathsroom.co.uk