©
2009 Mathematics Advanced Higher Finalised Marking Instructions
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Solutions for AH maths 2009
1.
f (x) = (x + 1) (x − 2)3
(a)
f ′ (x) = (x − 2)3 + 3 (x + 1) (x − 2)2
1
= (x − 2)2 ((x − 2) + 3 (x + 1))
(b)
= (x − 2)2 (4x + 1)
1
1 = 0 when x = 2 and when x = − . 4
1
Method 1 x2 + x = y − 5 ⇒ x2 + xy = y2 − 5y y dy dy dy + y = 2y − 5 dx dx dx dy dy dy − 1 = −2 − 5 6 + 3 dx dx dx dy dy −1 ⇒ = 5 = −10 dx dx 2 dy Note: a candidate may obtain dx = 2y2x− +x −y 5 and then substitute. 2x + x
2E1
1
x2 + x = y − 5 y
Method 2
dy 2xy − x2 dx dy + 1 = dx y2 dy −6 − 9 dx + 1 = 1 dy + 1 = −6 − 9 dx dy 1 = − dx 2
Mathod 3
1
2E1
dy dx dy dx 2E1
()
x2 1 + x = y − 5 ⇒ x2 + x = y − 5 y y 2x
( )
1 1 dy dy + x2 − 2 + 1 = y y dx dx
dy dy dy 1 + 1 = ⇒ = − dx dx dx 2 dy 2xy + y2 Note: a candidate may obtain dx = y2 + x2 (in 2 and 3) and then substitute. −6 − 9
2E1 2E1
2.
(a)
det
(
t + 4 3t 3 5
)
= 5 (t + 4) − 9t = 20 − 4t
−1
A
(
1 5 −3t 20 − 4t −3 t + 4
=
1
)
20 − 4t = 0 ⇒ t = 5
(b)
(
(c)
t + 4 3 3t 5
) ( ) 6 3 6 5
=
⇒ t = 2
1,1
1
1
dy = 1 dx dy = x−2 ey dx
3.
eyx2
∫ e dy y
=
∫x
−2
dx
1
ey = −x−1 + c
1
y = 0 when x = 1 so 1 = −1 + c ⇒ c = 2 ey = 2 −
4.
When n = 1, LHS = n = 1.
(
1 1 ⇒ y = ln 2 − x x
)
1 1 1 1 = , RHS = 1 − = . So true when 1 × 2 2 2 2
1 1
1
k
1 1 = 1 − Assume true for n = k , ∑ . k + 1 r = 1 r (r + 1) Consider n = k + 1 k+1
1 ∑ r (r + 1) = r=1
k
∑ r (r
r=1
1 1 + + 1) (k + 1) (k + 2)
= 1 −
1 1 + k + 1 (k + 1) (k + 2)
= 1 −
k + 2 − 1 k + 1 = 1 − (k + 1) ((k + 1) + 1) (k + 1) (k + 2)
1 ((k + 1) + 1) Thus, if true for n = k , statement is true for n = k + 1, and, since true for n = 1, true for all n ≥ 1. = 1 −
1
1
1 1
5.
Method 1 ex + e−x ∫ln 23 ex − e−x dx Let u = ex − e−x, then du = (ex + e−x) dx. When x = ln 23 , u = 23 − 23 = 65 and when x = ln 2, u = 2 − ln 2
ln 2
∫ln
3 2
ex + e−x dx = ex − e−x
3/2
∫5/6
1 2
1 1
du u
= [ ln u] 3/2 5/6 = ln
=
3 2.
1
3 5 9 − ln = ln 2 6 5
1
Method 2 ln 2
∫ln
6.
3 2
ex + e−x x −x ln 2 ( )] ln 23 dx = [ ln e − e ex − e−x 1 3 2 − = ln 2 − − ln 2 2 3 3 5 9 = ln − ln = ln 2 6 5
(
(
)
1,1
)
(1 + 2i)2 1 + 4i − 4 = 7 − i 7 − i −3 + 4i 7 + i × = 7 − i 7 + i (−3 + 4i) (7 + i) = 50 1 1 = − + i 2 2
1 1
1 1
1
1
|z| =
arg z = tan−1
1 2 − 12
1 1 1 + = 2 4 4 2
= tan−1 (−1) =
3π (or 135°) . 4
1
1
x = 2 sin θ ⇒ dx = 2 cos θ dθ
7.
∫0
2
x2 dx = 4 − x2 =
∫0
π/4
∫0
= 2
4 sin 2 θ (2 cos θ) dθ 2 cos θ
π/4
π/4
∫ 0 (1
− cos 2θ ) dθ
{
(b)
1 1
}
1
(1 + x)5 = 1 + 5x + 10x2 + 10x3 + 5x4 + x5
(a)
1
2 ∫0 (2 sin θ ) dθ
π/4 1 = 2 θ − sin 2θ 0 2 π 1 = 2 − − 0 4 2 π − 1 = 2
8.
1
4 sin 2 θ (2 cos θ) dθ 4 − 4 sin 2 θ
π/4
= 2
1 π ⇒ θ = 4 2
2 ⇒ sin θ =
x = 0 ⇒ θ = 0; x =
1
Let x = −0·1, then
1
1
0·95 = (1 + (−0·1))5 = 1 − 0·5 + 0·1 − 0·01 + 0·0005 − 0·00001
1
= 0·5 + 0·09 + 0·00049 = 0·59049
9.
1
∫0
1
x tan−1 x2 dx = tan−1 x2 ∫ x dx − 0 1 1 = x2 tan−1 x2 − 2 0
1
∫0
2x x2 dx 1 + x4 2
1,1
x3 ∫0 1 + x4 dx 1
1 1 1 1 = x2 tan−1 x2 − ln (1 + x4) 2 0 0 4 1 1 1 tan−1 1 − 0 − ln 2 − ln 1 = 4 2 4 π 1 = − ln 2 8 4
1 1 1,1
10.
14654 = 11 × 1326 + 68
1
1326 = 19 × 68 + 34 68 = 2 × 34
1
34 = 1326 − 19 × 68
1
= 1326 − 19 (14654 − 11 × 1326) = 210 × 1326 − 19 × 14654
11.
1
When x = 1, y = 1.
1 y = x
2x2 + 1
⇒ ln y = ln (x2x + 1) = (2x2 + 1) ln x 1 dy 2x2 + 1 = + 4x ln x y dx x Hence, when x = 1, y = 1 and dy = 3 + 0 = 3. dx 2
12.
aj = pj ⇒ Sk = p + p2 + … Sn = S2n
+ pk =
1
1,1
1
p (pk − 1) p − 1
p (pn − 1) p − 1
1
p (p2n − 1) = p − 1
p (p2n − 1) p − 1 n (p + 1) (pn − 1) pn + 1 ⇒ pn
1
65p (pn − 1) p − 1 = 65 (pn − 1) = 65 = 64 =
1 1
a2 = p2 ⇒ a3 = p3 but a3 = 2p so p3 = 2p ⇒ p2 = 2 ⇒ p =
2 since p > 0.
pn = 64 = 26 = n = 12
1
( 2)
12
1
13.
x2 + 2x x2 + 2x = (x − 1) (x + 1) x2 − 1 Hence there are vertical asymptotes at x = −1 and x = 1. f (x) =
f (x) =
1 + x + 2x = x2 − 1 1 − 2
2x x2 1 x2
=
1 + 1 −
1
2 x 1 x2
1
→ 1 as x → ∞. So y = 1 is a horizontal asymptote. f (x) = f ′ (x) = =
1
x + 2x x2 − 1 2
(2x + 2) (x2 − 1) − (x2 + 2x) 2x (x2 − 1)2
1
2x3 − 2x + 2x2 − 2 − 2x3 − 4x2 −2 (x2 + x + 1) = (x2 − 1)2 (x2 − 1)2
1
−2 ((x + 12 )2 + 34 ) < 0 (x2 − 1)2 Hence f (x) is a strictly decreasing function. =
x2 + 2x = 0 ⇒ x = 0 or x = −2 x2 − 1
1
x2 + 2x 1 = 1 ⇒ x2 + 2x = x2 − 1 ⇒ x = − 2 x − 1 2
1
f (x) = f (x) =
1
1 for shape 1 for position
Alternatively for the horizontal asymptote: 1 x2
− 1
x2 x2
+ 2x
− 1 2x + 1
⇒ f (x) = 1 +
2x + 1 → 1 as x → ∞ x2 − 1
x2 + 6x − 4 A B C + = + 2 2 (x + 2) (x − 4) (x + 2) x + 2 x − 4
14.
Let x Let x Let x −4 = Thus
x2 + 6x − 4 = A (x − 4) + B (x + = −2 then 4 − 12 − 4 = −6A ⇒ A = = 4 then 16 + 24 − 4 = 36C ⇒ C = = 0 then −4A − 8B + 4C ⇒ −4 = −8 − 8B +
M1
2) (x − 4) + C (x + 2)2 2. 1.
1 1
4 ⇒ B = 0.
1
x2 + 6x − 4 2 1 . = + 2 2 (x + 2) (x − 4) (x + 2) x − 4 Let f (x) = 2 (x + 2)−2 + (x − 4)−1 then f (x) = 2 (x + 2)−2 + (x − 4)−1 f ′ (x) = −4 (x + 2)−3 − (x − 4)−2 f ″ (x) = 12 (x + 2)−4 + 2 (x − 4)−3
⇒ ⇒ ⇒
f (0) = 12 − 14 = 14 f ′ (0) = − 12 − 161 = − 169 f ″ (0) = 34 − 321 = 23 32
1 1 1
Thus x2 + 6x − 4 1 9x 23x2 − + +… = (x + 2)2 (x − 4) 4 16 64
2E1
15.
dy − 3y = (x + 1)4 dx dy 3 − y = (x + 1)3 dx x + 1
(x
(a)
+ 1)
1
Integrating factor: −3 dx = −3 ln (x + 1) . + 1 Hence the integrating factor is (x + 1)−3. since
∫x
1 dy 3 − y = 1 3 (x + 1) dx (x + 1)4
1 1 1
d ((x + 1)−3 y) = 1 dx y = ∫ 1 dx (x + 1)3
1
= x + c y = 16 when x = 1, so 2 = 1 + c ⇒ c = 1. Hence y = (x + 1)4
1
(x + 1)4 = (1 − x)4
(b)
x + 1 = 1 − x ⇒ x = 0
1
or x + 1 = −1 + x which has no solutions.
y = f (x)
y = (1 − x)4
Area =
0
∫−1
= 2 =
(x + 1)4 dx + 0
∫−1 (x +
1
∫0 (1
− x)4 dx
1)4 dx
0 2 [ (x + 1)5] −1 = 2 − 0 = 2 5 5 5
M1 1 1
16.
x + y − z = 6 2x − 3y + 2z = 2 −5x + 2y − 4z = 1
(a)
1 2 −5
1 −3 2
−1 2 −4
1 6 2 ⇒ 0 0 1
1 −5 7
−1 4 −9
1 6 −10 ⇒ 0 0 31
1 −5 0
−1 4 − 175
6 −10 17 1,1,1
z = 17 ÷
( −175) = −5
1
−5y − 20 = −10 ⇒ y = −2 x − 2 + 5 = 6 ⇒ x = 3 (b)
Let x = λ. Method 1 In first plane: x + y − z = 6. λ + (4λ − 14) − (5λ − 20) = 5λ − 5λ + 6 = 6. In the second plane: 2x − 3y + 2z = 2λ − 3(4λ − 14) + 2(5λ − 20) = 5λ − 5λ + 2 = 2. Method 2 y − z = 6 − λ ⇒ y = 6 + z − λ −3y + 2z = 2 − 2λ (−18 − 3z + 3λ) + 2z = 2 − 2λ −z = 20 − 5λ ⇒ z = 5λ − 20 and y = 4λ − 14 Method 2 x + y − z = 2x − 3y + 2z = 5x − z = 4x − y =
6 2 20 14
(1) (2) (2) + 3 (1) (2) + 2 (1)
1
1 1
1 1
1
y = 4x − 14 z = 5x − 20 x = λ, y = 4λ − 14, z = 5λ − 20 (c)
Direction of L is i + 4j + 5k, direction of normal to the plane is −5i + 2j − 4k. Letting θ be the angle between these then −5 + 8 − 20 cos θ = 42 45 −17 = 3 210 This gives a value of 113.0° which leads to the angle 113.0° − 90° = 23.0°.
1
1M,1
1,1