mi_AH_Mathematics_all_2012

©

2012 Mathematics Advanced Higher Finalised Marking Instructions

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These Marking Instructions have been prepared by Examination Teams for use by SQA Appointed Markers when marking External Course Assessments. This publication must not be reproduced for commercial or trade purposes.

Advanced Higher Mathematics 2012 Marks awarded for

1.

(3,4)

f (x) =

(a)

3x + 1 x2 + 1 1M for quotient rule (or product) 1 for two correct terms 1 for third correct term

3 (x2 + 1) − (3x + 1) 2x f ′ (x) = (x2 + 1)2

(b)

3x2 + 3 − 6x2 − 2x = (x2 + 1)2 −3x2 − 2x + 3 = (x2 + 1)2 g (x) = cos2 x etan x

g′(x) = 2 cosx(− sin x)e

tan x

+ (cos x)(sec x) e 2

2

tan x

1M 1 1 1

product rule first correct term second correct term

= − sin 2x etan x + etan x simplification tan x = (1 − sin 2x) e (b) alternative g (x) = cos2 x exp (tan x) ln (g (x)) = ln (cos2 x) + tan x 1M = 2 ln (cos x) + tan x Differentiating g′ (x) (− sin x) = 2 + sec 2 x 1 g (x) cos x 1 − 2 sin x cos x g′ (x) = cos2 x exp (tan x) 1 2 cos x = (1 − sin 2x) tan x 1

(

2.

(5)

)

a = 2048 and ar3 = 256 ⇒ r3 = 18 ⇒ r = 12 . a (1 − rn) Sn = 1 − r 1 n 1 − (2) 4088 ⇒ = 1 1 − 2 2048 511 = 256 n 1 511 1 511 ⇒ 1 − = × = 2 256 2 512 1 511 1 = 1 − = 2n 512 512 n ⇒ 2 = 512 ⇒ n = 9

1M valid approach 1

correct answer only, 2 marks

1M for sum formula

()

1 1

any valid method

Marks awarded for

3.

(6)

4.

(5)

5.

(5)



Since w is a root, w = −1 − 2i is also a root.

The corresponding factors are (z + 1 − 2i) and (z + 1 + 2i) from which ((z + 1) − 2i)((z + 1) + 2i) = (z + 1)2 + 4 = z2 + 2z + 5 z3 + 5z2 + 11z + 15 = (z2 + 2z + 5) (z + 3)

1

for conjugate

1 1

evidence needed

The roots are (−1 + 2i) , (−1 − 2i) and −3.

1

for stating roots together

1 1

for two correct points for third correct point

The general term is given by: 1 r 9 (2x)9 − r − 2 r x 9−r 9−r 2 x (−1)r 9 = × r x2r 9 = × (−1)r 29 − rx9 − 3r r The term independent of x occurs when 9 − 3r = 0, i.e. when r = 3.

9! (−1)3 26 The term is: 6! 3! = −5376.

() () ()

( )

1 1 1 1

1

Method 1 → → PQ = 3i + j + 4k and QR = 2i − 2j − 2k A normal to the plane: i j k → → PQ × QR = 3 1 4 2 −2 −2 3 4 3 1 1 4 = i − j + k −2 −2 2 −2 2 −2 = 6i + 14j − 8k Hence the equation has the form: 6x + 14y − 8z = d.

The plane passes through P (−2, 1, −1) so d = −12 + 14 + 8 = 10 which gives an equation 6x + 14y − 8z = 10 i.e. 3x + 7y − 4z = 5.

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1

1M

1 1

1

→

PR could be used

Marks awarded for Method 2 A plane has an equation of the form ax + by + cz = d . Using the points P, Q, R we get −2a + b − c = d 1M a + 2b + 3c = d 3a + c = d Using Gaussian elimination to solve these we have or other valid method −2 1 −1 d −2 1 −1 d ⇒ 1 1 2 3 d 0 5 5 3d 3 0 1 d 0 6 8 2d

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−2 1 −1 d 0 5 5 3d 0 0 2 − 85 d 4 7 3 ⇒ c = − d, b = d, a = d 5 5 5 These give the equation ( 35 d) x + ( 75d ) y + (− 45 d ) z = d i.e.

3x + 7y − 4z = 5 ⇒

|

|

1 1

1

Marks awarded for

6.

(5)

Method 1 ex = 1 + x +

+

x2 2

+ …

x3 6

1

(1 + e ) = 1 + 2ex + e2x x 2

= 1 + 2 (1 + x +

1M

+ … )  3 2 + (1 + 2x + (2x) + (2x) 2 6 + … )   2 2 1 3 4 3 = 1 + 2 + 2x + x + 3 x + 1 + 2x + 2x + 3 x +… +

x2 2

x3 6

= 4 + 4x + 3x2 + 53 x3 + …

1 1 1

Method 2 ex = 1 + x +

x2 2

+

+ …

x3 6

1

( 1 + e x) = 2 + x + 2 + 6 + …

(1 + ex)2 = (2 + x + x2 + x6 + … )(2 + x + x2 + x6 + … ) x2

2

x3

3

2

3

1 1M

= 4 + 4x + 3x2 + 13 x3 + 12 x3 + 12 x3 + 13 x3 + …

1

= 4 + 4x + 3x2 + 53 x3 + …

1

Method 3 ex = 1 + x +

x2 2

f (x) = (1 + ex)2 f ′ (x) = 2ex (1 + ex) = 2ex + 2e2x f ″ (x) = 2ex + 4e2x f ′′′ (x) = 2ex + 8e2x

+

+ …

1

f (0) = 4 f ′ (0) = 4

1

x3 6

f ″ (0) = 6 f ′′′ (0) = 10

1 1

can award marks for correct columns.

f (x) = f (0) + f ′(0)x + f ″ (0) x2 + f ′′′(0) x6 +…

2

3

(1 + ex)2 = 4 + 4x + 3x2 + 53 x3 + …

1

y = |x + 2|

7.

(4)

y

(a)

(0,2)

(−2,0)

1 1

for shape for coordinates

1 1

for both horizontal lines for values: 1, −1, −2

x

y

(b)

1 −2 −1

x

Marks awarded for x = 4 sin θ ⇒ dx = 4 cos θ dθ

8.

(6)

x = 0 ⇒ θ = 0 x = 2 ⇒ θ = π6 

1 1

2

∫0 16 − x2 dx π/6

16 − (4 sin θ )2 . 4 cosθ dθ

= ∫0

π/6

1

16 (1 − sin 2 θ ) . 4 cosθ dθ

= ∫0

π/6

= ∫0

16 cos2 θ . 4 cos θ dθ

π/6

= ∫0 16 cos2 θ dθ = 8

π/6 ∫0

(1

= 8 [θ +

9.

(4)

1

+ cos 2θ ) dθ π/6

sin 2θ ] 0

1 2

1

for applying trig. identity and integrating

1

numerical approx. allowed

1

for multiplying by A

1

for rearranging A + A−1 = I

A2 = −A−1

1

for subtracting I

A3 = −I, i.e. k = −1

1

for multiplying by A

1

for rearranging

1

for squaring

1

for multiplying by A

=

8π 6

+ 4 sin π3

=

4π 3

+ 2 3 (≈ 7·65)

Method 1 A + A−1 = I A2 + I = A −1

A + I = I − A 2

Hence

Method 2 A + A−1 = I A = I − A−1 A2 = I − 2A−1 + (A

)

−1 2

A3 = A − 2I + A−1 A = (A + A

−1

3

Hence

)

− 2I = I − 2I

A = −I, i.e. k = −1 3

1

Method 3 A + A−1 = I A = I − A−1

1

for rearranging

A = A − A

1

for multiplying by A2

A3 = (A − I) − A

1

using A2 = A − I

3

2

= −I, i.e. k = −1 Plus other valid methods.

1

Marks awarded for

10.

(3)

Method 1 1234 = 7 × 176 + 2 176 = 7 × 25 + 1 25 = 7 × 3 + 4

1

123410 = 34127

1

1

Hence

Method 2 1234 = 7 × 176 + 2

answer only, 1 of 3

1

= 7 × (7 × 25 + 1) + 2

11.

(1,4)

= 7 × (7 × (7 × 3 + 4) + 1) + 2

1

= 3 × 73 + 4 × 72 + 1 × 7 + 2 Hence 123410 = 34127

1

d (sin −1 x) = dx

(a) (b) ∫ sin −1 x.

12.

(5)

x 1 − x2

dx =

sin −1 x ∫

x 1 − x2

dx − ∫

= sin −1 x ∫

x 1 − x2

dx − ∫

(

(

d dx

1 1 − x2

(sin −1 x) ∫

1 1 − x2

∫

x 1 − x2

= sin −1 x (− 1 − x2) − ∫

(

1 1 − x2

x 1 − x2

1

dx dx

)

1

)

1

)

dx dx

(−

1 − x2) dx

= sin −1 x (− 1 − x2) − ∫ (−1) dx

1

= x − sin −1 x. 1 − x2 + c

1

dr = −0.02; dt

answer only, 1 of 3

dh = 0.01 dt

for ∫

x dx = − 1 − x2 1 − x2

1

V = πr2h

( )

dV dr dh = π 2r h + πr2 dt dt dt = π (2 × 0.6 × (−0.02) × 2 + 0.36 × 0.01)

1M for implicit differentiation 1 for accuracy 1

= π (−0.048 + 0.0036) = −0.0444π (≈ −0.14) The rate of change in the volume is −0.0444π m3 s−1.

1

units required

Marks awarded for

13.

(10)

x = 2t + 12 t 2

⇒

y =

⇒

1 3 3t

− 3t

dx dt dy dt

= 2 + t

1

= t − 3

1

2

dy t − 3 = dx 2 + t d dy 2t (2 + t) − (t 2 − 3) t 2 + 4t + 3 = = dt dx (2 + t)2 (2 + t)2 2

1

( )

1

d 2y t 2 + 4t + 3 1 t 2 + 4t + 3 = × = dx2 (2 + t)2 2+t (2 + t)3

1

Stationary points when

dy dx

= 0, i.e.

t2 − 3 = 0 ⇒ t = ± 3 d 2y 3 + 4 3 + 3 = >0 dx2 (2 + 3)3 which gives a minimum.

When t =

1

3,

d 2y 3 − 4 3 + 3 = <0 dx2 (2 − 3)3 which gives a maximum.

1

no marks for using a nature table

1

no marks for using a nature table

When t = − 3,

2

d y At a point of inflexion, dx 2 = 0.

1

In this case, that means t 2 + 4t + 3 = (t + 1)(t + 3) = 0 and this has exactly two roots.

Note that this is a slimmed-down version of the complete story of points of inflexion.

1

need to show 2 values exist

Marks awarded for

14.

(5, 1,3)

(a)

| | |

4 0 6 2 −2 4 −1 1 λ 4 0 6 0 4 −2 0 4 6 + 4λ

1 −1 2 1 3 9

|

1

| |

4 0 6 1 0 4 −2 3 0 0 8 + 4λ 6 6 3 z = = 8 + 4λ 2 (2 + λ) 18 + 6λ 4y = 3 + 2z ⇒ 4y = 4 + 2λ 3λ + 9 ⇒ y = 4 (2 + λ) 2λ − 14 4x = 1 − 6z ⇒ 4x = 4 + 2λ λ − 7 ⇒ x = 4 (2 + λ) (b) When λ = −2, the final row gives 0 = 6 which is inconsistent.

There are no solutions.

for augmented matrix

1

1

triangular form needed

1

first root

1

other two roots

1

(c) λ = −2.1; x = 22.75; y = −6.75; z = −15 1,1 1 for first 2 values; 1 for third Although the values of λ are close, the values of x, y and z are quite different. The system is ill-conditioned near λ = −2.

1

Marks awarded for

15.

(4,7)

(a)

1 A B C = + + (x − 1)(x + 2)2 x − 1 x + 2 (x + 2)2

1M

1 = A(x + 2)2 + B(x − 1)(x + 2) + C (x − 1) x = 1 ⇒ A =

1 9

1

x = −2 ⇒ C = − 13 x = 0 ⇒ 1 = ∴

(b)

4 9

− 2B +

(

1 3

1

⇒ B = − 19

1 1 1 1 3 = − − 2 (x − 1)(x + 2) 9 x − 1 x + 2 (x + 2)2

(x − 1)

)

1

dy x − 1 − y = dx (x + 2)2

dy 1 1 − y = dx x − 1 (x + 2)2

1M for rearranging

Integrating factor: exp (∫ − x −1 1 dx)

1

= exp (− ln (x − 1) = (x − 1)−1

1

1 dy 1 1 − y= 2 (x − 1) dx (x − 1) (x − 1) (x + 2)2

( )

d y 1 = dx x − 1 (x − 1)(x + 2)2 =

(

1 1 1 3 − − 9 x − 1 x + 2 (x + 2)2

(

1

)

1

)

y 1 3 = ln|x − 1 | − ln|x + 2 | + +c 1 x−1 9 x+2 x−1 3 y= ln|x − 1 | − ln|x + 2 | + + c(x − 1) 1 9 x+2 x − 1 |x − 1 | 3 = ln + + c(x − 1) |x + 2 | x + 2 9

(

(

)

)

constant of integration needed.

Marks awarded for

16.

(6,4)

(a) For n = 1, the LHS = cos θ + i sin θ and the RHS = cos θ + i sin θ . Hence the result is true for n = 1.

1

Assume the result is true for n = k , i.e. (cos θ + i sin θ )k = cos kθ + i sin kθ .

1

Now consider the case when n = k + 1: (cosθ + i sinθ )k + 1 =(cosθ + i sinθ )k (cosθ + i sinθ ) = (cos kθ + i sin kθ) (cos θ + i sin θ )

1 1

= (coskθ cosθ − sinkθ sinθ) + i (sinkθ cosθ + coskθ sinθ) 1 = cos(k + 1)θ + i sin (k + 1)θ Thus, if the result is true for n = k the result is true for n = k + 1.

Since it is true for n = 1, the result is true for all n ≥ 1.

1

11π cos 11π (cos 18π + i sin 18π )11 18 + i sin 18 (b) = cos π9 + i sin π9 (cos 36π + i sin 36π )4

=

11π cos 11π cos π9 − i sin π9 18 + i sin 18 × cos π9 + i sin π9 cos π9 − i sin π9

π 11π π cos 11π 18 cos 9 + sin 18 sin 9 = + imaginary term cos2 π9 + sin 2 π9 11π π = cos − + imaginary term 18 9 π = cos + imaginary term 2 Thus the real part is zero as required.

(

)

END OF SOLUTIONS

1

working with n is penalised.

for applying the inductive hypothesis multiplying and collecting

using result from above

1

1

1

or equivalent