mi_AH_Mathematics_all_2011 by colin Mitchell

2011 Maths Advanced Higher Finalised Marking Instructions

Scottish Qualifications Authority 2011 The information in this publication may be reproduced to support SQA qualifications only on a noncommercial basis. If it is to be used for any other purposes written permission must be obtained from SQA’s NQ Delivery: Exam Operations Team. Where the publication includes materials from sources other than SQA (secondary copyright), this material should only be reproduced for the purposes of examination or assessment. If it needs to be reproduced for any other purpose it is the centre’s responsibility to obtain the necessary copyright clearance. SQA’s NQ Delivery: Exam Operations Team may be able to direct you to the secondary sources. 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.

General Marking Principles

These principles describe the approach taken when marking Advanced Higher Mathematics papers. For more detailed guidance please refer to the detailed Marking Instructions. 1

The main principle is to give credit for the skills demonstrated and the criteria met. Failure to have a correct method may not preclude a candidate gaining credit for their solution.

The answer to one part of a question, even if incorrect, can be accepted as a basis for subsequent dependent parts of the question.

The following are not penalised: working subsequent to a correct answer (unless it provides firm evidence that the requirements of the question have not been met) legitimate variation in numerical values / algebraic expressions.

Full credit will only be given where the solution contains appropriate working. Where the correct answer might be obtained by inspection or mentally, credit may be given.

Sometimes the method to be used in a particular question is explicitly stated; no credit will be given where a candidate obtains the correct answer by an alternative method.

Where the method to be used in a particular question is not explicitly stated in the question paper, full credit is available for an alternative valid method. (Some likely alternatives are included but these should not be assumed to be the only acceptable ones.)

In the detailed Marking Instructions which follow, marks are shown alongside the line for which they are awarded. There is one code used M. This indicates a method mark, so in question 1(a), 1M means a method mark for the product rule.

Advanced Higher Mathematics 2011 Marks awarded for

1. (5)

13 − x 13 − x = + 4x − 5 (x − 1) (x + 5)

A B + x − 1 x + 5 13 − x = A (x + 5) + B (x − 1) x = 1 ⇒ 12 = 6A ⇒ A = 2 x = −5 ⇒ 18 = −6B ⇒ B = −3 13 − x 2 3 Hence 2 = − x + 4x − 5 x − 1 x + 5 13 − x 2 3 ∫ x2 + 4x − 5 dx = ∫ x − 1 dx − ∫ x + 5 dx =

= 2 ln | x − 1 | − 3 ln | x + 5 | + c

2. (3)

( 12 x − 3)4 = 4C0 ( 2x )4 + 4C1 ( 2x )3 (−3) +   4C ( x )2 (−3)2 + 4C ( x ) (−3)3 + 4C (−3)4  2 2 3 2 4

1 1 1

for first value for second value

1 1

for logs for moduli

1  1

for powers for coefficients

= ( 2x ) + 4 ( 2x ) (−3) + 6 ( 2x ) (−3)2 4

+4 ( 2x ) (−3)3 + (−3)4 =

3. (6)

x4 3x3 27x2 − + − 54x + 81. 16 2 2

for simplifying

(a) Method 1 y + ey = x2 dy dy + ey = 2x dx dx dy (1 + ey) = 2x ⇒ dy = 2x y (1 + e ) dx dx Method 2 ln (y + ey) = 2 ln x dy (1 + ey) dx 2 = y y + e x y dy 2 (y + e ) = dx x (1 + ey) Method 3 y + ey = x2 ⇒ ey = x2 − y ⇒ y = ln (x2 − y) dy 2x − dx dy = 2 dx x − y dy 2 dy dy 2 dx (x − y) = 2x − dx ⇒ dx (x − y + 1) = 2x dy 2x = 2 dx x − y + 1

1M for applying implicit differentiation 1 for accuracy 1

1 1 1 1 1

Marks awarded for

(b) Method 1 f (x) = sin x cos3 x M1 for using the product rule 1 for first term 1 for second term

f ′ (x) = cos x + sin x (−3 cos x sin x) 4

= cos4 x − 3 cos2 x sin 2 x Method 2 3

f (x) = sin x cos x ln (f (x)) = ln sin x + ln (cos3 x) f ′ (x) cos x 3 cos2 x sin x = − f (x) sin x cos3 x cos x 3 sin x = − sin x cos x 3 cos x 3 sin x f ′ (x) = − sin x cos x sin x cos x 4 = cos x − 3 sin 2 x cos2 x

(

4. (6)

(a)

)

M1 1

Singular when the determinant is 0.

1 det

( )

0 2 3 0 3 2 − 2 det + (−1) det = 0 M1 −1 6 λ 6 −1 λ

−2λ − 2 (18 + 2) − 1 (3λ − 0) = 0

−5λ − 40 = 0 when λ = −8

)

7α = −7 ⇒ α = −1 and β = −1.

(

2 3α + 2β (b) From A, A′ = 2α − β 4 −1 3 Hence 2α − β = −1 and 3α + 2β

−1 3 . 2 = −5.

for transpose

4α − 2β = −2 3α + 2β = −5 for both values

Marks awarded for

5. (6)

  1  x)2 ⇒ f (0) = 1  1 x)− 2 ⇒ f ′ (0) = 12   3 x)− 2 ⇒ f ″ (0) = − 14   5 x)− 2 ⇒ f ″′ (0) = 83  1

Let f (x) = (1 + x)2 , then f (x) = (1 + 1 2

(1 +

− 14

(1 +

f ′ (x) = f ″ (x) =

f ″′ (x) =

3 8

(1 +

{

1 1

for derivatives for values

Hence 1 1 x2 3 x3 1 + x = 1 + x − × + × −… 2 4 2 8 6

1 x2 x3 x − + −… 2 8 16 and replacing x by x2 gives = 1 +

1 +

1 2 x4 x6 = 1 + x − + −… 2 8 16

Thus

(

(1 + x) (1 + x2) =

)(

)

1 x2 x3 1 2 x4 x6 1+ x− + −… 1 + x − + −… 1M for multiplying 2 8 16 2 8 16 1 1 1 1 1 = 1 + x + x2 − x2 + x3 + x3 + … 2 2 8 4 16 1 3 5 3 = 1 + x + x2 + x +… 1 2 8 16

6. (4)

Reflect in the line y = x to get y

x (a, 0) (0, −1)

1 1

for position for coordinates

1 1

for shape for coordinates

Now apply the modulus function

( 0, 1) ( a, 0)

Marks awarded for

7. (4)

Method 1 esin x (2 + x)3 1 − x ⇒ ln y = ln (esin x (2 + x)3) − ln ( 1 − x) = sin x + 3 ln (2 + x) − 12 ln (1 − x) 1 dy 3 1 ⇒ = cos x + + y dx 2 + x 2 (1 − x) dy 3 1 = y cos x + + dx 2 + x 2 (1 − x) When x = 0, y = 8 ⇒ 3 1 gradient = 8 1 + + = 24. 2 2 y =

(

)

(

)

Method 2 esin x (2 + x)3 dy y = ⇒ = 1 − x dx sin x d (2 + x)3) 1 − x − esin x (2 + x)3 (− 12 dx (e

1 1−x

(1 − x) [ cosx esin x (2 + x)3 + 3esin x (2 + x)2] (1 − x) = (1 − x)3/2 esin x (2 + x)3 + 2 (1 − x)3/2 When x = 0, ( 23 + 3 × 22) 23 gradient = + = 20 + 4 = 24 1 2 Method 3 sin x 3 y = e 1(2−+x x)

1M for use of logs 1 for preparing to differentiate

)

for final value

M1 for use of quotient rule M1 1

y 1 − x = esin x (2 + x)3 1 −1/2 dy 1 1 − x dx − 2 y(1 − x) = cosx esin x (2 + x)3 + 3esin x (2 + x)2 1,1 0 3 when x = 0, y = e 12 = 8. This leads to dy 1 dx = 24

Marks awarded for

8. (4)

∑r − 3

r=1

( ) (∑ ) n

∑r =

r=1

(

)

(

)

n2 (n + 1)2 n(n + 1) 2 − =0 4 2

n2 (n + 1)2 n(n + 1) + 4 2 r=1 r=1 2 2 2 2 n (n + 1) n (n + 1) = + 4 4 n2 (n + 1)2 = 2 n

∑r + 3

9. (5)

r =

Method 1 dy = 3 (1 + y) 1 + x dx 1 dy ∫ 1 + y = 3 ∫ (1 + x)2 dx

1 1

M1 separating variables

ln (1 + y) = 2 (1 + x)2 + c

1 1

1 + y = exp (2 (1 + x)2 + c) 3 y = exp (2 (1 + x)2 + c) − 1. 3 = A exp (2 (1 + x)2 ) − 1.

for LHS for term in x

1 1

for the constant

Method 2 dy − 3 1 + xy = 3 1 + x dx Integrating Factor

exp (−3 ∫ 1 + x dx) = exp (−2 (1 + x)3/2) d dx

(y exp (−2 (1

+ x)

))

3/2

3 1 + x (exp (−2 (1 + x)3/2)) 1

y (exp (−2 (1 + x)3/2)) =

− ∫ (−3 1 + x) exp (−2 (1 + x)3/2)) dx = − exp (−2 (1 + x)3/2) + c

y = −1 + c exp (2 (1 + x)

)

3/2

10. (5)

Let z = x + iy, so z − 1 = (x − 1) + iy. 2 | z − 1 | = (x − 1)2 + y2 = 9. The locus is the circle with centre (1, 0) and radius 3.

1 1 1

y x

1 1

Can subsume the first two marks. for circle for shading or other indication

Marks awarded for

11. (7)

(a) ∫0 (secx − x) (secx + x) dx = ∫0 (sec2 x − x2) dx π/4

π/4

π 4

x3   = tan x −   3 0 3 1π   = 1 −  − [0]  3 64  π3 = 1 − . 192 (b) Method 1 Let u = 7x2, then du = 14x dx. x 1 ∫ 1 − 49x4 dx = 14 1 = sin −1 u + c 14 1 = sin −1 7x2 + c 14

Exact value only

M1 1

∫

du 1 − u2

must be in terms of x

1 1 1

for fraction

must be in terms of x

Method 2

∫

x 1 dx = 4 14 1 − 49x =

12. (5)

∫

14x dx 1 − (7x2)2

1 sin −1 7x2 + c 14

for numerator for (7x2)2

For n = 2, 82 + 30 = 64 + 1 = 65. True when n = 2.

Assume true for k , i.e. that 8k + 3k − 2 is divisible by 5, i.e. can be expressed as 5p for an integer p.

for the inductive hypothesis

1 1

for replacing 8k

Now consider 8k + 1 + 3k − 1 = 8 × 8k + 3k − 1 = 8 × (5p − 3k − 2) + 3k − 1 = 40p − 3k − 2 (8 − 3) = 5 (8p − 3k − 2) which is divisible by 5. So, since it is true for n = 2, it is true for all n ≥ 2.

Marks awarded for

13. (9)

Method 1 Let d be the common difference. Then 1 u3 = 1 = a + 2d and u2 = = a + d a 1 1 = a + 2 − a a a = a2 + 2 − 2a2 a2 + a − 2 = 0 (a + 2)(a − 1) = 0 ⇒ a = −2 since a < 0. a = −2 gives 2d = 3 and hence d = 23 .

(

)

1 1 1 1 1

Method 2 1 , u3 = 1 a 1 1 ⇒ − a = 1 − a a ⇒ 1 − a2 = a − 1 ⇒ a2 + a − 2 = 0 (a + 2)(a − 1) = 0 ⇒ a = −2 since a < 0. 1 3 d = u3 − u2 = 1 − = a 2 u1 = a, u2 =

n [ 2a + (n − 1) d ] 2 n 3 3 = −4 + n −  2 2 2 = 14 [ 3n2 − 11n] ∴ 3n2 − 11n > 4000 11 4000 n2 − n > 3 3 2 11 48000 121 48121 n − > + = 6 36 36 36 11 48121 n − > 6 6 48121 + 11 n > ≈ 38.39 6

M1 1 1 1 1

Sn =

(

1 1

)

So the least value of n is 39.

1 1

for value for suitable justification

Marks awarded for

14. (10)

Auxiliary equation m2 − m − 2 = 0 (m − 2) (m + 1) = 0 m = −1 or 2 Complementary function is: y = Ae−x + Be2x

1 1 1

The particular integral has the form y = Cex + D 1 dy y = Cex + D ⇒ = Cex dx d 2y ⇒ = Cex 1 2 dx Hence we need: d 2y dy − − 2y = ex + 12 2 dx dx [ Cex] − [ Cex] − 2 [ Cex + D] = ex + 12 −2Cex − 2D = ex + 12 1 1 Hence C = − 2 and D = −6. So the General Solution is y = Ae−x + Be2x − 12 ex − 6. 1 x = 0 and y = − 23 ⇒ A + B − 12 − 6 dy x = 0 and dx = 12 ⇒ −A + 2B − 12 3B − 7 = −1 ⇒ B = So the particular solution is y = 3e−x + 2e2x −

= − 23 1 = 2 ⇒ A = 3

− 6.

1 2

1 x 2e

Setting up the equations

Marks awarded for

15. (10)

(a) In terms of a parameter s, L1 is given by x = 1 + ks, y = −s, z = −3 + s

In terms of a parameter t , L2 is given by x = 4 + t, y = −3 + t, z = −3 + 2t

Equating the y coordinates and equating the z coordinates: −s = −3 + t −3 + s = −3 + 2t Adding these −3 = −6 + 3t ⇒ t = 1 ⇒ s = 2.

}

From the x coordinates 1 + ks = 4 + t Using the values of s and t 1 + 2k = 5 ⇒ k = 2

The point of intersection is: (5, −2, −1).

(b) L1 has direction 2i − j + k. L2 has direction i + j + 2k. Let the angle between L1 and L2 be θ , then (2i − j + k) . (i + j + 2k) cos θ = |2i − j + k| |i + j + 2k| 2 − 1 + 2 3 1 = = = 6 6 6 2 θ = 60° The angle between L1 and L2 is 60°.

1 1 1

For both directions.

Marks awarded for

16. (11)

(a) In = ∫ 1

1 n dx 0 ( 1 + x2) 1

−n

= ∫ 1 × ( 1 + x2) 0

[

]

= (1 + x2) ∫1 dx 0 + ∫0 (2nx (1 + x2) −n

−n − 1

∫ 1 dx) dx 1

= [ x (1 + x2) ] 0 + ∫0 2nx2 (1 + x2) dx 1 −n − 1 1 = n − 0 + 2n ∫ x2 (1 + x2) dx 0 2 1 1 x2 = n + 2n ∫ n + 1 dx. 0 ( 1 + x2) 2 −n 1

(b)

−n − 1

A B x2 + = (1 + x2)n (1 + x2)n + 1 (1 + x2)n + 1 ⇒ A ( 1 + x2) + B = x2 ⇒ A = 1, B = −1 1 −1 x2 + = (*) (1 + x2)n (1 + x2)n + 1 (1 + x2)n + 1 1 1 x2 In = n + 2n ∫ n + 1 dx. 0 ( 1 + x2) 2 1 1 1 1 −1 = n + 2n ∫ dx + 2n n n + 1 dx ∫ 2 0 (1 + x ) 0 ( 1 + x2) 2 1 = n + 2nIn − 2nIn + 1 2 1 2nIn + 1 = n + (2n − 1) In 2 1 2n − 1 In + 1 = + I n. n + 1 n × 2 2n

(

(c)

)

1 dx = I3 + x2)3 1 3 + I2 16 4 1 3 1 1 + + I1 16 4 4 2 1 3 1 1 + ∫ dx 4 8 0 1 + x2 1 1 3 + [ tan−1 x] 0 4 8 1 3π 1 3π + = + . 4 84 4 32

1 1

Using (*)

Recognising In and In + 1

∫ 0 (1 = = = = =

(

for showing that 1 is integrated

)

1 1

END OF SOLUTIONS