2008 H2 Maths 9740 Paper 1

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GCE 'A' Level October/November 2008 Suggested Solutions. Mathematics H2 ( 9740/01) version 1.1. For tuition, exam papers & Last-Minute Buddha FootΒ ...
GCE β€˜A’ Level October/November 2008 Suggested Solutions

Mathematics H2 (9740/01)

version 1.1

MATHEMATICS (H2)

9740/01

Paper 1 Suggested Solutions

October/November 2008

2.

Topic: Summation of Series (Mathematical Induction) 1

Let Pn denotes the statement β€œπ‘†π‘› = 𝑛(𝑛 + 1)(4𝑛 + 5), βˆ€ 𝑛 ∈ β„€+ ”. 6

When n = 1, 1.

Topic: Definite Integrals

L.H.S.

= S1 = u1 = 3

From the diagram:

R.H.S.

= (1)(1+1)(4+5) = 3

2 2 ∫1 𝑦 dπ‘₯ = ∫1 π‘₯ 2 dπ‘₯…… 4 4 βˆ«π‘Ž π‘₯ d𝑦 = βˆ«π‘Ž �𝑦 d𝑦

Area A = Area B =

=

Equating (1) & (2): 2

4 1 βˆ«π‘Ž 𝑦 2 d𝑦……

(2)3 3

2

π‘₯3

οΏ½ οΏ½

βˆ’ 8 3

3 1

(1)3 3

βˆ’

1 3

3 2

3 4 2

2

=οΏ½ 𝑦 οΏ½ 3

π‘Ž

3

2

=

π‘Ž =

16 3

9

(1)

∴ P1 is true. 1

Assume Pk is true i.e. π‘†π‘˜ = π‘˜(π‘˜ + 1)(4π‘˜ + 5), for some π‘˜ ∈ β„€+ 6

(2)

2

3

2

βˆ’ π‘Ž 3

3 2

3

1

To show that Pk+1 is also true i.e. π‘†π‘˜+1 = (π‘˜ + 1)(π‘˜ + 2)(4π‘˜ + 9), L.H.S. = Sk+1 1

= k (k + 1) (4k + 5) + (k + 1) [2(k+1) + 1]

π‘₯ 𝑛+1 οΏ½ π‘₯ dπ‘₯ = +𝑐 𝑛+1

6

𝑛

1

= (k + 1) [k (4k + 5) + 6(2k + 3)] 6

1

∴a = ( )3

1

= (k + 1) (4k2 + 5k + 12k + 18) Bring out the (k + 1) 6 factor since it’s found on the R.H.S.

2 9 2

6

1 π‘†π‘˜ = π‘˜(π‘˜ + 1)(4π‘˜ + 5) 6

= Sk + uk+1

= (4)2 βˆ’ (π‘Ž)2 3

6

β‡’ L.H.S. = R.H.S.

1

4

∫1 π‘₯ 2 dπ‘₯ = βˆ«π‘Ž 𝑦 2 d𝑦

1

2

6

uk+1 = (k + 1) [2(k+1) + 1]

1

= (k + 1) (4k2 + 17k + 18) 6 1

= (k + 1) (k + 2) (4k + 9) 6

β‰ˆ 2.73 (3 sig. fig.)

= R. H. S. ∴Pk+1 is also true if Pk is true. Since P1 is true and Pk+1 is true if Pk is true, by mathematical induction, Pn is true βˆ€ 𝒏 ∈ β„€+

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Mathematics H2 (9740/01) 3.

Topic: Vectors

= =

1 5 οΏ½ 4 οΏ½βˆ™οΏ½βˆ’1οΏ½ βˆ’3 0 οΏ½12 +4 2 +(βˆ’3)2 οΏ½52 +(βˆ’1)2 +02 (1)(5)+(4)(βˆ’1)+(βˆ’3)(0) 26

√26√26

4.

Topics: Differentiation, Differential Equations (i)

d𝑦 dπ‘₯

=

3π‘₯

π‘₯2+ 1

3π‘₯

Express in

∫ d𝑦 = ∫ π‘₯ 2+ 1 dπ‘₯ 3

y =

O

οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½βƒ— 𝑂𝐴 βˆ™ οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½βƒ— 𝑂𝐡 ������⃗��𝑂𝐡 οΏ½οΏ½οΏ½οΏ½οΏ½οΏ½βƒ—οΏ½ �𝑂𝐴

1

P

οΏ½οΏ½οΏ½οΏ½οΏ½βƒ— = 𝐴𝐴 οΏ½οΏ½οΏ½οΏ½οΏ½βƒ— 𝑂𝑂 ∡ same magnitude & direction

1 5 = οΏ½ 4 οΏ½ + οΏ½βˆ’1οΏ½ βˆ’3 0 πŸ” =οΏ½ πŸ‘ οΏ½ βˆ’πŸ‘

=

B

Equal Vectors:

οΏ½οΏ½οΏ½οΏ½οΏ½βƒ— + οΏ½οΏ½οΏ½οΏ½οΏ½βƒ— (i) οΏ½οΏ½οΏ½οΏ½οΏ½βƒ— 𝑂𝐴 = 𝑂𝐴 𝐴𝐴 οΏ½οΏ½οΏ½οΏ½οΏ½βƒ— οΏ½οΏ½οΏ½οΏ½οΏ½βƒ— = 𝑂𝐴 + 𝑂𝑂

(ii) cos ∠AOB =

version 1.1

A Scalar Product of two vectors a and b: 𝐚 βˆ™ 𝐛 = |𝐚||𝐛| cos πœƒ b

ΞΈ

a

1

2

πŸ‘

=

𝟐

2π‘₯

∫ π‘₯ 2+ 1 dπ‘₯

π₯𝐧|π’™πŸ + 𝟏| + 𝒄

3

οΏ½

fβ€² (π‘₯) f(π‘₯)

f β€² (π‘₯) dπ‘₯ = ln|f(π‘₯)| + 𝑐 f(π‘₯)

(ii) Sub y = 2, x = 0 into 𝑦 = ln|π‘₯ 2 + 1| + 𝑐, 3

2

2 = ln|02 + 1| + 𝑐 β‡’ c = 2 2

πŸ‘

∴ π’š = π₯𝐧|π’™πŸ + 𝟏| + 𝟐 𝟐

(iii) As x β†’ +∞, As x β†’ βˆ’βˆž,

d𝑦

dπ‘₯ d𝑦 dπ‘₯

β†’ 0+

β†’ 0βˆ’

∴the gradient of every solution curve tends to be horizontal as x β†’ Β± ∞.

∠AOB = cos βˆ’1 ( ) 26

= 87.795Β° β‰ˆ 87.8Β° (3 sig. fig.)

(iv)

(iii) Area of parallelogram OAPB οΏ½οΏ½οΏ½οΏ½οΏ½βƒ— Γ— οΏ½οΏ½οΏ½οΏ½οΏ½βƒ— = �𝑂𝐴 𝑂𝑂 οΏ½ 1

5

= οΏ½οΏ½ 4 οΏ½ Γ— οΏ½βˆ’1οΏ½οΏ½ βˆ’3

βˆ’3

= οΏ½οΏ½βˆ’ 15οΏ½ οΏ½

0

βˆ’21

𝐒 𝐣 οΏ½1 4 5 βˆ’1

𝐀 4 βˆ’3 1 βˆ’3 1 4 �𝐒 βˆ’ οΏ½ �𝐣+ οΏ½ �𝐀 βˆ’3οΏ½ = οΏ½ βˆ’1 0 5 0 5 βˆ’1 0 = βˆ’3𝐒 βˆ’ 15𝐣 βˆ’ 21𝐀

TI-84 Plus

= οΏ½(βˆ’3)2 + (βˆ’15)2 + (βˆ’21)2 = √675

= πŸπŸ“βˆšπŸ‘ units2

Casio fx-9860G

N.B. Final answer expressed in surd form as question asks for exact area.

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GCE β€˜A’ Level October/November 2008 Suggested Solutions

Mathematics H2 (9740/01) 5.

version 1.1

Topic: Integration 1 √3

(i) ∫0

1

1+9π‘₯ 2

6. 1 √3

dπ‘₯ = ∫0 = = = = =

1 3

1

1

dπ‘₯

1+(3π‘₯)2 1

Express in

3

∫0√3 1+(3π‘₯)2 dπ‘₯

οΏ½

1 √3

[tanβˆ’1 (3π‘₯)]0 3 1 βˆ’1 3 3 1 3 Ο€

πŸ—

12

οΏ½tan ( ) βˆ’ tanβˆ’1 (0)οΏ½ Ο€

οΏ½ βˆ’0οΏ½ 3

√3

e

π‘₯ 𝑛+1

e

e π‘₯ 𝑛+1

ILATE/LIATE Rule: Sub u = ln π‘₯ (L) d𝑒 1 = dπ‘₯ π‘₯

Sub

d𝑣

dπ‘₯

= π‘₯ 𝑛 (A)

π‘₯ 𝑛+1 𝑣= 𝑛+1

= οΏ½ln e βˆ™ οΏ½ = = = =

e𝑛+1 𝑛+1

e𝑛+1 𝑛+1 1

βˆ’ βˆ’

(𝑛+1)2

𝑛+1 1

𝑛+1

1

1

οΏ½ βˆ’ ln 1 βˆ™ οΏ½

οΏ½

π‘₯ 𝑛+1

e

οΏ½

𝑛+1 1

(𝑛+1)2

1𝑛+1 𝑛+1

(a) By Cosine Rule, cos ∠ABC =

fβ€²(π‘₯)

1+[f(π‘₯)]2

cos ΞΈ =

β€²

f (π‘₯) dπ‘₯ = tanβˆ’1 [f(π‘₯)] + 𝑐 + [f(π‘₯)]2 Integration by parts: d𝑣 d𝑒 οΏ½ 𝑒 dπ‘₯ = 𝑒𝑒 βˆ’ οΏ½ 𝑣 dπ‘₯ dπ‘₯ dπ‘₯ 1

(ii) ∫1 π‘₯ 𝑛 ln π‘₯ dπ‘₯ = οΏ½ln π‘₯ βˆ™ οΏ½ οΏ½οΏ½ βˆ’ ∫1 οΏ½ οΏ½ οΏ½ οΏ½ dπ‘₯ 𝑛+1 𝑛+1 π‘₯ e𝑛+1

Topic: Maclaurin’s Series

οΏ½οΏ½ βˆ’

e π‘₯𝑛 ∫1 �𝑛+1οΏ½ dπ‘₯

= 2

6

1

1 2

β‰ˆ 4 οΏ½1 +

2 3 οΏ½ ΞΈ2 οΏ½ & 4

β‰ˆ 2 [1 + οΏ½ ΞΈ2 οΏ½ + …]

above

β‰ˆ2 β‰ˆ2

2 4 3 [1 + ΞΈ2 8 3 2 + ΞΈ 4 πŸ‘

∴ a = 2, b =

πŸ’

1

+ …]

1

B

Small Angle Approximation: 1 cos πœƒ β‰ˆ 1 βˆ’ πœƒ 2 2

Express in (1 + x)n since n is not a positive integer

β‡’ f(0) = tan ( Ο€) = 1

(b) f(x) = tan (2x + Ο€) 4

1

3 2 2 ΞΈ οΏ½ 4

1 3

ΞΈ

A

∴ AC β‰ˆ οΏ½4 + 3ΞΈ2 οΏ½2 (Shown)

Maclaurin’s expansion: (1 + π‘₯)𝒏 = 1 + 𝑛𝑛 + β‹―

1

1

f β€²(x) = 2 sec2 (2x + Ο€)

(𝒏+𝟏)𝟐

4

4

1

β‡’ f β€²(0) = 2sec2 ( Ο€) = 4

1

1

4

1

f β€³(x) = 2 β‹… 2 sec(2x + Ο€) [2 sec(2x + Ο€) tan (2x + Ο€)] 2

1

4

1

= 8 sec (2x + Ο€) tan (2x + Ο€) 4 4 β‡’ f β€³(0) = 8(2)(1) = 16 f(x) = f(0) + xf β€²(0) + π‘₯2

π‘₯2 2!

f β€³(0) + …

= 1 + x(4) + οΏ½ οΏ½16 + … 2!

= 1 + 4x + 8x2 + …

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10βˆ’ 𝐴𝐢 2

2

ignored since ΞΈ is small

π’πžπ’+𝟏 +𝟏

3

2 (1)(3)

β‰ˆ 10 – 6 + 3ΞΈ 2 β‰ˆ 4 + 3ΞΈ 2

x term i.e.

[(𝑛 + 1)e𝑛+1 βˆ’ e𝑛+1 + 1]

C

2 (𝐴𝐡)(𝐡𝐢)

12 + 32 βˆ’ 𝐴𝐢 2

β‡’ AC = 10 – 6 cos ΞΈ 1 β‰ˆ 10 – 6 (1 – ΞΈ 2)

2

(e𝑛+1 βˆ’ 1𝑛+1 )

𝐴𝐡2 + 𝐡𝐢 2 βˆ’ 𝐴𝐢 2

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4

4

d sec 𝑛 [f(π‘₯)] dπ‘₯

= 𝑛 sec π‘›βˆ’1 [f(π‘₯)] βˆ™

d sec[f(π‘₯)] dπ‘₯

d sec[f(π‘₯)] dπ‘₯ = f β€² (π‘₯) βˆ™ sec[f(π‘₯)] tan[f(π‘₯)]

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GCE β€˜A’ Level October/November 2008 Suggested Solutions

Mathematics H2 (9740/01) 7.

version 1.1

Topic: Differentiation

8. π‘₯

Total time taken: 180 = (2y + x)(3) + Ο€οΏ½ οΏ½ (9) 2

π‘₯

Total time taken = length of straight part Γ— 3 hrs/m + length of semicircular part Γ— 9 hrs/m

60 = 2y + x + Ο€οΏ½ οΏ½ (3) 2y = 60 – x βˆ’ π‘₯

y = 30 βˆ’ βˆ’ 2

2 3Ο€π‘₯ 2 3Ο€π‘₯ 4

……… (1)

1 2

π‘₯ 2

= 30π‘₯ βˆ’

When A is maximum,

π‘₯+

2

= π‘₯ οΏ½30 βˆ’ βˆ’

Sub y expression from (1)

β‡’ 30 βˆ’ π‘₯ βˆ’

π‘₯

10Ο€π‘₯

8 10Ο€π‘₯ 8

= 30π‘₯ βˆ’

d𝐴 dπ‘₯

= 0.

2

π‘₯2 2 π‘₯2 2

βˆ’ βˆ’

3Ο€π‘₯ 4

4 5Ο€π‘₯ 2

π‘₯2

1

8

+

4

2

Ο€π‘₯ 2 8

=0

= 30

x =

30

1+

10Ο€ 8

β‰ˆ 6.0889 β‰ˆ 6.09 (3 sig. fig.) Second derivative test:

d2 𝐴 dπ‘₯ 2

= βˆ’1 βˆ’

5Ο€ 4

< 0 β‡’ A is maximum when x = 6.09

6.0889

3Ο€(6.0889)

Sub x β‰ˆ 6.0889 into (1): y = 30 βˆ’ βˆ’ 2 β‰ˆ 12.608 β‰ˆ 12.6 (3 sig. fig.)

Alternative Method 2 3 Ο€ = 1 + 3√3i + 3�√3iοΏ½ + �√3iοΏ½ 𝑧 = 1 + √3i = 2ei 3 Ο€ 3 Ο€ = 1 + 3√3i βˆ’ 3(3) βˆ’ 3√3i 𝑧 3 = οΏ½2ei 3 οΏ½ = 23 ei 3 = βˆ’8 = βˆ’8 (ii) Given that 1 + √3i is a root, sub 𝑧 = 1 + √3i into 2𝑧 3 + π‘Žπ‘§ 2 + 𝑏𝑧 + 4 = 0: 2(βˆ’8) + π‘ŽοΏ½1 + √3iοΏ½ + 𝑏�1 + √3iοΏ½ + 4 = 0 βˆ’16 + π‘ŽοΏ½1 + 2√3i βˆ’ 3οΏ½ + 𝑏 + π‘βˆš3i + 4 = 0 z3= βˆ’8 βˆ’16 + 2π‘Žβˆš3i βˆ’ 2π‘Ž + 𝑏 + π‘βˆš3i + 4 = 0 from βˆ’12 βˆ’ 2π‘Ž + 𝑏 + 2π‘Žβˆš3i + π‘βˆš3i = 0 Comparing real parts: βˆ’12 βˆ’ 2π‘Ž + 𝑏 = 0 βˆ’2π‘Ž + 𝑏 = 12 𝑏 = 12 + 2π‘Ž … … … (1) Comparing imaginary parts:

οΏ½ + Ο€οΏ½ οΏ½

3Ο€π‘₯ 2

3

(i) z13 = �1 + √3i�

2

Total area of rectangular & semicircular parts: A = π‘₯𝑦 + Ο€ οΏ½ οΏ½

Binomial Expansion: (1 + π‘₯)3 = 1 + 3π‘₯ + 3π‘₯ 2 + π‘₯ 3

Topic: Complex Numbers

4

Given 1 + √3i is a root β‡’ 1 βˆ’ √3i is also a root. β‡’

Non-real roots occur in conjugate pairs in polynomial equations �𝑧 βˆ’ οΏ½1 + √3i���𝑧 βˆ’ οΏ½1 βˆ’ √3iοΏ½οΏ½ = 𝑧 2 βˆ’ 2𝑧 + 4 with real coefficients.

A quadratic factor of the equation is

β‡’ 2𝑧 3 βˆ’ 3𝑧 2 + 6𝑧 + 4 ≑ (𝑧 2 βˆ’ 2𝑧 + 4)(2𝑧 + 1) = 0

∴x = 6.09 & y = 12.6 gives the flower-bed a maximum area.

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2π‘Žβˆš3 + π‘βˆš3 = 0 𝑏 = βˆ’2π‘Ž … … … . … (2) Equating (1) & (2): 12 + 2π‘Ž = βˆ’2π‘Ž 4π‘Ž = βˆ’12 β‡’ π‘Ž = βˆ’πŸ‘ 𝑏 = βˆ’2(βˆ’3) = πŸ” (iii) Since the equation 2𝑧 3 βˆ’ 3𝑧 2 + 6𝑧 + 4 = 0 has real coefficients,

β‡’ 𝒛 = 𝟏 + βˆšπŸ‘π’, 𝟏 βˆ’ βˆšπŸ‘π’, βˆ’ facebook.com/JossSticksTuition

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𝟏 𝟐

By inspection or long division.

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GCE β€˜A’ Level October/November 2008 Suggested Solutions

Mathematics H2 (9740/01)

version 1.1

9.

ALTERNATE APPROACH (ii) Since the equation 2𝑧 3 + π‘Žπ‘§ 2 + 𝑏𝑧 + 4 = 0 has real coefficients, Given 1 + √3i is a root β‡’ 1 βˆ’ √3i is also a root. β‡’ A quadratic factor of the equation is

�𝑧 βˆ’ οΏ½1 + √3i���𝑧 βˆ’ οΏ½1 βˆ’ √3iοΏ½οΏ½ = 𝑧 2 βˆ’ 2𝑧 + 4

Non-real roots occur in conjugate pairs in polynomial equations with real coefficients.

β‡’ 2𝑧 3 + π‘Žπ‘§ 2 + 𝑏𝑧 + 4 ≑ (𝑧 2 βˆ’ 2𝑧 + 4)(𝐴𝑧 βˆ’ 𝑂) where A, B ∈ ℝ

Comparing coefficients of z3, A = 2 Comparing coefficients of constant, 4 = βˆ’4B β‡’ B = βˆ’1 Comparing coefficients of z2, a = βˆ’B βˆ’ 2A = βˆ’(βˆ’1) βˆ’ 2(2) = βˆ’3 Comparing coefficients of z, b = βˆ’2(βˆ’B) + 4A = βˆ’2[βˆ’(βˆ’1)] + 4(2) = 6

Topic: Graphing Techniques (i)

f(π‘₯) =

Quotient Rule: 𝑒 β€² 𝑣𝑒′ βˆ’ 𝑒𝑒′ (𝑐π‘₯ (π‘Žπ‘₯ + 𝑑)(π‘Ž) βˆ’ + 𝑏)(𝑐) οΏ½ οΏ½ = f β€² (π‘₯) = 𝑣 𝑣2 (𝑐π‘₯ + 𝑑)2 π‘Žπ‘π‘₯ + π‘Žπ‘‘ βˆ’ π‘Žπ‘π‘₯ βˆ’ 𝑏𝑐 = (𝑐π‘₯ + 𝑑)2 π‘Žπ‘‘ βˆ’ 𝑏𝑐 β‰  0, βˆ€π‘₯ ∡ π‘Žπ‘‘ βˆ’ 𝑏𝑐 β‰  0 (Given) = (𝑐π‘₯ + 𝑑)2

∴By differentiation, fβ€²(x) β‰  0, βˆ€x β‡’ the graph of y = f(x) has no turning points. π‘Ž 𝑐

π‘Ž

= +

2𝑧 3 βˆ’ 3𝑧 2 + 6𝑧 + 4 ≑ �𝑧 βˆ’ οΏ½1 + √3i���𝑧 βˆ’ οΏ½1 βˆ’ √3iοΏ½οΏ½(2𝑧 + 1) = 0 𝟏

√3 𝑧3 = βˆ’ 1 βˆ’ 2

1 2

0

𝑧1 = 1 + √3i

1

𝑐π‘₯+𝑑

π‘π‘βˆ’π‘Žπ‘‘

𝑐 (𝑐π‘₯+𝑑)

𝑐

𝑐(𝑐π‘₯+𝑑)

π‘Ž

By long division:

π‘Žπ‘‘ 𝑐

cx + d ax + b ax +

π‘Žπ‘‘βˆ’π‘π‘

π‘Ž

When ad – bc = 0, 𝑦 = f(π‘₯) = , π‘₯ β‰  βˆ’ 𝑐

𝑑 𝑐

a c

ad c ad bβˆ’ c

𝒂

𝒅

∴ The graph is a horizontal line at π’š = , but undefined at 𝒙 = βˆ’ 𝒄 𝒄 y

Re (z)

(iii) 𝑦 =

π‘Ž 𝑐

3π‘₯βˆ’7

2π‘₯+1

Sub a = 3, b = βˆ’7, c = 2, d = 1 into f β€²(x) in (i):

π‘Žπ‘Ž + 𝑏 , 𝑐𝑐 + 𝑑 π‘Žπ‘Ž = 𝑏𝑏

𝑦=

𝑑 x 𝑐 ∴ the graph has a positive gradient at all points since (2x + 1)2 > 0, x β‰  βˆ’πŸπŸ.

𝑧2 = 1 βˆ’ √3i

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π‘βˆ’

𝑐

= –

β‡’ βˆ’βˆš3

𝑐π‘₯+𝑑

(ii) f(x) = +

(iii) Sub a = βˆ’3, b = 6, A = 2, B = βˆ’1 in (ii), β‡’ 𝒛 = 𝟏 + βˆšπŸ‘π’, 𝟏 βˆ’ βˆšπŸ‘π’, βˆ’ 𝟐 Im (z)

π‘Žπ‘₯+𝑏

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d𝑦 dπ‘₯

=

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3(1)βˆ’ (βˆ’7)(2) (2π‘₯+1)2

=

17

(2π‘₯+1)2

>0

βˆ’

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GCE β€˜A’ Level October/November 2008 Suggested Solutions

Mathematics H2 (9740/01)

version 1.1

10. Topic: Arithmetic & Geometric Series

(iv)

(i) Since the student saves $3 more than the previous month in each subsequent month, the monthly savings = 10, 13, 16, 19, … β‡’ Arithmetic series with the 1st term, a = 10, and common difference, d = 3. To save over $2000 in total, 𝑆𝑛 > 2000 𝑛 [2π‘Ž + (𝑛 βˆ’ 1)𝑑] > 2000 2 Sub a = 10, d = 3 into Sn:

Casio fx-9860G TI-84 Plus

(a) y

𝑦= 1 π‘₯=βˆ’ 2

2

3π‘₯ βˆ’ 7 𝑦= 2π‘₯ + 1

3 2

x

7 οΏ½ , 0οΏ½ 3

(0, βˆ’7)

𝑛

Sum of A. P.: 𝑆𝑛 𝑛 = [2π‘Ž + (𝑛 βˆ’ 1)𝑑] 2

3 𝑦=οΏ½ 2 π‘₯=βˆ’

1 2

7

οΏ½3, 0οΏ½

𝑦2 =

Casio fx-9860G

n > 33.79 or n < βˆ’39.46 (reject) β‡’ n = 34 (earliest no. of months since 1 Jan 2009)

∴ she will first have saved over $2000 on the 1st of October 2011. (ii) (a) Taking into account only the original $10 deposit:

(b) y

[2(10) + (𝑛 βˆ’ 1)(3)] > 2000 𝑛 [17 + 3𝑛] > 2000 2 17𝑛 + 3𝑛2 > 4000 3𝑛2 + 17𝑛 βˆ’ 4000 > 0

3π‘₯ βˆ’ 7 2π‘₯ + 1

Month 1 2 3

Balance at Start of Month ($) 10 10+(0.02)(10) = (1.02)(10) (1.02)(10)+(0.02)(1.02)(10) = (1.02)(1.02)(10)

Interest Earned End of Month ($) (0.02)(10) (0.02)(1.02)(10) (0.02)(1.02)(1.02)(10)

n

(1.02)nβˆ’1(10)

(0.02)(10)(1.02)nβˆ’1

Total compound interest earned after n months from the original $10 x

= (10) (0.02) + (10) (0.02) [1.02 + 1.022 + … + 1.02n-1] = (10)(0.02) + (10)(0.02) οΏ½

3 𝑦 = βˆ’οΏ½ 2

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1.02π‘›βˆ’1.02 0.02

οΏ½

1.02οΏ½1.02π‘›βˆ’1 βˆ’1οΏ½ 1.02βˆ’1

οΏ½

Sum of G.P. where r > |1|: π‘Ž(π‘Ÿ 𝑛 βˆ’ 1) 𝑆𝑛 = π‘Ÿβˆ’1

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GCE β€˜A’ Level October/November 2008 Suggested Solutions

Mathematics H2 (9740/01)

version 1.1

∴total interest earned in 2 years (24 months) from original $10 = 10(1.0224 βˆ’ 1) β‰ˆ $6.08 (3 sig. fig.)

(b) Account balance including $10 deposited monthly: Month

Balance at Start of Month ($)

Balance at End of Month ($)

1

10

(1.02)(10)

2

(1.02)(10) + 10

3

1.022(10) + 1.02(10) + 10

1.02[1.02(10) + 10] = 1.022(10) + 1.02(10) 1.02[1.022(10) + 1.02(10) + 10] = 1.023(10) + 1.022(10) + 1.02(10) (1.02 + 1.022 + … + 1.02n)10

n

11. Topic: Three-Dimensional Geometry π‘₯ Let �𝑦�be the point of intersection of p1, p2, p3. Solving for p1, p2 and p3, 𝑧 πŸ’ Cartesian equation of plane βˆ’ 𝟏𝟏 π‘₯ 𝑛1 2 βˆ’5 3 3 π‘₯ βŽ› πŸ’βŽž �𝑦 οΏ½ = οΏ½3 2 βˆ’5οΏ½ βˆ’5 οΏ½ = βŽœβˆ’ 𝟏𝟏⎟ 𝐫 βˆ™ 𝐧 = �𝑦� βˆ™ �𝑛2 οΏ½ = 𝑑: 𝑧 𝑛3 5 βˆ’20.9 17 16.6 πŸ• 𝑧 𝑛1 π‘₯ + 𝑛2 𝑦 + 𝑛3 𝑧 = 𝑑 ⎝ 𝟏𝟏 ⎠

1.02(1.02𝑛 βˆ’1)

= 10 οΏ½ οΏ½ 1.02βˆ’1 = 510(1.02𝑛 βˆ’ 1) ……… (1)

Total in account at end of 2 years (24 months) = 510(1.0224 βˆ’ 1) Sub n = 24 into (1) β‰ˆ 310.3029972 β‰ˆ $310 (3 sig. fig.)

TI-84 Plus

(c) Let n be the number of complete months taken for the total in the account to first exceed $2000. Using (1) from (b)(ii), 510(1.02𝑛 βˆ’ 1) > 2000 1.02𝑛 > 4.9215 𝑛 ln 1.02 > ln 4.9215 𝑛 > 80.475 ∴ n = 81 months.

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GCE β€˜A’ Level October/November 2008 Suggested Solutions

Mathematics H2 (9740/01)

version 1.1

(ii) Given that l lies on p3 β‡’ all points on l lies on p3.

(i) Since l lies on p1 and p2 β‡’ solve for p1 and p2, 2π‘₯ βˆ’ 5𝑦 + 3𝑧 = 3………… (1) 3π‘₯ + 2𝑦 βˆ’ 5𝑧 = βˆ’5 ……… (2)

Eliminate y (1) Γ— 2 + (2) Γ— 5: 19π‘₯ βˆ’ 19𝑧 = βˆ’19 β‡’ 𝑧 = π‘₯ + 1

Eliminate x (1) Γ— 3 βˆ’ (2) Γ— 2: βˆ’19𝑦 + 19𝑧 = 19 β‡’ 𝑧 = 𝑦 + 1 β‡’

π‘₯+1 1

=

𝑦+1 1

=

𝑧

1

Cartesian equation of π‘Ž1 𝑏1 line οΏ½π‘Ž2 οΏ½ + 𝑠 �𝑏2 οΏ½: π‘Ž3 𝑏3

π‘₯ βˆ’ π‘Ž1 𝑦 + π‘Ž2 𝑧 βˆ’ π‘Ž3 = = 𝑏1 𝑏2 𝑏3

βˆ’πŸ 𝟏 ∴vector equation of l is: 𝐫 = οΏ½βˆ’πŸοΏ½ + 𝒔 �𝟏� , 𝒔 ∈ ℝ 𝟎 𝟏 ALTERNATE APPROACH

Picking two points on l from the vector equation of l in (i), βˆ’1 βˆ’1 5 Let s = 0: 𝐫 = οΏ½βˆ’1οΏ½ β‡’ οΏ½βˆ’1οΏ½ βˆ™ οΏ½ πœ† οΏ½ = πœ‡ 0 0 17 βˆ’5 βˆ’ πœ† = πœ‡ ……………… (1)

0 0 5 Let s = 1: 𝐫 = οΏ½0οΏ½ β‡’ οΏ½0οΏ½ βˆ™ οΏ½ πœ† οΏ½ = πœ‡ 1 1 17 πœ‡ = 17 ……….……… (2)

Sub (2) into (1): βˆ’5 βˆ’ πœ† = 17 β‡’ πœ† = βˆ’22 ∴λ = βˆ’22, Β΅ = 17

Direction vector of l, 𝐝�𝑙 = 𝐧𝑝1 Γ— 𝐧𝑝2 (where 𝐧𝑝1 and 𝐧𝑝2 are the normal vectors of p1 and p2 respectively)

(iii) For the three planes to have no point in common, a.

2 3 1 = οΏ½βˆ’5οΏ½ Γ— οΏ½ 2 οΏ½ = 19 οΏ½1οΏ½ 3 βˆ’5 1

3

Using the common point obtained earlier, πŸ’

βˆ’ 𝟏𝟏 𝟏 βŽ› πŸ’βŽž vector equation of l: 𝐫 = βŽœβˆ’ 𝟏𝟏⎟ + 𝒔 �𝟏� , 𝒔 ∈ ℝ 𝟏 πŸ• 𝐒 ⎝ 𝟏𝟏 ⎠

b.

β‡’ π‘₯ βˆ’ 𝑧 = βˆ’1 β‡’ 𝑦 βˆ’ 𝑧 = βˆ’1

2 3 βˆ’5 3 2 βˆ’5 =οΏ½ �𝐒 βˆ’ οΏ½ �𝐣 + οΏ½ �𝐀 3 βˆ’5 2 βˆ’5 3 2 = 19𝐒 + 19𝐣 + 19𝐀

1 β‡’ 𝐝�𝑙 = οΏ½1οΏ½ 1

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5 1 οΏ½ πœ† οΏ½ βˆ™ οΏ½1οΏ½ = 0 1 17 5 + πœ† + 17 = 0 πœ† = βˆ’22

p3 must not contain l β‡’ al ⋅𝐧𝑝3 β‰  πœ‡

𝐣 𝐀 𝐝𝑙 = οΏ½2 βˆ’5 3 οΏ½ 3 2 βˆ’5

TI-84 Plus

p3 must be parallel to l β‡’ 𝐧𝑝 βŠ₯ 𝐝̂𝑙 direction vector of 𝑙

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βˆ’1 5 β‡’ οΏ½βˆ’1οΏ½ βˆ™ οΏ½βˆ’22οΏ½ β‰  πœ‡ 0 17 πœ‡ β‰  17

np3 dl

p1 p2 p3

aβŠ₯b⇔aβ‹…b=0 al: Pick a point on l obtained from (i)

∴Given that the three planes have no point in common, Ξ» = βˆ’22 and Β΅ can be any real number but not 17.

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GCE β€˜A’ Level October/November 2008 Suggested Solutions

Mathematics H2 (9740/01)

version 1.1

(iv) Let p and q be two distinct direction vectors that lie on the plane (required to get the equation of the plane). n p q

l

(-1, -1, 0)

O

(1, -1, 3)

1 Since plane contains l β‡’ 𝐩 = οΏ½1οΏ½ 1

Direction vector of l, 𝐝�𝑙 obtained in (i).

1 Since plane contains the point (1, βˆ’1, 3) β‡’ πͺ = οΏ½βˆ’1οΏ½ βˆ’ πšπ‘™ 3 1 βˆ’1 2 Position vector of point on l, = οΏ½ οΏ½ βˆ’ οΏ½ οΏ½ = οΏ½ βˆ’1 βˆ’1 0οΏ½ al obtained in (i). 3 0 3 2 1 3 Normal vector to the plane, 𝐧 = 𝐩 Γ— πͺ = οΏ½1οΏ½ Γ— οΏ½0οΏ½ = οΏ½βˆ’1οΏ½ 1 3 βˆ’2 1 3 Vector equation of plane: 𝐫 βˆ™ 𝐧 = οΏ½βˆ’1οΏ½ βˆ™ οΏ½βˆ’1οΏ½ 3 βˆ’2 π‘₯ 3 �𝑦� βˆ™ οΏ½βˆ’1οΏ½ = βˆ’2 𝑧 βˆ’2 ∴ Cartesian equation of plane: πŸ‘π’™ βˆ’ π’š βˆ’ πŸπ’› = βˆ’πŸ

Cartesian equation of plane 𝑛1 π‘₯ 𝐫 βˆ™ 𝐧 = �𝑦� βˆ™ �𝑛2 οΏ½ = 𝑑: 𝑛3 𝑧 𝑛1 π‘₯ + 𝑛2 𝑦 + 𝑛3 𝑧 = 𝑑

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