2010 H2 Maths 9740 Paper 1

GCE ‘A’ Level October/November 2010 Suggested Solutions

Mathematics H2 (9740/01)

version 2.1

MATHEMATICS (H2)

Paper 1 Suggested Solutions 1.

|𝐛| =

∴p =

𝟑

+

(−2)2

Since|𝐚| = |𝐛| 7p = 3

(ii)

𝟕

+

(2)2

2.

= 3

1

= �1 + 𝑥 + 𝑥 2 + ⋯ � [1 + 2𝑥 − ⋯ ]

6

4

𝟐

𝑛

3

1 3 4

8

𝑛(𝑛−1) 4 2!

( 𝑥)2 + ⋯ 3

From MF15: 𝑛(𝑛−1) 2 (1 + 𝑥)𝑛 = 1 + 𝑛𝑛 + 𝑥 +⋯

= 1 + 𝑛𝑥 + 𝑛(𝑛 − 1)𝑥 2 + ⋯ ………………… (2) 3

6

9

2!

Comparing coefficients of x in (1) and (2) 4 3

𝑛 = 3

𝒏 =

𝟗 𝟒

∴ Third term in series (2)

= ⎜7⎟∙⎜7⎟

=

4

⎝7⎠ ⎝7⎠ 13 115 128 = − − +

=

49

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𝑛 4

= 1 + ( 𝑥)1 +

7 ⎛ 23 ⎞

= 0 ∴ (a +b) ∙(a – b) = 0 (shown)

2

𝟓

(ii) (1 + 𝑥)

−1

49

Sub 2x into sin x expansion

= 𝟏 + 𝟑𝒙 + 𝒙𝟐 + ⋯………….………………… (1)

⎡ 7 ⎤ ⎡ 7 ⎤ 1 1 ⎢⎛ 9 ⎞ ⎥ ⎢⎛ 9 ⎞ ⎥ (a +b)∙(a – b) = ⎢⎜ 7 ⎟ + �−2�⎥ ∙ ⎢⎜ 7 ⎟ − �−2�⎥ 2 ⎥ ⎢ 18 2 ⎥ ⎢ 18 ⎣⎝ 7 ⎠ ⎦ ⎣⎝ 7 ⎠ ⎦

49

1

= 1 + 2𝑥 + 𝑥 + 2𝑥 2 + 𝑥 2 + ⋯

7 2 ⎛9⎞ a = �3� = ⎜ 7 ⎟ 7 6 18 ⎝7⎠

32

𝑥2 𝑥3 𝑥𝑟 + +⋯+ + ⋯ 2! 3! 𝑟! (−1)𝑟 𝑥 2𝑟+1 𝑥3 𝑥5 sin 𝑥 = 𝑥 − + − ⋯ +⋯ (2𝑟 + 1)! 3! 5! e𝑥 = 1 + 𝑥 +

Topic: Maclaurin’s Series

2

3

7 ⎛−5⎞

𝐚 ∙ 𝐚 = |𝐚|2

a∙a –a∙b + a∙b–b∙b |𝐚|2 − |𝐛|2 |𝐚|2 − |𝐚|2 (since |𝐚| = |𝐛|) 0 From MF15:

(i) e𝑥 (1 + sin 2𝑥)

6

13

(a + b) ∙(a – b) = = = =

October/November 2010

Topic: Vectorsin Three Dimensions(Points & Lines) 2𝑝 1 (i) Given a = �3𝑝�, b = �−2�, where p > 0. 6𝑝 2 |𝐚| = p√22 + 32 + 62 = 7p �12

ALTERNATIVE APPROACH

9740/01

𝟖 𝟗 𝟓 𝟐

𝟗

� � � − 𝟏� 𝒙𝟐

𝟗 𝟒

𝒙

𝟐

𝟒

= Third term in series (1) (shown)

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Mathematics H2 (9740/01) 3.

version 2.1

(ii)

Topic: Arithmetic Progression, Recurrence (i) Given Sn = n(2n + c)

un+1 = 4(n + 1) + c – 2

un = Sn − Sn−1

= 4n + 4 + c – 2

= n (2n + c) – (n − 1) [2(n – 1) + c] 2

= (4n + c – 2) + 4

2

= 2n + nc – 2(n – 1) – c (n – 1)

= un + 4

= 2n2 + nc – 2 (n2 – 2n + 1) – cn + c


= 4n + c – 2

From un in (i):


u1 = 2 + c

S1 = u1

= 1[2(1) + c]

= 2+c

S 2 = u2 + u1

= 2[4 + c]

= 8 + 2c

= 8 + 2c − 2 – c

= 6+c

= 3(6 + c)

= 18 + 3c

= 18 + 3c − 8 − 2c

= 10 + c

⇒ u2 = S2 − S1 S3 = u3 + u2 + u1 ⇒ u 3 = S3 – S2

S4 = u4 + u3 + u2 + u1 = 4[8 + 2c] ⇒ u4 = S4 – S3

un = 4n + c – 2

u2 = 6 + c = 4 + (2 + c) = 4 + u1 u3 = 10 + c = 4 + (6 + c) = 4 + u2 u4 = 14 + c = 4 + (10 + c) = 4 + u3 ∴un+1 = 4 + un

f(un) = 4 + un

= 32 + 4c

= 32 + 4c − 18 − 3c = 14 + c

Since u2 − u1 = u3 − u2 = u4 − u3 = 4 ⇒ un is an arithmetic progression (A. P.).

1st term, a = 2 + c Common difference, d = 4

∴un = 2 + c + (n – 1) 4 = 2 + c + 4n – 4 = 4n + c– 2

nth term in an A.P.: Tn = a + (n – 1) d



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Mathematics H2 (9740/01) 4.

version 2.1

Topic: Differentiation (Tangents &Normals) (i)

2

5.

2

= f(x) (i) Given y = x3 f(x – 2) Translate 2 units in the positive x- direction

Given x – y + 2xy + 4 = 0………………… (1) Using implicit differentiation, 2𝑥 − 2𝑦

d𝑦 d𝑥

+ �2𝑦 + 2𝑥

d𝑦 d𝑥

� = 0

𝑥 + 𝑦 = (𝑦 − 𝑥) ∴

𝐝𝒚 𝐝𝒙

=

𝒙+𝒚 𝒚−𝒙

y = (x – 2)3 = g(x)

d𝑦

1

d𝑥

2

⇒

d𝑥

𝑥+𝑦

𝑦−𝑥

y =

= 0 = 0

y =

1 2

2

(x – 2)3 = h(x)

𝟏 𝟐

(x – 2)3− 6

When x = 0, y =

Sub (2) into (1)

1

(−2)3 – 6

1

(x – 2)3 − 6

2

= − 10

x2 – (− x)2 + 2x(− x) + 4 = 0 x2 – x2 – 2x2 + 4 = 0

When y = 0, 0 =

2

2x = 4

Sub x= √2 into (2)

1

h(x)– 6 Translate 6 units in the negative y -direction

y = − x ……………… (2)

Sub x = − √2 into (2)

g(x)

Stretch with scale factor parallel to the y- axis

(ii) For the tangent of the curve to be parallel to the x-axis, d𝑦

Topic: Functions, Graphs (Transformations)

2

(x – 2)3 = 12

x = ±√2

3

x = 2 + √12

𝟑

∴Curve crosses the y-axis at (0, −10) and the x-axis at (2 + √𝟏𝟐, 0).

y = √2

y = − √2

The coordinates are (−√𝟐, √𝟐) and (√𝟐, −√𝟐).



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version 2.1

(ii) Using G. C. (refer to Appendix for detailed steps),

6.

(i) Using G.C. for y = x3 − 3x + 1 (refer to Appendix for detailed steps and various methods that can be used),

y = f−1(x)

y=x

Topic: Integration

y = f(x)

β = 0.3473, γ = 1.5321 ∴β = 0.347 andγ= 1.532(3d.p.) (ii) Using G.C. (refer to Appendix for detailed steps and various methods used),

The graph of y = f−1(x) is obtained by reflecting the graph of y = f(x) in the line y = x.

For y = f−1(x), when y = 0, x = −10 𝟑 x = 0, y = 2 + √𝟏𝟐

The x-coordinate in f(x) becomes the y-coordinate in f−1(x) and vice versa.



Area ofthe region bounded by the curve and the x- axis (between x = 𝛽 and x 𝛾 = 𝛾) = �∫𝛽 (𝑥 3 − 3𝑥 + 1) d𝑥� = | (−0.7814168)| ≈ 0.781 units2(3sig.fig.)

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Remember to obtain the absolute value for the answer as we are finding the area.


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version 2.1

(iii) Obtain the x-coordinates of the intersection points for 3

y = x – 3x + 1 y = 1

(1) = (2)

(iv)

………………… (1) ………………… (2)

x3– 3x + 1 = 1 x(x2 – 3) = 0 x = 0, x = ±√3

From the diagram, the lower and upper limit of integration to find the region is −√3 and 0 respectively.

Hence, area of region (shaded region A) between the curve and the line = = = = =

0

∫−√3(𝑥 3 − 3𝑥 + 1)d𝑥 − �1 × √3� 1

3

0

� 𝑥 4 − 𝑥 2 + 𝑥� 4

1

2

3

−√3

− √3

Area of rectangle

�0 − � × 9 − (3) − √3�� − √3 − 𝟗 𝟒

9

4

4 9

2

Sincex = −1 and x = 1 are stationary points, When x = −1, ymax = (−1)3 – 3(−1) + 1 = 3 When x = 1, ymin = (1)3 – 3(1) + 1 = −1 For x3 – 3x + 1 = k to have three real distinct roots, ymin