Add Maths 1990 Paper 1

2000-CE A MATH

HONG KONG EXAMINATIONS AUTHORITY

PAPER 1

HONG KONG CERTIFICATE OF EDUCATION EXAMINATION 2000

ADDITIONAL MATHEMATICS PAPER 1 8.30 am – 10.30 am (2 hours) This paper must be answered in English

1.

Answer ALL questions in Section A and any THREE questions in Section B.

2.

All working must be clearly shown.

3.

Unless otherwise specified, numerical answers must be exact.

4.

In this paper, vectors may be represented by bold-type letters such as u, but & candidates are expected to use appropriate symbols such as u in their working.

5.

The diagrams in the paper are not necessarily drawn to scale.

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2000-CE-A MATH 1–1

FORMULAS FOR REFERENCE

sin ( A ± B ) = sin A cos B ± cos A sin B cos ( A ± B) = cos A cos B # sin A sin B tan ( A ± B) =

tan A ± tan B 1 # tan A tan B

sin A + sin B = 2 sin

A+ B A− B cos 2 2

sin A − sin B = 2 cos

A+ B A− B sin 2 2

cos A + cos B = 2 cos

cos A − cos B = −2 sin

A+ B A− B cos 2 2 A+ B A− B sin 2 2

2 sin A cos B = sin ( A + B) + sin ( A − B) 2 cos A cos B = cos ( A + B) + cos ( A − B) 2 sin A sin B = cos ( A − B) − cos ( A + B)

2000-CE-A MATH 1–2

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Section A (42 marks) Answer ALL questions in this section.

1.

Solve

1 >1. x (3 marks)

2.

Find

(a)

d sin 2 x , dx

(b)

d sin 2 (3 x + 1) . dx (4 marks)

3.

(a)

Show that

(b)

Find

1 x + ∆x

−

1 x

=

−∆ x x ( x + ∆x ) ( x + x + ∆x )

.

d 1 ( ) from first principles. dx x (5 marks)

4.

P (−1, 2) is a point on the curve ( x + 2) ( y + 3) = 5 . Find dy at P, dx

(a)

the value of

(b)

the equation of the tangent to the curve at P. (5 marks)

5.

(a)

Solve | 1 − x | = 2.

(b)

By considering the cases | 1 − x | = x −1.

x ≤ 1 and x > 1 , or otherwise, solve (5 marks)

2000-CE-A MATH 1–3

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6.

Express the complex number

1+ 3i 3+ i

in polar form.

 1+ 3i   Hence find the argument θ of   3+ i   principal values −π < θ ≤ π . 7.

2000

, where θ is limited to the (6 marks)

α and β are the roots of the quadratic equation x 2 + ( p − 2) x + p = 0 , where p is real. (a)

Express α + β and α β in terms of p.

(b)

If α and β are real such that α 2 + β 2 = 11 , find the value(s) of p. (7 marks)

8.

B D

C

O

Figure 1

A In Figure 1, OA = i , OB = j .

C is a point on OA produced such that

AC = k , where k > 0 . D is a point on BC such that BD : DC = 1 : 2 . 1+ k 2 i + j. 3 3

(a)

Show that OD =

(b)

If OD is a unit vector, find (i)

k,

(ii)

∠ BOD , giving your answer correct to the nearest degree. (7 marks)

2000-CE-A MATH 1–4

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Section B (48 marks) Answer any THREE questions in this section. Each question carries 16 marks. 9.

F E C

A

D

B

Figure 2

O

In Figure 2, OAC is a triangle. B and D are points on AC such that AD = DB = BC. F is a point on OD produced such that OD = DF . E is a point on OB produced such that OE = k (OB) , where k > 1. Let OA = a and OB = b . (a)

(i)

Express OD in terms of a and b.

(ii)

Show that OC = −

(iii)

Express EF in terms of k, a and b.

1 3 a+ b. 2 2

(5 marks) (b)

It is given that OA = 3 , OB = 2 and ∠AOB = 60° . (i)

Find a . b and b .b .

(ii)

Suppose that ∠OEF = 90° .

2000-CE-A MATH 1–5

(1)

Find the value of k.

(2)

A student states that points C, E and F are collinear. Explain whether the student is correct. (11 marks) –4–

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10.

Let f ( x) = (a)

7 − 4x x2 + 2

.

(i)

Find the x- and y-intercepts of the curve y = f ( x) .

(ii)

Find the range of values of x for which f ( x) is decreasing.

(iii)

Show that the maximum and minimum values of f ( x) are 4 and −

1 respectively. 2 (9 marks)

(b)

In Figure 3, sketch the curve y = f ( x) for −2 ≤ x ≤ 5 . (3 marks)

(c)

Let p =

7 − 4 sin θ sin 2θ + 2

, where θ is real.

From the graph in (b), a student concludes that the greatest and 1 least values of p are 4 and − respectively. Explain whether 2 the student is correct. If not, what should be the greatest and least values of p ? (4 marks)

2000-CE-A MATH 1–6

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Candidate Number

Centre Number

Seat Number

Total Marks on this page

If you attempt Question 10, fill in the first three boxes above and tie this sheet into your answer book. 10.

(b)

(continued) y

–2

O

5

x

Figure 3

2000-CE-A MATH 1–7

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This is a blank page.

2000-CE-A MATH 1–8

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11.

(a)

Let w = cos θ + i sin θ , where 0 < θ < π . It is given that the 5 complex number w 2 + − 2 is purely imaginary. w Show that 2 cos 2θ + 5 cos θ − 3 = 0 . Hence, or otherwise, find w. (8 marks)

(b)

A and B are two points in an Argand diagram representing two distinct non-zero complex numbers z1 and z2 respectively. Suppose that z 2 = w z1 , where w is the complex number found in (a). z2 z and arg ( 2 ) . z1 z1

(i)

Find

(ii)

Let O be the point representing the complex number 0. What type of triangle is ∆OAB ? Explain your answer. (8 marks)

2000-CE-A MATH 1–9

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12.

Consider the function f ( x) = x 2 − 4 m x − (5m 2 − 6 m + 1) , where m > (a)

1 . 3

Show that the equation f ( x) = 0 has distinct real roots. (3 marks)

(b)

Let α, β be the roots of the equation f ( x) = 0 , where α < β . (i)

Express α and β in terms of m.

(ii)

Furthermore, it is known that 4 < β < 5. 6 . 5

(1)

Show that 1 < m <

(2)

Figure 4 shows three sketches of the graph of y = f ( x ) drawn by three students. Their teacher points out that the three sketches are all incorrect. Explain why each of the sketches is incorrect. y

y = f (x) –1

O

x 4 5

Sketch A Figure 4

2000-CE-A MATH 1–10

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12.

(b)

(ii)

(2)

(continued) y y = f (x)

O –1

x 4 5

Sketch B

y

y = f (x) O –1

x 4 5

–1 Sketch C

Figure 4 (continued)

2000-CE-A MATH 1–11

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(13 marks)

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13. B N

R

ym Q

)θ

100 m A xm P 100 m

Figure 5

Two boats A and B are initially located at points P and Q in a lake respectively, where Q is at a distance 100 m due north of P. R is a point on the lakeside which is at a distance 100 m due west of Q. (See Figure 5.) Starting from time (in seconds) t = 0, boats A and B sail northwards. At time t, let the distances travelled by A and B be x m and y m respectively, where 0 ≤ x ≤ 100 . Let ∠ARB = θ. (a)

Express tan ∠ ARQ

in terms of x.

Hence show that tan θ = (b)

100 (100 − x + y ) . 10000 − 100 y + x y

(4 marks)

Suppose boat A sails with a constant speed of 2 m s–1 and B adjusts its speed continuously so as to keep the value of ∠ARB unchanged. 100 x . 200 − x

(i)

Using (a), show that y =

(ii)

Find the speed of boat B at t = 40.

(iii)

Suppose the maximum speed of boat B is 3 m s–1. Explain whether it is possible to keep the value of ∠ARB unchanged before boat A reaches Q. (12 marks) END OF PAPER

2000-CE-A MATH 1–12

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2000 Additional Mathematics Paper 1 Section A 1.

0 < x <1

2.

(a)

2 sin x cos x

(b)

6 sin (3 x + 1) cos(3x + 1)

3.

(b)

− x 2x 2

4.

(a)

–5

(b)

5x + y + 3 = 0

(a)

x = −1 or 3

(b)

x ≥1

5.

6.

cos

π π 2π + i sin , − 3 6 6

7.

(a)

2− p, p

(b)

–1

(b)

(i)

8.

(ii)

5 −1 48°

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Section B Q.9

(a)

(i) OD =

(ii) OB =

a+b 2

OA + 2OC 1+ 2

a + 2OC 3 1 3 OC = − a + b 2 2 b=

(iii)

EF = OF − OE = 2OD − k OB a+b = 2( ) − kb 2 = a + (1 − k ) b

(b)

(i) a . b = | a | | b | cos ∠ AOB = 3(2)cos60° =3 b . b = | b |2 = 4 (ii) (1) OE . EF = 0 k b . [ a + (1 − k ) b ] = 0 ka . b + k (1 − k ) b . b = 0 3k + 4k (1 − k ) = 0 7 k − 4k 2 = 0 k = 0 (rejected) ∴ k=

or k =

7 4

7 . 4

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(2) Put k =

7 : 4

7 3 EF = a + (1 − )b = a − b 4 4 CE = OE − OC 7 1 3 = b − (− a + b) 4 2 2 1 1 = a+ b 2 4 Since CE ≠ µ EF , C, E, F are not collinear. The student is incorrect.

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Q.10 (a)

7 ∴ the y-intercept is 2 7 ∴ the x-intercept is Put y = 0 , x = 4

(i) Put x = 0 , y =

(ii) f ( x) is decreasing when f ′( x) = =

7 . 2 7 . 4

f ′( x) ≤ 0 .

2

− 4( x + 2) − ( 7 − 4 x ) 2 x ( x 2 + 2) 2 4 x 2 − 14 x − 8 ( x 2 + 2) 2

4 x 2 − 14 x − 8

≤0 ( x 2 + 2) 2 (2 x + 1) ( x − 4) ≤ 0 −

1 ≤x≤4 2

(iii) f ( x) is increasing when f ′( x) ≥ 0, i.e. x ≥ 4 or x ≤ −

1 . 2

f ′ ( x) = 0 when x = 4 or −

1 . 2

As f ′( x ) changes from positive to negative as x increases through − maximum at x = −

1 , so f ( x) attains a 2

1 . 2

1 , y=4 2 ∴ the maximum value of f ( x) is 4. As f ′( x ) changes from negative to positive as At x = −

x increases through 4, so f ( x) attains a minimum at x = 4 . At x = 4 , y = −

1 2

∴ the minimum value of f ( x) is −

1 . 2

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(b)

y

1 (− , 4) 2 7 2 (–2,

5 ) 2

y = f(x)

7 4 –2

5

O (4, – (5, −

(c)

1 ) 2

13 ) 27

Put x = sinθ , f (sinθ ) = 7 − 24 sin θ = p . sin θ + 2

The range of possible value of sinθ is −1 ≤ sinθ ≤ 1. From the graph in (b), the greatest value of f(x) in the range −1 ≤ x ≤ 1 is 4. ∴ the greatest value of p is 4 and the student is correct. From the graph in (b), f(x) attains its least value at one of the end-points. f (1) = 1 , f ( − 1) =

11 . 3

∴ the least value of p is 1 and the student is incorrect.

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x

Q.11 (a)

w = cos θ + i sin θ w 2 = cos 2θ + i sin 2θ 1 1 = cos θ + i sin θ w = cos(−θ ) + i sin( −θ ) = cos θ − i sin θ 5 w2 + − 2 w = cos 2θ + i sin 2θ + 5(cos θ − i sin θ ) − 2 = cos 2θ + 5 cos θ − 2 + i(sin 2θ − 5 sin θ ) 5 − 2 is purely imaginary, w cos 2θ + 5 cos θ − 2 = 0 (2 cos 2 θ − 1) + 5 cos θ − 2 = 0

Since w 2 +

2 cos 2 θ + 5 cos θ − 3 = 0 1 cos θ = or cosθ = –3 2

(rejected)

π ( 0 < θ < π ) 3 Imaginary part π 2π = sin − 5 sin ≠ 0 3 3

θ=

∴ w = cos

(b)

(i)

π π + i sin 3 3

z2 =|w| z1 =1 z2 arg( ) = arg( w) z1 =

π 3

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(ii)

z2 |z | = 2 =1 z1 | z1 | ∴ | z 2 | = | z1 | i.e. OA = OB. ∠AOB = arg( z 2 ) − arg( z1 ) = arg( =

z2 ) z1

π 3

Since OA = OB , ∆OAB is isosceles.

π π 1 (π − ) = 2 3 3 ∴ ∆OAB is equilateral. ∠OAB = ∠OBA =

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Q.12 (a)

f ( x) = x 2 − 4 m x − (5m 2 − 6 m + 1) Discriminant ∆ = (−4m) 2 + 4(5m 2 − 6 m + 1)

= 36 m 2 − 24 m + 4 = 4(9 m 2 − 6 m + 1) = 4(3 m − 1) 2 > 0

1 ( m > ) 3

∴ the equation f ( x) = 0 has distinct real roots. (b)

(i)

4m ± ∆ 2 = 2 m ± (3m − 1) Since α < β , α = 2 m − (3 m − 1) = − m + 1 x=

β = 2 m + (3 m − 1) = 5m − 1 (ii) (1) Since 4 < β < 5 , 4 < 5m − 1 < 5 5 < 5m < 6 1< m <

6 5

(2) Sketch A : Since the coefficient of x2 in f(x) is positive, the graph of y = f(x) should open upwards. However, the graph in sketch A opens downwards, so sketch A is incorrect. Sketch B : Since α = 1 − m and 1 < m < 1 −1 > 1− m > 1 −

6 , 5

6 5

1 5 In sketch B, α is less than –1, so sketch B is incorrect.

0 >α > −

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Sketch C :  y = x 2 − 4mx − (5m 2 − 6m + 1)  y = −1 

− 1 = x 2 − 4mx − (5m 2 − 6m + 1) x 2 − 4mx − (5m 2 − 6 m) = 0 – – – – (*) Discriminant ∆ = (−4m) 2 + 4(5m 2 − 6m) = 36m 2 − 24m = 12m(3m − 2) 6 , ∆ >0. 5 As ∆ > 0 , equation (*) has real roots, i.e. y = f(x) and y = –1 always have intersecting points. However, the line and the graph in sketch C do not intersect, so sketch C is incorrect. Since 1 < m <

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Q.13 (a)

100 − x 100 tanθ = tan (∠ARQ + ∠QRB) tan ∠ARQ + tan ∠QRB = 1 − (tan ∠ARQ ) (tan ∠QRB )

tan ∠ARQ =

y 100 − x + 100 100 = y 100 − x )( ) 1− ( 100 100 100(100 − x + y ) = 10000 − 100 y + xy (b)

(i) At t = 0 , tan θ =

PQ RQ

100 = 1. 100 Since ∠ARB remains unchanged, 100(100 − x + y ) =1 10000 − 100 y + xy 10000 − 100 x + 100 y = 10000 − 100 y + xy 200 y − xy = 100 x =

y=

(ii)

100 x 200 − x

d y (200 − x ) (100) − 100 x(−1) d x = dt dt (200 − x ) 2 = =

dx (200 − x) d t 20000

2

40000

(200 − x) 2 At t = 40 , x = 40 × 2 = 80. dy 40000 = d t (200 − 80) 2 =

25 9

∴ the speed of boat B at t = 40 is

25 m s −1 . 9

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(iii)

From (ii),

dy 40000 = d t (200 − x) 2 dy ≤3 dt 40000 (200 − x) 2 200 − x ≥

≤3

200

x ≤ 200(1 −

3 1 3

When x > 200(1 −

) 1

),

dy > 3. dt

3 So it is impossible to keep ∠ARB unchanged before boat A reaches Q.

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