Add Math (4047/01) - korlinang · PDF fileAdd Math (4047/01) Prepared by Mr Ang, Nov 2016 3 3. (a) State the values between which each of the following must lie:

Add Math (4047/01)

Prepared by Mr Ang, Nov 2016 1

Requirement : Answer all questions

Total marks : 80

Duration : 2 hours

1. The equation of a curve is 22 4y x kx , where k is a constant, and the equation of a line

is 2 12y x .

(i) In the case where k = 6, find the coordinates of the points of intersection of the line

with the curve. [3]

(ii) Show that, for all values of k, the line intersects the curve at two distinct points. [2]

Solution:

(i) Solving 22 4y x kx and 2 12y x ,

22 2 16 0x k x ----- (1)

For k = 6,

2 2 8 0x x

2 4 0x x

2x or 4x

When 2x , 12 2 2 16y

When 4x , 12 2 4 4y

the coordinates of the points of intersection (‒2, 16) and (4, 4)

(ii) From equation (1),

The discriminant

2

2 4 2 16k

2

2 128 128 0k

Since the discriminant > 0, the equation 22 2 16 0x k x will always have two distinct

roots, for all values of k.

Add Math (4047/01)


2. A rectangular block has a square base of side 6 2 cm and a height of h cm.

The volume of the rectangular block is 16 4 3 cm3. Without using a calculator, show

that h can be expressed as 3a b , where a and b are integers. [5]

Solution:

2

16 4 3

6 2

h

2

2 8 2 3

2 3 1

h

2

2 4 3

3 1

h

2

2 4 3

3 1

h

2 4 3

2 2 3h

4 3 2 3

2 3 2 3h

2

2

8 2 3 4 3 3

2 3

h

5 2 3h

Add Math (4047/01)


3. (a) State the values between which each of the following must lie:

(i) the principal value of 1sin x ,

(ii) the principal value of 1cos x .

(b)

The figure shows part of the graph of cosx

y a cb

.

Find the value of each of the constants a, b and c. [3]

Solution:

(a) (i) 190 sin 90x

(ii) 10 cos 180x

(b)

3 1

22

c

1 3

12

a

Period = 12π

12

62

b

y

x O ‒3π

3

2

1

‒6π ‒9π ‒12π 3π 6π 9π 12π

Add Math (4047/01)


4. (i) On the same axes sketch the curves 2 256y x and 22y x . [3]

(ii) Find the equation of the line passing through the points of intersection of the two

curves. [2]

Solution:

(i)

(ii) Solve 2 256y x and 22y x

2

22 256x x

44 256x x

4 64 0x x

3 64 0x x

0x or 3 64x

4x

Coordinates of the points of intersection are (0, 0) and (4, 32).

Gradient of the line = 8

Equation of the line, 8y x

y

x O

22y x

2 256y x

Add Math (4047/01)


5. Express 2

2

2 4 31

6

x x

x x

in partial fractions. [5]

Solution:

2

2 2

2 4 31 2 192

6 6

x x x

x x x x

.

2

2 19 2 19

6 3 2

x x

x x x x

Let

2 19

3 2 3 2

x A B

x x x x

2 19 2 3x A x B x

When 3x , 25 5A

5A

When 2x , 15 5B

3B

2

2

2 4 31 5 32

6 3 2

x x

x x x x

.

Add Math (4047/01)


6.

The diagram shows part of the curve 2y ax bx c , where 0a .

The curve touches the x-axis at A(p, 0) and at B(6,0) and has a maximum point at M(4, 6).

(i) Explain why p = 2. [1]

(ii) Determine the value of each of a, b and c. [4]

(iii) State the set of values of q for which the line y = q intersects the curve at four distinct

points. [2]

Solution:

(i) Since 0a , consider 2y ax bx c

Line of symmetry, 4x

4 6 4 2p

(ii)

Let 2 6y a x x

For 4, 6 ,

6 4 2 4 6a

6

4a

3

2a

(iii) 0 6q

y

x O A (p, 0)

2y ax bx c

B (6, 0)

M (4, 6)

y

x O A (p, 0)

2y ax bx c

B (6, 0)

M (4, ‒6)

Add Math (4047/01)


7. At a given instant, a cyclist is at a cross roads O, travelling due west at a constant speed

of 5 m/s.

At the same instant a second cyclist is 100 m from O, travelling due north towards O at a

constant speed of 10 m/s. This situation is shown in Fig. 1.

The position, t seconds later, when the cyclists have reached points P and Q, is shown in

Fig. 2.

(i) Express OP and OQ in terms of t and hence show that the distance, s m, between the

two cyclist at time t is given by 2125 16 80s t t . [3]

(ii) Obtain an expression for d

d

s

t. [2]

(iii) Find the least distance between the two cyclists. [2]

Solution:

(i) 5OP t

100 10OQ t

2 2

5 100 10s t t

N • O 5 m/s

Q • 10 m/s

P •

Fig. 2

N • O 5 m/s

100 m

• 10 m/s

Fig. 1

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2 225 10000 100 2000s t t t

2125 2000 10000s t t

2125 16 80s t t

(ii) 2 2125 16 80s t t

d

2 125 2 16d

ss t

t

d 125

8d

st

t s

2

125 8d

d 125 16 80

ts

t t t

(iii)

Let d

0d

s

t ,

2

125 80

125 16 80

t

t t

0 125 8t

8t

When 7t , d

0d

s

t , s is downward sloping.

When 8t , d

0d

s

t , s is a stationary point.

When 9t , d

0d

s

t , s is upward sloping.

When 8t , s is a locally minimum point.

2125 8 16 8 80 20000 44.7s (3 s.f.)

the least distance between the two cyclists is 44.7 m when 8t s.

Add Math (4047/01)


8.

The diagram shows a triangle ABC in which the point A is (‒2, 6), the point C lies on

the x-axis and angle ABC is 90°. The equation of BC is 2 3 45y x .

(i) Find the coordinates of B [5]

(ii) Given that M is the midpoint of AC and that ABCD is a rectangle,

find the coordinates of M and of D. [3]

Solution:

(i) 2 3 45y x , gradient of BC = 3

2

Gradient of AB =2

3

The equation of AB ,

26

2 3

xy

2 22

3 3y x

2 22

2 3 453 3

x x

13 91

3 3x

7x

2 3 7 45y

12y

the coordinates of B (7, 12)

y

x O

A

(‒2, 6)

2 3 45y x

B

C

Add Math (4047/01)


(ii) A(‒2, 6). Let 0y , 15x . C(15, 0)

M(2 15

2

,

6 0

2

)

M(6.5, 3)

Let D(a, b), B (7, 12)

7

6.52

a

6a

12

32

b

6b

D(6, ‒6)

Add Math (4047/01)


9. The equation of a curve is 2

2

162y x

x .

(i) Find the coordinates of the stationary points of the curve. [5]

(ii) Use the second derivative test to determine the nature of each of these points. [3]

Solution:

(i) 3

d 322

d

yx

x x

Let d

0d

y

x ,

3

320 2x

x

4 16x

2x

When 2x , 2

2

162 2 6

2y

When 2x ,

2

2

162 2 6

2y

the coordinates of the stationary points are (2, ‒6) and (‒2, ‒6)

(ii) 3

d 322

d

yx

x x

2

2 4

d 962

d

y

x x

When 2x , 2

2

d0

d

y

x , it is a locally maximum point.

When 2x , 2

2

d0

d

y

x , it is a locally maximum point.

Add Math (4047/01)


10. (i) Show that 2 3 3

d ln 1 2ln

d

x x

x x x x

. [3]

(ii) Integrate 3

ln x

x with respect to x. [3]

(iii) Given that the curve fy x passes through the point 3

1,4

and is such that

3

lnf

xx

x , find f x . [2]

Solution:

(i) let 2

ln xy

x , we get

2

22

12 ln

d

d

x x xy x

x x

4

d 2 ln

d

y x x x

x x

4

1 2lnd

d

x xy

x x

3

d 1 2ln

d

y x

x x

Therefore,2 3 3

d ln 1 2ln

d

x x

x x x x

(ii) since 2 3 3

d ln 1 2ln

d

x x

x x x x

2 3 3

d ln 1 2lnd d

d

x xx x

x x x x

2 3 3

d ln 1 2lnd d d

d

x xx x x

x x x x

3 3 2

2ln 1 d lnd d d

d

x xx x x

x x x x

3 3 2

2ln 1 d ln2 d d d

2 d

x xx x x

x x x x

3 2 2

ln 1 lnd

4 2

x xx c

x x x

Add Math (4047/01)


(iii) Given that 3

lnf

xx

x and the curve fy x passes through the point

31,

4

.

3

lnf d d

xx x x

x

2 2

1 lnf

4 2

xx c

x x

2 2

1 ln

4 2

xy c

x x

At 3

1,4

,

3 1 ln1

4 4 2c

1c

2 2

1 ln1

4 2

xy

x x

Add Math (4047/01)


11. (a) Prove that 2 1 sin

sec tan1 sin

. [4]

(b)

A public building has a large clock set on an exterior wall. The distance from the

centre of the clock, O, to the tip, M, of the minute hand is 80 cm.

The distance, d cm, of M from the vertical line through O is given by 80sind kt ,

where t is the time in minutes past the hour.

(i) Find the value of k in radians per minute. [1]

(ii) For how long in each hour is d > 40? [3]

Solution:

(a) R.H.S

21 sin

1 sin 1 sin

2

2

1 sin

1 sin

2

2

1 sin

cos

21 sin

cos

21 sin

cos cos

2

sec tan

XII

80 cm

O III

VI

IX •

d cm M

Add Math (4047/01)


(b) (i) When t = 15, kt = 2

k = 30

(ii) For 80sin 40kt ,

80sin 40kt or 80sin 40kt

sin 0.5kt

1sin 0.5kt

From the sketch of the graph of siny kt ,

Principal angle, 1sin 0.56

kt

6

kt

or 5

6 6kt

5 2 30

206 6 30 3

t

By symmetry, d > 40, 2 20 2 40t minutes

y

kt O

‒1

1

0.5

‒π π 2π

Add Math (4047/01)


12. It is given that f sin cos 2x dx x k x c , where c is a constant of integration,

and that 6

0

3f

4x dx

.

(i) Show that 1

2k . [1]

(ii) Find f x . [3]

(iii) Find the equation of the normal to the curve fy x at the point where 6

x

. [5]

Solution:

(i) 6

0

3f

4x dx

3

sin cos 2 64

0

x k x c

3

sin cos sin 0 cos06 3 4

k c k c

1 3

2 2 4

kc k c

1 3

2 2 4

k

1

2k

(ii) 1

f sin cos 22

x dx x x c

d 1

f sin cos 2d 2

x x x cx

f cos sin 2x x x

(iii) d

sin 2cos 2d

yx x

x

When 6

x

,

d

sin 2cosd 6 3

y

x

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d 1 1 1

2d 2 2 2

y

x

cos sin 36 3

y

Gradient of the normal at 6

x

is ‒2.

3

2 6

yx

2 33

y x

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Add Math (4047/01) - korlinang · PDF fileAdd Math (4047/01) Prepared by Mr Ang, Nov 2016 3 3. (a) State the values between which each of the following must lie: