Modul2 Jpnt

KEM SPM TERBILANG TERENGGANU 2008 (MODUL 2)

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ADDITIONAL MATHEMATICS PAPER 2

 MODULE 2  PAPER 2 (3472/2)

 DIFFERENTIATION / INTEGRATION



5 cm

1.

In the diagram above, water is poured into a cylindrical container at a constant rate of p cm3 s1. The height of the water level in the container is increasing at a rate of 2 cms1. Given that the radius of the cylinder is 5 cm, find (a) the value of p, (b) the rate of change of the curved surface area of the container filled with water, (c) the time taken for the container to be fully filled with water if the volume of the cylinder is 1000 cm3.

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 2)

2.

(a) Given that y =

64 x

3

, find the value of

the approximate value of

64 (1  97) 3

ADDITIONAL MATHEMATICS PAPER 2

dy dx

when x = 2. Hence, by using differential method, find

.

10 cm

30 cm

.

(b) The diagram above shows a conic container with rounded surface radius of 10 cm and the height of 30 cm. Water is poured into the container at a rate of 15 cm3 s1. Find in terms of , the rate of increase of water height when the height is 9 cm from its vertex.

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 2)

3.

(a) Find



( x  2)( x  2) x2

ADDITIONAL MATHEMATICS PAPER 2

dx .

4 cm h cm

perfume (b) The diagram above shows a pyramid-shaped perfume container with square base of area 36 cm2. The height of the pyramid is 4 cm. The perfume is poured into the container such that the square-shaped surface area of the perfume is 4p2 cm2 and its height from the vertex is h cm. 3 (i) Show that the volume of the space not occupied by the perfume is V = (64 – h3). 4 (ii) Given that the rate of change of perfume height is 0·2 cms1, calculate the rate of change of unoccupied space volume when h = 2 cm. 1 [Volume of pyramid =  area of base  height] 3

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 2)

ADDITIONAL MATHEMATICS PAPER 2

B

A E D

C 4.

(a) Ren has a piece of wire of 120 cm long. He intends to form a shape as shown in the diagram above. Given that AB = DC and AED is a semi circle with radius x cm. 3 (i) Show that the area of the enclosed region, A cm2, is given by A = 120r – 4r2 – r2. 2 (ii) Find the value of x when A is maximum and show that the value of A is maximum. (b) The volume of a sphere increases at a rate of 20 cm3s1. Find the rate of change of the surface area of the sphere when the radius is 8 cm. [Volume of a sphere =

4 3

 r 3]

[Surface area of a sphere = 4 r 2]

10 cm Water surface

h cm

5.

(a) The diagram above shows a hemispheric bowl with radius 10 cm. Water is poured into the bowl such that the height of the water surface from its base, h cm, increases at a rate of 02 cms1. (i) Show that the area of the water surface, L cm2, is given by L = (20h – h2). (ii) Find the rate of increase area of the water surface when h = 4 cm.

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 2)

ADDITIONAL MATHEMATICS PAPER 2

y B 

A

O

x

(b) The diagram above shows the curve y = 6x  3x2. AB is a tangent to the curve at the maximum point, B. Calculate the volume of the solid block generated when the shaded area is revolved 360 about x-axis. b

6.



(a) Give the geometrical meaning of the expression

f ( x) dx .

a

b

3

Hence, if f (x) = 2x , find the value of b when



f ( x) dx = 312.

1

y

y=

x2 2

1

x+y=5

O

x x2

 1 and a straight line x + y = 5. 2 Find the volume generated when the shaded region is rotated through 360º about y-axis.

(b) The diagram above shows the curve y =

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 2)

7.

ADDITIONAL MATHEMATICS PAPER 2

The diagram above shows a block which consists of a hemisphere placed on top of the cylinder. Given that the hemisphere and the cylinder have the same radius, r cm, and the volume of the cylinder is 81 cm3. (a) Find the total surface area of the block in terms of r. (b) Given that the radius increases at the rate of 08 cms – 1, find the rate of change of the surface area of the block when its radius is 4 cm. (c) If the radius of the cylinder increases from 4 cm to 4008 cm, find the approximate value of the changes in the surface area of the block.

y

y = (x – 1)(x – 4) A

O

8.

C

x

B

(a) The diagram above shows the shaded region bounded by the curve y = (x – 1)(x – 4) and A straight line AB. Find (i) the coordinates of A, B and C, (ii) the area of the shaded region.

y

y = x2 – 2 x 3

O 2008 Hak Cipta Jabatan Pelajaran Negeri Terengganu



y 5

=1

x 6

KEM SPM TERBILANG TERENGGANU 2008 (MODUL 2)

ADDITIONAL MATHEMATICS PAPER 2

(b) The diagram above shows the graph y = x2 – 2 and the straight line

x



y

= 1. 3 5 Find the volume generated when the shaded area is rotated 360 about y-axis.

9.

d 

2

 2x . Hence, evaluate (a) Show that   = dx  1  5 x  (1  5 x )3 x

y

3

4x

 (1  5 x)

3

.

1

y=k

J

y=x2+2



0

1

1

x

2

2

(b) Based on the graph above, (i)

find the area of the shaded region.

(ii) J is a block produced when the region bounded by the curve y = x2 + 2 and a line y = k 1 is rotated through 180 about y-axis. If the volume produced is  unit3, find the value 2 of k.

10. (a) Given that y =

(3 x 2  5)3 (2 x  5)

3

, find

dy dx

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 2)

ADDITIONAL MATHEMATICS PAPER 2

Q

R

P

O

(b) The diagram above shows a quadrilateral PQRS enclosed in a circle. Given that PQ = 2x cm and QR = 5 cm. (i)

Show that the area of the shaded region, A cm2, is given by A = πx 2 – 10x +

25

π.

4

(ii) Find the value of x such that A is minimum. (iii) If x increases from 6 cm to 605 cm, find the approximate increase in the area of the shaded region.

11. (a) Find the equation of a curve which has gradient (2x + 1)3 and passes through the point 1   ,  3 . 2 

y

y = x(x – 1)(x + 4)

O

x

(b) The diagram above shows the graph y = x(x – 1)(x + 4). Calculate the area bounded by the curve and the x-axis.

y Q 2008 Hak Cipta Jabatan Pelajaran Negeri Terengganu

O

R(2, 0)

8

P

x

KEM SPM TERBILANG TERENGGANU 2008 (MODUL 2)

ADDITIONAL MATHEMATICS PAPER 2

12. (a) The diagram above shows part of the curve y2 = 4x. The point P lies on the x-axis and the point O lies on the curve. PQ is parallel to y-axis and PQ = p unit. Given that R is a fixed point (2, 0), express the area of PQR, A, in terms of p. dA . (ii) Find dp

(i)

The point P moves along x-axis and point Q moves along the curve such that PQ is always parallel to y-axis. Given that p increases at the rate of 02 unit per second, find (iii) the rate of change of A when p = 6 unit, (iv) the value of A when p = 6 + k unit where k is a small value.

y 1 O

p

x

2y =  4 x  1 (b) The diagram above shows the shaded region bounded by the curve 2y =  4 x  1 , the line x = 1 and the line x = p. When the shaded region is rotated through 360 about x-axis, 9 the volume generated is  unit3. Find the value of p. 2

13. (a) Point M and point N lie on the sides BC and CD of a square ABCD respectively. The sides of the square are p cm such that BM = x cm and CN = 2x cm. 2008 Hak Cipta Jabatan Pelajaran Negeri Terengganu

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 2)

ADDITIONAL MATHEMATICS PAPER 2

1 1 Show that the area of AMQ is x2  px + p2. 2 2 (ii) Find, in terms of p, the value of x such that the area of AMQ is minimum.

(i)

(iii) Find, in terms of p, the minimum area of AMQ.

y

A

P

y=3

B O

x y=4–x2

(b) The diagram above shows part of the curve y = 4 – x 2. The straight line y = 3 intersects the curve at point P. (i) Determine the coordinates of P. (ii) Find the ratio of the area of A to the area of B.

ANSWERS FOR

MODULE 2

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KEM SPM TERBILANG TERENGGANU 2008 (MODUL 2)

20 cm2s1

1.

(a)

50

(b)

2.

(a)

2, 8·36

(b)

3.

(a) x +

4.

(a)

(ii)

5.

(a)

(ii) 24 cm2 s1

4 x

5 3

(c)

ADDITIONAL MATHEMATICS PAPER 2

20 s

cm s1

(b) 1·8 cm3s1

+c

(b) 21

(b)

 cm2 s1

5

6.

(a)

k= 4

(b)

6

2

 unit3

3 162

7.

(a)

L = 3  j2 +

8.

(a)

(i) A(0, 4), B(4, 0), C(1, 0)

(b)

13 unit3.

9.

(a)

17

(b)

j

(b)

(i)

1152

11.

(a)

1

(ii)

1067 unit2

(ii)

3

2

2

(2 x  5)

(b)

(ii)

(a)

y=

(a)

4

x = 1591 cm

(2 x  1)

(iii) 1385

4

5

(b)

10

(i) A =

1

(a)

(ii)

x=

2

unit2

3

3

p p

(ii)

3p

2

1

(iii) 25 unit per second

8

8

13.

0111 cm2

6(3 x  5) (3 x  15 x  5)

8

12.

(c)

12

2

10.

2

111 cm2s – 1

1 4

p

(iii) A =

7

p2

(iv)

25

k

(b)

p=3

2

(b)

(i)

P(1, 3)

(ii)

1:7

16

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