Mathematics CSSA 1

2 UNIT MATHEMATICS (HSC) – COORDINATE METHODS IN GEOMETRY – CSSA

Coordinate Methods in Geometry 2U96-3)! In the diagram AB = BC and CD is perpendicular to AB. CD intersects the y axis at P. Copy the diagram onto your answer sheet. y

4

A

D P C

B -3

0

x

NOT TO SCALE a. b. c.

Find the length of AB. Hence show the co-ordinates of C are (2, 0). Show the equation of CD is 3x + 4y = 6.

d.

Show the co-ordinates of P are (0,

e.

Use Pythagoras’ Theorem on POC to show the length of CP is 2

f. g.

Prove that ADP is congruent to COP. Hence calculate the area of the quadrilateral DPOB.† « a) 5 units b) c) d) e) f) Proof g) 4.5 units2 »

3 ). 2

1 units. 2

2U95-2)! A(-2, 2), B(2, 7) and C(8, 1) are the vertices of ABC and line L passes through point D, as shown in the diagram. y B(2, 7) L D A(-2, 2)

C(8, 1) O

a. b. c. d. e. †©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

x

NOT TO SCALE Copy the diagram onto your answer page. D is the midpoint of BC. Show the co-ordinates of D are (5, 4). What is the gradient of line AC? Line L was drawn through point D and parallel to AC. Show the equation of line L is x + 10y = 45. Line L and side AB intersect at point E. i. Show the position of E on your diagram.

2 UNIT MATHEMATICS (HSC) – COORDINATE METHODS IN GEOMETRY – CSSA

f. g.

ii. Prove ABC and EBD are similar. Hence, or otherwise, write down the co-ordinates of point E. Explain how you know the interval AC is twice as long as the interval ED.† y B(2, 7) L

E

A(-2, 2)

« b) Proof c)

1 d) Proof e) i) 10

D

C(8, 1) O

x

1 2

ii) Proof f) E(0, 4 ) g) The ratio of the corresponding sides is 1 : 2. »

2U94-2)! The line L has equation x + 2y = 5 and P is the point (2, 4). a. On a number plane, mark the origin O, the point P and draw the line L. b. Find the midpoint M, of the interval OP. c. Show M lies on the line L. d. Find the gradients of the line OP and the line L. e. Show the line L is the perpendicular bisector of the interval OP. f. Line L meets the x axis at Q. Find the co-ordinates of Q. g. A line is drawn through O parallel to PQ and it meets line L in R. Find the equation of OR. h. Explain why PQOR is a rhombus.† y R(-3, 4)

P(2, 4)

Q O

« a)

5

x

b) M(1, 2) c) Proof d) 2, 

1 f) Q(5, 0) g) 4x + 3y = 0 h) 2

The two diagonals are perpendicular and bisect each other. All sides are equal and opposite sides are parallel. » 2U90-4)! A line, L, is inclined at an angle of 45° to the positive direction of the x-axis and passes through the point X(0, 5). i. Show that the equation of the line L is x - y + 5 = 0. ii. Line P is perpendicular to line L. Show that the gradient of line P is -1. iii. Show that the equation of the line P, through Y(12, 5) is x + y = 17. iv. Find the shortest distance between the line L and the point Y(12, 5). Leave your answer in surd form with a rational denominator. v. The point Z(6, 11) lies on the line L. Show that (6, 11) is the point of intersection of the lines L and P. vi. Show that the distance between Z and X can be expressed in the form a 2 units. vii. What type of triangle is XYZ?† « i) Proof ii) Proof iii) Proof iv) d = 6 2 v) Proof vi) Proof vii) Right-angled isoceles. » 2U88-3)! A(2, -2), B(-2, -3) and C(0, 2) are the vertices of a triangle ABC. i. Draw a sketch diagram of the triangle. ii. Find the length of the line AC and the gradient of AC. iii. Find the equation of the line AC in the general form. †©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – COORDINATE METHODS IN GEOMETRY – CSSA

Calculate the perpendicular distance of B from the side AC and hence find the area of ABC. Find the co-ordinates of D such that ABCD is a parallelogram.†

iv. v.

y 2

-1

C(0, 2)

1

2

x

-1 -2

A(2, -2)

B(-2, -3) « i)

ii) AC = 2 5 units, m = -2 iii) 2x + y - 2 = 0 iv)

9 5 , ABC = 9 units2 v) D(4, 3) » 5

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – APPLICATION OF GEOMETRICAL PROPERTIES – CSSA

Application of Geometrical Properties 2U96-1f)! D

A x°

30°

35° B

50° E

C

NOT TO SCALE

Find the value of x.† « 115 » 2U96-2b)! In the diagram, AB || CD, AD = CD and BAC = 120°. Copy the diagram onto your answer sheet.

A

B

120°

C

D NOT TO SCALE

i. ii.

Explain why ACD = 60°. Show that ADC is equilateral, giving reasons.†

« Proof » 2U96-9a)! ABC is right-angled at A and AD is drawn perpendicular to BC. AB = 15cm and AD = 12cm. Copy the given diagram onto your answer sheet.

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – APPLICATION OF GEOMETRICAL PROPERTIES – CSSA

B

15cm D 12cm C A NOT TO SCALE

i. ii. iii.

Show that BD = 9cm. Prove that ABC is similar to DBA. Hence find the length of AC.† « i) ii) Proof iii) 20 cm »

2U95-1d)! In the diagram AB || CE, ABF = 75° and BFE = 35°.

A

C

75° B

D

35°

F

 E NOT TO SCALE Find the size of  giving reasons.† « 40 » 2U95-5d)!

E y 2y

F L

96° H

G

NOT TO SCALE The diagram shows a rhombus EFGH. A line EL is drawn through E so that HEL = 2  FEL. i. Copy the diagram onto your answer page. ii. FGH = 96°, find the size of ELF giving reasons.† « 106 » 2U95-9b)! †©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – APPLICATION OF GEOMETRICAL PROPERTIES – CSSA

In the diagram ABCD is a square. AB is produced to E so that AB = BE and BC is produced to F so that BC = CF.

A

B

E

C

D

F i. ii. iii.

NOT TO SCALE Copy the diagram onto your answer page. Prove AED  BFA. Hence prove AED = BFA.† « Proof »

2U94-3b)! In the diagram AE || BD, AC || ED, AED = 130° and ABC = 90°. A B

C

E

130

D

i. ii.

NOT TO SCALE Copy this diagram onto your answer sheet. Find the size of BAC giving reasons.† « 40 »

2U94-7c)! In the figure triangles ACB and APO are equilateral.

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – APPLICATION OF GEOMETRICAL PROPERTIES – CSSA

A P

O

B

i. ii. iii. iv.

C

NOT TO SCALE Copy this diagram onto your answer sheet and include all the given information. Explain why BAO = PAC. Prove AOB  APC. Hence prove OB = CP.† A P

O

« i)

B

C

ii) Each angle is equal to 60 - OAC iii) Proof iv) Proof »

2U93-4a)! In the diagram CA = AD = DB and EBD = 20°. Copy this diagram onto your answer sheet. C

D 20° E

i. ii.

A FIGURE NOT TO SCALE Show ADC = 40°, giving reasons. Hence find the size of CAE, giving reasons.†

B

« i) Proof ii) 60 » 2U93-5c)! In the diagram CT bisects ACB, AE is perpendicular to CT and M is the midpoint of AB. AE produced meets BC at the point P.

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – APPLICATION OF GEOMETRICAL PROPERTIES – CSSA

C

P E A i. ii. iii. iv.

B

T M

FIGURE NOT TO SCALE Copy this diagram onto your answer sheet and mark in all the given information. Prove that ACE is congruent to PCE. Explain why AE = EP. Hence prove that EM is parallel to PB.† C

P E

« i) 2U92-5a)!

A

B

T M

ii) Proof iii) Corresponding sides in congruent 's iv) Proof »

N

L 123° K



2°

M

J NOT TO SCALE In the diagram above JKLM is a quadrilateral and LMN is a triangle. JM || LN, JK = KL, JM = ML = MN, KLM = 123°, JKL = 2° and JML = °. i. Copy this diagram onto your answer sheet. ii. Show that JML = 38° giving reasons. iii. Determine the size of LNM giving reasons.† « ii) Proof iii) 38 »

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – APPLICATION OF GEOMETRICAL PROPERTIES – CSSA

2U92-7a)! P

Q

N Z

M

R

S

In the given diagram PQ || RS. MQ bisects PQR, NR bisects QRS and MQ = NR. i. Copy this diagram onto your answer sheet and mark on it all the given information. ii. Explain how you know that MQZ = NRZ. iii. Prove that QMZ  RNZ. iv. Hence prove that the intervals QR and MN bisect each other.† P

Q

N Z

M

« i)

ii) iii) iv) Proof » 2U91-4c)! In the diagram given below, ABC is a right angle triangle with BAC = 90°, CQ = CR, PB = RB and ACB = 40°. R

S

A Q

P

40° B

i. ii.

C

R

NOT TO SCALE Copy this diagram onto your answer booklet. Write down the size of PRQ. (No reasons are required in your solution).† « 45 »

2U91-7a)! P

Q

V T

S

R

NOT TO SCALE PQRS is a parallelogram. TQ bisects PQR and VS bisects PSR. i. Copy this diagram onto your answer booklet. ii. State why PQR = PSR. †©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – APPLICATION OF GEOMETRICAL PROPERTIES – CSSA

iii. iv.

Prove that PVS and RTQ are congruent. Hence find the length of TV if PR = 20cm and TR = 8cm.† « ii) Opposite 's in a parallelogram iii) Proof iv) 4 cm »

2U90-2c)! L

N M

K

NOT TO SCALE KLM is an isosceles triangle with KL = LM, LKM = 80°, LN bisects KLM and KMN = 20°. i. On your answer sheet, draw a neat sketch of the diagram above, showing all the given information. ii. Find the size of LMN, giving reasons for your answer. iii. Find the size of LNM, giving reasons.† L

N 20 80

« i)

M

ii) LMN = 60 iii) LNM = 110 »

K

2U90-5d)! PQRS is a quadrilateral with PR = QS, PQ  PS and SR  PS. i. On your answer sheet, draw a neat sketch and mark on it all the given information. ii. Prove that QPS and RSP are congruent. iii. Hence prove that PQRS is a parallelogram.†

« i)

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

P

Q

S

R

ii) Proof iii) Proof »

2 UNIT MATHEMATICS (HSC) – APPLICATION OF GEOMETRICAL PROPERTIES – CSSA

2U89-3b)! L

N

K

M

NOT TO SCALE In the diagram above, KN = NM, KL = LM, KNM = 110° and NKL = 45°. i. Reproduce a neat sketch and mark on it all the given information. ii. Find the size of MKN and KLM, giving reasons.† L

N 45 110

« i)

K

M

ii) MKN = 35, KLM = 20 »

2U89-6c)! P

Q

T

R U

S NOT TO SCALE In the diagram above, SQ  PQ, RU SQ and PS || QR. i. Prove that RQU ||| PSQ. ii. If RU = x units, QR = y units and PS is four times the length of RU, find the length of PQ in terms of x and y.†

« i) Proof ii) 2U88-6i)! †©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

4x 2 » y

2 UNIT MATHEMATICS (HSC) – APPLICATION OF GEOMETRICAL PROPERTIES – CSSA

F I

E

G J H NOT TO SCALE In the figures FG = 10cm, EG = 15cm, EF = 12.5cm, IJ = 4cm, HJ = 6cm and HI = 5cm. a. Draw a neat sketch and mark on it all the given information. b. Show that EFG ||| HIJ giving reasons.†

« Proof » 2U88-6ii)! L

M

P

N

K

NOT TO SCALE The figure above shows quadrilateral KLMN with diagonals KM and LN intersecting at P. a. Reproduce this diagram on your answer sheet. b. If the diagonals KM an LN bisect each other at right angles, prove that KLMN is a rhombus.† « Proof » 2U87-5i)! In the diagram AB || CD and GH  AB. If y = 25 find the size of GMH. Hence or otherwise find the size of MFD.† y G

A

C

H

B

M

F

D

« GMH = 65, MFD = 115 » 2U87-5ii)! PQRS is a trapezium with PQ || SR. Diagonals PR and SQ intersect at T.

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – APPLICATION OF GEOMETRICAL PROPERTIES – CSSA

P

Q

T

S

a. b. c.

R

NOT TO SCALE Reproduce this diagram on your answer sheet. Prove, giving reasons, that PQT ||| RST. Hence, find PQ, given that SR = 36cm, PT = 5cm and RT = 15cm.† « a) Proof b) 12 cm »

2U87-5iii)! In the diagram below, UXY = UYX and XZ = YZ. U

Y

X

Z

a. b. c.

V Copy this diagram on your answer sheet. Prove that UVY  UWX, giving reasons. Hence prove that VZW is isosceles.†

W

« Proof » 2U86-5i)! List three properties of a rhombus.† « All sides are equal. Opposite sides are parallel. Diagonals bisect each other at right angles. The diagonals bisect the angles through which they pass. » 2U86-5ii)! A, B, C are collinear points. BD || AE, BA || DE, BC = BD and BCD = 58°. Reproduce this diagram on your answer sheet and find the size of DEA.

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – APPLICATION OF GEOMETRICAL PROPERTIES – CSSA

C 58°

B

D

A

E NOT TO SCALE

†

« 116 » 2U86-5iii)! In the triangle PSU, QR || SU, SP || TR, ST = 7.5cm, PQ = 10cm, PR = 12cm and UT = 15cm. Find the length of SQ giving reasons.

P

R

Q

S

U

T

NOT TO SCALE† « 20 cm » 2U86-5iv)! GL is a median in HFG and HJ || FK. a. Draw a neat sketch of this diagram on your answer sheet. b. Prove, giving reasons, that KL = LJ.

F K L J

H †©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

G

2 UNIT MATHEMATICS (HSC) – APPLICATION OF GEOMETRICAL PROPERTIES – CSSA

NOT TO SCALE† « Proof » 2U85-7i)! Q

P

T

S

R NOT TO SCALE PQRS is a square with PQ = 1 unit. Find the perimeter of PTRS.†

« 4 +

2 units »

2U85-7ii)! A

Z D

X

Y

C

B NOT TO SCALE ABC and ABD are two triangles, X, Y and Z are points such that XY CB and YZ BD. Prove that XY : YZ = CB : BD.† « Proof » 2U84-2i)! Find the area of the rhombus ABCD given AB = 10cm and EB = 8cm.

A

10cm m

B 8cm

E

D

C

† « 96 cm2 »

2U84-8i)!

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – APPLICATION OF GEOMETRICAL PROPERTIES – CSSA

W

Y

Z

V

X NOT TO SCALE

In the triangle WXV, YZ = 9cm, VX = 12cm, WX = 8cm and YZ || VX. Prove that WZY is similar to WXV and find the length of WZ.† « 6 cm » 2U84-9iv)!

A

B C

F NOT TO SCALE

D

E

In the figure, AB = AC, BD || FE, BF || DE and CAB = 54. Find the size of FED giving reasons.† « 63 »

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – GEOMETRICAL APPLICATIONS OF DIFFERENTIATION – CSSA

Geometrical Applications of Differentiation 2U96-8c)! Consider the curve given by the equation y = 9x(x - 2)2. i. Find the co-ordinates of the stationary points and determine their nature. ii. Find the co-ordinates of any points of inflexion. iii. Sketch the curve in the domain -1  x  3. iv. What is the maximum value of 9x(x - 2)2 in the domain -1  x  3?†

2 3

 1  3

2 3

1 3

« i) Maximum turning point at  , 10  and minimum turning point at (2, 0) ii) 1 , 5  iii) y

(3, 27) 2 2  ,10  3 3

x

(2, 0)  1 1 1 , 5   3 3

(-1, -81)

iv) 27 » 2U95-4b)! The function y = x3 - 3x2 - 9x + 1 is defined in the domain -4  x  5. i. Find the co-ordinates of any turning points and determine their nature. ii. Find the coordinates of any points of inflexion. iii. Draw a neat sketch of the curve. iv. Determine the minimum value of the function y, in the domain -4  x  5.† « i) Maximum turning point at (-1, 6) and minimum turning point at (3, -26) ii) (1, -10) iii) y (-1, 6)

(5, 6) x (1, -10)

(3, -26)

(-4, -75)

iv) -75 »

2U94-6a)! For the function y = x3 - 6x2 + 9x + 1 find the: i. stationary points and determine their nature. ii. co-ordinates of any points of inflection. iii. values of x for which the curve is increasing. Hence sketch the curve y = x3 - 6x2 + 9x + 1.† « Maximum turning point at (1, 5), Minimum turning point at (3, 1) ii) (2, 3) iii) x < 1 and x > 3 y (1, 5) 5 (2, 3) 1

(3, 1) x

2U93-9a)! The curve f(x) = x3 + 3x2 - 9x - 1 is defined in the domain -4  x  2. i. Find the co-ordinates of the two stationary points and determine their nature. ii. Show a point of inflexion occurs at x = -1. iii. Sketch this curve.

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

»

2 UNIT MATHEMATICS (HSC) – GEOMETRICAL APPLICATIONS OF DIFFERENTIATION – CSSA

« i) Maximum turning point at (-3, 26) and minimum turning point at (1, -6) ii) Proof iii) y

(-3, 26) (-4, 19)

(2, 1) x (1, -6)

»

2U92-9c)! The gradient function of a curve y = f(x) is given by: f (x) = x2(3 - x). i. Show that the curve has two stationary points and determine their nature. ii.

3 4

If f(0) = 2 and f(x) has a maximum value of 8 , sketch y = f(x).† 3   3, 8   4

y

2 O

1

2

3

4

« i) Minimum at x = 0, Maximum at x = 3 ii)

x

»

2U91-7b)! The gradient function of a curve is given by i. ii.

dy = 3x2 - 12. dx

For what values of x does the curve increase with downward concavity? If this curve passes through (-3, 2), find the equation of the curve.† « i) x < -2 ii) y = x3 - 12x - 7 »

2U90-10b)! The gradient function

dy of a curve is illustrated by the graph below: dx dy dx

dy 2 dx  3(x  1)

3

0

i. ii.

iii.

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

1

x

A stationary point is located at x = 1. Justify this statement by reference to the graph.

dy for all x, x  1. dx

.

Comment on the sign of

.

What does this imply about the curve y = f(x)?

The graph of

d2 y is given below. dx2

2 UNIT MATHEMATICS (HSC) – GEOMETRICAL APPLICATIONS OF DIFFERENTIATION – CSSA

d2y dx 2

0

1

x

-6 Copy and complete this table: x

0

1

2

2

sign of d y2 dx

iv. v.

What is the nature of the stationary point at x = 1? If the curve y = f(x) passes through (0, 0), find the equation of the curve.†

« i) At x = 1,

dy dy  0 ii) )  0 for all x, except x = 1 ) The curve is increasing for all x, except dx dx x

0 –

2 sign of d y2

x = 0. iii) 2 +

1 0

dx

iv) An horizontal point of inflexion. v) y = x3 - 3x2 + 3x » 2U89-5b)! Consider the function f(x) = 1 - 3x + x3, in the domain -2  x  3. i. There are two turning points for f(x). Find their co-ordinates and determine their nature. ii. Draw a sketch of the curve y = f(x) in the domain -2  x  3, clearly showing all its essential features. iii. What is the maximum value of the function f(x) in the domain -2  x  3 ?† y (3, 19)

(-1, 3) x

« i) Minimum at (1, -1), Maximum at (-1, 3) ii)

(-2, -1)

(1, -1)

iii) 19 »

2U88-5iii)! Determine the greatest value of the function f(x) = 17 + 4x - x2.† « 21 » 2U85-5iii)! Find the values of x for which the curve y = 2x3 - 3x2 - 12x + 6 rises with downward concavity.† « x  -1 »

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – INTEGRATION – CSSA

Integration 2U96-2c)! Find i. ii.

 sec 3xdx ;  (5x  3) dx ; 2

5

0

iii.

dx

 2x  3 .†

1

« i)

1 1 1 tan 3x + C ii) (5x  3)6  C iii) Ln 3 » 3 30 2

2U96-5a)! Wasteland bordering a river bank and a straight road was fenced off and used as a recreational park. Perpendicular distances from the road to the river bank are shown on the diagram. Use Simpson’s Rule, with 5 function values, to approximate the area of the recreational park.†

RIVER 13m

10m

9m

F E 13m N C E

ROAD 24m

NOT TO SCALE « 242 m2 » 2U96-6b)! The region bounded by the curve y = x3, the y-axis and the line y = 8 is rotated about the y-axis. Find the volume of the solid formed.† « 2U95-3c)! i.

96 units3 » 5

Find  (2  x )dx .  4

ii.

Evaluate

 sec

2

3x dx .†

0

2 23 1 « i) 2x + x + C ii)  » 3 3

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – INTEGRATION – CSSA

2U95-5c)! Use Simpson’s Rule with the five function values given in the table below to evaluate correct to 1 5

decimal place:  f ( x)dx .† 1

x f(x)

1 0

2 1.34

3 3.30

4 5.55

5 8.05 « 14.1 »

2U95-6a)!

5m

2m

5m

The diagram shows a sketch of a skateboard ramp which is 2 metres high and 5 metres wide. The cross section of the ramp was determined by using the graph of y = f(x). y 2 1 0

3

4

x

NOT TO SCALE

2 f (x)   1  cos x i. ii.

0 x3 3 x  4

Find the area of the cross section of the ramp, the shaded area in the diagrams. The ramp is solid concrete. How much concrete was used to make the ramp? Give your answer to the nearest m3.† « i) 7 m2 ii) 35 m3 »

2U95-8a)! A bowl was designed by rotating the section of the curve y = centimetres, about the y axis.

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

1 2 x between x = 2 and x = 12 4

2 UNIT MATHEMATICS (HSC) – INTEGRATION – CSSA

y

O

i. ii.

12 x

2

NOT TO SCALE Calculate the volume of the bowl, leaving your answer in terms of . Hence calculate the capacity of the bowl, correct to the nearest litre. (1 litre = 1000 cm3)† « i) 2590 m3 ii) 8 L »

2U94-5c)! Below is the graph of y = f(x) for -2  x  4. y

4

(1, 4)

(2, 4) y = f(x)

3

(3, 2)

2 1

-2

i. ii.

-1

O

1

2

3

x

4

Write down an expression for the exact area bounded by the curve and the x axis. (You are not required to find the equation of this curve.) Use Simpson’s Rule with 5 function values to approximate the area enclosed by the curve, the x axis and the lines x = 0 and x = 4.† « i)

0

4

2

0

2U94-6b)! A woodturner made the wooden bowl shown in the diagram below.

holding capacity wood †©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2

 f (x)dx   f (x)dx ii) 10 3 units2 »

2 UNIT MATHEMATICS (HSC) – INTEGRATION – CSSA

She designed the solid shape by rotating the area bounded by the curves y = x2 - 1, and y =

x2 + 1, in 2

the first quadrant about the y axis. The area she rotated is shown in the diagram below: y y = x2 - 1 2 y  x 1 2

0

x

NOT TO SCALE Show the co-ordinates of A are (2, 3). Calculate the holding capacity of the bowl. Find the volume of wood in the finished bowl.†

i. ii. iii.

« i) Proof ii) 4 units3 iii)

7 units3 » 2

2U93-4c)! The logo for the company “Top Hats” was designed using parts of the curves y = 2x2 and y = 12 - x2.

y y = 2x2

y = 12 - x2

0 DIAGRAM 1

x

DIAGRAM 2

Diagram 1 shows a drawing of the logo and Diagram 2 shows a sketch of the logo related to the coordinate axes. i. Show the curves intersect at (-2, 8) and (2, 8). ii. Hence find the area of the shaded part of the logo.† †©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – INTEGRATION – CSSA 2

« i) Proof ii) 32 units » 2U93-6c)! The diagram shows a block of land 60 metres long. At intervals of 10 metres, the width of the block was measured.

12m

15m 19.5m

22m

17m

15m

13m

60m DIAGRAM NOT TO SCALE Approximate the area of this block of land using the Trapezoidal Rule with the seven heights shown.† « 1010 m2 » 2U92-6b)! The diagram below shows a paddock with one side bounded by a river.

RIVER

20m

0

14m

12m

7m

8m

40m

NOT TO SCALE Use Simpson’s Rule with five function values shown on the diagram to approximate the area of the paddock.† « 453 2U92-8a)!

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

1 2 m » 3

2 UNIT MATHEMATICS (HSC) – INTEGRATION – CSSA

y

y = x2

P

0

x y = 3 - 2x

NOT TO SCALE The diagram shows the parabola y = x2 and the line y = 3 - 2x intersecting at the point P, in the first quadrant. i. Show that the co-ordinates of the point P are (1, 1). ii. The shaded region is rotated about the x axis. Find the volume of the solid formed.† « i) Proof ii) 2U91-6d)! The table below gives the values of f(t) for 0  t  2. t 0 0.5 1 f(t) 0 0.30 0.37

1.5 0.33

11 units3 » 30

2 0.27

2

Use the Trapezoidal Rule with 5 function values to evaluate:  f ( t )dt correct to 1 decimal place.† 0

« 0.6 » 2U90-5a)! Find: i. ii.

 (3x - 2)dx;  cos 3x dx.† 2

« i) x3 - 2x + C ii) 2U90-6d)!

A

3m 5.6

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

7.0 8.5 9.2 10.8 9.6 8.8 6.5

6.0 5m

1 sin 3x  C » 3

2 UNIT MATHEMATICS (HSC) – INTEGRATION – CSSA

FIGURE NOT DRAWN TO SCALE The diagram given above represents the cross-section of a river which is 50 metres wide. A depthgauge was used to measure the depth of the river at intervals of 5 metres from one bank A. Use Simpson’s Rule to approximate the area of cross-section of this river. Give your answer correct to 4 significant figures.† « 385.7 m2 » 2U90-6e)! To calculate the area of the region bounded by the curve y = x2 - 2x and the axis x between the 4

ordinates x = 0 and x = 4, Ernie used



(x2 - 2x)dx.

0

i. ii.

Explain why Ernie’s method of calculating this area is incorrect. Find the area of the required region.† « 8 units2 »

2U90-8d)! The diagram shows a region bounded by y = k2 - x2 and the x axis. If the area of this region is

256 3

square units, find the value of k.† y k2

-k

0

k x

NOT TO SCALE «k » 2U89-6a)! Find: i. ii. iii.

 x x dx;  sin 3x dx; dx  (2x  3) .† 2

« i)

2 25 1 1 x + C ii)  cos3x + C iii) C » 5 3 2(2x  3)

2U88-8i)! Find a primitive function of the following: a. x2 - x ; b. 3sec2 2x; c.

x2 .† x3  2 x3 2 23 3 1 « a)  x  C b) tan 2x + C c) Ln(x3 - 2) + C » 3 3 2 3

2U88-8ii)! †©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – INTEGRATION – CSSA

y E

D B 70m 50m

A

x

C

100m NOT TO SCALE The diagram represents the span of a bridge, 70 metres high and 100 metres wide. The curved part of the span is a parabola with vertex 50 metres above the ground. Using the axes shown in the diagram, find: a. the equation of the arc ABC; b. the shaded area ABCDE.† « a) y = -x2 + 50 b) 6528.60 m2 » 2U87-7ii)! Use Simpson’s Rule with 3 function values (2 subintervals) to approximate the area enclosed between

the curve y =

1 and the lines x = 0 and x = 4 correct to 2 significant figures.† (x  1)2 « 0.99 »

2U87-7iv)! y

new expressway

1km

x

old highway

NOT TO SCALE The straight new expressway and a parabolic arc of the old highway are the boundaries of a property. Taking the axes shown in the diagram, the parabolic arc has equation y2 = 9x. The new expressway is perpendicular to the x axis. The greatest distance between the old highway and the new expressway is 1 km as shown in the diagram. Use integration to find the area of the property.† « 4 km2 » 2U86-7i)! Find: †©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

2 UNIT MATHEMATICS (HSC) – INTEGRATION – CSSA

a. b. c.

 

2

(3x + 5) dx. sin(3x - 1) dx.

ex  ex  5 dx .† « a)

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

1 1 (3x + 5)3 + C b) cos(3x - 1) + C c) Ln(ex + 5) + C » 9 3

2 UNIT MATHEMATICS (HSC) – INTEGRATION – CSSA

2U86-7ii)! Find the volume of the solid formed when the area bounded by the curve y = 5 - x2 for x  0, the y axis and the line y = 1 is rotated about the y axis.† « 8 units3 » 2U86-7iii)! A builder wishes to install a window which is in the shape of a parabola as shown in the diagram. a. If the parabola is symmetrical about the vertical axis, has a window sill 4 metres wide and height 4 metres, find the equation of the parabola which satisfies these conditions. b. Hence calculate the area of wall this window will occupy.

4m

4m NOT TO SCALE† « a) y = 4 - x2 b)

32 2 m » 3

2U85-5ii)! Find the primitive function of the following: a. (x2 - 2)2; b. c.

x2  2 ; x x .† 2 x 2 « a)

1 1 5 4 3 1 x - x + 4x + C b) x2 - 2Lnx + C c) Ln(x2 - 2) + C » 3 2 5 2

2U84-6i)! Find the following indefinite integrals: a.

 3cos 2xdx ;

b.



c.



4x  5 dx ; 2x2  5x x 1 dx.† x « a)

†©CSSA OF NSW 1984 - 1996 ©EDUDATA: DATAVER1.0 1996

3 2 sin 2x + C b) Ln(2x2 + 5x) + C c) x x  2 x  C » 2 3