Answers [EXTENSION 1 MATHEMATICS HSC MATHS IN FOCUS]

562

Maths In Focus Mathematics Extension 1 HSC Course

Answers Chapter 1: Geometry 2

8.

Exercises 1.1 1.

2.

+ABE = 180° −+ABD +CBE = 180° −+CBD = 180° −+ABD = +ABE

(straight angle 180° ) (straight angle)

(corresponding sides in congruent ∆s)

(+ABD =+CBD —given)

+DFB = 180° − (180 − x) ° (+AFB is a straight angle) =x ` +AFC = x (vertically opposite angles) +CFE = 180° − (x + 180° − 2x)

` OC bisects AB 9.

(+AFB is a straight angle)

=x ` +AFC = +CFE ` CD bisects +AFE 3.

+WBC + +BCY = 2x + 115 + 65 − 2x = 180° These are supplementary cointerior angles. `VW < XY

4.

x + y = 180° (given) ` +A + +D = 180° These are supplementary cointerior angles. ` AB < DC Also +A = +B (similarly) These are supplementary cointerior angles. ` AD < BC ` ABCD is a parallelogram.

5.

+ADB = +CDB = 110° (given) (BD bisects +ABC) +ABD = +CBD BD is common ` by AAS, ∆ ABD / ∆ CBD

6.

(a)

AB = AE ( given) +B = +E ( base angles of isosceles ∆) ( given) BC = DE ` by SAS, ∆ ABC / ∆ AED

10.

(corresponding angles in congruent ∆s)

(corresponding angles in congruent ∆s)

` AC bisects +DCB 11. (a) +NMO = +MOP (alternate angles, MN < PO) (alternate angles, PM < ON) +PMO = +MON MO is common ` by AAS, ∆ MNO / ∆ MPO (b) +PMO = +MON (alternate angles, PM < ON) (given) MN = NO (base angles of isosceles ∆) +MON = +NMO ` +PMO = +NMO i.e. +PMQ = +NMQ (c)

MN = NO PM = NO

(given)

(corresponding sides in congruent ∆s)

PM = MN +PMQ = +NMQ MQ is common ` by SAS, ∆PMQ / ∆NMQ `

+ACD = 180° − +BCA (BCD is a straight angle) = 180° − +EDA = +ADC ` since base angles are equal, ∆ ACD is isosceles

(corresponding sides in congruent ∆s)

AB = AD (given) (given) BC = DC AC is common ` by SSS, ∆ABC / ∆ADC ` +DAC = +BAC So AC bisects +DAB Also +BCA = +DCA

(corresponding angles in congruent ∆s)

DC = BC ( given) ( given) +B = +D = 90° 1 ( given) DM = AD 2 1 and BN = AB 2 ` DM = BN ` by SAS, ∆MDC / ∆NBC ` MC = NC

+CDB = +BEC = 90° (altitudes given) (base angles of isosceles ∆) +ACB = +ABC CB is common ` by AAS, ∆CDB / ∆BEC ` CE = BD (corresponding sides in congruent ∆s)

(b) +BCA = +EDA

7.

+OCA = +OCB = 90° (given) (equal radii) OA = OB OC is common ` by RHS, ∆OAC / ∆OBC ` AC = BC

(d)

(from (b))

+MQN = +MQP

(corresponding angles in congruent ∆s)

But +MQN ++MQP = 180° (+PQN straight angle) +MQN = +MQP = 90° ` 12.

ANSWERS

Let ABCD be a parallelogram with diagonal AC. +DAC = +ACB (alternate angles, AD < BC) (alternate angles, AB < DC) +BAC = +ACD AC is common ` by AAS, ∆ADC / ∆ABC

19.

13.

∆ ADC / ∆ ABC (see question 12) ` +ADC = +ABC (corresponding angles in congruent ∆s) Similarly, by using diagonal BD we can prove +A = +C ` opposite angles are equal 14.

15.

17.

(corresponding sides in congruent ∆s)

20.

AB = DC (opposite sides in < gram) (given) BM = DN ` AB − BM = DC − DN i.e. AM = NC (ABCD is a < gram) Also AM < NC Since one pair of sides is both parallel and equal, AMCN is a parallelogram. AD = BC BC = FE ` AD = FE

Let ABCD be a rectangle with +D = 90° `

+DEC = +DCE Also, +DEC = +ECB ` +DCE = +ECB ` CE bisects +BCD

= 90° +B = 180° − 90° (+B and +C cointerior angles, AB < DC)

= 90° +A = 180° − 90° (+A and +B cointerior angles, AD < BC)

= 90° ` all angles are right angles 21.

(base angles of isosceles ∆) (alternate angles, AD < BC)

AB = CD (given) (given) +BAC = +DCA AC is common ` by SAS, ∆ABC / ∆ADC ` AD = BC

+C = 180° − 90°

(+D and +C cointerior angles, AD < BC)

(opposite sides in < gram) (similarly)

Also AD < BC (ABCD is a < gram) (BCEF is a < gram) and BC < FE ` AD < FE Since one pair of sides is both parallel and equal, AFED is a parallelogram. 16.

Let ABCD be a rectangle AD = BC (opposite sides in < gram) +D = +C = 90° DC is common ` by SAS, ∆ ADC / ∆ BCD ` AC = DB

AD = CD (given) (opposite sides of < gram) AD = BC Also AB = CD (similarly) ` AB = AD = BC = CD ` all sides of the rhombus are equal

22.

(corresponding sides in congruent ∆s)

Construct BE < AD Then AD = BE (opposite sides of < gram) But AD = BC (given) Then BE = BC (base angles of isosceles ∆) ` +BCD = +BEC (corresponding angles, AD < BE) Also, +ADC = +BEC ` +ADC = +BCD

Since two pairs of opposite sides are equal, ABCD is a parallelogram. 18. (a)

AE = EC (diagonals bisect each other in < gram)

+AEB = +CEB = 90° EB is common ` by SAS, ∆ ABE / ∆CBE ` AB = BC

(given)

(corresponding sides in congruent ∆s)

(b) +ABE = +CBE

(corresponding angles in congruent ∆s)

23.

AD = AB (given) (given) DC = BC AC is common ` by SSS, ∆ADC / ∆ΑΒC ` +ADC = +ABC (corresponding angles in congruent ∆s)

563

564

Maths In Focus Mathematics Extension 1 HSC Course

24.

AD = BC (opposite sides of < gram) (given) +D = +C = 90° DE = EC (E is the midpoint—given) ` by SAS, ∆ADE / ∆BCE ` AE = BE (corresponding sides in congruent ∆s)

Exercises 1.3 1.

2.

AD = BC (opposite sides of < gram) AB = DC (similarly) DB is common ` by SSS, ∆ADB / ∆BCD

25. (a)

3.

(a) AB = AC =

4.

Show m XY = mYZ =

5.

(a) Show AB = AD =

(b) +ABE = +CBE

(corresponding angles in congruent ∆s)

+AEB = +BEC

(corresponding angles in congruent ∆s)

But +AEB + +BEC = 180° (AEC is a straight angle) ` +AEB = +BEC = 90°

Exercises 1.2 1.

(a) 14 452 mm2 (b) 67 200 mm3

3.

V = 2x3 + 3x2 − 2x

4.

V = π r2 h 250 = π r2 h 250 = r2 πh 250 =r πh 6. A =

5.

V = 5π r3

8.

S = 24π h2

9.

V = x ] 3 − 2x g2 = 4x3 − 12x2 + 9x

2.

3

3b2 2

7. V = ] x + 2 g3 = x3 + 6x2 + 12x + 8

16. V = 19. l = 20. y =

h2 + r2 4π

400 17. h = πr 2

850 − πr2 πr 810 000 − x2

r =1

7.

(a) 2x − 3y + 13 = 0 (b) Substitute (7, 9) into the equation (c) Isosceles

8.

+AOB OD OB OC OA OD ` OB

= +COD = 90° 4 = =2 2 14 = =2 7 OC = OA

(a) OB is common OA = BC = 5 AB = OC = 20 ` by SSS ∆OAB / ∆OCB (b) Show mOA = m BC = 1

1 (6x3 − 5x2 − 34x − 15) 3

x2 y

72 = 6 2,

6.

9.

14. (a) V = 18x3 − 12x2 + 2x (b) S = 54x2 − 30x + 4 15. l =

26 ,

Since 2 pairs of sides are in proportion and their included angles are equal, ∆ OAB 0 for all x ! 0

(b)

12 > 0 for all x ! 0 x4 So the function is concave upward for all x ! 0.

So

15. (a) (0, 7), (1, 0) and ^ −1, 14 h (b) (0, 7) 16. (a)

d2 y dx2

= 12x2 + 24

x2 ≥ 0 for all x So 12x2 ≥ 0 for all x 12x2 + 24 ≥ 24 So 12x2 + 24 ≠ 0 and there are no points of inflexion. (b) The curve is always concave upwards.

(c)

17. a = 2 18. p = 4 19. a = 3, b = − 3 20. (a) ^ 0, − 8 h, ^ 2, 2 h (b)

dy dx

= 6x5 − 15x4 + 21

At ^ 0, − 8 h:

dy dx

= 6 ] 0 g5 − 15 ] 0 g4 + 21 = 21 ≠0

At

^ 2, 2 h :

dy dx

= 6 ] 2 g5 − 15 ] 2 g4 + 21

= − 27 ≠0 So these points are not horizontal points of inflexion.

(d)

dy dx dy dx dy dx

> 0, > 0, > 0,

d2 y dx

2

d2 y dx2 d2 y dx2

> 0 (b) < 0 (d)

dy dx dy dx

< 0, < 0,

d2 y dx2 d2 y dx2

0

>0

dP d2 P > 0, < 0 (b) The rate is decreasing. dt dt2

575

576

Maths In Focus Mathematics Extension 1 HSC Course

5.

13. (a) a = − 14. m = − 5

2 3

1 2

Exercises 2.7 1. 6.

7.

dM d2 M < 0, >0 dt dt2

8.

(a) The number of fish is decreasing. (b) The population rate is increasing. (c)

9.

The level of education is increasing, but the rate is slowing down.

10. The population is decreasing, and the population rate is decreasing.

2.

3.

Exercises 2.6 1.

(1, 0) minimum 2. (0,1) minimum

3.

(2, − 5), y m = 6 > 0 4. (0.5, 0.25), y m < 0 so maximum

5.

(0, − 5); f m(x) = 0 at (0, − 5), f m(x) < 0 on LHS, f m(x) > 0 RHS

6.

Yes—inflexion at (0, 3)

7.

^ − 2, −78 h minimum, ^ − 3, −77 h maximum

8.

(0, 1) maximum, ^ − 1, − 4 h minimum, ^ 2, − 31 h minimum

9.

(0, 1) maximum, (0.5, 0) minimum, ^ − 0.5, 0 h minimum

10. (a) (4, 176) maximum, (5, 175) minimum (b) (4.5, 175.5) 11. (3.67, 0.38) maximum 12. ^ 0, −1 h minimum, ^ − 2, 15 h maximum, ^ − 4, −1 h minimum

4.

(b) maximum, as y m < 0 15. a = 3, b = − 9

ANSWERS

9.

5.

6.

10.

7.

11.

8.

(a) ^ 0, −7 h minimum, ^ − 4, 25 h maximum (b) ^ − 2, 9 h (c) 12.

577

578

Maths In Focus Mathematics Extension 1 HSC Course

13.

2.

y

3. 14.

dy dx

=

−2 ≠0 (1 + x) 2 (0, 0)

-1 (-3, -3)

4.

15.

5.

Exercises 2.8 6. 1.

1 2

x

ANSWERS

y

12.

7.

⎛ 1⎞ -1, ⎝ 2⎠

x ⎛1, - 1⎞ 2⎠ ⎝

8.

13.

y

(2.16, 30.24)

x −2

9.

2

(−2.16, −30.24)

f(t)

14.

10. (2.41, 4.83)

t (−0.41, −0.83)

y

11. 6 5 4 3 2 1

(0.24, 0.66) x

4 ( 4.24, 0.1)

3

2

1

1 2 3 4 5

1

2

3

4

1 −1

579

580

Maths In Focus Mathematics Extension 1 HSC Course

x

15.

6 5 4 3 2 1 −4 −3 −2 −1

⎛ 1⎛ 0, 2 − ⎝ 4⎝

h −1

1

2

3

4

−2 −3 −4

5.

Maximum value is −2.

6.

1 Maximum value is 5, minimum value is − 16 . 3

−5

Exercises 2.9 1.

2.

Maximum value is 4.

7.

Absolute maximum 29, relative maximum −3, absolute minimum −35, relative minimum −35, −8

8.

Minimum −25, maximum 29

9.

Maximum 3, minimum 1

Maximum value is 9, minimum value is −7.

3.

Maximum value is 25.

4.

Maximum value is 86, minimum value is −39.

ANSWERS

10. Maximum ∞, minimum −∞

1 2 1 2 xn + d yn 4 4 y2 x2 = + 16 16 x2 + y2 = 16 x2 + (30 − x)2 = 16 x2 + 900 − 60x + x2 = 16 2x2 − 60x + 900 = 16 2 (x2 − 30x + 450) = 16 x2 − 30x + 450 = 8

A =d

Problem 30 cm. (This result uses Stewart’s 7 theorem—check this by research.)

The disc has radius

6.

(a) x2 + y2 = 2802 = 78 400 y2 = 78 400 − x2 y=

Exercises 2.10 1.

A 50 50 ` x P

= x 78 400 − x2

78 400 − x2

7.

= xy = xy =y = 2x + 2y = 2x + 2 ´ = 2x +

2.

2x + 2y 2y y A

3.

xy = 20 20 y= x S =x+y 20 =x+ x

4.

V 400 400 π r2 S

V = x (10 − 2x) (7 − 2x) = x (70 − 20x − 14x + 4x2) = x (70 − 34x + 4x2) = 70x − 34x2 + 4x3

50 x

100 x

= 120 = 120 − 2x = 60 − x = xy = x (60 − x) = 60x − x2

8.

Profit per person = Cost − Expenses = (900 − 100x) − (200 + 400x) = 900 − 100x − 200 − 400x = 700 − 500x For x people, P = x (700 − 500x) = 700x − 500x2

9.

= π r2h = π r2 h =h = 2π r2 + 2π rh = 2π r 2 + 2π r e = 2π r 2 +

5.

(b) A = xy

400 o π r2

800 r

(a) x + y = 30 ` y = 30 − x (b) The perimeter of one square is x, so its side is 1 1 x. The other square has side y. 4 4

After t hours, Joel has travelled 75t km. He is 700 − 75t km from the town. After t hours, Nick has travelled 80t km. He is 680 − 80t km from the town. d= =

(700 − 75t)2 + (680 − 80t)2 490 000 − 105 000t + 5625t2 + 462 400 − 108 800t + 6400t2 = 952 400 − 213 800t + 12 025t2

581

582

Maths In Focus Mathematics Extension 1 HSC Course

10. The river is 500 m, or 0.5 km, wide Distance AB: d= =

x + 0.5 x2 + 0.25 distance Speed = time distance ` Time = speed x2 + 0.25 t= 5 2

2

Distance BC: d =7−x distance Time = speed 7−x t= 4 So total time taken is: t=

10. V = πr2h = 54π 54π h= πr 2 54 = 2 r S = 2πr (r + h) 54 = 2πr d r + 2 n r 108π 2 = 2πr + r Radius is 3 m. 11. (a) S = 2π r2 +

13. (a)

Exercises 2.11 2 s, 16 m 2. 7.5 km

3.

2x + 2y 2y y A

= 60 = 60 − 2x = 30 − x = xy = x (30 − x) = 30x − x2

(b) 100 cm2 from (1)

(1)

= 2x + 2 c

from (1)

17. 301 cm2 18. 160

5.

4 m by 4 m 6. 14 and 14 7. -2.5 and 2.5

8.

x = 1.25 m, y = 1.25 m

9.

(a) V = x (30 − 2x) (80 − 2x) = x (2400 − 220x + 4x2) = 2400x − 220x2 + 4x3 (b) x = 6

2 cm 3

(c) 7407.4 cm3

15. 1.12 m2

14. 20 cm by 20 cm by 20 cm 16. (a) 7.5 m by 7.5 m

(a) A = xy = 4000 4000 ` y= x P = 2x + 2y 4000 m x 8000 = 2x + x (b) 63.2 m by 63.2 m (c) $12 332.89

xy = 400 400 ` y= x A = (y − 10) (x − 10) = xy − 10y − 10x + 100 400 400 m − 10 c m − 10x + 100 = xc x x 4000 = 400 − − 10x + 100 x 4000 = 500 − 10x − x

(1)

Max. area 225 m2 4.

(b) 2323.7 m2

12. 72 cm2

x2 + 0.25 7 − x + 5 4

1.

17 200 r

(b) 2.4 m 1 cm2 6

19. 1.68 cm, 1.32 cm

20. d2 = (200 − 80t) 2 + (120 − 60t) 2 = 40 000 − 32 000t + 6400t2 + 14 400 − 14 400t + 3600t2 = 10 000t2 − 46 400t + 54 400 24 km 1 21. (a) d = ] x2 − 2x + 5 g − ] 4x − x2 g (b) unit 2 2 2 = x − 2x + 5 − 4x + x = 2x 2 − 6 x + 5

22. (a)

y 1 (2πr) where r = 2 2 y 1 1200 = 2x + y + e 2π ´ o 2 2 πy = 2x + y + 2 πy = 2x 1200 − y − 2 y πy =x 600 − − 2 4 2400 − 2y − πy =x 4 Perimeter = 2x + y +

ANSWERS

1 2 πr 2 2400 − 2y − πy

(b) A = xy +

y 2 oy + 1 π e o 4 2 2 2400y − 2y2 − πy2 πy2 = + 4 8 2 2 πy 2 4800y − 4y − 2πy = + 8 8 4800y − 4y2 − πy2 = 8 =e

2.

3

(e) f (x) = 3.

2x 2 3

+C x− 3 + 2x− 1 + C 3 2 (d) y = − + C x

(b) y = −

(a) y = x5 − 9x + C

x4 x3 − +C 20 3 x4 x2 − +x+C (e) y = 4 3

(c) y =

4.

(a)

2 x3 +C 3

(b) −

x− 2 +C 2 1

1

Substitute ^ −1, 2 h

(c) −

1 +C 7x7

b (−1) + b a b =− +b a 2a = − b + ab = b ] −1 + a g = b ]a − 1 g 2a =b a−1

(e) −

x− 6 + 2x− 1 + C 6

10. y = 4x2 − 8x + 7 11. y = 2x3 + 3x2 + x − 2

(b) a = 2, b = 4

12. f (x) = x3 − x2 − x + 5 13. f (2) = 20.5

2 =

24. 26 m 25. (a)

d t d So t = s 1500 = s

Exercises 2.12

(c)

x6 − x4 + C 6

(e) 6x + C

x4 1 − x3 + 5x − 4 4

y=

7.

f (1) = 8 8. y = 2x − 3x2 + 19 9. x = 16

x3 + 4x2 + x + C 3 x3 − x2 + x + C (d) 3 (b)

6. f (x) = 2x2 − 7x + 11

14. y =

x3 x2 1 + − 12x + 24 3 2 2

16. y =

x3 2 − 2x2 + 3x − 4 3 3

15. y =

1 3

4x3 1 − 15x − 14 3 3

17. f ] x g = x4 − x3 + 2x2 + 4x − 2

19. f ] − 2 g = 77

18. y = 3x2 + 8x + 8

(b) 95 km/h (c) $2846

(a) x2 − 3x + C

(d) 2x 2 + 6x 3 + C

5.

s=

Cost of trip taking t hours: C = ] s2 + 9000 g t 1500 = ] s2 + 9000 g s 9000 ´ 1500 = 1500s + s 9000 c m = 1500 s + s

1.

(b) f (x) =

x2 − 2x + C 2

(c) f (x) =

(c) x = 168 m, y = 336 m 23. (a) Equation AB: y = mx + b b = x+b a

x5 − x3 + 7x + C 5 x3 (d) f (x) = − x2 − 3x + C 3

x2 +C 2

(a) f (x) = 2x3 −

20. y = 0

Test yourself 2 1.

(− 3, −11) maximum, (−1, −15) minimum

2.

x >1

4.

(a) -8

6.

(0, 0) minimum

8.

(a) (0, 1) maximum, (− 4, − 511) minimum, (2, −79) minimum (b)

1 6

3. y = 2x3 + 6x2 − 5x − 33 (b) 26

(c) -90

5. 50 m

7. x > −1

583

584

Maths In Focus Mathematics Extension 1 HSC Course

9.

5x3 1 + 6x2 − 49x + 59 c , −1 m 10. f (x) = 2 2

11. (a)

V 375 375 πr 2 S

18. (a)

= πr 2 h = πr 2 h =h = 2πr2 + 2πrh = 2πr2 + 2πr d = 2πr2 +

375 n πr 2

(b)

750 r

(b) 3.9 cm 12. (a) (0, 0) and (−1, 1) (b) (0, 0) minimum, (−1, 1) point of inflexion (c)

(c)

13. (a)

19. (a) x2 + y2 = 52 = 25 y2 = 25 − x2 y=

S = 2x2 + 4xh 250 = 2x2 + 4xh 250 − 2x2 = 4xh 250 − 2x2 =h 4x 2 (125 − x2) =h 4x 2 125 − x =h 2x 2 V =x h 125 − x2 n = x2 d 2x x (125 − x2) = 2 125x − x3 = 2

25 − x2

1 xy 2 1 = x 25 − x2 2

A=

(b) 6.25 m2 20. (a) (4, −171) minimum, (− 6, 329) maximum (b) (−1, 79) (c)

(b) 6.45 cm ´ 6.45 cm ´ 6.45 cm 14. x < 3

15. y = x3 + 3x2 + 3

17. For decreasing curve,

dy dx

16. 150 products

0 on LHS and RHS (b) y m < 0 on LHS, y m > 0 on RHS 21. 21

1 cm3 3

22. (a) (0, 1)

b3 b2 + +C 3 2

8.

x 2 + 4x + C

11.

x3 + x2 + 5x + C 3

10.

x 8 3x 7 − − 9x + C 7 8

14.

x6 x4 + + 4x + C 4 6

15.

x4 x3 x2 + − − 2x + C 2 3 2

17.

4x3 5x2 − − 8x + C 3 2

18.

19.

3x4 5x3 + − 4x + C 2 3

20. − x− 3 −

23.

a4 a2 − −a+C 4 2

12. x4 − x3 + 4x2 − x + C

x5 x4 + +C 5 2

13. x6 +

(b) k = 2, 4, 6, 8, . . .

9.

16.

3x5 x4 x2 − + +C 5 2 2 x− 2 − 2x − 1 + C 2

4

1 24. minimum −1; maximum − 5

Exercises 3.1 2.5

6.

0.39

2.

10

7.

11. 0.94

3.

2.4 4.

0.41 8.

0.225 5.

1.08 9.

12. 0.92 13. 75.1

24. x − 2x2 +

4x3 +C 3

1 +C 2x2

28. −

29.

Chapter 3: Integration

1.

22.

27. −

25. 87 kmh−1

y3 3

+

8.

16

2 3 2 20. 6 9 14. 4

35.

1.

(b)

3.

125 4. -1

5.

10 6.

54 7.

(x + 1)

1 3 3

(f)

5

3 +C x

3 3 x4 +C 4

33.

2 x3 +x+C 3

x6 +C 2. 2 6.

2x5 +C 3. 5

(7x + 8)13

h8 + 5h + C 8

7.

y2 2

− 3y + C

(5x − 1)

(4 + 3x)5

(e)

+C

(g) −

15 (1 − x)7 7

(i) −

(j) − 3 (x + 7) − 1 + C

(k) −

3

33 (4x + 3)4 16

(g) 4

2 3

9

+C

(b) − 1

(h) −

1 8

1 4

+C +C

2 (3x + 1)− 3

+C

9

1 +C 16 (4x − 5) 2 1

(m) − 2 (2 − x) 2 + C

+C

5 2 (t + 3) +C 5

(a) 288.2

(3x − 4)3

+C

50

+C

(n)

2.

(c)

+C

(2x − 5)3

(h)

m2 4. +m+C 2

+C

24 91

(ii) 10

(3y − 2)8

Exercises 3.4

t3 − 7t + C 3

26. −

t4 t3 − − 2t2 + 4t + C 4 3

1 +C 2t4

5

(l)

5.

x 3 3x 2 + − 10x + C 3 2

(a) (i) 3x3 − 12x2 + 16x + C

25. 0.0126

1.

30.

2 x5 +C 5

1 2 2 9. 50 10. 52 11. 12. 21 13. 0 4 3 3 1 1 1 15. 1 16. 4 17. 0 18. 2 19. 0 4 3 3 3 1 2 1 21. 101 22. −12 23. 22 24. 2 4 4 3 3

x3 +C 3

+ 5y + C

34.

(d) 10

x4 − x3 + x2 + C 4

3 4 7 −x+ 2 − 4 +C x 4x 2x

32. −

Exercises 3.3 2.

23.

Exercises 3.5

1. 48.7 2. 30.7 3. 1.1 4. 0.41 5. (a) 3.4475 (b) 3.4477 6. 2.75 7. 0.693 8. 1.93 9. 72 10. 5.25 11. 0.558 12. 0.347 13. 3.63 14. 7.87 15. 175.8

8

6

+C

25.

2 x3 +C 3

14. 16.5 15. 650.2

Exercises 3.2

1.

y− 6

4

31. (a) 28 (b) 22

0.75 10. 0.65

3x 3

1 21. − 7 + C 7x

(o)

7 2 (5x + 2) +C 35

(c) −

(i) 1

1 5

1 8

(j)

(d) 60 3 5

2 3

(e)

1 6

(f)

1 7

ANSWERS

Exercises 3.6 1. 5. 9.

1

1 units2 3

1 units2 6

2. 36 units2 3. 4.5 units2 4. 10

2 units2 3

4.

7.

2π units3 3

992π units3 5

14.

1 units2 6

15.

2 units2 3

1 12. 9 units2 3 16.

1 units2 3

8.

18. 18 units2 19. π = 3.14 units2

4

a units2 2

π units3 7

5.

39π units3 2

6.

5π units3 3

10.

9.

12.

64π units3 3

13.

16 385π units3 7

14.

25π units3 2

15.

65π units3 2

16.

1023π units3 5

5π units3 3 3π 20. units3 5

344π units3 27 72π 22. units3 5

18. 13π units3

19.

2π units3 5

21.

y = r2 − x2 ` y2 = r2 − x2 b

V = π # y2 dx a r

= π # ] r2 − x2 g dx

Exercises 3.7

−r

1.

2

4.

1.5 units

7.

10

2. 20 units

2 3. 4 units2 3

1 5. 1 units2 4

1 6. 2 units2 3

2

2 units2 3

8.

1 units2 6

9. 3

1 units2 4

12. 60 units2

13. 4.5 units2 14. 1

1 units2 3

15. 1.9 units2

1 2. 1 units2 3

1. 4.

2 10 units2 3

7.

2 units2 3

(− r)3 r3 oH n − e − r3 − 3 3 2r3 − 2r3 n =πd − 3 3 3 4π r = units3 3

Exercises 3.10

8. 166

2 units2 3

2 10. units2 3

1 11. units2 12

13. 36 units2

14. 2

2 units2 3

+C

(b)

+C

(d)

24 (x2 − 9)6 12

6. 8 units

3

+C

(x2 + 4x + 1)5

+C 10 3 2x 2 − 1 (f) +C 2

9. 0.42 units2 2. 1 12. units2 3 15. (π − 2) units2

u = ax + b du = a dx # (ax + b)n dx = 1a # (ax + b)n a dx 1 = # un du a 1 un + 1 n+C = d a n +1 n +1 1 (ax + b) o+ C = e a n +1 =

Exercises 3.9

3. 2.

2 (x + 3)3

1 +C 2 (x2 + 3x − 1)2 1 +C (g) − 9 (x3 + 1)3

206π units3 15

243π units3 5

(3x − 4)8

(e) −

2

Problem

1.

(a) (c)

1 3. units2 6

5 5. 20 units2 6

x3 r F 3 −r

= π > d r3 −

1.

Exercises 3.8 1 1 units2 3

= π < r2 x −

7 units2 9

10. 2 units2 11. 11

485π units3 3

9π units3 2

27π units3 2

23.

1 21 units2 3

758π units3 3

11.

17.

1 17. 5 units2 3 20.

376π units3 15

6. 14.3 units2 7. 4 units2 8. 0.4 units2

8 units2 10. 24.25 units2 11. 2 units2

13. 11

2 units2 3

3.

4.

(ax + b)n + 1 a (n + 1)

+C

x2 − 4 + C (2x + 7)5 10

−

7 (2x + 7)3 6

+C

587

588

Maths In Focus Mathematics Extension 1 HSC Course

5.

6.

7.

2 ] x + 2 g3 −8 x+2 +C 3 −10 (5 − x)3

(2x − 5)3 6

2 (5 − x)5

+

3

+C

5

5 2x − 5 +C 2

+

40 (x + 4)3

8.

2 (x + 4)5 −

9.

5 1 − y=− 2 (x2 − 1) 6

10. y =

(b)

3

6.

+C

3 units2 4 206π 12. units3 + C 13. 3 units2 14. (a) 15 84 π 2 (b) units3 15. (a) x = ± y − 3 (b) 3 units2 2 3

8.

(3x2 − 1) 2 + 11 12 (x − 2)5 3

11. f (x) =

16. 36

12. (a) Domain: all real x 2 −1 2 (b) y = 4 3

20. (a)

Exercises 3.11 1.

(a) 1

3 4

(c) 7 − 2 (c) 0

(b)

(f) 24.51 (g) − 3 2.

(a) 63.05 (b) 4 2

(e) 2 2 + 3.

6.

2 2

10.2 units

211π units3 15

2 units2 3 (7x + 3)12 2

(c)

+1

15

3 units2 7. 1.1 units2

(d)

1 8

(e)

15 256

1 3 (c) −

1 4

(d) −

19 72

−4=3 2 −4

4.

3 units2 8

5. 65 536

40π units3 7. 3

8. 2

2 3

1 units2 4 9. 6

2 units2 5

1. 2.

3. 5.

(2x + 5)8

14.83

16

11.

17. 10.55 (2x − 1)5 2

18. 85

+C

6 (x + 1)5

(b)

1 units2 3

19.

3π units3 5

x6 +C 8

− 2 (x + 1)3 + C

5

Challenge exercise 3 1 units2 12

(a)

2.

(a) Show f (− x) = − f (x) (b) 0

3.

27.2 units3 4. 9 units2 5. (a) 36x3 (x4 − 1)8 (b)

(x4 − 1)9

(b)

2π units3 35

1.

+C

36 − 22x (3x2 − 4)2

6.

(a)

9.

f (0) =

(c) 12 units2

(b)

1 8

7. 7.35 units2 8.

2π units3 3

1 =∞ 0

10. (a)

(a) 0.535 (b) 0.5 3x2 5x2 + x + C (b) −x+C (a) 2 2 (d)

1 2

10. 4

5π units3 2

10. 64 units2

Test yourself 3

9. 9π units3

(c)

2 x3 +C 3

+C

4. (a) 2

(b) 0 (c) 2

1 5

(b) 3.08 units2

(a) 11.

17 17 units2 6

13. (a)

3x + 6 2 x+3

=

12.

215π units3 6

3 (x + 2) 2 x+3

(b)

2x x + 3 +C 3

ANSWERS

14. (a) 6

15.

2 2 (b) 6 3 3

5 units2 16. (a) 12

589

8.

3 units2 (b)

58π units3 15

17. (a)

(b) 1.69 units2 18. 0

9.

(a)

2

8a units2 3 (b) 2πa3 units3

19. (a)

10. 1

Practice assessment task set 1 1.

198π units3 7

1 units2 3

(b)

96π units3 5

11. f l(3) = 20, f m(− 2) = −16 12. 68

13. AB = AC (given) (given) BD = CD AD is common. ` by SSS ΔABD / ΔACD ` +ADB = +ADC (corresponding +s in congruent Δs)

But +ADB + +ADC = 180° (+BDC is a straight +)

` +ADB = +ADC = 90° ` AD = BC Let ABCD be a parallelogram with diagonal AC. (alternate +s, AB < DC) +ACD = +BAC +DAC = +BCA (alternate +s, AD < BC) AC is common ` by AAS ΔACD / ΔACB ` AB = DC and AD = BC

7 9 16. f (1) = 3, f l(1) = − 2, f m(1) = 18; curve is decreasing and concave upwards at (1, 3)

14. (a) 78.7 units3 (b) 1.57 units3 15. x > −

(corresponding sides in congruent Δs)

` opposite sides are equal 1 2

2.

x<

6.

(opposite sides of < gram equal) AC = FD BC = FE (given) ` AB = AC − BC = FD − FE = ED

3. x3 − x2 + x + C

4. 24

5. 8 m

(since ACDF is < gram) Also AB < ED ` since AB = ED and AB < ED, ABDE is a parallelogram

7.

x9 + 2x 2 + C 3

17. P = 8x + 4y = 4 4y = 4 − 8x y = 1 − 2x A = 3x2 = 3x2 = 3x2 = 7x2

+ y2 + ] 1 − 2x g2 + 1 − 4x + 4x2 − 4x + 1 3 6 2 Rectangle m ´ m, square with sides m 7 7 7 18. f (−1) = 0

Ans_PART_1.indd 589

19. 12

20. 0.837

6/24/09 9:43:55 AM

590

Maths In Focus Mathematics Extension 1 HSC Course

21.

AB2 = 242 = 576 BC2 = 322 = 1024 AC2 = 402 = 1600 AB2 + BC2 = 576 + 1024 = 1600 = AC2

42. y =

2 (x − 4)5 5

+

8 (x − 4)3 3

43.

` ∆ABC is right angled at +B (Pythagoras’ theorem) 22.

(3x + 5)8 24

+C

23. − 5

1 3

24. (1, 1)

25. (a) 1.11 (b) 1.17 26. ^ 0, 3 h maximum, (1, 2) minimum, (−1, 2) minimum 27. 2

8 15

29. 12

2 3

28. (a) 1.58 units2

30. 10

(b)

44.

5π units3 2

2 units2 3

31. +B is common +BDC = +ACB = 90° ` ∆ABC 0 on both LHS and RHS of (0, 0) 33. 2

36.

2 3 m 3

34. f (2) = − 16 35. 9 units2

2 (x3 − 5)3 3

37. 119.3 m2

(diagonals perpendicular in rhombus)

1 DE = BE = y 2 (diagonals bisect each other)

1 1 1 ´ x ´ y = xy 4 2 2 1 1 1 ∆ADC has area ´ x ´ y = xy 4 2 2 1 1 1 ` ABCD has area xy + xy = xy 4 4 2

∆ACB has area

+C

38. f (x) = x3 − 4x2 − 3x + 20

39. (0, −1) maximum 45.

AB AG = AC AD AG AF = AD AE AB AF = ` AC AE

46. f (2) = 1

(corresponding sides in congruent ∆s)

` OC bisects AB 1 41. 20 2

(equal ratios of intercepts, GF < DE)

2 3

xn + 1 +C n+1 d (b) Since (C) = 0, the primitive function dx could include C.

47. (a) (given) 40. +OCA = +OCB = 90° (equal radii) OA = OB OC is common ` by RHS ∆OAC / ∆OBC ` AC = BC

(equal ratios of intercepts, BG < CD)

48.

3AB 2

49. 1249

7 units2 8

ANSWERS

50. Let +ADB = x Then+BCD = 2x (given) +A = 180 − 2x

3.

(e) 3ex ] ex + 1 g 2

+ABD = 180 − (180 − 2x + x) (angle sum of triangle)

(m) So ADB is an isosceles triangle.

3

55. (d)

56. (b)

52. (c), (d) 57. (a)

ex (x − 1)

(g) 4ex ] 2ex − 3 g

(j) xex ] x + 2 g

x2

(k) (2x + 1)ex + 2ex = ex (2x + 3) (l)

=x = +ADB

51.

(c) ex + 2x (d) 6x2 − 6x + 5 − ex (f) 7 ex (ex + 5)6

(h) ex ] x + 1 g (i)

(opposite angles in cyclic quadrilateral)

4 (6 6 − 1)

(b) − ex

(a) 9ex

53. (a), (b)

58. (b)

60. (b)

(7x − 3) 2

5ex − 5xex 5 (1 − x) = ex e 2x

4.

f l(1) = 6 − e; f m(1) = 6 − e

7.

19.81

54. (c)

59. (d)

ex (7x − 10)

10. c −1, −

5. e

6. − e− 5 = −

1 e5

8. ex + y = 0 9. x + e3 y − 3 − e6 = 0 1 m min e

Chapter 4: Exponential and logarithmic functions

Exercises 4.1 1.

(a) 4.48 (b) 0.14 (c) 2.70 (d) 0.05 (e) -0.14

2.

(a)

11.

12.

(b)

dy dx

= 7ex; dy

d2 y dx2

= 7ex = y d2 y

= 2ex;

dx dx2 y = 2ex + 1 ` y − 1 = 2ex d2 y ` =y−1 dx2

= 2ex

Exercises 4.2 1.

(a) 7e7x

(b) − e− x

(e) (3x2 + 5) ex (i) 2e2x + 1

3

(c) 6e6x − 2

+ 5x + 7

(d) 2xex

2

+1

(g) − 2e− 2x

(f) 5e5x

(j) 2x + 2 − e1 − x

(h) 10e10x

(k) 5 (1 + 4e4x) (x + e4x)4

e (3x − 2) 3x

(l) e2x (2x + 1) (m) (c)

(o)

x3

(n) x2 e5x (5x + 3)

4e2x + 1 (x + 2) (2x + 5)2

2.

28e2x (e2x + 1)5 (7e2x + 1)

3.

f (1) = 3e; f m(0) = 9e− 2

5.

x + y − 1 = 0 6.

7.

y = 2ex − e

8.

−

4.

5

1 3e3

f m(−1) = −18 − 4e2

591

592

Maths In Focus Mathematics Extension 1 HSC Course

9.

10.

(0, 0) min; (−1, e − 2) max

dy dx d2 y dx2

= 4e 4x − 4e − 4 x = 16e4x + 16e− 4x

15.

Exercises 4.3 1.

= 16 (e4x + e− 4x) = 16y 11.

12.

y = 3e2x dy = 6e 2 x dx d2 y = 12e2x dx2 d2 y dy −3 + 2y LHS = dx dx2 2x = 12e − 3 (6e2x) + 2 (3e2x) = 12e2x − 18e2x + 6e2x =0 = RHS d2 y dy ` −3 + 2y = 0 dx dx2 y = aebx dy = baebx dx d2 y = b2aebx dx2 = b2 y

1 2x e +C 2

(b)

1 4x e +C 4

(c) − e− x + C

(d)

1 5x e +C 5

3 1 1 − 2x e + C (f) e4x + 1 + C (g) − e5x + C 4 5 2 x2 1 7x 1 2t x−3 +C (h) e + C (i) e − 2x + C (j) e + 7 2 2 (e) −

2 1 1 5 (e − 1) (b) e− 2 − 1 = 2 − 1 (c) e7 (e9 − 1) 5 3 e 1 4 2 1 4 1 1 (d) 19 − e (e − 1) (e) e + 1 (f) e2 − e − 1 2 2 2 2 1 1 (g) e6 + e− 3 − 1 2 2

2.

(a)

3.

(a) 0.32 (b) 268.29 (e) 755.19

4.

e4 − e2 = e2 (e2 − 1) units2

6.

2.86 units2 7. 29.5 units2

9.

4.8 units3 10. 7.4

12.

1 x e 3

15.

1 4 (e − 5) units2 2

13. n = − 15 14.

(a)

3

+1

+C

(c) 37 855.68

5.

(d) 346.85

1 (e − e− 3) units2 4 π 8. (e6 − 1) units3 2

11. (a) x (2 + x) ex

(b) x2 ex + C

1 4 (e − 1) 14. πe units3 2

13.

Exercises 4.4 1.

(a) 4

2.

(a) 9 (b) 3 (i) 1 (j) 2

3.

(a) −1 (h)

4.

1 3

(b) 2

1 2 1 (i) 1 2 (b)

(c) 3

(d) 1

(c) −1

(c)

1 2

(e) 2

(d) 12

(d) −2

(j) − 1

(f ) 1

(e) 8

(e)

1 4

(g) 0

(f ) 4

(f ) −

(h) 7

(g) 14

(h) 14

1 3

1 2

(g) −

1 2

(a) 3.08 (b) 2.94 (c) 3.22 (d) 4.94 (e) 10.40 (f ) 7.04 (g) 0.59 (h) 3.51 (i) 0.43 (j) 2.21

ANSWERS

5.

(a) log3 y = x (b) log5 z = x (c) logx y = 2 (d) log2 a = b (e) logb d = 3 (f ) log8 y = x (g) log6 y = x (h) loge y = x (i) loga y = x (j) loge Q = x

4.

(a) x + y (b) x − y (c) 3x (d) 2y (e) 2x (g) x + 1 (h) 1 − y (i) 2x + 1 (j) 3y − 1

5.

(a) p + q (b) 3q (c) q − p (d) 2p (e) p + 5q (f ) 2p − q (g) p + 1 (h) 1 − 2q (i) 3 + q (j) p − 1 − q

(f ) x + 2y

6.

(a) 3x = 5 (b) ax = 7 (c) 3b = a (d) x9 = y (e) ay = b (f ) 2y = 6 (g) 3y = x (h) 10y = 9 (i) ey = 4 (j) 7y = x

6.

(a) 1.3 (b) 12.8 (c) 16.2 (d) 9.1 (g) -3.7 (h) 3 (i) 22.2 (j) 23

7.

(a) x = 1 000 000 (b) x = 243 (c) x = 7 (d) x = 2 (e) x = − 1 (f ) x = 3 (g) x = 44.7 (h) x = 10 000 (i) x = 8 (j) x = 64

7.

(a) x = 4 (b) y = 28 (c) x = 48 (d) x = 3 (e) k = 6

8.

y = 5 9. 44.7 10. 2.44 11. 0

Exercises 4.6

13. (a) 1 (b) (i) 3 (ii) 2

(iii) 5

(vii) 3 (viii) 5 (ix) 7

(iv)

(x) 1

12. 1 1 2

(v) -1

(vi) 2

(xi) e

(e) 6.7

(f ) 23.8

1.

(a) 1.58 (b) 1.80 (c) 2.41 (d) 3.58 (g) 1.40 (h) 4.55 (i) 4.59 (j) 7.29

2.

(a) x = 1.6 (b) x = 1.5 (c) x = 1.4 (d) x = 3.9 (e) x = 2.2 (f ) x = 2.3 (g) x = 6.2 (h) x = 2.8 (i) x = 2.9 (j) x = 2.4

3.

(a) x = 2.58 (b) y = 1.68 (c) x = 2.73 (d) m = 1.78 (e) k = 2.82 (f ) t = 1.26 (g) x = 1.15 (h) p = 5.83 (i) x = 3.17 (j) n = 2.58

4.

(a) x = 0.9 (b) n = 0.9 (c) x = 6.6 (d) n = 1.2 (e) x = − 0.2 (f ) n = 2.2 (g) x = 2.2 (h) k = 0.9 (i) x = 3.6 (j) y = 0.6

5.

(a) x = 5.30 (b) t = 0.536 (c) t = 3.62 (d) x = 3.81 (e) n = 3.40 (f ) t = 0.536 (g) t = 24.6 (h) k = 67.2 (i) t = 54.9 (j) k = − 43.3

14. Domain: x > 0; range: all real y

15. Curves are symmetrical about the line y = x.

(e) 2.85

Exercises 4.7 1.

(a) 1 +

1 x

(b) −

1 x

(c)

3 3x + 1

(d)

2x x2 − 4

10x2 + 2x + 5 15x2 + 3 5 (f) + 2x = 3 5x + 1 5x + 1 5x + 3x − 9 8 6x + 5 1 (g) 6x + 5 + (h) (i) x 8x − 9 (x + 2) (3x − 1) − 30 4 2 (j) − = 4x + 1 2x − 7 (4x + 1) (2x − 7) (e)

5 1 (1 + loge x)4 (l) 9 c − 1 m (ln x − x)8 x x 4 1 3 (m) (loge x) (n) 6 c 2x + m (x2 + loge x)5 x x 1 − loge x (o) 1 + loge x (p) x2

(k) 16. x = ey

Exercises 4.5 1.

(a) loga 4y

(b) loga 20 (c) loga 4 (d) loga

x5 (e) logx y3 z (f ) logk 9y3 (g) loga 2 y p3 q (i) log10 ab4 c3 (j) log3 2 r

b 5

xy (h) loga z

2.

(a) 1.19 (b) -0.47 (c) 1.55 (d) 1.66 (e) 1.08 (f ) 1.36 (g) 2.02 (h) 1.83 (i) 2.36 (j) 2.19

3.

(a) 2 (b) 6 (c) 2 (d) 3 1 (h) (i) -2 (j) 4 2

(e) 1

(f ) 3 (g) 7

x3 2x + 1 + 2 loge x (r) + 3x2 loge (x + 1) x x+1 x − 2 − x loge x 1 (s) (t) x loge x x (x − 2) 2

(q)

(u)

e2x (2x loge x − 1) x (loge x) 2

1 (v) ex c + loge x m x (w)

10 loge x x

(f ) 2.66

593

594

Maths In Focus Mathematics Extension 1 HSC Course

1 4. x − 2y − 2 + 2 loge 2 = 0 x loge 10 2 5. y = x − 2 6. − 7. 5x + y − loge 5 − 25 = 0 5 1 1 1 1 8. 5x − 19y + 19 loge 19 − 15 = 0 9. d , loge − n 2 2 2 4 1 10. c e, m maximum e 2.

f l(1) = −

1 2

3.

11. (a)

2.

(a) ln (4x − 1) + C

(b) loge (x + 3) + C

1 1 loge (2x6 + 5) + C (c) ln (2x3 − 7) + C (d) 6 12 1 (e) loge (x2 + 6x + 2) + C 2 3.

(a) 0.5

(b) 0.7

(c) 1.6

4.

loge 3 − loge 2 = loge 1.5 units2

6.

(0.5 + loge 2) units2

8.

π loge 3 units3

=

(e) 0.5

5. loge 2 units2

7. 0.61 units2

9. 2π loge 9 units3

10. 47.2 units2 11. 12. (a) RHS =

(d) 3. 1

π 2 4 e (e − 1) units3 2

1 2 + x+3 x−3 1 (x − 3)

(x + 3) (x − 3) 2 (x + 3) + (x + 3) (x − 3) x − 3 + 2x + 6 = x2 − 9 3x + 3 = 2 x −9 = LHS 3x + 3 1 2 = + ` 2 x −9 x+3 x−3

(b)

(b) loge (x + 3) + 2 loge (x − 3) + C (c)

5 x−1 x−1 5 − = x−1 x−1 x−6 = x−1 = LHS

13. (a) RHS = 1 −

`

12.

2 13. (a) 3x ln 3 (b) 10x ln 10 (2x + 5) loge 3 (c) 3 ln 2 ´ 23x − 4

14. 4 ln 4 $ x − y + 4 = 0 15. 3 loge 3 $ x + y − 1 − 9 loge 3 = 0

(b) x − 5 loge (x − 1) + C 14.

(a) loge (2x + 5) + C

(b) loge (2x2 + 1) + C 1 1 (c) ln (x5 − 2) + C (d) loge x + C or loge 2x + C 2 2 5 (e) 2 ln x + C (f) loge x + C (g) loge (x2 − 3x) + C 3 3 1 2 (h) ln (x + 2) + C (i) loge (x2 + 7) + C 2 2 1 (j) loge (x2 + 2x − 5) + C 2

3 2x − 1 +C 2 loge 3

15. 1.86 units2

Test yourself 4 1.

(a) 6.39 (b) 1.98 (c) 3.26 (d) 1.40 (e) 0.792 (f) 3.91 (g) 5.72 (h) 72.4 (i) 6 (j) 2

2.

(a) 5e5x

Exercises 4.8 1.

x−6 5 =1− x−1 x−1

(b) − 2e1 − x

(c)

1 x

(d)

4 4x + 5

1 − ln x (g) 10ex (ex + 1)9 x2 1 1 (a) e4x + C (b) ln (x2 − 9) + C 4 2

(e) ex (x + 1)

(f)

3.

(d) ln (x + 4) + C

(c) − e− x + C

ANSWERS

e2 e +1

1 4 6 e (e − 1) units2 2

4.

3x − y + 3 = 0 5. −

7.

π 6 6 e (e − 1) units3 6

8.

(a) 0.92 (b) 1.08 (c) 0.2 (d) 1.36

9.

e (e2 − 1) units2

6.

2

10.

1 x e 3

11.

d 2 3x +C (x loge x) = x (1 + 2 loge x); 18 loge 3 12. loge 3 dx

(e) 0.64

−1

+C

(b) x − y − 1 = 0; x − loge 10 $ y − 1 = 0

13. (a) (1, 0)

10. (a) 2.16 units2 (b) x = ey

1 (c) f 1 − p units 14. 0.645 units2 loge 10

(c) 2.16 units2

11. (a) x = 1.9 (b) x = 1.9 (c) x = 3 (d) x = 36 (e) t = 18.2

15.

ex (1 + loge x) − xex loge x

=

16.

3 12. (a) (e2 − 1) 2 1 (b) ln 10 3 1 (c) 8 + 3 ln 2 6

17.

13. e4 x − y − 3e4 = 0 14. 0.9 15. (a) e (e − 1) units2 π (b) e2 (e2 − 1) units3 2

18. 19.

16. (a) loga x5 y3 (b) logx

3

k2 p 3

17. 2x + y − ln 2 − 4 = 0 18. (0, 0) point of inflexion, (− 3, − 27e− 3) minimum 19. 5.36 units2 20. (a) 0.65 (b) 1.3

e 2x 1 + loge x − x loge x ex

1 3x e 6

2

+1

+C

y = ex + e− x dy = ex − e− x dx d2 y = ex − (− e− x) dx2 = ex + e− x =y e8 − 1 1 6 (e − e− 2) = 3 3e 2 y = 3e5x − 2 dy = 15e5x dx d2 y = 75e5x dx2 d2 y dy −4 − 5y − 10 LHS = dx dx2 5x = 75e − 4 (15e5x) − 5 (3e5x − 2) − 10 = 75e5x − 60e5x − 15e5x + 10 − 10 =0 = RHS

20. f (x) = 3e2x − 6x

Challenge exercise 4 (e2x + x) 1.

1 − (2e2x + 1) loge x x (e2x + x) 2

4.

(a) 2.8 (b) 1.8 (c) 2.6

5.

9 c 4e4x +

7.

−

2 2x − 3

21.

1 m 4x (e + loge x) 8 x 8.

12 units3

2.

1 x e +C 2 2

3.

2e

6. 0.42 units2 9. 5x loge 5 22. y =

1 x e 2

2

595

596

Maths In Focus Mathematics Extension 1 HSC Course

Chapter 5: Trigonometric functions

5.

π π π π cos + sin sin 4 4 3 3 3 1 1 1 = ´ + ´ 2 2 2 2

LHS = cos

Exercises 5.1 1.

2.

(a) 36° (b) 120° (c) 225° (d) 210° (e) 540° (g) 240° (h) 420° (i) 20° (j) 50° 3π (a) 4 (h)

π (b) 6

5π 2

(i)

4π (d) 3

5π (c) 6

5π 4

(j)

5π (e) 3

7π (f) 20

(f) 140°

3 1 + 2 2 2 2 π π π π RHS = sin cos + cos sin 4 4 6 6 3 1 1 1 + = ´ ´ 2 2 2 2 3 1 = + 2 2 2 2 LHS = RHS π π π π π π So cos cos + sin sin = sin cos + 4 4 4 6 3 3 π π cos sin 4 6 =

π (g) 12

2π 3

3.

(a) 0.98 (b) 1.19 (c) 1.78 (d) 1.54 (e) 0.88

4.

(a) 0.32 (b) 0.61 (c) 1.78 (d) 1.54 (e) 0.88

5.

(a) 62° 27’ (b) 44° 0’ (c) 66° 28’ (d) 56° 43’ (e) 18° 20’ (f) 183° 21’ (g) 154° 42’ (h) 246° 57’ (i) 320° 51’ (j) 6° 18’

6.

(a) 0.34 (b) 0.07 (c) 0.06 (d) 0.83 (e) −1.14 (f) 0.33 (g) −1.50 (h) 0.06 (i) −0.73 (j) 0.16

6.

(a)

7.

(a)

8.

(a)

9.

(a)

Exercises 5.2 1. π 3

π 4

π 6

sin

3 2

1

1 2

cos

1 2

1

tan

2

1

1

3 2

cosec

3 2

2

3 2

2

10. (a)

3 2

sec

1

cot

2.

(a) (g)

1 2

(c)

2− 3

(h)

(b)

3 3 8

3

1

3

1 3

2

2

(d)

2+2 2

4 3 3

(i)

3

(e) 0

3−2 2 2

(f) (j)

2 3+1 2

3π 4π π = − 4 4 4 π =π− 4 5π 6π π = − 6 6 6 π =π− 6 7 π 8π π = − 4 4 4 π = 2π − 4 4π 3π π = + 3 3 3 π =π+ 3 5 π 6π π = − 3 3 3 π = 2π − 3

11. (a) − 1

12. (a) (i)

2+ 3 2 (b) (i)

3.

(a) 1

1 4

(b)

6− 2 4

(c)

3 2

(d) 1

(e) 4

1 4 13. (a)

4.

(a)

6+ 4

2

(b)

3−2

(b)

3 2

(c) −

(b) 2nd

(c)

(b) 4th

(c) − 1

(b) 3rd

(c) −

(b) 4th

(c) −

3 2

(c) − 3

(d) −

1

13π 12π π = + 6 6 6 π = 2π + 6 1 2

π 5π , 3 3

14. (a) sin θ

(ii)

(b)

1

(b) 2nd

3

2

1 2

1 2

(ii) 1st

(iii) −

5π 7 π , 4 4

(c)

1 2

(iv)

1

(d)

1 3

3 2

(iii)

π 5π , 4 4

(b) − tan x (c) − cos α

(e)

2

3

(v) −

π 4π , 3 3

(d) sin x

3 2

(e)

5π 7π , 6 6

(e) cot θ

ANSWERS

π π 2 π + 3 π 5π + = = 6 4 12 12

15. (a)

sin θ +

16. (a)

3 cos θ 2

1 − tan x (c) 1 + tan x

(d)

(b)

(b)

3 −1 2 2

cos θ − sin θ

2 cos y − 3 sin y

=

=

(e)

2

2 ( 3 − 1) 4 2 (cos θ − sin θ )

1.

2 cos θ +

3 sin θ 2

6π − 9 3 2 125π − 75 cm2 m (c) 4 3 49 (π − 2 2 ) 3 (π − 3) (d) mm2 cm2 (e) 4 8 (a) 8π cm2

(b)

2.

(a) 0.01 m2 (b) 1.45 cm2 (c) 3.65 mm2 (d) 0.19 cm2 (e) 0.99 m2

3.

0.22 cm2 4.

5.

134.4 cm

2π 4π π 3π 5 π 7 π , , , , 2π (b) x = , 4 4 4 4 3 3 π π π 3π (c) x = (d) x = (e) x = 0, π , 2π , , 3 2 2 2

7.

(a) 10.5 mm

8.

(a)

π π π (b) x = nπ − (c) x = nπ + (−1)n 4 3 3 π n π (d) x = nπ ± (e) x = πn + ^ − 1 h , 2πn 6 2 π π n (f) x = n π, n π + (−1) , n π − (−1)n 2 2 π (g) x = 2πn ± 6

9.

(a) 77° 22l (b) 70.3 cm2 (c) 26.96 cm2 (d) 425.43 cm2

17. (a) sin2 θ

(c) tan d θ +

(b) sin α

(d) cos2 θ − sin2 θ = cos 2θ

π n 3

(e) tan θ

3 4 18. cos x = − ; sin x = 5 5 19. (a) x = 0,

20. (a) x = 2nπ ±

Exercises 5.3 π cm 2

7π mm 4

2.

(a) 0.65 m (b) 3.92 cm (c) 6.91 mm (d) 2.39 cm (e) 3.03 m 1.8 m

8.

13

(b) π m

25π cm 3

(a) 4π cm

3.

4. 7.5 m

7 mm 9

V=

5.

(c)

2π 21

(d)

(e)

6. 25 mm 7. 1.83

9. 25.3 mm 10. SA =

9π cm2 14

(c) 0.07 cm2

6. (a) 2.6 cm

(b)

5π cm 6

(c) 0.29 cm2

(b) 4.3 mm2

25π cm2 4

(b) 0.5 cm2

11π cm 9

(c) 22 +

(b) 22 −

11π 11 (18 + π ) = cm 9 9

(b)

3π 2 m 2

(c)

125π cm2 3

(d)

3π cm2 4

49π mm2 8

(a) 0.48 m2 (e) 3.18 m2

(b) 6.29 cm2

3.

16.6 m2

7.

6845 mm2 8. 75 cm2 9. 11.97 cm2 8π

4. θ = 4

π , r = 3 cm 15

4 9

13. (a) 10π cm 14. (a) 8 +

121π 396 − 121π cm2 = 18 18

(c) 15.6 cm

(b) 24π cm2

20π 4 (18 + 5π ) = cm 9 9

225π cm3 2

(b)

(b) 3:7

15 (7π + 24) 105π cm2 + 180 = 2 2

Exercises 5.6

2.

10. θ =

11. (a)

175π cm2, 36

Exercises 5.4

(e)

(b)

10. 9.4 cm2

15. (a)

125 35 π cm3 648

(a) 8π cm2

3π cm 7

(a)

12. (a) 5 cm2 (b) 0.3%

1.

1.

Exercises 5.5

(c) 24.88 mm2

5. 6 m

6. (a)

7π cm 6

(d) 7.05 cm2

(b)

49π cm2 12

1.

(a) 0.045

2.

1 4

4.

(a) sin d

3.

(b) 0.003

(c) 0.999

(d) 0.065

1 3 π π π + x n = sin cos x + cos sin x 3 3 3 3 1 cos x + sin x = 2 2 3 1 ´1 + ´ x Z 2 2 1 = ( 3 + x) 2

(e) 0.005

597

598

Maths In Focus Mathematics Extension 1 HSC Course

(b) cos d

π π π − x n = cos cos x + sin sin x 4 4 4 1 1 = cos x + sin x 2 2 1 1 Z ´1 + ´x 2 2 2 1 = ( 1 + x) ´ 2 2 2 = ( 1 + x) 2

(c)

2

1

π tan x + tan π 4 (c) tan d x + n = π 4 1 − tan x tan 4 tan x + 1 = 1 − tan x ´ 1 x +1 Z 1− x 5.

1 343 622 km

y

3π 2

π 2

π

π 2

π

2π

x

-1 y

(d)

3

6. 7367 m

2

Exercises 5.7 1.

1

y

(a)

1

x 3π 2

2π

y

(e) x π 2

π

3π 2

2π

3

-1

x

y

(b)

π 2

π

2π

3π 2

-3

2 y

(f) x π 2

π

3π 2

2π

4

-2 π 2 -4

π

x 3π 2

2π

ANSWERS

y

(g)

(b)

y

4 3 2

π 4

1

7π 4

5π 4

3π 2

4π 3

5π 3

2π

2π

3π 2

y

(h)

π

3π 4

x

π

π 2

π 2

(c)

y

1 5 π 2

x

π

2π

3π 2

π 3

2π 3

π

x

2π

–1 (i)

y

y

(d)

3

3

(j)

x

π

π 2

3π 2

2π

π 4

y

π 2

3π 4

π

5π 4

3π 2

7π 4

2π

x

–3

(e)

1

y

6 π 2

π

2π

3π 2

x π 3

2π 3

π

4π 3

5π 3

2π

x

-6 2.

(a)

(f)

y

y

1 π

π 4

-1

π 2

3π 4

π

5π 4

3π 2

7π 4

2π

x

x 2π

x

599

600

Maths In Focus Mathematics Extension 1 HSC Course

(g)

3.

y

(a)

y

1 π 3

2π 3

π

4π 3

5π 3

2π

x

-π (h)

π - 3π - 2 4

-

π 4

y

3π 4

π

π 2

3π 4

π

π 4

π 2

3π 4

π 4

π 2

3π 4

x

-1 y

(b)

7

3

π 2

π

3π 2

-π

x

2π

-

3π - π 2 4

-

π 4

π 4

x

-7 (c)

-3 (i)

π 2

π 4

y

y

-π

-

2

3π - π 2 4

-

π 4

π

x

y

(d) x π

π 2

2π

3π 2

5

−2 -π

-

3π π 4 2

-

π 4

-5

y

(j)

x

π

(e)

y

4 2

π 2

-4

π

3π 2

2π

x

-π

-

3π - π 2 4

-

π 4

π 4

-2

π 2

3π 4

π

x

ANSWERS

(e)

y

4.

y

8

2

2π

π

3π

x

4π

π 2

-8

2π

x

-2 (f)

5.

3π 2

π

(a)

y

y

4 1

π 2

3π 2

π

x

2π

π 4

π 2

π 4

π 2

3π 4

π

5π 4

3π 2

7π 4

2π

x

-4 -1 (g)

y

(b) y

1

π

π 2

3π 2

2π

x

3π 4

π

5π 4

3π 2

7π 4

x

2π

-1 (c)

(h)

y

y

1

π 2

π

1

x 3π 2

2π

π 4

-1

π 2

3π 4

π

5π 4

3π 2

7π 4

x

2π

y

(d)

(i)

y

3 3

π 2

π

x 3π 2

2π

2 1

-3 -1

π 2

π

3π 2

2π

x

601

602

Maths In Focus Mathematics Extension 1 HSC Course

(j)

8.

y

(a)

y

5

5

4

4

3

3

2

2

1

1

−1 π 2

-1

π

3π 2

2π

y = 2 cos x π 2

π

−2

x

2π

3π 2

y = 3 sin x

−3 −4 −5

6.

(a)

y (b)

y 5

1

4

-2

2

1

-1

3

x

2 1

-1

π 2

−1

(b)

π

−2

y

−3 −4

3

3π 2

2π

x

y = 2 cos x + 3 sin x

−5

x

-2

-1

1

2

9.

y

-3 y = cos 2x - cos x

2 7.

(a)

y 1

π 2

y = sin x

1 π 2

-1

π y = sin 2x

3π 2

2π

x

-2

y

(b)

2 1

-1 -2

π 2

π

3π 2

-1

2π

x

π

3π 2

2π

x

x

ANSWERS

10. (a)

y

y

(e) 3

2 2

y = cos x + sin x 1

1

π

π 2

2π

3π 2

x

π 2

-1

π

3π 2

y = sin x - sin

-2

-1

x

2π x 2

-3 -2

-4

y

(b)

Exercises 5.8 1.

2

(a)

y y= x 2

y = sin 2x – sin x

1 1

π

π 2

3π 2

2π

x

–1

1 π 2 2

3 π

4 3π 5 2

–2

x

2π

6

y = sin x -1 y

(c)

There are 2 points of intersection, so there are 2 solutions to the equation. y (b)

y = sin x + 2 cos 2x

2

y= 1

x 2

1

π 2

π

2π

3π 2

–1

x

y = sin x π -π -3 -2 - -1 2

1

π 2 2

x 3π

–2

-1 –3

(d)

y

There are 3 points of intersection, so there are 3 solutions to the equation.

3 2

y =3 cos x − cos 2x

1

−1 −2 −3 −4

π 2

π

3π 2

2π

x

603

Maths In Focus Mathematics Extension 1 HSC Course

2.

x =0

6.

x = 0.8, 4

(m) 3 sin 2x sec2 3x + 2 tan 3x cos 2x

4. x = 0, 4.5 5. x = 0, 1

3. x = 1.5

(n)

y

2x cos x − 2 sin x x cos x − sin x = 4x 2 2x2 3 sin 5x − 5 (3x + 4) cos 5x

(o)

sin2 5x

(p) 9 (2 + 7 sec2 7x) (2x + tan 7x)8 (q) 2 sin x cos x 1

π 2

π

3π 2

2π

(r) − 45 sin 5x cos2 5x

1 cos (1 − loge x) x cos x = cot x (u) (ex + 1) cos (ex + x) (v) sin x 3x 3x (w) − 2e sin 2x + 3e cos 2x = e3x (3 cos 2x − 2 sin 2x) (s) ex + 2 sin 2x

y = cos x

x

y = sin x

7.

(a) Period 12 months, amplitude 1.5 (b) 5.30 p.m.

8.

(a) 1300 (b) (i) 1600 (ii) 1010 (c) Amplitude 300, period 10 years

2.

9.

(a)

3. 12

18

4 cos2 x sin3 x − sin5 x = sin3 x (4 cos2 x − sin2 x)

16 14

(t) −

2e2x tan 7x − 7e2x sec2 7x tan2 7x 2x e (2 tan 7x − 7 sec2 7x) = tan2 7x

(x)

-1

4. 6 3 x − 12y + 6 − π 3 = 0

5.

− sin x = − tan x cos x

7.

sec2 xetanx

9.

y = 2 cos 5x dy = − 10 sin 5x dx d2 y = − 50 cos 5x dx2 = − 25 (2 cos 5x) = − 25y

10.

f (x) = − 2 sin x f l(x) = − 2 cos x f m(x) = 2 sin x = − f (x)

12

6. −

2 3 3

=−

2 3 9

10 8

8. 8 2 x + 48y − 72 2 − π 2 = 0

6 4 2 0 January February March

April

May

June

July

August

(b) It may be periodic - hard to tell from this data. Period would be about 10 months. (c) Amplitude is 1.5 10. (a)

4 3.5 3 2.5 2

d [loge (tan x)] dx sec2 x = tan x tan2 x + 1 = tan x tan2 x 1 = + tan x tan x = tan x + cot x = RHS

11. LHS =

1.5 1 0.5

am 6.1 5p m 11 .48 pm

m

.55 11

pm

6.2

0a

m

.48 11

am

6.1

5p

am

.55 11

6.2 0

pm

m

.48 11

am

6.1

.55 11

5p

am

0 6.2 0

604

(b) Period 24 hours, amplitude 1.25 (c) 2.5 m

Exercises 5.9 1.

(a) 4 cos 4x (b) − 3 sin 3x (c) 5 sec2 5x (d) 3 sec2 (3x + 1) (e) sin (− x) (f) 3 cos x (g) − 20 sin (5x − 3) (h) − 6x2 sin (x3) (i) 14x sec2 (x2 + 5) (j) 3 cos 3x − 8 sin 8x (k) sec2 (π + x) + 2x (l) x sec2 x + tan x

`

d 7 loge (tan x) A = tan x + cot x dx

π 12. d , 3

d

3−

π n maximum, 3

5π 5π n minimum ,− 3 − 3 3

ANSWERS

13. (a) 14.

π sec2 x° 180

(b)

−π sin x° 60

π cos x° 900

(c)

(b)

y = 2 sin 3x − 5 cos 3x dy = 6 cos 3x + 15 sin 3x dx d2 y = −18 sin 3x + 45 cos 3x dx2 = − 9 (2 sin 3x − 5 cos 3x) = − 9y

14. (a)

1.

(b) − cos x + C

(c) tan x + C

− 45 1 cos x° + C (e) − cos 3x + C (d) π 3 1 (g) tan 5x + C (h) sin (x + 1) + C 5 (i) −

1 cos (2x − 3) + C 2

(j)

(f)

1 cos 7x + C 7

1 sin (2x − 1) + C 2

(a) 1

(b)

1 π

(f)

(e)

3 3 (g) 4

1 2

1

3.

4 units2 4.

6.

0.51 units3 7.

3−

10.

12. (a)

1

3−

3 2

=

=

2 3 3 (h) −

(c)

2 2

=

2. (d) −

2

1 5

#

1 3 3.

2 units2 6

5. 0.86 units2

π units3 8. etan x + C 4

9.

5.

1 24

π 3 3− π units2 11. 2 2 units2 = 3 3

V =π

(b)

π (π − 2) units3 8

x3 1 1 1 1 sin 2x − x + C (b) x − sin 2x + +C 4 4 2 2 3 7x 3 1 + C (d) x − sin 2x − sin x + C (c) sin 2x − 4 2 2 7x 7 1 (e) + sin 2x + tan 7x + C 4 7 2 5x 1 +C (f) sin 2x + sin x + 4 2 1 1 1 (g) − cos 2x − x + sin 2x + C 4 2 2 (a)

(h) tan x +

(k) cos (π − x) + C (l) sin (x + π ) + C x 2 (m) tan 7x + C (n) − 8 cos +C 7 2 x (o) 9 tan + C (p) − cos (3 − x) + C 3 2.

1 (1 − cos 2x) 2

Exercises 5.11

Exercises 5.10 (a) sin x + C

1 1 sin 2x + x + C 4 2

15. y = − 2 sin 3x

15. a = − 7, b = − 24

1.

605

7.

(i)

1 1 1 sin 3x + x + sin 2x + C 4 3 2

(a)

π 4

(e)

3 2π − 3 3 π − = 12 8 24

y 2 dx

(b)

3π 2

(c)

π (2 + 3π ) 8

(j)

(d)

1 1 x + sin 4x + C 2 8

π 3

π−2 π units2 (π − 2) units3 4. 8 8 π 1 1 (3π + 8) units3 6. (a) (x − sin 6x) + C 4 6 2 90 1 1 sin (2x)° E + C (b) (x + sin x) + C (c) ; x − π 2 2 3 4x 5 1 1 n +C sin 2ax) + C (e) d x − sin (d) (x + 4 5 2 2a 2 (a) sin 3x = 3 sin x − 4 sin3 x (b)

b

1 1 x − sin 2x + C 4 2

1e8 3 3 o 1 − = (16 − 9 3 ) 4 3 2 24

a

=π

#

π 2

cos x dx

8.

(a) sin (7x + 3x) + sin (7x − 3x) = sin 7x cos 3x + cos 7x sin 3x + sin 7x cos 3x − cos 7x sin 3x = 2 sin 7x cos 3x 1 ` [sin (7x + 3x) + sin (7x − 3x)] 2 = sin 7x cos 3x 1 (b) 10

9.

(a) −

0

= π ; sin x E

π 2 0

π − sin 0) 2 = π (1 − 0) = π units 3 = π (sin

(b) 3.1 units3 13. (a)

cos 2x = cos x − sin x = cos2 x − (1 − cos2 x) = 2 cos2 x − 1 cos 2x + 1 = 2 cos2 x 2

1 (cos 2x + 1) = cos2 x 2

2

(b) sin x +

(c)

1 1 (θ − sin 4θ ) + C 4 8

(e)

1 1 c sin 3θ + 3 sin θ m + C 4 3

10. (a)

Ans_PART_2.indd 605

1 cos 2x + C 4

1 1 (x − sin 2x) + C 2 2

(d) x −

1 cos 2x + C 2

π 1 (2 2 − 3 − 1) units2 (b) (2 − 3 ) units3 4 2

6/30/09 12:09:23 PM

606

Maths In Focus Mathematics Extension 1 HSC Course

Test yourself 5 25π cm2 12 3 3 (b) (c) 2 2

5π cm 6

1.

(a)

2.

(a)

3.

(a) x =

4.

(a)

3

15. (a)

(b)

3π 7π , 4 4

(b) x =

(c) 0.295 cm2 (d) −

1 2

π 5π , 6 6

(b) x = 0.6

16. (a)

(π − 6 + 3 3 ) 3− 2 units2 (b) π units3 2 24

17. 2 units2 18. 4x + 8y − 8 − π = 0

19. y = − 3 cos 2x

20. (a) (b)

(b) x = 0.9, 2.3, 3 5.

(a) − sin x (e)

6.

(b) 2 cos x

x sec2 x − tan x x2

(a) −

1 cos 2x + C 2

(c) sec2 x

(f) − 3 sin 3x (b) 3 sin x + C

(d) x cos x + sin x

(c)

1 tan 5x + C 5

(d) x − cos x + C 1

(a)

8.

3x + 2 y − 1 −

9.

x = cos 2t dx = − 2 sin 2t dt 2 d x = − 4 cos 2t dt2 = − 4x

10.

1 2

2

(b)

14. (a)

0.27

4.

(a) Period = 2, amplitude = 3

(c)

4π − 3 3 12

(d)

π 12

3π =0 4

units2 11.

13. − 3 3

2.

3− 3 1 1 e1 − o= 6 2 3

1.

(b)

2 3 3

7.

Challenge exercise 5

(g) 5 sec2 5x

5.

(a) y = − sin 3x (b) LHS =

π 3

units3 12. (a) 5

8π cm2 7

(b) 0.12 cm2

(b) 2

d2 y

+ 9y dx2 = 9 sin 3x + 9 (− sin 3x) = 9 sin 3x − 9 sin 3x =0 = RHS

3. r = 64 units, θ =

π 512

ANSWERS

π 5π 9π 13π 20. d , 0 n, d , 0 n, d , 0 n, d , 0n 8 8 8 8

6.

21. (a)

1 2

−

1 = 2

22. x = nπ + (−1)n 24. (a) 7.

π sec2 x° 180

9.

(a)

8. 2 sec2 x tan x 1 cos2 x = sin x cos x cos x 1 = ´ cos2 x sin x 1 1 ´ = cos x sin x = sec x cosec x = LHS

RHS =

(b) loge 3 =

1.

10. (2x cos 2x + sin 2x) ex sin 2x

12. −

14. −

16. 0.204 units

19.

(c) Amplitude = 1

15. −

π− 2 64

f (x) = 2 cos 3x f l(x) = − 6 sin 3x f m(x) = −18 cos 3x = − 9(2 cos 3x) = − 9f (x)

1 2

cos x − sin x 17. sin x + cos x

9 (π − 2) 9 π d − 1n = cm2 4 2 2

(a) R = 20 − 8t

(b) R = 15t2 + 4t

(c) R = 16 − 4x

(d) R = 15t − 4t + 2 (e) R = e (f) R = −15 sin 5θ 100 400 x (g) R = 2π − 3 (h) R = (i) R = 800 − 2 r r x2 − 4 (j) R = 4π r2 2.

180 cos x° + C π

cos3 x − cos x + C 3

sin 2x = sin (x + x) = sin x cos x + cos x sin x = 2 sin x cos x 1 sin 2x = sin x cos x 2

4

4 (2π − 3 3 ) 3 π o cm2 = − cm2 3 2 3

3

18.

3π π , 2n 11. (a) d , 4 n and d 4 4

1 cos3 x + C 3

13. 8 e

23.

Exercises 6.1

1 loge 3 2

(b) Maximum = 4

25.

π π , nπ ± (−1)n 4 2

Chapter 6: Applications of calculus to the physical world

sec2 x tan x

` sec x cosec x =

(b)

2−1 π units2 (b) (π + 6 − 3 3 ) units3 2 24

3

t

(a) h = 2t2 − 4t3 + C (b) A = 2x4 + x + C 4 (c) V = π r3 + C (d) d = − 7 cos t + C 3 (e) s = 4e2t − 3t + C

3. 9.

20

4. 1

5. 6e12 6. 13

8. 2e3 + 5

7. 900

dM = 1 − 4t; R = − 19 dt [i.e. melting at the rate of 19 g per minute (g min-1)] y = x3 − x2 + x + 6

10. R =

11. − 11 079.25 cm per second (cms-1) 13. 165 cm2 per second (cm2s-1)

12. 21 000 L

14. − 0.25

15. 41 cm2 per minute (cm2 min-1)

16. 31 cm3

17. 108 731 people per year 18. (a) 27 g (b) − 2.7 (i.e. decaying at a rate of 2.7 g per year) 19.

y = e 4x dy = 4e4x dx = 4y

20.

S = 2e2t + 3 dS = 2 (2e2t ) dt = 2 (2e2t + 3 − 3) = 2 (S − 3)

607

608

Maths In Focus Mathematics Extension 1 HSC Course

Exercises 6.2 1.

3

(a) 8x

35. 2.55 m per minute

(b) 54x

(c) − 6x + 3

2

2x

(d) 10e

(e) − sin x 36. (a)

2. 3.

9.

(a) 297 6 2 1 1 4

(c) − 6084

4

(b) 4e

=3 2

4.

6 17

5. −

5 9

(d) 3 ln 4 + 3

(e) − 20

5 D D tan α = 5 tan α =

D=

6. 426 7. 289 8. 44

5 tan α

dD 5 =− dα sin2 α (c) -39 (decreasing by 39 radians/hour) (d) No, the angle will reach zero after a few seconds. (b)

10. 6

3 -1

11. 8100 mm s

13. 205.84 cm2s-1

3 -1

12. 0.287 mm s

14. 159.79 cm3s-1

15. 40 units per second 16. − 34 560, i.e. decreasing at the rate of 34 560 mm3s-1

Exercises 6.3 1.

(a) 80 (d)

2.

(a) 99 061

(b) 7 hours

3.

(a) When t ` M When t ` 95 0.95 ln 0.95 0.01 So M

= 0, M = 100 = 100e− kt = 5, M = 95 = 100e− 5k = e − 5k = − 5k =k = 100e− 0.01t

17. -614, i.e. decreasing by 614 radios per week

(b) 146

(c) 92 days

18. 2411.5 cms-2 19. -11.12, i.e. decreasing by 11.12 mm3s-1 20. -2.5, i.e. decreasing by 2.5 ms-1 21. -2765, i.e. decreasing by 2765 rabbits per day 22. 1.02 cms-1

23. 2.14 houses per year 24. 0.92 mh-1

25. -0.01, i.e. decreasing at the rate of 0.01 cms-1 1 2 x sin 60° 2 3 x2 = ´ 2 2 3 x2 = 4

26. (a) A =

(b)

2 cms-1 3

27. (a) 663.5 mm2s-1

(b) 29 194.2 mm3s-1

28. 0.66 mm2s-1 29. 0.57 mm3s-1 30. − 30.48, i.e. decreasing at the rate of 30.48 cm2s-1

(b) 90.25 kg (c) 67.6 years

31. − 0.52 ms-1, i.e. moving down at the rate of 0.52 ms-1 32. 458 kmh-1

33. 2.6 ms-1

34. Rate of volume decrease is proportional to surface area. i.e.

`

dV = − kS dt = − k (4π r2) 4 V = π r3 3 dV = 4π r2 dr dr 1 = dV 4π r2 dr dr dV = ´ dt dV dt 1 = ´ − k (4π r2) 4π r2 = −k

` radius will decrease at a constant rate of k

4.

(a) 35.6 L

(b) 26.7 minutes

5.

(a) P0 = 5000 (d) 8.8 years

6.

2.3 million m2 7. (a) P = 50 000e0.069t (c) 4871 people per year (d) 2040

8.

(a) 65.61° C

9.

(a) 92 kg (b) Reducing at the rate of 5.6 kg per hour (c) 18 hours

(b) k = 0.157

(c) 12 800 units

(b) 70 599

(b) 1 hour 44 minutes

10. (a) M0 = 200; k = 0.00253 (b) 192.5 g (c) Reducing by 0.49 g per year (d) 273.8 years 11. (a) B = 15 000e0.073t 12. 11.4 years

(b) 36 008

13. (a) 19%

(c) 79.6 hours

(b) 3200 years

ANSWERS

14. (a)

P (t) = P (t0) e− kt dP (t) = − kP (t0) e− kt dt = − kP (t)

(b) 23%

5.

(c) 2% decline per year (d) 8.5 years

15. 12.6 minutes

16. 12.8 years

17. (a) 76.8 mg/dL

(b) 9 hours 18. 15.8 s

19. 8.5 years

20. (a) Q = Aekt dQ = kAekt dt = kQ (b)

dQ

(a) When `

T t 80 62 ` T When t 68 50 50 62 50 ln 62

1

`

dx = 2Ae2t dt = 2 (100 + Ae2t − 100) = 2 (x − 100)

(b) A = 0.198 (c) t = 2.76 2.

(a)

dN = 0.14Ae0.14t dt = 0.14 (45 + Ae0.14t − 45) = 0.14 (N − 45)

(b) A = 27.96 (c) N = 101.3 (d) t = 7.05 3.

dv (a) = kAekt dt = k (5000 + Aekt − 5000) = k (v − 5000) (b) A = 82 000, k = 0.0414 (c) 1 615 609.47 kL (d) 3 days, 23 h

4.

(a)

dN = kAekt dt = k (P + Aekt − P) = k (N − P)

(b) t = 6.27

50 62 =k −15 0.0143 = k T = 18 + 62e− 0.0143t

(b) After 114.5 minutes (c) As t " ∞, e− 0.0143t " 0 ` T " 18 6.

(a) 25 464 (b) After 68.8 weeks

7.

(a) − 5.2c C (b) After 8 minutes

8.

(a)

Exercises 6.4 (a)

= ln e −15k

ln

1

1.

= e −15k

= −15k ln e = −15k

= kQ

dt dt 1 So = dQ kQ 1 t =# dQ kQ 1 1 dQ = # k Q 1 = ln Q + C k kt = ln Q + C1 kt − C1 = ln Q ekt − C = Q kt e ´ e− C = Q Aekt = Q

= 18 + Ae − kt = 0, Τ = 80 = 18 + Ae0 =A = 18 + 62e − kt = 15, Τ = 68 = 18 + 62e −15k = 62e −15k

dv = − kAe− kt dt = − k (P + Ae− kt − P) = − k (v − P)

(b) A = − 500, k = 0.008 58 (c) 78.85 ms-1 (d) 500 ms-1 9.

(a) 7.2c C (b) 68 s or 1 m 8 s

10. (a) 3738 (b) After 17.1 years 11. 5.1 12. (a) 8.4 years (b) 7.5 years 13. 125.7 years 14. 16 minutes 15. 50.1% 16. (a) 2800 (b) 17.7 years

609

610

Maths In Focus Mathematics Extension 1 HSC Course

T = 28 + 172e−kt dT So = − k (172e−kt ) dt = − k (28 + 172e− kt − 28) = − k (T − 28)

(d)

17. (a)

(b) 28° C (c) 40 minutes 18. (a) 17.6% (b) 156 years (e) 19. (a) 32.5% (b) 80.8 years 20.

dN = k (N − P) dt dt 1 So = dN k (N − P) 1 t =# dN k (N − P) 1 1 dN = # k N−P 1 = ln (N − P) + C k kt = ln (N − P) + C1 kt − C1 = ln (N − P) ekt − C = N − P kt e ´ e− C = N − P Aekt = N − P P + Aekt = N

2.

(a) t2, t4, t6 (b) 0 to t1, t3, t5 (c) 0 to t1 (d) t5

3.

(a)

1

1

Exercises 6.5 1.

(a)

v

a (b)

t

t

(b)

v

a

t

(c)

v

(a) O, t2, t4, t6 (b) t1, t3, t5 (c) t5 (d) (i) At rest, accelerating to the left. (ii) Moving to the left with zero acceleration.

5.

(a)

6.

(a) At the origin, with positive velocity and positive constant acceleration (moving to the right and speeding up). (b) To the right of the origin, at rest with negative constant acceleration. (c) To the left of the origin, with negative velocity and positive acceleration (moving to the left and slowing down). (d) To the right of the origin, with negative velocity and acceleration (moving to the left and speeding up). (e) To the left of the origin, at rest with positive acceleration.

t

a

t

4.

t

π 3π 5 π , , ,… 4 4 4

(b) 0,

3π π , π, ,… 2 2

ANSWERS

Exercises 6.6 1.

2.

(a) 18 cms (b) 12 cms (d) After 5 s -1

-2

(c) The particle is on the RHS of the origin, travelling to the right and accelerating.

(c) When t = 0, x = 0; after 3 s 9.

(a) − 8 ms− 1 (b) a = 4; constant acceleration of 4 ms-2 (c) 13 m (d) after 2 s (e) − 5 m (f)

(a) v = 5 − 10t

10. v =

(b) − 95 ms− 1

(c) a = − 10 = g

− 102 17 ,a= (3t + 1) 2 (3t + 1)3

1 1 cms− 2 cms− 1 (c) − 6 36 (d) The particle is moving to the right but decelerating (e) (e3 − 1) s

11. (a) At the origin

12. (a) 3 ms-1

(b)

(b) When t = 0 s, 1 s, 3 s

(c) 10 ms-2

13. (a)

3.

(a) 4 m (b) 40 ms-1

(c) 39 m (d) 84 m

(e) (b) ox = 6 cos 2t, px = − 12 sin 2t (c) − 6 3 cms-2 (d) px = − 12 sin 2t = − 4 (3 sin 2t) = − 4x 14. (a) 7 m (d)

4.

(a) 2 cm (b) After 1 s (c) -4 cm (e) -7 cms-1

5.

(a) 2 ms-1 (d)

(b) 4e2 ms− 2

(b) 16 m

(c) After 7 s

(d) 6 cm

(c) a = 4e2t = 2 (2e2t ) = 2v

(e) 10 m 15. (a) 18.75 m

6.

(a) v = − 2 sin 2t

(b) a = − 4 cos 2t

3π π (d) 0, , π , ,...s 2 2

(e) ±1 cm

(c) 1 cm

π 3π 5 π (f) , , ...s 4 4 4

(g) a = − 4 cos 2t = − 4x 7.

(a) v = 3t2 + 12t − 2; a = 6t + 12 (d) 42 ms-1

8.

(a) x = 20 (4t − 3)4 , x = 320 (4t − 3)3 (b) x = 1, cm, ox = 20 cms− 1, px = 320 cms− 2 •

(b) 266 m (c) 133 ms-1

(b) -15 ms-1

(c) 5 s

16. (a) At the origin: x = 0 2t3 − 3t2 + 42t = 0 t (2t2 − 3t + 42) = 0 t = 0, 2t2 − 3t + 42 = 0 Since t = 0, the particle is initially at the origin. 2t2 − 3t + 42 = 0 b2 − 4ac = (−3) 2 − 4 (2) (42) = −327 0 Also 2t > 0 and 3t + 9 > 0 when t > 0 So 2t < 3t + 9 2t 0 displacement is always positive v = 5x Since x > 0, 5x > 0 ` v >0 So velocity is always positive.

`

11. (a) 12.

3s

Exercises 6.10 1.

(a) x = 2 cos t

(b) 2 s

ln 21 m 3

Exercises 6.9 1.

v=

2.

v = − 2x4 − 4x2 + 2x − 68

3.

8 ms-1 4.

(b) v = − 2 sin t

2x2 + 10x + 9

(a) 2.8 ms-1

(b) For v = 0, i.e.

πx 12 +4=0 sin π 2 πx 12 = −4 sin π 2 π x 4π = sin 2 12 = − 1.05

(c) a = − 2 cos t

This has no solution, so v ≠ 0 (particle is never at rest). v 2 = n 2 (a 2 − x 2 )

5.

7.9 cms-1

8.

0.37 m

9.

(a) v = − 2x2 + 4x + 16 (b) 3 2 ms− 1 (c) Between -2 m and 4 m

10. v =

6.

1.55 cms-1

7.

(x2 − 3) 5 + 79 5

11. (a) 4 kms-1

2.

(a) x = 5 sin t

(b) At rest, v = 0 400 i.e. =0 x 400 = 0 This is impossible. ` the rocket never comes to rest 12. v =

k (6400 − x) 3200x (b) v = 5 cos t

613

614

Maths In Focus Mathematics Extension 1 HSC Course

(c) a = −5 sin t

3.

(a) x = 4 cos 2t

(b) t = 0,

(c) x = cos 2t

5.

(a) x = 2 cos 3t ox = − 6 sin 3t px = − 18 cos 3t = − 9 (2 cos 3t) = − 9x (b) ± 2 (c) v = 0 (d) v = ± 6; a = 0

6.

(a) x = 7 cos 5t ox = − 35 sin 5t px = − 175 cos 5t = − 25 (7 cos 5t) = − 25x

3π π , π, , 2π , . . . ; x = ± 4 2 2

(b) t = 0,

(c) v = − 8 sin 2t 7.

(d) v = 0 (e) a = − 16 cos 2t

(c) Period =

(a) x = 3 sin 4t ox = 12 cos 4t px = − 48 sin 4t = −16 (3 sin 4t) = −16x (b) t =

8.

π 2π 3 π , , , . . . ; x = ±7 5 5 5

π 3π 5 π , , ,... 8 8 8

(c) x = ±3; px = ±48

(a) px = − 36x (b) 12 ms-1 π π π (c) t = 0, , , , . . . ; ox = ±12 ms− 1 6 3 2 (d) v = ± 144 − 36x2

9.

(a) x = 2 cos b t + ox = − 2 sin b t +

π l; 4

π π l ; px = − 2 cos b t + l 4 4

= − x, ` SHM π 5π 9 π (b) t = , , ,... 4 4 4 (f) a = 0 4.

(a) x = cos 2t ox = − 2 sin 2t px = − 4 cos 2t = − 4x (b) amplitude = 1, period = π

(c) 2π

(d) x = ±2

10. (a) x = 5 cos 3t + 2 sin 3t ox = − 15 sin 3t + 6 cos 3t px = − 45 cos 3t − 18 sin 3t = − 9 (5 cos 3t + 2 sin 3t) = − 9x (b) 16.2 ms-1 11. (a) x = 4 cos (3t + π ) ox = − 12 sin (3t + π ) px = −36 cos (3t + π ) = − 9 [4 cos (3t + π )] = − 9x (b) 2 3 cm

(c) Amplitude = 4, period =

2π 3

2π 5

ANSWERS

(c) 4 ms-1 (d) px = 2 − x

12. (a) 0 m, 4 m (b) 2 m 13. (a) ±1.5 m 14. (a) v =

5 ms-1

(b)

± 81 − x2 3

(b) x = ±3 5 cm

15. v = − 6.5 cms–1; a = − 5.5 cm− 2 16. Period

24. (a) v2 = 4x − x2 − 3 x2 3 1 2 − v = 2x − 2 2 2 d 1 2 c v m=2−x dx 2 px = 2 − x So = −x + 2 = − (x − 2)

4 3 2π , amplitude ; 3 3

This is in the form px = − n2 (x − x0) so SHM. (b) Centre x = 2, endpoints x = 1, x = 3 (c) a = 1, n = 1

4 3 π cos c 3t − m or 6 3 4 3 π x= sin c 3t + m 3 3 x=

17. (a) Between x = −1 and x = 9 (b) Yes—centre of motion is x = 4 Let X = x − 4: d2 x = 8 − 2x dt2 = − 2 (x − 4) = − 2X 18. (a) a = −1600x

(n =

2)

(b) Period =

π 20

(c) 30 cms-1

19. (a) px = − an2 sin nt − bn2 cos nt = − n2 (a sin nt + b cos nt) = − n2 x (b) Amplitude 20. (a) − (b)

a2 + b2 ; period =

This is in the form px = − n2 (x − x0) so SHM.

Exercises 6.11 2π n

(c) n a2 + b2

1.

5 3π cms− 1 6

5π 2 cms− 2 9

(b) Amplitude 45 = 3 5

5 , period

23. (a) Centre x = 3, endpoints x = 0, x = 6 (b) x = 3 − 3 cos 3t ox = − 3 (− 3 sin 3t) = 9 sin 3t px = 9 (3 cos 3t) = 27 cos 3t = − 9 (− 3 cos 3t) = − 9 (3 − 3 cos 3t − 3) = − 9 (x − 3)

(ii) y = − 5t2 +

15 2 t 2

3 2 s 2

(a) 6 3 s

4.

(a) 2.59 m

6.

(a) 4 minutes

7.

(a) px = 0, ox = u cos a, x = ut cos a py = − g, oy = − gt + u sin a, gt2 y=− + ut sin a 2

(b) 540 m

3. (a) 2.3 s

(b) 7.3 m

5.

(b) 5.8 m

2.8 s

(b) 102 km

(b) 15 m 5x2 (1 + tan2 θ ) + x tan θ 256

8.

y=−

9.

x = vt cos β , y = −

2π 6

15 2 t 2

2.

2π 3

22. (a) Equilibrium x = 1, endpoints x = 0, x = 2 (b) Period

(a) (i) x = (b)

21. (a) x = 2 sin 3t − cos 3t ox = 2 (3 cos 3t) − (− 3 sin 3t) = 6 cos 3t + 3 sin 3t px = 6 (− 3 sin 3t) + 3 (3 cos 3t) = −18 sin 3t + 9 cos 3t = − 9 (2 sin 3t − cos 3t) = − 9x

(c)

25. x = cos2 2t = (cos 2t) 2 ox = 2 (cos 2t)1 (− 2 sin 2t) = − 4 sin 2t cos 2t = − 2 (2 sin 2t cos 2t) = − 2 sin 4t px = − 2 (4 cos 4t) = − 8 cos 4t = − 8 (cos2 2t − sin2 2t) = − 8 (cos2 2t − 5 1 − cos2 2t ? ) = − 8 (cos2 2t − 1 + cos2 2t) = − 8 (2 cos2 2t − 1) = − 8 (2x − 1)

10. 20° 34l

gt2 2

+ vt sin β + h

11. 1.8 ms-1

12. 28.86 ms-1

13. 2° 45l, 87° 15l 14. 9.3 ms-1 15. (a) 6 s

(b) 91.7 m

16. 63° 6l or 50° 52l 17. 2 m

18. 8 m

19. 0.12 m

20. (a) 8 ms-1

21. (a) 4.8 s

(b) 100.9 m

(b) 53° 8l (c) 3.2 m

615

616

Maths In Focus Mathematics Extension 1 HSC Course

22. (a) Second stone is first by 1.46 s (b) 1st stone: x = 10t

(b) Particle is 28 m to the left of the origin, travelling at 15 ms− 1 to the right, with 18 ms− 2 acceleration (to the right), so the particle is speeding up.

= 10 (2 3 ) = 20 3 m

11. (a) t1, t3, t5

(b) t2, t4

(c) t3 and after t5

(d) (i)

2nd stone: x = 10 3 t = 10 (10 3 (2)) = 20 3 m So both land at the same place. 23. 63° 26l 39° 27l

(ii) 24. 2.7 s 25. (a) 29.5 m (b) x = 11.5, y = 8.2 or (11.5, 8.2); No they will not collide—they reach this point at different times.

Test yourself 6 1.

-2 ms-1

2.

(a) 0 m, 0 ms-1, 8 ms-2

3.

(a)

(b) 0, 0.8 s

(c) 0.38 m

T = 25 + Ae− kt dT = − kAe− kt dt = − k (Ae− kt + 25 − 25) = − k (T − 25)

(b) A = 295, k = 0.042 (c) 108.4c (d) 97 min or 1 h 37 min 4.

-76 m, -66 ms-2

6.

(a) 6 cms− 1 (b) 145 855.5 cms-2 (c) x = 2e3t ox = 6e3t px = 18e3t = 9 (2e3t ) = 9x

7.

1m

8.

x = 2 sin 3t ox = 6 cos 3t px = −18 sin 3t = − 9 (2 sin 3t) = − 9x

9.

(a) 2, 6 s (b) (i) 16 cm (ii) 15 cms− 1 (iii) −18 cms− 2 (c) Particle is 16 cm to the right of the origin, travelling at 16 cms− 1 to the right. Acceleration is −18 cms− 2 (to the left), so the particle is slowing down.

10. (a) (i) 18 ms− 2 (ii) 15 ms− 1 (iii) -28 m

5.

39.6 years

12. (a) 48.2% 13. (a)

(b) 1052.6 years

π 5π 13π 17π , , , ,...s 6 6 6 6

(c) −

1

2 14. (a) 15 m

(b)

π 5π 7π 11π , , , ,...s 3 3 3 3

ms− 2 (b) 20 m

(c) 4 s

15. (a) 16 941 (b) 1168 birds/year (c) 18.3 years 16. 0.0193 mms-1 17. (a) (i)

(ii)

(b) t1, t3 2π (b) 7 ms− 1 7 π 5π 13π 17π , , , , . . . seconds (c) 21 21 21 21

18. (a) Period

19. 80° 58l, 16° 38l 20.

55 033 m

ANSWERS

Challenge exercise 6 1.

2.

3.

4.

8 2 (4 − cos 3t) 2 cos 3t + = 3 3 3 8 (b) px = − 9 c x − m 3 2π 2 (c) Yes. Amplitude = , period = 3 3 (a) x = −

7.

(a) 36 208

8.

88° 35l, 10° 52l

9.

(a) 7.85 mm2 per hour

10. (a) x = cos 4t

`

px = 0, ox = V , x = 2

,

2

gt2 Vt py = − g, oy = − gt + V , y = − + 2 2 2 Range: when y = 0, − t f−

gt2 2 gt 2

+ +

Vt 2 V 2

`

When t =

2V g 2

,x =

V

=

V2 g

2

´

V g 2

t = 0 or 2V t= g 2

14. (a) 16.1 ms-1 (b) If v = 0, 3x2 + 2x + 225 = 0 ∆ = 22 − 4 ´ 3 ´ 225 0, range: all real y 2 (d) f − 1 (x) = ; domain: all real x ≠ 0, range: all real y ≠ 0 x 1 (e) f − 1: y = − 1; domain: all real x ≠ 0, range: all real x y ≠ −1 x (a) y = ; domain: x ≥ 0, range: y ≥ 0 2 (b) y =

x − 2 ; domain: x ≥ 2, range: y ≥ 0

(c) y =

x + 3; domain: x ≥ 0, range: y ≥ 3

(d) y =

x + 1 + 1; domain: x ≥ −1, range: y ≥ 1

(e) f − 1 [f (x)] = f − 1 (3x + 1) (3x + 1) − 1 = 3 3x = 3 =x x−1 n f [f − 1 (x)] = f d 3 x−1 n+ 1 = 3d 3 =x

(e) y = 6 x ; domain: x ≥ 0, range: y ≥ 0 (f) y = − 1 − x ; domain: x ≤ 1, range: y ≤ 0 (g) y = 4 x + 1 ; domain: x ≥ −1, range: y ≥ 0 1 ; domain: x > 0, range: y < 0 (h) y = − x 3.

4.

(a) x ≥ − 3 (b) y = x + 9 − 3; domain: x ≥ − 9, range: y ≥ − 3 (c) x ≤ − 3 (d) y = − x + 9 − 3; domain: x ≥ − 9, range: y ≤ − 3 (a) y = −

3 x

(d) y = − 5.

(a) (i) y = (b) (i) y =

(b) y = −

x

(e) y = − 4

x+1 3 2 x

(c) f 1 (x) = − 4 x + 2

x + 1 + 1 (ii) y = − x + 1 + 1 4

x+2

(ii) y = − 4 x + 2

(a) f − 1 [f (x)] = f − 1 (x + 7) = (x + 7) − 7 =x f [f − 1 (x)] = f (x + 7) = (x − 7) + 7 =x (b) f − 1 [f (x)] = f − 1 (3x) 3x = 3 =x x f [f − 1 (x)] = f d n 3 x = 3d n 3 =x

25. y = ± x + 16 − 4

1.

4

Exercises 7.5

f − 1 (x) = x2 − 2

24. y = ± 6 x + 3

26. y = ± 4 − x + 2

y=

6.

12. y =

14. f − 1 (x) =

ln (x − 1)

y =3 x−5

9.

ln x 17. y = 2

16. y = ln x 20. f − 1 (x) =

3.

9 11. y = 2 x

10. y = x + 7 5

x+3 2

y = 2x − 1

8.

3

13. y =

(c) (i) y =

2.

(a) Domain: all real x ≠ 1, range: all real y ≠ 0 2 (b) f − 1 (x) = + 1 x (c) Domain: all real x ≠ 0, range: all real y ≠ 1

3.

(a)

ANSWERS

(b) y = ex (c) y = ln x — domain: x > 0, range: all real y; y = ex — domain: all real x, range: y > 0 4.

(ii) x = y

(a) (i) 5x4

(iii) (c) (i)

1 x2

dx = − x2 dy

1 x+1

(iii)

(e) (i)

5.

´

dx

(iv)

dy

´

dy

´

dx

1 3

(iv) 3

(iv) 15 d (c) (i) x = (iv) −

(b)

4

x 5 n 3

(v)

(v)

dy dx

´

1 x −5 d n 15 3 4

´

2 −3 x

(iii) −

(iii)

(d) 1

3 3 (x + 7) 2

2

(v)

(ii) y = dy dx

´

ln x − 1 5

(iii)

dx 1 ´ 5x = 1 = dy 5x

π π π (b) 0 (c) 0 (d) (e) − 4 6 2 π π π π (f) − (g) (h) (i) 4 6 2 2 5π π π π (k) (l) (m) − (j) − 4 6 6 3 (a)

2 x2

dx x2 2 =− 2 ´− =1 2 dy x

Exercises 7.6 1.

4

dx x 5 1 x −5 d n ´ 15 d n = 1 = 3 dy 15 3

dx 1 ´ 3 3 (x + 7) 2 = 1 = dy 3 3 (x + 7) 2

(e) (i) x = e5y + 1 (iv) 5x

´

(c) 4

(iii)

(ii) y = 3 x + 7

(iv) 3 3 (x + 7) dx

dx

dx 1 = ´3 =1 dy 3

1

x 5 n 3

(ii) y =

(d) (i) x = y3 − 7

dy

dy

x−1 3 ´

dx

(ii) y = d

2 y+3

x2 2

dy

(v)

2

1 5x (e)

(e)

3 (i) 2

(a)

dx 1 = ´2 x− 3 =1 dy 2 x − 3 (ii) y =

3 (g) 0 (h) − 3

1

5.

dx = − e− x (− ex) = 1 dy

(ii) x = y2 + 3

(d)

(a) 0.67 (b) − 0.14 (d) 0.97 (e) − 0.90

dx 1 (x + 1) = 1 = dy x + 1

dx

π 2

4.

dx =2 x−3 dy

(b) (i) x = 3y5

(v)

dy

(iv)

dx 1 = − − x = − ex e dy

(a) (i) x = 3y + 1 (iii)

dx 1 = − 2 (− x2) = 1 dy x

´

dx

(c)

(a) 0.41 (b) 1.04 (c) 0.97 (d) − 0.64 (e) −1.31

(ii) x = − ln y

2 x−3

(iv)

dy

(iv)

dx =x+1 dy

1

(iii)

1 y

(b) − 1

3.

(ii) x = ey − 1

(d) (i) − e − x (iii)

dx 1 ´ = 5x4 ´ x− 4 = 1 (iv) 5 dx dy

(ii) x =

(a) 0 (f)

1 5

dy

dx 1 − 4 (iii) = x dy 5 (b) (i) −

2.

(c) 1.64

1 2 (j) − 1

621

622

Maths In Focus Mathematics Extension 1 HSC Course

(f)

(l)

(g)

(m)

(h)

(n)

(i)

6.

(a) − 1 ≤ x ≤ 1 (b) − 1 ≤ x ≤ 1 (c) −

7.

(a) π (b) 0 (c) 0 π π π (d) (e) (f) 2 2 2

8.

(a) (d)

(j) 9.

4 5

(b) 3 58

3 5

(c)

5 12

(e)

π 4

(f) −

(a) 2 sin θ cos θ

(b)

π 4

24 25

10. (a) odd (b) odd (c) even (d) neither (e) odd (f) odd

(k)

π π ≤x≤ 4 4

π 11. (a) tan− 1 (−1) = − 4 = − (tan− 1 1) π (b) sin− 1 (−1) = − 2 = − (sin− 1 1) (c) tan− 1 (− 3) Z − 1.249 = − (tan− 1 3)

(g) odd

ANSWERS

(d) LHS = cos− 1 c − =

2π 3

RHS = π − cos− 1

1 2

π 3

=π− =

18. (a)

1 m 2

π 3 3 + cos− 1 = θ + − θ 7 7 2 π = 2

sin− 1

2π 3

(e) sin− 1 e −

1

o=−

2

π 4

(b) sin− 1 d −

= − e sin− 1

1 2

o

= − sin− 1 d

13. (a) θ = π n − (−1)n

π 2

π 14. (a) θ = 2π n ± 3

5π 7π (b) θ = ± ,± 3 3

π 3

(b) θ = −

(b) θ =

2 2 So cos− 1 d − n = π − cos− 1 d n 5 5

π 3π , 2 2

(d) tan− 1 d −

16. (a) x = 2π n ± 1.53 (b) x = π n + (−1)n 0.39 (c) x = π n + 0.92 (d) x = π n − (−1)n 1.04 (e) x = π n − 0.75 (f) x = 2π n ± 1.80

1

(a) −

1−x 3

(d) − (g)

17. (a) and (b) y

40

=

1 − (3x + 2) −x

+ cos x −1

1 − x2

(b) 1

ex 1 + e 2x

(e)

1 9 − x2

2 x (1 − x)

2

1

y = sin-1 x π 2

2

−3

(l)

15

(a) − 1 (d)

1 − 4x

(i) −

1 − 64x2 3

(o) 2.

(e)

2

36 − x2

(n)

x

1−x 8

(c)

2

1 1 + x2

(f)

2x 1 − x4

2 1 = 1 + (2x − 1)2 2x2 − 2x + 1

(k)

π 2

2

(b)

2

1 − 9x

(h) −

π

-1

7 n 10

Exercises 7.7 1.

y = sin-1 x + cos-1 x

7 n = − 0.61 10 = − tan− 1 d

2π 5π 8π , , 3 3 3

y = cos-1 x

5 n 9

2 (c) LHS = cos− 1 d − n 5 = 1.9823 2 RHS = π − cos− 1 d n 5 = 1.9823

π π (b) θ = 2πn (c) θ = π n + 3 3 π π π (d) θ = π n + (−1)n (e) θ = π n − (f) θ = 2π n ± 4 4 6 π π n (g) θ = 2π n ± (h) θ = π n + (−1) 6 2 π π (i) θ = π n − (−1)n (j) θ = π n − 4 6

12. (a) θ = π n + (−1)n

15 (a) θ = π n −

5 n = − 0.589 9

(c)

(j)

2 4 + x2

(m) −

1 49 − x2

15 − 9x − 12x − 3 2

(p)

5 (tan− 1 x + 1)4 1 + x2 1

x 1 − (loge x)2 1

1 − x2 sin− 1 x

−1 1 (g) (1 + x2) (tan− 1 x)2 1 − x2 6 1 + (cos− 1 x + 1)2 @ 1 1 1 = (h) − (i) 2 1 + x2 − x 2 − 4x x 2 1 − d + 1n 2

(f) −

(j)

− ecos

−1

x

1 − x2

623

624

Maths In Focus Mathematics Extension 1 HSC Course

3.

(a) (i) − 1 (b) (i) 1

(ii) 1

3 5

(ii) −

π2

(c) (i)

11. (a) − 5 8

(ii) −

6 3 1 (d) (i) − (ii) 3 3 1 (e) (i) (ii) − 5 5

4. 6.

7.

(c) 6 3 π2

1 − x2

1−x

2

1.

+ sin− 1 x = 0 sin− 1 x =

2. −x

9.

−x (9 − x2)3

− 1 for −

(b)

− 1 − 2x tan− 1 x

6 (1 + x2) (tan− 1 x) @

h 6 h n 6

(a) sin− 1 x + C

(b) 2 cos− 1 x + C or − 2 sin− 1 x + C x 1 (c) tan− 1 x + C (d) tan− 1 + C 3 3 x x 5 (e) sin− 1 + C (f) tan− 1 + C 2 2 2

(a)

(i)

(b)

1 sin− 1 2x + C 2

3 3x o+C tan− 1 e 6 2

1.1 units2

5.

(a)

6.

(a)

7.

0.1

9.

3x 2 n +C cos− 1 d 5 15

π π π (b) (c) 4 2 3 π π π (d) (e) (f) 6 12 3 π 3π (g) π (h) (i) 9 28 π (j) 3 5

4.

8.

(j)

(a)

2

π π π π < x < , 1 for − π ≤ x < − , < x ≤ π 2 2 2 2

x 1 tan− 1 + C 6 6

t 1 + C (d) tan− 1 3x + C 3 3 5 4x x 1 1 n + C (f) tan− 1 d n +C (e) sin− 1 d 5 4 2 3 5 2t (g) sin− 1 d n + C (h) tan− 1 ( 5 x) + C 5 5

10. (a) 0 (b)

etan x 1 + x2

(c) sin− 1

3.

(a)

1 − x2

(e)

3 x 1 sin− 1 x + C (h) sin− 1 + C 5 4 2 x x 1 (i) tan− 1 + C (j) sin− 1 +C 3 3 5

` (0, 0) is a stationary point π (b) Domain: − 1 ≤ x ≤ 1, range: 0 ≤ y ≤ 2 (c)

8.

x (1 + 5 ln x ? 2)

−1

1

(g)

1 − x2 x = 0 satisfies this equation x = 0, y = 0 sin− 1 0 =0

When

(d) −

Exercises 7.8

d x (x sin− 1 x) = + sin− 1 x dx 1 − x2 dy = 0, dx x

1

1 (1 + x2) tan− 1 x

(b) 0° 0l 33m per second −1

d 1 (sin− 1 x + cos− 1 x) = + dx 1 − x2 =0 π −1 −1 (b) sin x + cos x = 2 d π d n=0 ` dx 2

For

(b)

` θ = sin− 1 d

(a)

(a)

1 − e 4x

12. (a) sin θ =

5. 40x + 100y − 25π − 8 = 0

y = 2x

2e2x

1 2

π units2 6

(b)

π (7 3 − 12) units3 6

1 1 − x2 sin− 1 x

units2

1 sin− 1 x3 + C 3

10. y = sin− 1 d

5π x n− 3 14

(b) 0.3

ANSWERS

1 1 + 1 + x2 4 + x2 4 + x2 + 1 + x2 = (1 + x2) (4 + x2) 2x2 + 5 = (1 + x2) (4 + x2) = LHS

11. (a) RHS =

12. LHS = cos− 1 e −

(b) 0.48

=π−

14. d

3 π o units2 (b) e + 1 − 6 2

2 tan− 1 x

(a)

π 4

(d) 0

13. (a) Domain: all real x ≠ − 2; range: all real y ≠ 0 1 (b) f −1 (x) = − 2 x (c) Domain: all real x ≠ 0; range: all real y ≠ − 2 14. (a)

π units2 6

15. (a)

π 2

(b)

3π π (c) 4 3 π (e) − 6

(a) 3 tan− 1 x + C

3.

(a)

4.

5π units2 12

1

(b)

1 − x2

(b) sin− 1 3 1 + 9x 2

x +C 4 (c) −

10 1 − 25x2 17. 4 3 x − 18y − 6 3 + 3π = 0 18. (a) x > 2

3− x 2

5.

f −1 (x) =

6.

π θ = 2πn ± 4

7.

f −1 (x) =

8.

(a) 0

19. (a) 2 20. (a)

ln (x − 1) 3

x + tan− 1 x 1 + x2

10. (a) Domain: all real x ≥ 1; range: all real y ≥ 0 (b) f − 1 (x) = x2 + 1 Domain: all real x ≥ 0

(b) f −1 (x) =

x+4 +2

(b) 10.5

π 3

(b)

7π 14

Challenge exercise 7 1.

(a) y = − 1 − x2 , domain: 0 ≤ x ≤ 1, range: −1 ≤ y ≤ 0

d π π d n=0 (b) sin− 1 x + cos− 1 x = and 2 dx 2

11.

4 3

(b)

2.

9.

(b)

(b) 1.73 units3

16. (a) Domain: −1 ≤ x ≤ 1; range: − π ≤ y ≤ π

Test yourself 7 1.

3 o 2

= RHS

π − 1 n units2 2

15. v =

π 6

= π − cos− 1 e

π2 12. units3 12 13. (a) cos− 1 x

3 o 2

1 + 1 ; domain: x > 0, x ≤ −1, x range: y ≤ 0, y ≠ −1 (b) y = −

2.

3.

x−

1 − x2 sin− 1 x x

2

1 − x2

1 tan− 1 (x2) + C 2

; no: x ≠ 0

625

626

Maths In Focus Mathematics Extension 1 HSC Course

4.

(a)

9.

LHS = tan− 1 =θ+

5 4 + tan− 1 5 4

π −θ 2

π 2 = RHS =

10. (a) 0.9 units3 (b) π units3

(b) 0 (c)

Chapter 8: Series

Exercises 8.1

1 x x 1 + tan− 1 x 1

LHS = tan− 1 x + tan− 1 = tan− 1

π =θ+ −θ 2 π = 2 = RHS 5.

6. 7.

14, 17, 20

2.

23, 28, 33

4.

85, 80, 75

5.

1, −1, − 3

7.

1 2, 2 , 3 2

8.

1 1 1 , , 16 32 64

3. 6.

3.1, 3.7, 4.3

44, 55, 66 87, 83, 79

9.

16, 32, 64

11. 16, − 32, 64

10. 108, 324, 972 13.

12. 48, − 96, 192

16 32 64 , , 135 405 1215

14.

15. 36, 49, 64

16. 125, 216, 343

17.

35, 48, 63

18. 38, 51, 66

19. 126, 217, 344

20. 21, 34, 55

1.4 ms− 1 3π 1 − = 6 2

3π − 3 units2 6

Exercises 8.2 1.

(a) T1 = 3, T2 = 11, T3 = 19 (c) u1 = 5, u2 = 11, u3 = 17 (e) t1 = 19, t2 = 18, T3 = 17 (g) Q1 = 9, Q2 = 11, Q3 = 15 (i) T1 = 8, T2 = 31, T3 = 70

2.

(a) T1 = 1, T2 = 4, T3 = 7 (b) t1 = 4, t2 = 16, t3 = 64 (c) T1 = 2, T2 = 6, T3 = 12

3.

(a) 349 (b) 105 (c) 248 (d) -110

Let sin− 1 x = θ x Then sin− 1 = θ 1

By Pythagoras’ theorem, BC =

`

1− x 1 − x2 cos θ = 1 = 1 − x2 θ = cos− 1 1 − x2

` sin− 1 x = cos− 1

8.

1.

1 sin− 1 (3x2) + C 6

1 − x2

(e) -342

4.

(a) 1029 (b) -59 039 (d) 53 (e) 1002

5.

(a), (b), (d)

9.

7th term

2

12. (a) -572 13. n = 33

6.

(b) T1 = 5, T2 = 7, T3 = 9 (d) T1 = 3, T2 = − 2, T3 = −7 (f) u1 = 3, u2 = 9, u3 = 27 (h) t1 = 2, t2 = 12, t3 = 58 (j) T1 = 2, T2 = 10, T3 = 30

(c) 1014

(b), (d)

7.

10. 23rd term

16th term

11. (a) 1728

(b) 17th term 14. n = 9

15. n = 88, 89, 90, . . .

16. n = 41, 42, 43 17. n = 501 19. (a) n = 14

(b) -4

20. -6

18. n = 151

8.

Yes

(b) 25th term

ANSWERS

27. (a) T2 − T1 =

Exercises 8.3 1.

(a) 128 (b) 54 (c) 70 (d) 175 (e) 220 47 (g) 40 (h) 21 (i) 126 (j) 1024 (f) 60

2.

(a) 65 (b) 99 (c) 76 (d) 200 (e) 11 (f) 39 (g) 97 (h) 66 (i) 75 (j) 45

3.

(a)

6

10

/ 2n − 1

(b)

n =1 n

(f)

k =3

n =1

/ (−n)

(g)

n =1

/ a + (k − 1) d

/ 3.2k

(h)

k =0 n

(j)

k =1

/ ar

/ 6k − 4

(a) y = 13 (f ) x = 42

k =1

Since T2 − T1 = T3 − T2 it is an arithmetic series with d =

/ n =0

1 2n

28. 26

k =1

(b) x = − 4 1 2

(g) k = 4

(c) x = 72 1 2

(d) b = 11

(h) x = 1

(e) x = 7

(i) t = − 2

(j) t = 3

3.

(a) 590

4.

(a) -110

5.

Tn = 2n + 1

6.

(a) Tn = 8n + 1 (b) Tn = 2n + 98 (c) Tn = 3n + 3 (d) Tn = 6n + 74 (e) Tn = 4n − 25 (f) Tn = 20 − 5n n+6 (g) Tn = (h) Tn = − 2n − 28 (i) Tn = 1.2n + 2 8 3n − 1 (j) Tn = 4

(b) 12.4 (c) -8.3

8.

10. 15th term

17. 103

20. d = 9

(d) 37 (e) 15

54th term 9.

11. Yes

14. n = 13

19. (a) d = 8

(e) 67

(b) -850 (c) 414 (d) 1610 (e) -397

28th term

16. -2

29. 122b

30. 38th term

Exercises 8.5

(a) 46 (b) 78 (c) 94 (d) -6

13. Yes

3.

(b) 50 3

k−1

2.

7.

3

9

Exercises 8.4 1.

3

= 9 ´ 3 − 4 ´ =3 3 −2 3 = 3

n

(d)

n

n

(i)

5

/ n3

(c)

n =1 50

/ k2

(e)

/ 7n

12 −

= 4 ´ 3 − 3 =2 3 − 3 = 3 T3 − T2 = 27 − 12

4 5

30th term

12. No

15. n = 30, 31, 32 . . .

1.

(a) 375

2.

(a) 2640

(b) 4365

3.

(a) 2050

(b) − 2575

4.

(a) − 4850

5.

(a) 28 875 (b) 3276 (c) −1419 (e) 6604 (f) 598 (g) − 2700 (h) 11 704 (i) − 290 (j) 1284

21. a = 12, d = 7

22. 173

23. a = 5

24. 280

25. 1133

26. (a) T2 − T1 = log5 x2 − log5 x = 2 log5 x − log5 x = log5 x T3 − T2 = log5 x3 − log5 x2 = 3 log5 x − 2 log5 x = log5 x Since T2 − T1 = T3 − T2 it is an arithmetic series with d = log5 x. (b) 80 log5 x or log5 x80 (c) 8.6

(c) 480 (c) 240

(b) 4225 (d) 6426

6.

(a) 700 (b) − 285 (c) −1170 ] 18 terms g (d) 6525 (e) − 2286

7.

21

8.

8

9. 11

11. a = − 3, d = 5

10. a = 14, d = 4

12. 2025

14. 8 and 13 terms

13. 3420

15. 1010

16. (a) (2x + 4) − (x + 1) = (3x + 7) − (2x + 4) =x+3 (b) 25 (51x + 149)

18. 785

(b) 87

(b) 555

17. 1290

18. 16

19. Sn = Sn − 1 + Tn

` Sn − Sn − 1 = Tn

20. 4234

Exercises 8.6 1.

(a) No (d) No

3 2 (c) Yes, r = 7 4 (f ) No (g) Yes, r = 0.3

(b) Yes, r = −

(e) No 3 (h) Yes, r = − (i) No 5

2.

(j) Yes, r = − 8

(a) x = 196 (b) y = − 48 (c) a = ±12 2 (d) y = (e) x = 2 (f ) p = ±10 3 (g) y = ±21 (h) m = ±6 (i) x = 4 ! 3 5

627

628

Maths In Focus Mathematics Extension 1 HSC Course

(j) k = 1 ± 3 7 3.

(a) Tn = 5n − 1

(k) t = ±

1 6

(l) t = ±

(b) Tn = 1.02n − 1 (e) Tn = 6 . 3n − 1

(d) T = 2 . 5n − 1 n

(f) Tn = 8 . 2n − 1

(a) 1944 (b) 9216 (c) -8192 64 (d) 3125 (e) 729

n +2

14. (a)

1. 2.

(a) 3 ´ 219 (b) 719 (c) 1.0420 3 20 1 1 19 1 (d) d n = 21 (e) d n 4 2 4 2

3. 4.

6th term

10. 5th term

11. No

5.

12. 7th term

13. 11th term

10.

14. 9th term

15. n = 5

12.

16. r = 3

17. (a) r = − 6 (b) −18

13.

20. 208

2 7

14. 15.

Exercises 8.7

17.

1.

(a) 2 097 150 (b) 7 324 218

19.

2.

(a) 720 600 (b) 26 240

3.

(a) 131 068

4.

(a) 7812 (b) 35 (c) 8403

5.

127 128

6.

(a) 1792

7.

148.58

9.

n=9

32 769 65 536

20. LS − Sn = =

1 2

(b) No

(c) Yes LS = 12

a ^ 1 − r nh a − 1−r 1−r a − a ^ 1 − r nh

1−r a − a + ar n = 1−r ar n = 1−r

55 64

364 729

(c) 97 656.2

(e) 87 376

Exercises 8.9 (b) 3577

1.

(a) 210

133.33

2.

(a) 39

10. 10 terms

3.

(a) 3n + 3

8.

3

4 5

(d) No

25 (e) Yes LS = 3 (f) Yes LS = (g) No 32 3 5 (h) Yes LS = − 1 (i) No (j) Yes LS = 1 7 22 2 2 7 (a) 80 (b) 426 (c) 66 (d) 12 (e) (f) 54 3 3 10 9 16 2 (g) − 10 (h) (i) 48 (j) − 7 20 39 3645 7 4 1 1 (a) (b) (c) (d) (e) 64 12 27 12500 4096 1 1 1 2 2 (b) (c) (d) 2 (e) 3 (f) 5 (g) (a) 1 4 5 5 2 48 5 4 1 (h) − 5 (i) 1 (j) 5 6 3 3 2 1 7 a = 4 6. r = 7. a = 5 8. r = 9. r = − 5 4 5 8 2 2 1 r=− 11. a = 3, r = and a = 6, r = 3 3 3 3 1 a = 192, r = − , LS = 153 4 5 3 2 2 a = 1, r = , LS = 3, a = − 1, r = − , LS = − 4 3 3 3 a = 150, r = , LS = 375 5 3 2 2 1 2 a = , r = , LS = 1 16. a = 3, r = and a = 2, r = 5 5 5 5 3 3 21 2 x= 18. (a) − 1 1 k 1 1 (b) − (c) k = 4 5 32 1 5 1 1 (a) − 1 p 1 (b) (c) p = 7 2 2 14

(d) 273 (e) 255

(a) 255 (b) (d) 1

(b)

1 − (− 5) n

Choice 1 gives $465.00. Choice 2 gives $10 737 418.23! 382 apples

1.(a) Yes LS = 13

7.

19. n = 7

3

=

Exercises 8.8

2.

1 , r = ±2 10

(− 5) n − 1

Puzzles

(a) 234 375 (b) 268.8 (c) -81 920 2187 (d) (e) 27 156 250

18. a =

(b) Sn = −

15. 2146

6.

9.

/ 2 (− 5) k − 1

k =1

(a) 256 (b) 26 244 (c) 1.369 3 (d) -768 (e) 1024

1149

(b) $178 550.21

n

5.

8.

12. 10 terms

13. (a) $33 502.39

(c) Tn = 9n – 1

=2 1 n −1 n−1 ] g (g) Tn = · 4 (h) Tn = 1000 −10 4 = − ] −10 gn + 2 n−2 =4 1 2 n −1 (i) Tn = − 3 ] −3 gn − 1 (j) Tn = d n 3 5 = ] −3 gn 4.

11. a = 9

2 3

(b) 13th (b) 29th

(c) 57 (c) 32

ANSWERS

(b) Sn = = = = =

4.

14. $4543.28

n [2a + (n − 1) d ] 2 1 n [2 ´ 6 + (n − 1) ´ 3] 2 1 n (12 + 3n − 3) 2 1 n (3n + 9) 2 3 n (n + 3) 2

18. $7.68

Exercises 8.11

(a) (i) 93% (ii) 86.49% (iii) 80.44% (b) 33.67% (c) 19 weeks

6.

(a) 0.01 m (b) 91.5 m

7.

(a) 49 (b) 4 mm

8.

(a) 3k m

9.

(a) 96.04%

(c) 9

(b) 34 (c) 68.6

10. (a) 77.4% (b) 13.5 (c) 31.4

12. 0.625 m

1.

$27 882.27

2. $83 712.95

3.

$50 402.00

4. $163 907.81

5.

$40 728.17

6. $29 439.16

7.

$67 596.72

8. $62 873.34

9.

$164 155.56 (28 years)

10. $106 379.70

11. $3383.22

12. $65 903.97

13. $2846.82

14. $13 601.02

15. $6181.13

16. $4646.71

18. (a) $26 361.59

17. $20 405.74

(b) $46 551.94

19. $45 599.17 20. (a) $7335.93 361 (j) 2 999

13. 15 m 14. 20 cm

15. 3 m

16. (a) 74.7 cm (b) 75 m 17. (a) 4.84 m (b) After 3 years 18. 300 cm

19. Kate $224.37

20. Account A $844.94

5.

2 4 7 11. (a) (b) (c) 1 9 9 9 9 25 7 (d) (e) 2 (f) 99 11 30 43 131 7 (g) 1 (h) 1 (i) 450 90 990

16. 8 years

17. (a) x = 7 (b) x = 5 (c) x = 8 (d) x = 6.5 (e) x = 8.5

(a) (i) $23 200 (ii) $26 912 (iii) $31 217.92 (b) $102 345.29 (c) 6.2 years

(b) k (3k + 3) m

15. 4 years

(b) $1467.18

21. $500 for 30 years

22. Yes, $259.80 over

23. No, shortfall of $2013.75 24. (a) $14 281.87 (b) $9571.96 (c) No, they will only have $23 853.83. 25. $813.16

19. 3.5 m 20. 32 m

21. (a) 1, 8, 64, … (b) 16 777 216 people (c) 19 173 961 people

Exercises 8.10

Exercises 8.12 1.

$1047.62

2. $394.46 3. $139.15

4.

(a) $966.45

(a) $740.12 (b) $14 753.64 (c) $17 271.40 (d) $9385.69 (e) $5298.19

5.

$2519.59

2.

(a) $2007.34 (b) $2015.87 (c) $2020.28

6.

(a) $592.00

3.

(a) $4930.86 (b) $4941.03

7.

(a) $77.81

4.

$408.24

8.

$78 700

6.

$1733.99

9.

(a) Get Rich $949.61, Capital Bank $491.27 (b) $33 427.80 more through Capital Bank

8.

$22 800.81

1.

5.

$971.40

7. 9.

$3097.06

(b) $39 319.89 (b) $2645.42

$691.41 10. $43 778.80

10. $1776.58

(b) $1265.79

11. $61 292.20

11. $14 549.76 12. $1 301 694.62

13. (a) $4113.51 (b) $555.32 (c) $9872.43 (d) $238.17 (e) $10 530.59

12. NSW Bank $175.49 a month ($5791.25 altogether) Sydney Bank $154.39 a month ($5557.88 altogether) Sydney Bank is better

629

630

Maths In Focus Mathematics Extension 1 HSC Course

13. (a) $249.60 (b) $13 485.12

Step 2: Assume true for n = k So 5 + 11 + 19 + … + ] 8k – 3 g = k ] 4k + 1 g

14. (a) $13 251.13 (b) $374.07 (c) $20 199.77

Step 3: Prove true for n = k + 1 Prove 5 + 11 + 19 + … + ] 8k − 3 g + ] 8 5 k + 1 ? − 3 g = ]k + 1 g]4 5k + 1 ? + 1 g i.e. 5 + 11 + 19 + … + ] 8k – 3 g + ] 8k + 5 g = ] k + 1 g ] 4k + 5 g

15. (a) $1835.68 (b) $9178.41

Exercises 8.13 1.

LHS = 5 + 11 + 19 + … + ] 8k − 3 g + ] 8k + 5 g = k ] 4k + 1 g + ] 8k + 5 g = 4k2 + k + 8k + 5 = 4k2 + 9k + 5 = ] k + 1 g ] 4k + 5 g = RHS

Step 1: Prove true for n = 1 LHS = 7n – 4 = 7 ]1g – 4 =3 n RHS = (7n − 1) 2 1 = (7 ´ 1 − 1) 2 =3 LHS = RHS

So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1.

So the statement is true for n = 1. 3.

Step 2: Assume true for n = k So 3 + 10 + 17 + … + ] 7k – 4 g =

k (7k − 1) 2

Step 3: Prove true for n = k + 1 Prove 3 + 10 + 17 + … + ] 7k – 4 g + ^ 7 [k + 1] − 4 h k+1 = (7 [k + 1] − 1) 2 i.e. 3 + 10 + 17 + … + ] 7k – 4 g + ] 7k + 3 g k+1 = (7k + 6) 2

So the statement is true for n = 0. Step 2: Assume true for n = k So 5 + 10 + 20 + … + 5.2k = 5 (2k + 1 − 1) Step 3: Prove true for n = k + 1 Prove 5 + 10 + 20 + … + 5 · 2k + 5 · 2k + 1 = 5(2k + 1 + 1 − 1) i.e. 5 + 10 + 20 + … + 5 · 2k + 5 · 2k + 1 = 5(2k + 2 − 1) LHS = 5 + 10 + 20 + … + 5 · 2k + 5 · 2k + 1 = 5 ] 2k + 1 − 1 g + 5 · 2k + 1 = 5 · 2k + 1 − 5 + 5 · 2k + 1 = 10 · 2k + 1 − 5 = 5 · 2 · 2k + 1 − 5 = 5 · 2k + 2 − 5 = 5 ] 2k + 2 − 1 g = RHS

LHS = 3 + 10 + 17 + … + ] 7k – 4 g + ] 7k + 3 g k = (7k − 1) + ] 7k + 3 g 2 2 (7k + 3) k = (7k − 1) + 2 2 k (7k − 1) 2 (7k + 3) = + 2 2 2 7k − k + 14k + 6 = 2 7k2 + 13k + 6 = 2 (k + 1) (7k + 6) = 2 (k + 1) = (7k + 6) 2 = RHS So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1. 2.

Step 1: Prove true for n = 1 LHS = 8n − 3 = 8 ]1 g −3 =5 RHS = n ] 4n + 1 g = 1 (4 ´ 1 + 1) =5 LHS = RHS

Step 1: Prove true for n = 0. LHS = 5 · 20 =5 RHS = 5 (20 + 1 − 1) = 5 ]2 − 1 g =5 LHS = RHS

So the statement is true for n = k + 1. The statement is true for n = 0 so it must be true for n = 1. It is true for n = 1 so it must be true for n = 2 and so on. The statement is true for all integers n ≥ 0. 4.

Step 1: Prove true for n = 1 1 LHS = 1 − 1 2 1 = 0 2 =1 1 RHS = 2 d 1 − 1 n 2 1 =2´ 2 =1 LHS = RHS So the statement is true for n = 1.

So the statement is true for n = 1.

Ans_PART_2.indd 630

6/18/09 12:39:49 PM

ANSWERS

Step 2: Assume true for n = k 1 1 1 1 So 1 + + + … + k − 1 = 2 d 1 − k n 2 4 2 2

Step 3: Prove true for n = k + 1

Prove 2 + 5 + 8 + … + ] 3k – 1 g + ] 3 5 k + 1 ? –1 g 3 (k + 1)2 + (k + 1) = 2 i.e. 2 + 5 + 8 + … + ] 3k – 1 g + ] 3k + 2 g 3(k2 + 2k + 1) + (k + 1) = 2 3k2 + 6k + 3 + k + 1 = 2 3k2 + 7k + 4 = 2

Step 3: Prove true for n = k + 1 1 1 1 1 1 Prove 1 + + + … + k − 1 + k + 1 − 1 = 2 d 1 − k + 1 n 2 4 2 2 2 1 1 1 1 1 i.e. 1 + + + … + k − 1 + k = 2 d 1 − k + 1 n 2 4 2 2 2 1 1 1 1 LHS = 1 + + + … + k − 1 + k 2 4 2 2 1 1 = 2d1 − k n + k 2 2 1 1 = 2 − 2· k + k 2 2 1 =2− k 2 1 2 =2− k ´ 2 2 2 = 2 − k+1 2 1 = 2e1 − k+1 o 2 = RHS

LHS = 2 + 5 + 8 + … + ] 3k – 1 g + ] 3k + 2 g 3k2 + k = + ] 3k + 2 g 2 3k2 + k 2 (3k + 2) = + 2 2 2 3k + k 6k + 4 = + 2 2 3k2 + k + 6k + 4 = 2 3k2 + 7k + 4 = 2 = RHS So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1.

So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1.

n

6. n

5.

/ (3r − 1) = (3 ´ 1 − 1) + (3 ´ 2 − 1) + (3 ´ 3 − 1) + . . .

r =1

+ (3n − 1) = 2 + 5 + 8 + . . . + ] 3n − 1 g

Step 1: Prove true for n = 1 LHS = 3n −1 = 3 ] 1 g −1 =2 3n 2 + n RHS = 2 2 3 (1) + 1 = 2 =2 LHS = RHS

r =1

= 2 + 4 + 8 + . . . + 2n Step 1: Prove true for n = 1 LHS = 21 =2 RHS = 2 ] 21 − 1 g = 2 ]1 g =2 LHS = RHS So the statement is true for n = 1. Step 2: Assume true for n = k So 2 + 4 + 8 + … + 2k = 2 ] 2k − 1 g

So the statement is true for n = 1. Step 2: Assume true for n = k So 2 + 5 + 8 + … + ] 3k − 1 g =

/ (2r) = 21 + 22 + 23 + . . . + 2n

3k2 + k 2

Step 3: Prove true for n = k + 1 Prove 2 + 4 + 8 + … + 2k + 2k + 1 = 2 ] 2k + 1 – 1 g LHS = 2 + 4 + 8 + … + 2k + 2k + 1 = 2 ] 2k – 1 g + 2k + 1 = 2k + 1 – 2 + 2k + 1 = 2 · 2k + 1 – 2 = 2 ] 2k + 1 – 1 g = RHS So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1.

631

632

Maths In Focus Mathematics Extension 1 HSC Course

Step 3: Prove true for n = k + 1 Prove −2 − 4 − 6 − … − 2k − 2 ] k + 1 g = − ] k + 1 g ] k + 1 + 1 g i.e. −2 − 4 − 6 − … − 2k − 2 ] k + 1 g = − ] k + 1 g ] k + 2 g LHS = − 2 − 4 − 6 − … − 2k − 2 (k + 1) = − k (k + 1) − 2 (k + 1) = − k 2 − k − 2k − 2 = − k 2 − 3k − 2 = − ] k 2 + 3k + 2 g = − ]k + 1g]k + 2g = RHS

n

7.

/ (5r) = (5 ´ 1) + (5 ´ 2) + (5 ´ 3) + . . . + (5 ´ n)

r =1

= 5 + 10 + 15 + . . . + 5n

Step 1: Prove true for n = 1 LHS = 5 ´ 1 =5 5 RHS = (1) (1 + 1) 2 5 = ´2 2 =5 LHS = RHS

So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1.

So the statement is true for n = 1. Step 2: Assume true for n = k So 5 + 10 + 15 + … + 5k =

5 k (k + 1) 2

9.

Step 3: Prove true for n = k + 1 Prove 5 + 10 + 15 + … + 5k + 5] k + 1 g 5 = (k + 1) (k + 1 + 1) 2 i.e. 5 + 10 + 15 + … + 5k + 5 ] k + 1 g =

5 (k + 1) (k + 2) 2

LHS = 5 + 10 + 15 + … + 5k + 5 ] k + 1 g 5 = k (k + 1) + 5 ] k + 1 g 2 5 (k + 1) ´ 2 5 = k (k + 1) + 2 2 5 5 = k (k + 1) + $ 2 (k + 1) 2 2 5 = [k (k + 1) + 2 (k + 1)] 2 5 = (k2 + k + 2k + 2) 2 5 = (k2 + 3k + 2) 2 5 = (k + 1) (k + 2) 2 = RHS

So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1. 8.

Step 1: Prove true for n = 1 LHS = − 2 ] 1 g = −2 RHS = −1 ] 1 + 1 g = −2 LHS = RHS So the statement is true for n = 1. Step 2: Assume true for n = k. So − 2 − 4 − 6 − . . . − 2k = − k ] k + 1 g

Step 1: Prove true for n = 1.

LHS = 5 ] 1 g + 4 =9 5(1)2 + 13 (1) RHS = 2 18 = 2 =9 LHS = RHS So the statement is true for n = 1.

Step 2: Assume true for n = k So 9 + 14 + 19 + … + ] 5k + 4 g =

5k2 + 13k 2

Step 3: Prove true for n = k + 1 Prove 9 + 14 + 19 + … + ] 5k + 4 g + ] 5 5 k + 1 ? + 4 g 5(k + 1)2 + 13(k + 1) = 2 i.e. 9 + 14 + 19 + … + ] 5k + 4 g + ] 5k + 9 g 5(k2 + 2k + 1) + 13k + 13 = 2 5k2 + 10k + 5 + 13k + 13 = 2 5k2 + 23k + 18 = 2

LHS = 9 + 14 + 19 + … + ] 5k + 4 g + ] 5k + 9 g 5k2 + 13k = + ] 5k + 9 g 2 5k2 + 13k 2(5k + 9) = + 2 2 2 5k + 13k 10k + 18 = + 2 2 5k2 + 13k + 10k + 18 = 2 5k2 + 23k + 18 = 2 = RHS So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1.

ANSWERS

Step 2: Assume true for n = k

n

10.

/ (3r) = 32 + 33 + 34 + . . . + 3n

r =2

So – 4 + 8 – 16 − … − 4 ] − 2 g k – 1 =

= 9 + 27 + 81 + . . . + 3n Step 1: Prove true for n = 2

i.e. – 4 + 8 – 16 − … − 4 ] − 2 gk – 1 – 4 ] − 2 gk 4 6 (− 2) k + 1 − 1 @ = 3

Step 2: Assume true for n = k 9(3k − 1 − 1) 2

Step 3: Prove true for n = k + 1 Prove 9 + 27 + 81 + . . . + 3k + 3k + 1 = =

9(3k + 1 − 1 − 1) 2 9(3k − 1) 2

LHS = 9 + 27 + 81 + … + 3k + 3k + 1 9(3k − 1 − 1) = + 3k + 1 2 9(3k − 1 − 1) 2 (3k + 1) = + 2 2 k−1 9 · 3 − 9 + 2 · 3k + 1 = 2 32 · 3k − 1 − 9 + 2 · 3k + 1 = 2 3k + 1 − 9 + 2 · 3k + 1 = 2 3 · 3k + 1 − 9 = 2 9 · 3k − 9 = 2 9(3k − 1) = 2 = RHS So the statement is true for n = k + 1. The statement is true for n = 2 so it must be true for n = 3. It is true for n = 3 so it must be true for n = 4 and so on. The statement is true for all integers n ≥ 2. 11. Step 1: Prove true for n = 1 LHS = − 4 ] − 2 g1 – 1 = − 4 ] − 2 g0 = −4 4 6 (− 2)1 − 1 @ RHS = 3 4 5 −2 − 1 ? = 3 4 (− 3) = 3 = −4 LHS = RHS So the statement is true for n = 1.

3

Step 3: Prove true for n = k + 1 Prove – 4 + 8 – 16 − … − 4 ] − 2 g k – 1 − 4 ] − 2 g k + 1 – 1 4 6 (− 2) k + 1 − 1 @ = 3

LHS = 32 =9 9 (32 − 1 − 1) RHS = 2 9(3 − 1) = 2 9(2) = 2 =9 LHS = RHS So the statement is true for n = 2.

So 9 + 27 + 81 + . . . + 3k =

4 6 (− 2)k − 1 @

LHS = − 4 + 8 − 16 − … − 4 ] − 2 gk – 1 − 4 ] − 2 gk 4 6 (− 2)k − 1 @ = − 4 ] − 2 gk 3 4 6 (− 2)k − 1 @ 3 ´ 4 (− 2)k = − 3 3 k 4 (− 2) − 4 12 (− 2)k = − 3 3 4 (− 2)k − 4 − 12 (− 2)k = 3 − 8 (− 2)k − 4 = 3 4 ´ − 2 (− 2)k − 4 = 3 4 (− 2)k + 1 − 4 = 3 4 6 (− 2)k + 1 − 1 @ = 3 = RHS So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1. 12. Step 1: Prove true for n = 1 LHS = 12 =1 1 RHS = 1 ] 1 + 1 g (2 ´ 1 + 1) 6 1 = ]2 g]3g 6 =1 LHS = RHS So the statement is true for n = 1. Step 2: Assume true for n = k So 12 + 22 + 32 + . . . + k2 =

1 k ] k + 1 g ] 2k + 1 g 6

Step 3: Prove true for n = k + 1 Prove 12 + 22 + 32 + . . . + k2 + (k + 1) 2 1 = (k + 1) ] k + 1 + 1 g ] 2 5 k + 1 ? + 1 g 6 i.e. 12 + 22 + 32 + . . . + k2 + (k + 1) 2 1 = ] k + 1 g ] k + 2 g ] 2k + 3 g 6

633

634

Maths In Focus Mathematics Extension 1 HSC Course

LHS = 12 + 22 + 32 + . . . + k2 + (k + 1)2 1 = k (k + 1) (2k + 1) + (k + 1)2 6 6 1 = k (k + 1) (2k + 1) + (k + 1)2 6 6 1 = 6 k (k + 1) (2k + 1) + 6 (k + 1)2 @ 6 1 = 7 (k + 1) " k (2k + 1) + 6 (k + 1) , A 6 Factorise by taking out a common factor of k + 1.

1 [] k + 1 g ] 2k2 + k + 6k + 6 g] 6 1 = [] k + 1 g ] 2k2 + 7k + 6 g] 6 1 = ] k + 1 g ] k + 2 g ] 2k + 3 g 6 = RHS =

So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1. n

13.

/ (2r − 1)3 = (2 ´ 1 − 1)3

r =1

+ (2 ´ 2 − 1)3 + (2 ´ 3 − 1)3

+ . . . + (2 ´ n − 1)3 = 13 + 33 + 53 + . . . + ] 2n − 1 g3 Step 1: Prove true for n = 1 LHS = (2 ´ 1 − 1) = 13 =1 RHS = 12 (2 ´ 12 − 1) = 2 −1 =1 LHS = RHS 3

So the statement is true for n = 1. Step 2: Assume true for n = k So 13 + 33 + 53 + . . . + ] 2k − 1 g3 = k2 ] 2k2 − 1 g Step 3: Prove true for n = k + 1 Prove 13 + 33 + 53 + . . . + (2k − 1)3 + (2 5 k + 1 ? − 1)3 = ] k + 1 g2 ^ 2 5 k + 1 ? 2 − 1 h i.e. 13 + 33 + 53 + . . . + (2k − 1)3 + (2k + 1)3 = (k + 1)2 (2 5 k2 + 2k + 1 ? − 1) = (k2 + 2k + 1) (2k2 + 4k + 2 − 1) = (k2 + 2k + 1) (2k2 + 4k + 1) = 2k4 + 4k3 + k2 + 4k3 + 8k2 + 2k + 2k2 + 4k + 1 = 2k4 + 8k3 + 11k2 + 6k + 1 LHS = 13 + 33 + 53 + . . . + (2k − 1) 3 + (2k + 1) 3 = k2 (2k2 − 1) + (2k + 1) 3 = 2k4 − k2 + 8k3 + 12k2 + 6k + 1 = 2k4 + 8k3 + 11k2 + 6k + 1 = RHS So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2.

It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1. 14. Step 1: Prove true for n = 1 71 − 1 = 7 − 1 = 6 which is divisible by 6 So the statement is true for n = 1. Step 2: Assume true for n = k So 7k –1 is divisible by 6. i.e. 7k − 1 = 6p where p is an integer Step 3: Prove true for n = k + 1 Prove 7k + 1 − 1 is divisible by 6 i.e. 7k + 1 –1 = 6q where q is an integer 7 k − 1 = 6p ` 7 ^ 7k − 1 h = 7 ´ 6p 7k + 1 − 7 = 42p k +1 7 − 7 + 6 = 42p + 6 7k + 1 − 1 = 42p + 6 = 6 ^ 7p + 1 h = 6q where q is an integer So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1. 15. Step 1: Prove true for n = 1 32 ´ 1 – 1 = 9 − 1 = 8 which is divisible by 8 So the statement is true for n = 1. Step 2: Assume true for n = k So 32k − 1 is divisible by 8. i.e. 32k − 1 = 8p where p is an integer Step 3: Prove true for n = k + 1 Prove 32 (k + 1) − 1 is divisible by 8 i.e. 32k + 2 − 1 = 8q where q is an integer 3 2 k − 1 = 8p ` 3 2 ] 3 2 k − 1 g = 3 2 ´ 8p 32k + 2 − 9 = 72p 2k + 1 3 − 9 + 8 = 72p + 8 32k + 2 − 1 = 72p + 8 = 8 ^ 9p + 1 h = 8q where q is an integer So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1. 16. Step 1: Prove true for n = 1 51 − 1 = 5 – 1 = 4 which is divisible by 4 So the statement is true for n = 1. Step 2: Assume true for n = k So 5k − 1 is divisible by 4. i.e. 5k − 1 = 4p where p is an integer

ANSWERS

Step 3: Prove true for n = k + 1 Prove 5k + 1 − 1 is divisible by 4 i.e. 5k + 1 − 1 = 4q where q is an integer 5k − 1 ^ ` 5 5k − 1 h 5k + 1 − 5 5k + 1 − 5 + 4 5k + 1 − 1

= 4p = 5 ´ 4p = 20p = 20p + 4 = 20p + 4 = 4 ^ 5p + 1 h = 4 q where q is an integer

So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1. 17. Step 1: Prove true for n = 2 2 ]2 + 2 g = 2 ]4 g = 8 which is divisible by 4 So the statement is true for n = 2. Step 2: Assume true for n = k where k is even So k ] k + 2 g is divisible by 4. i.e. k2 + 2k = 4p where p is an integer Step 3: Prove true for n = k + 2 (the next even integer) Prove ] k + 2 g ] k + 2 + 2 g is divisible by 4 i.e. ] k + 2 g ] k + 4 g = 4q where q is an integer ] k + 2 g ] k + 4 g = k 2 + 4k + 2k + 8 = k2 + 2k + 4k + 8 = 4p + 4k + 8 = 4^ p + k + 2h = 4q where q is an integer So the statement is true for n = k + 2. The statement is true for n = 2 so it must be true for n = 4. It is true for n = 4 so it must be true for n = 6 and so on. The statement is true for all even positive integers. 18. Step 1: Prove true for n = 1 1 + 3 = 4 which is divisible by 4 So the statement is true for n = 1. Step 2: Assume true for n = k where k is odd So k + ] k + 2 g is divisible by 4. i.e. k + k + 2 = 4p where p is an integer 2k + 2 = 4p Step 3: Prove true for n = k + 2 (the next odd integer) Prove 2 ] k + 2 g + 2 is divisible by 4 i.e. 2 ] k + 2 g + 2 = 4q where q is an integer 2k + 4 + 2 = 4 q 2k + 6 = 4q 2k + 6 = 2k + 2 + 4 = 4p + 4 = 4^ p + 1h = 4q where q is an integer So the statement is true for n = k + 2. The statement is true for n = 1 so it must be true for n = 3.

It is true for n = 3 so it must be true for n = 5 and so on. The statement is true for all odd positive integers. 19. Step 1: Prove true for n = 1 51 + 31 = 5 + 3 = 8 which is divisible by 2 So the statement is true for n = 1 Step 2: Assume true for n = k So 5k + 3k is divisible by 2. i.e. 5k + 3k = 2p where p is an integer 5 k = 2p − 3 k Step 3: Prove true for n = k + 1 Prove 5k + 1 + 3k + 1 is divisible by 2 i.e. 5k + 1 + 3k + 1 = 2q where q is an integer 5k + 1 + 3k + 1 = 5 · 5k + 3 · 3k = 5 ^ 2p − 3 k h + 3 · 3 k = 10p − 5 · 3k + 3 · 3k = 10p − 2 · 3k = 2 ^ 5p − 3 k h = 2q where q is an integer So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1. 20. Step 1: Prove true for n = 1 71 + 31 = 7 + 3 = 10 which is divisible by 10 So the statement is true for n = 1 Step 2: Assume true for n = k where k is odd So 7k + 3k is divisible by 10. i.e. 7k + 3k = 10p where p is an integer 7k = 10p − 3k Step 3: Prove true for n = k + 2 (next odd integer) Prove 7k + 2 + 3k + 2 is divisible by 10 i.e. 7k + 2 + 3k + 2 = 10q where q is an integer 7k + 2 + 3k + 2 = 72 · 7k + 32 · 3k = 49 · 7k + 9 · 3k = 49 ^ 10p − 3k h + 9 · 3k = 490p − 49 · 3k + 9 · 3k = 490p − 40 · 3k = 10 ^ 49p − 4 · 3k h = 10q where q is an integer So the statement is true for n = k + 2. The statement is true for n = 1 so it must be true for n = 3. It is true for n = 3 so it must be true for n = 5 and so on. The statement is true for all odd integers n ≥ 1. 21. Step 1: Prove true for n = 1 LHS = 12 =1 RHS = 1 − 5 = −4 LHS > RHS So the statement is true for n = 1.

635

636

Maths In Focus Mathematics Extension 1 HSC Course

since 5 ] 4k g > 4 ] 4k g and 112 > 20 The statement is true for n = 3 so it must be true for n = 4. It is true for n = 4 so it must be true for n = 5 and so on. The statement is true for all integers n ≥ 3.

Step 2: Assume true for n = k So k2 > k − 5 Step 3: Prove true for n = k + 1 ] k + 1 g2 > k + 1 − 5 Prove k 2 + 2k + 1 > k − 4 k 2 + 2k > k − 5 We know that k2 > k − 5 So k2 + 2k > k − 5 since 2k > 0 for k > 0 So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1.

24. Step 1: Prove true for n = 1 LHS = 31 =3 RHS = 21 + 1 =3 LHS ≥ RHS So the statement is true for n = 1. Step 2: Assume true for n = k So 3k ≥ 2k + k Step 3: Prove true for n = k + 1 Prove 3k + 1 ≥ 2k + 1 + k + 1 LHS = 3k + 1 = 3 ] 3k g ≥ 3 ] 2k + k g ≥ 3 ] 2 k g + 3k ≥ 2 ] 2k g + k + 1 ≥ 2k + 1 + k + 1

22. Step 1: Prove true for n = 2 LHS = 42 = 16 RHS = 3 ] 2 g + 7 = 13 LHS ≥ RHS So the statement is true for n = 2.

since 3 ] 2k g ≥ 2 ] 2k g and 3k ≥ k + 1 for k ≥ 1 So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1.

Step 2: Assume true for n = k So 4k ≥ 3k + 7 Step 3: Prove true for n = k + 1 Prove 4k + 1 ≥ 3 ] k + 1 g + 7 i.e. 4 k + 1 ≥ 3k + 3 + 7 4k + 1 ≥ 3k + 10 LHS = 4k + 1 = 4 ] 4k g ≥ 4 ] 3k + 7 g ≥ 12k + 28 ≥ 3k + 10 since 12k ≥ 3k and 28 ≥ 10 So the statement is true for n = k + 1. The statement is true for n = 2 so it must be true for n = 3. It is true for n = 3 so it must be true for n = 4 and so on. The statement is true for all integers n ≥ 2.

25. Step 1: Prove true for n = 1 LHS = 51 =5 RHS = 31 + 21 =5 LHS ≥ RHS So the statement is true for n = 1. Step 2: Assume true for n = k So 5k ≥ 3k + 2k Step 3: Prove true for n = k + 1 Prove 5k + 1 ≥ 3k + 1 + 2k + 1 LHS = 5k + 1 = 5 ] 5k g ≥ 5 ] 3k + 2k g ≥ 5 ] 3k g + 5 ] 2k g ≥ 3 ] 3k g + 2 ] 2k g ≥ 3k + 1 + 2k + 1

23. Step 1: Prove true for n = 3 LHS = 53 − 3 = 125 − 3 = 122 RHS = 43 + 20 = 64 + 20 = 84 LHS > RHS So the statement is true for n = 3.

since 5 ] 3k g ≥ 3 ] 3k g and 5 ] 2k g ≥ 2 ] 2k g So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1.

Step 2: Assume true for n = k So 5k − 3 > 4k + 20 i.e. 5k > 4k + 23 Step 3: Prove true for n = k + 1 Prove 5k + 1 − 3 > 4k + 1 + 20 5k + 1 − 3 = 5 ] 5k g − 3 > 5 ] 4k + 23 g − 3 > 5 ] 4k g + 115 − 3 > 5 ] 4k g + 112 > 4 ] 4k g + 20 > 4k + 1 + 20

Test yourself 8 1.

(a) Tn = 4n + 5 (c) Tn = 2:3

n −1

(e) Tn = ] − 2 gn

(b) Tn = 14 − 7n 1 n −1 (d) Tn = 200 d n 4

ANSWERS

2.

(a) 2 (b) 1185 (c) 1183 (d) T15 = S15 − S14

29. n = 11

S15 = S14 + T15

30. (a)

n

r =1

(e) n = 16 3.

(a) 11 125 (b) 1 (c) 3 985 785

13 140

(d) 34 750

1 (e) 2 4.

(a) Each slat rises 3 mm so the bottom one rises up 30 ´ 3 mm or 90 mm. (b) 87 mm (c) 90, 87, 84, . . . which is an arithmetic sequence with a = 90, d = − 3 (d) 42 mm (e) 1395 mm

5.

$3400.01

6.

(a) (i) (b) (ii) (c) (i) (d) (iii) (e) (i) (f) (ii) (g) (ii) (h) (i) (i) (i) (j) (i)

7.

n = 108

8.

(a) $24 050 (b) $220 250

9.

a = − 33, d = 13

10. (a) 59 (b) 80

12. (a)

4 9

(b)

13 18

19 33

13. x = 3 14. (a) 136 (b) 44 15. 121

1 2

(c) 6

16. $8066.42

17. (a) Tn = 4n + 1

(b) Tn = 1.07n − 1

18. (a) − 1 < x < 1

(b) 2

1 2

(c) x =

1 3

19. d = 5 20. (a) 39 words/min (b) 15 weeks 21. (a) $59 000

(b) $15 988.89

22. 4.8 m 23. x = −

2 ,2 17

24. (a) $2385.04 (b) $2392.03 25. 1300 26. (a) 735 (b) 4315 27. (a) $1432.86 (b) $343 886.91 28. n = 20

Step 2: Assume true for n = k So 15 + 45 + 135 + … + 5 ] 3 gk =

=

(c) 18th term

(c) 1

= 15 + 45 + 135 + . . . + 5] 3 gn Step 1: Prove true for n = 1 LHS = 5 ] 3 g1 = 15 15 (31 − 1) RHS = 2 15 (2) = 2 = 15 LHS = RHS So the statement is true for n = 1. 15 (3k − 1) 2

Step 3: Prove true for n = k + 1 Prove 15 + 45 + 135 + … + 5 (3) k + 5 (3) k + 1

(b) x = ± 15

11. (a) x = 25

/ 5(3)r = 5(3)1 + 5(3)2 + 5(3)3 + . . . + 5(3)n

15 (3k + 1 − 1) 2

LHS = 15 + 45 + 135 + … + 5 (3) k + 5 (3) k + 1 15 (3k − 1) = + 5 (3) k + 1 2 15 (3k − 1) 2 ´ 5 (3) k + 1 = + 2 2 k 15 (3 − 1) + 10 (3) k + 1 = 2 15 ´ 3k − 15 + 10 ´ 3k + 1 = 2 5 ´ 3 ´ 3k − 15 + 10 ´ 3k + 1 = 2 5 ´ 3k + 1 − 15 + 10 ´ 3k + 1 = 2 15 ´ 3k + 1 − 15 = 2 15 (3k + 1 − 1) = 2 = RHS

So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1. (b) Step 1: Prove true for n = 1 LHS = 3 ´ 1 + 1 =4 1 (3 ´ 1 + 5) RHS = 2 8 = 2 =4 LHS = RHS So the statement is true for n = 1.

637

638

Maths In Focus Mathematics Extension 1 HSC Course

Step 2: Assume true for n = k So 4 + 7 + 10 + … + ] 3k + 1 g =

k (3k + 5) 2

Step 3: Prove true for n = k + 1 Prove 4 + 7 + 10 + … + ] 3k + 1 g + ] 3 5 k + 1 ? + 1 g (k + 1) (3 [k + 1] + 5) = 2 i.e. 4 + 7 + 10 + … + ] 3k + 1 g + ] 3k + 4 g =

2 3k + 8k + 3k + 8 = 2 3k2 + 11k + 8 = 2

LHS = 4 + 7 + 10 + … + ] 3k + 1 g + ] 3k + 4 g k (3k + 5) = + ] 3k + 4 g 2 k (3k + 5) 2 ´ (3k + 4) = + 2 2 k (3k + 5) + 2 (3k + 4) = 2 3k 2 + 5k + 6k + 8 = 2 3k2 + 11k + 8 = 2 = RHS So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1. (c) Step 1: Prove true for n = 2 2 ´ 2 ]2 − 1 g = 4 ´ 1 = 4 which is divisible by 4 So the statement is true for n = 2. Step 2: Assume true for n = k So 2k (k − 1) is divisible by 4. i.e. 2k2 – 2k = 4p where p is an integer Step 3: Prove true for n = k + 1 Prove 2 ] k + 1 g ] 5 k + 1 ? – 1 g is divisible by 4 i.e. 2 ] k + 1 g ] 5 k + 1 ? – 1 g = 4q where q is an integer

2 ] k + 1 g k = 4q 2 ] k + 1 g k = 2k2 + 2k = 2k2 − 2k + 4k = 4p + 4k = 4^ p + kh = 4q So the statement is true for n = k + 1. The statement is true for n = 2 so it must be true for n = 3. It is true for n = 3 so it must be true for n = 4 and so on. The statement is true for all integers n > 1.

1.

(a) 8.1 (b) 19th term

2.

(a)

π 9π (b) 4 4

(c)

33π 4

(a) 2 097 170

4.

(a) $40

5.

6th term 6.

7.

5 terms

9.

− 56

(k + 1) (3k + 8) 2

Challenge exercise 8

3.

11. x =

(b) -698 775

(b) $2880

8.

17 823 n = 1, 2 , 3

10. $1799.79

3 12. $8522.53 13. k = 20 8

14. (a) $10 100 (b) $11 268.25 (c) $4212.41 (d) $2637.23 15. (a) cosec2 x (b) −1 ≤ cos x ≤ 1 So 0 ≤ cos2 x ≤ 1 | cos2 x | ≤ 1 So the limiting sum exists. 16. $240 652.62 17. Step 1: Prove true for n = 1 LHS = ar1 − 1 =a a (r1 − 1) RHS = r −1 =a LHS = RHS So the statement is true for n = 1. Step 2: Assume true for n = k a(r k − 1) So a + ar + ar2 + . . . + ar k – 1 = r −1 Step 3: Prove true for n = k + 1 Prove a + ar + ar2 + . . . + ar k – 1 + ar k + 1 – 1 = i.e. a + ar + ar2 + . . . + ar k – 1 + ar k =

a(r k + 1 − 1) r −1

a(r k + 1 − 1)

r −1 LHS = a + ar + ar2 + . . . + ar k – 1 + ar k a(r k − 1) = + ar k r −1 a(r k − 1) ar k(r − 1) = + r −1 r −1 a(r k − 1) + ar k (r − 1) = r −1 ar k − a + ar k + 1 − ar k = r −1 − a + ar k + 1 = r −1 a(r k + 1 − 1) = r −1 = RHS So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1.

ANSWERS

n

18.

/ xr − 1 = x1 − 1 + x2 − 1 + x3 − 1 + … + xn − 1

5.

3e 3 x 3e3x = 1 + (e3x + 1) 2 e6x + 2e3x + 2

6.

0.73

7.

(a)

r =1

= 1 + x + x2 + … + xn – 1 To prove 1 + x + x2 + … + xn − 1 =

1 − xn 1− x

Step 1: Prove true for n = 1 LHS = x1 – 1 =1 1 − x1 RHS = 1− x =1 LHS = RHS So the statement is true for n = 1. Step 2: Assume true for n = k 1 − xk So 1 + x + x2 + … + xk – 1 = 1− x Step 3: Prove true for n = k + 1 ]

Prove 1 + x + x2 + … + xk – 1 + x k + 1 i.e. 1 + x + x2 + … + xk – 1 + xk =

g −1

1 − xk + 1 = 1− x

1 − xk + 1 1− x

LHS = 1 + x + x2 + … + xk – 1 + xk 1 − xk + xk = 1− x k 1 − xk x (1 − x) + = 1− x 1− x 1 − xk xk − xk + 1 + = 1− x 1− x k 1 − x + xk − xk + 1 = 1− x 1 − xk + 1 = 1− x = RHS So the statement is true for n = k + 1. The statement is true for n = 1 so it must be true for n = 2. It is true for n = 2 so it must be true for n = 3 and so on. The statement is true for all integers n ≥ 1.

8.

(b)

1 unit2 2

4 9

9.

1.

1 − x2 −6

(c)

+

1 x

13. 0.22

14. ±104 + 52 ± 26 + …

1 tan− 1 2x + C 2

16.

7π Z 0.61 36 cos− 1 x +

=

4 (tan− 1 x) 3 1 + x2

12. $2929.08

15.

17.

(b)

1 − 9x 2

π2 units3 18

.

Practice assessment task set 3

1

10. (a)

11.

3 49

x

1 − x2 (cos− 1 x) 2 1 − x2 cos− 1 x + x 1 − x2 (cos− 1 x) 2

18. 9x − 12y + 2π − 3 3 = 0

19. 44th term

20. Step 1: Prove true for n = 0 LHS = 3 ] 20 g =3 RHS = 3 ] 20 + 1 − 1 g = 3 ]2 − 1 g =3 So true for n = 0 Step 2: Assume true for n = k 3 + 6 + 12 + . . . + 3 ] 2k g = 3 ] 2k + 1 − 1 g

2.

π 6

3.

−

4.

324 625

1 1 − x cos x 2

−1

Step 3: Prove true for n = k + 1 Prove 3 + 6 + 12 + . . . + 3 ] 2k g + 3 ] 2k + 1 g = 3 ] 2k + 1 + 1 − 1 g i.e. 3 + 6 + 12 + . . . + 3 ] 2k g + 3 ] 2k + 1 g = 3 ] 2k + 2 − 1 g

639

640

Maths In Focus Mathematics Extension 1 HSC Course

LHS = 3 + 6 + 12 + . . . + 3 ^ 2k h + 3 ^ 2k + 1 h = 3 ^ 2k + 1 − 1 h + 3 ^ 2k + 1 h = 3 ^ 2k + 1 h − 3 + 3 ^ 2k + 1 h = 6 ^ 2k + 1 h − 3 = 3 ] 2 g ^ 2k + 1 h − 3 = 3 ^ 2k + 2 h − 3 = 3 ^ 2k + 2 − 1 h = RHS So it is true for n = k + 1. Since it is true for n = 0, then it is true for n = 1. If true for n = 1, then it is true for n = 2 and so on. So it is true for all n ≥ 0.

23.

25.

26.

(b) 1410

24 7

9841a 6561

(b) −

π 4

(c)

π 3

−1

34. f (x) = 35. (a) −

38. (a)

(b)

3x 2 n +C tan− 1 d 3 2

27. (a) 3 000 000 (b) 3 000 336 (c) 146 insects per day 28. (2, 0), inflexion

(c)

29. (a)

(d)

d2 = (230 − 65t) 2 + (125 − 80t) 2 (Pythagoras) = 52 900 − 29 900t + 4225t2 + 15 625 − 20 000t + 6400t2 = 10 625t2 − 49 900t + 68 525 (b) 2.3 h (c) 109.7 km 30. sin− 1

3

1 2

x +C 2

31. 199; 5050

x + 1 ; domain: all real x, range: all real y (b) −

π 3

(c)

π 2

36. 2.98 units3

24. 2.4 m

1 x 7 1 + (loge x + 1) 2 A 1 = x 7 (loge x) 2 + 2 loge x + 2 A

(c) 29th term

3 2

33. −

37. x sin− 1 x +

21. $945 22. (a)

32. (a) T1 = 4, T2 = 11, T3 = 18, T12 = 81

1 − x2 + cos x + C

ANSWERS

(g) a = − 48 cos 4t = − 16 (3 cos 4t) = − 16x

(e)

45.

47.

8 45

46. $180.76

AC2 = 162 = 256 AB + BC2 = 9.62 + 12.82 = 256 Since AC2 = AB2 + BC2, ∆ ABC is right angled at +B. 2

39. Step 1: Prove true for n = 5 LHS = 25 − 1 = 31 RHS = 5 ] 5 g + 2 = 27

48. x =

LHS > RHS So true for n = 5

49.

Step 2: Assume true for n = k 2k − 1 > 5k + 2 i.e. 2k > 5k + 3 Step 3: Prove true for n = k + 1 Prove 2k + 1 − 1 > 5 ] k + 1 g + 2 i.e. 2 k + 1 − 1 > 5k + 7 2 k + 1 − 5k + 8 2k + 1 = 2 ^ 2k h > 2 ] 5k + 3 g > 10k + 6 > 5k + 8

since 10k + 6 > 5k + 8 for k > 0 So it is true for n = k + 1. Since it is true for n = 5, then it is true for n = 6. If true for n = 6, then it is true for n = 7 and so on. So it is true for all n > 4. 40. − 15 – 4 + 7 + . . . 1+ x 1 + 1 or y = ; domain: all real x ≠ 0, range: all x x real y ≠ 1

41. y =

42. $2851.52 43.

44. (a) v = − 12 sin 4t (b) a = − 48 cos 4t (c) 3 cm π π π 3 π 5π (d) t = 0, , , . . . s (e) ±3 cm (f) t = , , ,... s 4 2 8 8 8

5 6

n

/ 3r = 30 + 31 + 32 + … + 3n

r =0

= 1 + 3 + 32 + … + 3n Step 1: Prove true for n = 0 LHS = 30 =1 30 + 1 − 1 RHS = 2 3 −1 = 2 2 = 2 =1 LHS = RHS So true for n = 0 Step 2: Assume true for n = k 3k + 1 − 1 1 + 3 + 32 + … + 3k = 2 Step 3: Prove true for n = k + 1 3k + 1 + 1 − 1 1 + 3 + 32 + … + 3k + 3k + 1 = 2 k+2 3 −1 i.e. 1 + 3 + 32 + . . . + 3k + 3k + 1 = 2 LHS = 1 + 3 + 32 + … + 3k + 3k + 1 3k + 1 − 1 = + 3k + 1 2 k +1 3 k + 1 − 1 2 (3 ) = + 2 2 3 k + 1 − 1 + 2 (3 k + 1) = 2 3 (3 k + 1) − 1 = 2 3k + 2 − 1 = 2 = RHS So it is true for n = k + 1. Since it is true for n = 0, then it is true for n = 1. If true for n = 1, then it is true for n = 2 and so on. So it is true for all n ≥ 0.

641

642

Maths In Focus Mathematics Extension 1 HSC Course

60. (a) 1.3 s

50.

(b) 6 m

(c) 2.15 m

61. (a) log 3 + log 9 + log 27 + . . . = log 3 + log 32 + log 33 + . . . = log 3 + 2 log 3 + 3 log 3 + . . . Arithmetic series, since 2 log 3 − log 3 = 3 log 3 − 2 log 3 = log 3 Let ABCD be a rhombus with AB = BC.

(b) 210 log 3

AB = DC and AD = BC (opposite sides in a parallelogram)

62.

` AB = BC = DC = AD ` all sides are equal 51. (a) k Z 0.025 (c) 20.6 years 52. (a) x = 0, (c) x =

(b) after 42.4 years

2π , 2π 3

(b) x =

π 5π 3 π , , 6 6 2

π 4π , 3 3

53. 450 cm

63. (a) 12.6 mL

2

(b) 30 minutes

64. (a) A = 24 000, k Z 0.038 −1

4

(b) 48e ms 54. (a) 12 ms (c) x = 3e4t + 2

−2

. ..

x = 12e4t

(b) 37.4 years

65. −9 cms− 1 66. (a)

x = 48e4t = 4 (12e4t )

.

= 4x (d)

(b)

1 2

units2

67. (a) θ = nπ + (− 1) n 68. x + 3y = 0

55. n = 4 56. (a)

57.

π 3

(b)

3 2

1 10

70. (a) x = 4 71. (b)

(b) Amplitude 1

72. (c), (d)

73. (a)

59.

75. (d)

76. (a)

77. (c) 78. (c) 79. (b), (c), (d) 80. (d)

(b) θ = nπ ±

69. $277.33

74. (b), (d) 58. (a) Square 46.3 m ´ 46.3 m, rectangle 30.9 m ´ 92.7 m (b) $8626.38

π 6

π 6

ANSWERS

(b) P ] 2 g = − 4 < 0 P ] 3 g = 15 > 0 So the root lies between x = 2 and x = 3. (c) x = 2.25

Chapter 9: Polynomials 2

Exercises 9.1 1.

2.

(a) f (3) < 0, f (4) > 0 (b) f (3) > 0, f (4) < 0 (c) P (− 3) < 0, P (− 2) > 0 (d) P (0) > 0, P (1) < 0 (e) f (2) < 0, f (3) > 0 (a) f (0) < 0, f (1) > 0 ` root lies between 0 and 1 (b) f (0.5) > 0 ` root lies between 0 and 0.5

3.

f (0) > 0, f (1) > 0 and minimum turning point at (0.8, 0.9) ` no root lies between 0 and 1

4.

(a) P (1) < 0, P (2) > 0 (b) P (1.5) > 0, so the root lies between 1 and 1.5; P (1.25) = 0 ` 1.25 is a zero of P (x)

5.

6.

(a) f (1) < 0, f (2) > 0 (b) f (1.5) > 0, so the root lies between 1 and 1.5 f (− 5) > 0, f (− 4) < 0, so the root lies between − 5 and − 4; f (− 4.5) < 0, so the root lies between − 5 and − 4.5; f (− 5) = 1 and f (− 4.5) = −1.25, so x = − 5 is the better approximation

13. (a) f ] 2 g = 0.51 > 0 f ] 3 g = − 0.46 < 0 So the root lies between x = 2 and x = 3. (b) x = 2.5 14. (a) f ] 0.1 g = 0.026 > 0 f ] 0.2 g = − 0.058 < 0 So the root lies between x = 0.1 and x = 0.2. (b) f ] 0.15 g = − 0.014 < 0 So the root lies between x = 0.1 and x = 0.15. f ] 0.125 g = 0.0063 > 0 So the root lies between x = 0.125 and x = 0.15. 15. P ] 1 g = 1 > 0 P ]2 g = − 4 < 0 So the root lies between P ] 1.5 g = − 0.55 < 0 So the root lies between P ] 1.25 g = 0.43 > 0 So the root lies between P ] 1.375 g = − 0.006 < 0 So the root lies between

x = 1 and x = 2. x = 1 and x = 1.5. x = 1.25 and x = 1.5. x = 1.25 and x = 1.375.

Exercises 9.2 7.

x Z1

8.

x Z 3.25

9.

x Z −4

10. (a)

1.

2.39

2.

− 1.4

3.

0.772

4.

(a) f ] 0 g > 0, f ] 1 g < 0 (b) 0.448

5.

(a) f ] 0 g < 0, f ] 1 g > 0 (b) 0.379

6.

7.

(a) f (0) < 0, f (1) > 0 (b) x Z 0.75 (c) x Z 0.8 (d) x Z 0.78

8.

(a) x3 − 9 = 0 x3 = 9

(b) f (x) has 1 zero between − 1 and 0 (c) 0 11. (a) For f ] x g = ex – sin x – 3 f ] 1 g = − 1.1 < 0 f ] 2 g = 3.5 > 0 So the root lies between x = 1 and x = 2. (b) x = 1.375 12. (a) x3 – 12 = 0 x3 = 12 x = 3 12

(a) f ] − 2 g < 0, f ] − 1 g > 0 (b) x Z − 1 (c) x Z − 1.18

`

x =3 9

(b) Between x = 2 and x = 3 (c) Using x = 2.5 as 1st approximation, x Z 2.15

643

644

Maths In Focus Mathematics Extension 1 HSC Course

9.

(a) 1.74

(b) 3.33 (c) 2.76 (d) 1.91

10. x Z 2.31; no since f ] 2.31 g Z 9.2

3.

11. x Z 1.34

14. (a) x = 2 and x = 3

12. x Z 1.99 13. x Z 0.42 (b) x Z 2.75 15. x Z 0.74

(a) f ] 6 g = − 0.09 < 0 f ] 6.5 g = 0.06 > 0 So the root lies between x = 6 and x = 6.5. (b) f ] 6.25 g = − 0.01 < 0 So the root lies between x = 6.25 and x = 6.5. f ] 6.375 g = 0.027 > 0 So the root lies between x = 6.25 and x = 6.375. (c) x = 6.253

Test yourself 9 1.

x = 1.39 2. x = 1.87 (b) x = 2.05

3. (a) f (2) = − 1, f (3) = 49

4.

(a) f (− 2) = 9, f (− 3) = − 5

5.

(a)

Chapter 10: The binomial theorem

Exercises 10.1

(b) x = − 2.75

1.

2.

3.

(b) x = 1, − 0.9, 4.4 6. x = 1.375 8.

(a) x = 1.61

(b) x = 1.58

(b) f ’(1) = 0

(c) x =

10. (a) x = 2.25

1 3

7. x = 1.34

9. (a) f ] 0 g = 1, f ] 1 g = − 1

(d) x = 0.25

(b) x = 2.3

Challenge exercise 9 1.

2.

P ] x g = x3 – a Pl(x) = 3x2 P (b) x1 = b − Pl(b) b3 − a =b− 3b2 3b3 b3 − a = − 3b2 3b2 3b3 − b3 + a = 3b2 3 2b + a = 3b 2

(a) 15 (b) 45 (c) 84 (h) 792 (i) 2 (j) 45 9 9! (a) c m = 5 (9 − 5)!5! 9! = 4!5! 9 9! c m= 4 (9 − 4)!4! 9! = 5!4! 9 9 So c m = c m 5 4 7! 7 (b) C2 = (7 − 2)!2! 7! = 5!2! 7! 7 C5 = (7 − 5)!5! 7! = 2!5! So 7 C2 = 7C5

(a)

(b) x Z − 0.17, − 3.21

(a) 5040 (b) 40 320 (c) 6 (d) 120 (e) 1 (f) 3 628 798 (g) 72 (h) 3600 (i) 72 576

12 12! m= 5 (12 − 5)!5! 12! = 7!5! 12 12! c m= 7 (12 − 7)!7! 12! = 5!7! 12 12 So c m = c m 5 7 11! 11 (d) C3 = (11 − 3)!3! 11! = 8!3! 11! 11 C8 = (11 − 8)!8! 11! = 3 !8 ! So 11 C3 = 11 C8 (c) c

(c) x Z − 3.21

(d) x Z − 3

(d) 165

(e) 1

(f) 1

(j) 1680 (g) 4

ANSWERS

10! (10 − 1)!1! 10! = 9!1! 10! 10 C9 = (10 − 9)!9! 10! = 1 !9 !

(e) 10 C1 =

7

(h)

= 6

C4 + 6 C5 = =

So 10 C1 = 10 C9

=

9 9! (f) c m= 7 (9 − 7)!7! 9! = 2!7! 8 8 8! 8! + c m+ c m= 6 7 (8 − 6)!6! (8 − 7)!7! 8! 8! = + 2!6! 1!7! 8! 8! = + 2!6! 7! 8! ´ 7 8! ´ 2 = + 2!6! ´ 7 7! ´ 2 7(8!) 2 (8!) = + 2!7! 2 (7!) 7(8!) 2 (8!) = + 2! (7!) 2!7! (since 2! = 2) = =

= =

7! (7 − 5)!5! 7! 2!5! 6! 6! + (6 − 4)!4! (6 − 5)!5! 6! 6! + 2!4! 1!5! 5(6!) 2(6!) + 2!5! 2!5! 5(6!) + 2(6!) 2 !5 ! 7(6!)

2 !5 ! 7! = 2!5! = 7 C5 10 10! m= 6 (10 − 6)!6! 10! = 4!6! 9 9 9! 9! + c m+ c m = 5 6 (9 − 5)!5! (9 − 6)!6! 9! 9! = + 4!5! 3!6! 6(9!) 4(9!) = + 6(4!5!) 4(3!6!) 6(9!) 4(9!) = + 4!6! 4!6! 10(9!) = 4!6! 10! = 4!6! 10 =c m 6 7 7! (j) c m= 3 (7 − 3)!3! 7! = 4!3! 6 6 6! 6! + c m+ c m = 2 3 (6 − 2)!2! (6 − 3)!3! 6! 6! = + 4!2! 3!3! 3(6!) 4(6!) = + 3(4!2!) 4(3!3!) 3(6!) 4(6!) = + 4!3! 4!3! 7(6!) = 4!3! 7! = 4!3! 7 =c m 3 n n! c m= k (n − k)!k! n n! c m= n−k [n − (n − k)]!(n − k)! n! = (n − n + k)!(n − k)! n! = k!(n − k) n =c m k (i)

7(8!) + 2 (8!) 2!7! 9(8!)

2 !7 ! 9! = 2 !7 ! 9 =c m 7 (g)

C5 =

11 11! m= 6 (11 − 6)!6! 11! = 5!6! 10 10 10! 10! + c m+ c m = 5 6 (10 − 5)!5! (10 − 6)!6! 10! 10! = + 5!5! 4!6! 6(10!) 5(10!) = + 6(5!5!) 5(4!6!) 6(10!) 5(10!) = + 5 !6 ! 5 !6 ! 11(10!) = 5!6! 11! = 5 !6 ! 11 =c m 6

c

4.

c

645

646

Maths In Focus Mathematics Extension 1 HSC Course

5.

x =5

9.

k =6 10. (a) (2 + k)k! (b) (− r) r! (c) (n + 2) n! (d) (k2 + k + 1) (k − 1)! (e) (p2 + p − 1) (p − 1)! (f) (t2 + 2t − 1) (t − 1)!

1 11. (a) k 12. (a)

14.

7.

a =3

n = 11

8.

1 (e) k(k + 1) (k + 2)

1 (b) k(k + 1) (c) n – 1 (d) m+1

2 5!3!

13. (a) 1

y =9

6.

9 4!5!

(b)

(c)

(d)

3 (d) 7

1 (c) 2 4

(b) 1

6 2!4!

72 5!4!

(e)

3 (f) 4

7 (e) 8

90 7!3! 1 (g) 2

n−k+1 k

(c) (i) (1 + x) 3 (ii) 1 + 3x + 3x2 + x3 (d) (i) (1 + x) 6 (ii) 1 + 6x + 15x2 + 20x3 + 15x4 + 6x5 + x6 (e) (i) (1 + x) 8 (ii) 1 + 8x + 28x2 + 56x3 + 70x4 + 56x5 + 28x6 + 8x7 + x8 4.

(a) 35

(b) 126

(c) 15

(d) 70

5.

(a) 84

(b) 10

(c) 165

(d) 120

6.

(a) 3

7.

(a) 252

8.

(a) 5

9.

(a) 28x2 (b) 3x2

(d)

(n − 1)! (n − 1)! n−1 n−1 + m+c m= k−1 k [n − 1 − (k − 1)]!(k − 1)! (n − 1 − k)!k! (n − 1)! (n − 1)! = + (n − k)! (k − 1)! (n − 1 − k)!k! (n − k) (n − 1)! k(n − 1)! = + k (n − k)! (k − 1)! (n − k) (n − 1 − k)!k! k (n − 1)! (n − k) (n − 1)! = + (n − k)!k! (n − k)!k! (k + n − k) (n − 1)! = (n − k) !k! n(n − 1)! = (n − k)!k! n! = (n − k)!k! n =c m k

(a) (i) (1 + x) 3 = 1 + 3x + 3x2 + x3 (iv) (1 + x) = 3

3

/ k=0

3

Ck x

(ii) 8

(b) (i) (1 + x) = 1 + 4x + 6x2 + 4x3 + x4 (iv) (1 + x) 4 =

(c) 70

(ii) 16 (iii) 5

4

/ 4 Ck xk

k=0

(d) (i) (1 + x) 6 = 1 + 6x + 15x2 + 20x3 + 15x4 + 6x5 + x6 (ii) 64 (iii) 7

(iv) (1 + x) 6 =

(a)

Ck – 1 x

k –1

(d) 126

(e) 21

(e) 330

(c) 6x2 (d) 21x2 (e) 15x2 (b) 2n Ck – 1 xk − 1 (e)

/ c 12 m 612 − k (5x) k k

3n − 1

Ck – 1 x

(c) n − 1 Ck – 1 xk – 1 k –1

24 (c) / c m (3y) 24 − k 2k k k=0 12 12 12 − k (e) / c m 6 (5x) k k k=0

/ c 18 m a18 − k (− b) k k 18

(b)

k=0

k=0

/ c 16 m x16 − k (− 2y) k k 16

24

(d)

k=0

2.

(a) a4 + 4a3 x + 6a2 x2 + 4ax3 + x4 (b) a6 + 6a5 x + 15a4 x2 + 20a3 x3 + 15a2 x4 + 6ax5 + x6 (c) a5 + 5a4 x + 10a3 x2 + 10a2 x3 + 5ax4 + x5 (d) 8a3 + 12a2 + 6a + 1 (e) x7 – 14x6 + 84x5 – 280x4 + 560x3 – 672x2 + 448x − 128 (f) 256x8 + 768x6 + 864x4 + 432x + 81 (g) 729 – 2916x + 4860x2 − 4320x3 + 2160x4 − 576x5 + 64x6 (h) 64a3 –240a2 b + 300ab2 – 125b3 (i) 32 + 240m + 720m2 + 1080m3 + 810m4 + 243m5 (j) 1 − 16x + 112x2 − 448x3 + 1120x4 − 1792x5 + 1792x6 − 1024x7 + 256x8

3.

(a) (i) (b) (i) (ii) (c) (i) (ii)

2

(d) (i) (ii) (e) (i) (ii)

6

/ 6 Ck xk

k=0

(e) (i) (1 + x) = 1 + 5x + 10x + 10x + 5x + x 5 5 (ii) 32 (iii) 6 (iv) (1 + x) 5 = / c m xk k=0 k 5

2n + 1

(d) 3003

12

1.

(iii) 4

(c) (i) (1 + x) 7 = 1 + 7x + 21x2 + 35x3 + 35x4 + 21x5 + 7x6 + x7 7 7 (ii) 128 (iii) 8 (iv) (1 + x) 7 = / c m xk k=0 k

3.

(b) 35

(c) 56

(e) 91

k

4

2.

(b) 792

(d) 21

(e) 4

Exercises 10.3

Exercises 10.2 1.

(c) 45

10. (a) n + 1 Ck – 1 xk − 1

n n! 15. c m = k (n − k)!k!

c

(b) 15

(e) 5

3

(a) (i) 225 (ii) 26 (b) (i) 234 (ii) 35 (d) (i) 263 (ii) 64 (e) (i) 240 (ii) 41

4

5

(f) (i) (g) (i) (ii) (h) (i) (ii)

(c) (i) 217 (ii) 18

(a) (i) (1 + x) 4 (ii) 1 + 4x + 6x2 + 4x3 + x4 (b) (i) (1 + x) 7 (ii) 1 + 7x + 21x2 + 35x3 + 35x4 + 21x5 + 7x6 + x7

(i) (i) (ii) (j) (i) (ii)

(a + x) 3 (ii) a3 + 3a2 x + 3ax2 + x3 (3x − 2) 5 243x5 – 810x4 + 1080x3 – 720x2 + 240x – 32 (x + 2y) 6 x6 + 12x5 y + 60x4 y2 + 160x3 y3 + 240x2 y4 + 192xy5 + 64y6 (2a + 5b) 4 16a4 + 160a3 b + 600a2 b2 + 1000ab3 + 625b4 (x + y) 7 x7 + 7x6 y + 21x5 y2 + 35x4 y3 + 35x3 y4 + 21x2 y5 + 7xy6 + y7 (4p − 3q) 3 (ii) 64p3 – 144p2 q + 108pq2 – 27q3 (3 + 2n) 5 243 + 810n + 1080n2 + 720n3 + 240n4 + 32n5 (2a − b) 6 64a6 − 192a5 b + 240a4 b2 − 160a3 b3 + 60a2 b4 − 12ab5 + b6 (3ab − 4c) 4 81a4 b4 − 432a3 b3 c + 864a2 b2 c2 − 768abc3 + 256c4 (2 − x) 7 128 − 448x + 672x2 − 560x3 + 280x4 − 84x5 + 14x6 − x7

ANSWERS

4.

5.

(a) (a + 7) 10 (b) (6 − y) 9 (c) (3a + 4b) 7 (d) (x2 + 3) 8 (e) (5p − q) 11 (f) (4x − 9) n (g) (3a2 + b) n (h) (3x − 2) n + 1 (i) (a2 + 6b) n − 1 (j) (8x2 + 7y) 2n (a) 41 + 29 2 (b) 208 − 120 3 (c) 70 2 − 99 (d) 124 + 32 15 (e) 89 3 − 109 2 27x2 3x3 x4 (f) 81 + 54x + + + 2 2 16 3x 3x 2 x 3 10 5 1 5 3 (g) x + 5x + 10x + + − + 3 + 5 (h) 1 − x 4 2 8 x x (i) x6 + 12x4 + 60x2 + 160 + 3

(j)

2

2

240 192 64 + 4 + 6 x x2 x

(a) 303 (e) 10

8.

(a) − 1280

6.

(d) − 2268 3 4

(b) 2

a = 45, b = 29

10 777 536

8.

a = 29 656, b = − 20 880 9.

15. n = 19

a = 161, b = − 72 x = − 76, y = 5808

11. a = 11, b = 9

16 27

(b) 13 608

(e) 12 (c) 279 936

9.

7654

(b) 216

1 2

10. (a) − 4

(c) 1120

(b) 8 660 520

12. (a) 0.9703 (b) 0.9606 (c) 0.9510 (d) 0.9415 (e) 0.9321

7−k k

3 8

(b) 2

(d) 489 888

13. − 945

3 16

(e) 210

14. 112 640

16. a = 2, b = − 1 or a = − 2, b = 1

17. a = 2, b = − 3 or a = − 2, b = 3

19. (a)

(d) 3 784 704

1 2

12. (a) 175 000

6.

10. p = 452, q = 165 888

7.

112 729 (c) 720

11. (a) 65 536

3

a a b ab b − + − 4 6 27 8 7.

5.

18. (a)

9−k k

(b) 70

(b) 240

14. (a) 1.0303 (b) 1.0510 (c) 1.0406 (d) 1.0721 (e) 1.0615

20. (a) 78 732 (b) 15 360 (c) 5760 (d) 1792 (e) 1 082 565 (f) 240 (g) 241 920 (h) 10 450 944 (i) 15 C11 24 511 (j) 12 C8 44 78

15. (a) 1215 (b) 40 (c) 65 625 (d) 314 928 (e) 103 680

21. (a) 672

13. (a) 0.9604 (b) 0.9039 (c) 0.9412 (d) 0.9224 (e) 0.8858

(b) ±80

(c) 78 732

(d) 1792

(e) −11 547 360

16. (a) 1120 (b) 280 (c) 8960 (d) 326 592 (e) 1215

Exercises 10.5 10 14 15 10 (b) − c m 55 45 (c) c m 29 35 m6 5 5 5 9 4 5 20 15 5 (d) − c m 3 5 (e) − c m 7 5 5 5

17. (a) c

1.

5 5 (b) c m + c m = 20 3 2 2 8 2 8 2 8 (c) c m c m + c m c m + c m c m = 120 0 3 1 2 2 1 4 7 4 7 4 7 4 7 (d) c m c m + c m c m + c m c m + c m c m = 165 0 3 1 2 2 1 3 0

12 6 8 18. (a) c m x8 − k 5k (b) c m (2a) 12 − k 3k (c) c m(5a) 6 − k (− b) k k k k 15 21 15 − k k 21 − k (d) c m (4x) (− y) (e) c m (3a) (− 2b) k k k 9 5 19. (a) c m 210 − k xk − 1 (b) c m (5x) 6 − k (2y) k − 1 k−1 k−1 8 13 (c) c m 39 − k (− 2d) k − 1 (d) c m m14 − k (− 6n) k − 1 k−1 k−1 20 (e) c m (3a) 21 − k (− 2b) k − 1 k−1 6 9 m 38 − k xk − 2 (b) c m a11 − k 2k − 2 k−2 k−2 6 18 (c) c m 58 − k (3a) k − 2 (d) c m 320 − k (− 4x) k − 2 k−2 k−2 8 (e) c m (3x2) 10 − k (− 2y) k − 2 k−2

2.

3.

4.

22 680y8 x3 (a)

2.

4. − 448x12

3.

21 879 32

243 405 + + 270x3 + 90x7 + 15x11 + x15 x x5

(b) 405

Coefficient of x4 from (1 + x)10 is c

10 m. 4

9 9 Coefficient of x4 from (1 + x) (1 + x) 9 is c m + c m . 4 3 10 9 9 ` c m = c m+ c m 4 4 3

21. 1 + 5 sin x + 10 sin2 x + 10 sin3 x + 4 sin4 x + sin5 x 22. 34

1.

6 From (1 + x)6 , the coefficient of x2 is c m . 2 From (1 + x)3 (1 + x)3 , the coefficient of x2 is 3 3 3 3 3 3 c mc m + c mc m + c mc m. 0 2 1 1 2 0 6 3 3 3 3 3 3 ` c m = c mc m + c mc m + c mc m 2 0 2 1 1 2 0 3 3 3 3 = 2 c mc m + c mc m 0 2 1 1

20. (a) c

Exercises 10.4

(a) 2 040 714

n+1 m. k k n Coefficient of x from (1 + x) (1 + x) is

Coefficient of xk from (1 + x) n + 1 is c

n n c m+c m. k−1 k n n+1 n `c m=c m+ c m k−1 k k

647

648

Maths In Focus Mathematics Extension 1 HSC Course

5.

(a) (2 + x) (1 + x)5

8.

(a) (1 + x)n =

(b) 30

6.

92

7.

n =9

n

/ n Ck xk

k=0

Substitute x = 6 (1 + 6)n = 7 = n

n

/ n Ck 6k

k=0 n

/ n Ck 6k

k=0

n

/ n Ck xk

(b) (1 + x)n =

k=0

Substitute x = 2 (1 + 2)n = 3n =

n

/ n Ck 2k

k=0 n

/ n Ck 2k

k=0

(c) (1 + x)n =

n

/ n Ck xk

k =0

Substitute x = 3 (1 + 3)n = 4n = (22)n = 22 n =

9.

(1 + x)n =

n

/ n Ck 3k

k =0 n

/ n Ck 3k

k =0 n

/ n Ck 3k

k =0 n

/ n Ck 3k

k =0

n

/ n Ck xk

k =0

n n n n n = c m + c m x + c m x2 + c m x3 + . . . + c m xn 0 1 2 3 n Differentiating: n n n n n(1 + x)n − 1 = c m + c m 2x + c m 3x2 + . . . + c m nxn − 1 1 2 3 n Substitute x = 1 n n n n(1 + 1)n − 1 = c m + c m 2 (1) + c m 3 (1) 2 1 2 3 n + . . . + c m n (1)n − 1 n n n n n = c m + 2c m + 3c m + . . . + n c m 1 2 3 n n n n2 n − 1 = / k c m k k =1 n n n 10. (a) (a + x)n = c m an + c m an − 1x + c m an − 2 x2 0 1 2 n n−3 3 n n + c ma x + . . . + c mx n 3 Substitute x = 0 n n n (a + 0)n = c m an + c m an − 1 (0) + c m an − 2 (0) 2 0 1 2 n n−3 3 n + c m a (0) + . . . + c m (0)n 3 n n an = c m an 0 n 1=c m 0 n n n (b) (a + x)n = c m an + c m an − 1 x + c m an − 2 x2 0 1 2 n n−3 3 n + c m a x + . . . + c m xn n 3

Substitute x = 1 n n n (a + 1)n = c m an + c m an − 1 (1) + c m an − 2 (1) 2 0 1 2 n n−3 3 n + c m a (1) + . . . + c m (1) n n 3 n n n n −1 n n − 2 = c ma + c ma + c ma 0 1 2 n n−3 n + c ma + ... + c m n 3 n n n−k = / c ma k=0 k n n n (c) (a + x)n = c m an + c m an − 1x + c m an − 2x2 0 1 2 n n−3 3 n n + c ma x + . . . + c mx 3 n Substitute x = − 1 n n n (a − 1)n = c m an + c m an − 1 (−1) + c m an − 2 (−1)2 0 1 2 n n−3 n 3 + c m a (−1) + . . . + c m (−1)n n 3 n n n−k k = / (−1) c m a k k=0 n n n (d) (a + x)n = c m an + c m an − 1 x + c m an − 2 x2 0 1 2 n n−3 3 n + c m a x + . . . + c m xn n 3 Differentiating: n n n (a + x)n − 1 = c m an − 1 + c m an − 2 2x 1 2 n n−3 2 n + c m a 3x + . . . + c m nxn − 1 n 3 n n −1 n n−2 = c ma + 2 c ma x 1 2 n n−3 2 n + 3 c m a x + . . . + n c m xn − 1 n 3 n n n − k k −1 = / k c ma x k k =1 (e) From (d): n n n(a + 1)n − 1 = c m an − 1 + 2 c m an − 2 x 1 2 n n −3 2 n + 3 c m a x + . . . + n c m xn − 1 3 n Substitute x = 0 n n n n(a + 0)n − 1 = c m an − 1 + 2 c m an − 2 (0) + 3 c m an − 3 (0) 2 1 2 3 n + . . . + n c m (0) n − 1 n n n −1 n−1 na = c ma 1 n n =c m 1 n n n n n 11. (1 + x)n = c m + c m x + c m x2 + c m x3 + . . . + c m xn 0 1 2 3 n Substitute x = − 2 n n n n (1 − 2)n = c m + c m (− 2) + c m (− 2) 2 + c m (− 2) 3 + . . . 0 1 2 3 n + c m (− 2) n n n n n n n n (−1) = c m − 2 c m + 4 c m − 8 c m + . . . + (− 2) n c m n 0 1 2 3

ANSWERS

12. (a) Let x = 1. n n n n Then (1 + 1)n = c m + c m 1 + c m 12 + . . . + c m 1n . 0 1 2 n n n n n 2n = c m + c m + c m + . . . + c m 0 1 2 n (b) Let x = −1 n n n Then (1 − 1)n = c m + c m (−1) + c m (−1) 2 0 1 2 n n + . . . + c m (−1) n n n n n 0 = c m − c m + c m − . . . + (−1)n c m n 0 1 2

n n x2 n x3 15. c m x + c m +c m 0 1 2 2 3 (1 + x) n + 1 − 1 n xn + 1 + ... + c m = n n+1 n+1 Let x =1 n n 12 n 13 c m1 + c m + c m + . . . 0 1 2 2 3 (1 + 1) n + 1 − 1 n 1n + 1 +c m = n n +1 n +1 n 1 n 1 n 1 n c m + c m + c m+ ... + c m 0 2 1 3 2 n +1 n 2n + 1 − 1 = n +1

x =1

13. (a) Let

(1 + 1) 2n = 2 2n =

/ c 2kn m 1k 2n

k=0

/ c 2kn m 2n

Test yourself 10

k=0

2n 4 = /c m k=0 k 2n 2n 2n (b) (1 + x)2n = c m + c m x + c m x2 0 1 2 2 n 2n + ... + c m x 2n Differentiating: 2n 2n 2n 2n(1 + x) 2n − 1 = c m + 2 c m x + . . . + 2n c m x2n − 1 1 2 2n Let x =1 2n 2n 2n 2n(1 + 1) 2n − 1 = c m + 2 c m 1 + . . . + 2n c m 12n − 1 1 2 2n 2n 2n 2n − 1 ) = / kc m 2n(2 k k =1 2n

n

r =0

2.

4.

k

= n4n 5. n # (1 + x)n dx = # / c r m xr dx n

14.

r =0

8.

(1 + x) n + 1

n n xr + 1 + C1 = / c m + C2 n+1 r+1 r=0 r n +1 n (1 + x) n xr + 1 ` =/c m +C n+1 r+1 r=0 r x = 0: Let n n 0r + 1 1 = /c m +C n + 1 r =0 r r + 1 1 ` =C n +1 n (1 + x) n + 1 n xr + 1 1 ` = /c m + n +1 n +1 r =0 r r + 1 n r +1 (1 + x) n + 1 1 ` − / c nr m x = r +1 n +1 n +1 r =0 (1 + x) n + 1 − 1 = n+1 Let x = −1 n (−1)r + 1 (1 − 1)n + 1 − 1 = / c nr m r +1 n +1 r =0 −1 = n +1 r n n (−1) 1 ` /c m = r +1 n +1 r =0 r

Differentiating: LHS = n (1 + x) n − 1 n n n n RHS = c m + c m 2x + c m 3x2 + . . . + c m nxn − 1 1 2 3 n Let x = 1: LHS = n (1 + 1) n − 1 = n 2n − 1 n n n n n n RHS = c m + c m 2 + c m 3 + . . . + c m n = / r c m 1 2 3 n r r =1

3. − 6048

/ k a 2 n k = n . 2 2n

k =1

r

3 / c 12 m x12 − r c m x r 12

1.

2n

`

(from question 14)

8 Coefficient of x3 in (1 + x)8 = c m 3 Coefficient of x3 in (1 + x)3 (1 + x)5 3 5 3 5 3 5 3 5 = c mc m + c mc m + c mc m + c mc m 3 0 2 1 1 2 0 3 (a) Let x = 0

6.

924

7. 152

n Coefficient of xk in (1 + x)n = c m k Coefficient of xk in (1 + x) (1 + x) n − 1

=c 9.

(b) Let x = − 1

n−1 n−1 m+c m k−1 k

32x5 + 240x4 y + 720x3 y2 + 1080x2 y3 + 810xy4 + 243y5

4 5 4 5 4 5 4 5 10. c m c m + c m c m + c m c m + c m c m = 84 3 0 2 1 1 2 0 3 11. 960 740 352

12. 15 360

13. a = 3, b = − 4

10 − k + 1 k (b) 295 245

14. (a) =

15. n = 8

16. 105

17. 489 888

18. (a) 243 + 810p + 1080p2 + 720p3 + 240p4 + 32p5 (b) 2889 − 1292 5 54 108 81 (c) x4 − 12x + 2 − 5 + 8 x x x (d) 8a3 − 12a2 b + 6ab2 − b3 (e) 184 592 − 130 728 2

649

650

Maths In Focus Mathematics Extension 1 HSC Course

19. a = 220, b = − 284 20. (a) (a + x) n =

an + 1 + C3 = 0 n +1 an + 1 n +1 (a + x) n + 1 n n an + 1 x2 So − = c m an x + c m an − 1 0 1 n +1 n +1 2 n n − 2 x3 n xn + 1 + c ma + ... + c m 2 n n+1 3 (a + x) n + 1 − an + 1 n n n an − 1 x2 = c ma x + c m 0 1 n +1 2 n an − 2 x3 n xn + 1 +c m + ... + c m 2 n n+1 3 n n an − k xk + 1 = /c m n+1 k =0 k

/ c n0 m an − k xk n

C3 = −

k=0

Substitute x = 2 n n (a + 2)n = / c m an − k 2k k=0 0 n n n (b) (a + x) n = c m an + c m an − 1 x + c m an − 2 x2 0 1 2 n n−3 3 n n + c ma x + . . . + c mx n 3 Differentiating: n n n n(a + x) n − 1 = c m an − 1 + c m an − 2 2x + c m an − 3 3x2 1 2 3 n n−1 + . . . + c m nx 4 n n−1 n n = c ma + 2 c m an − 2 x + 3 c m an − 3 x2 1 2 3 n n−1 + . . . + n c mx n Substitute x = 1 n n n n(a + 1) n − 1 = c m an − 1 + 2 c m an − 2 (1) + 3 c m an − 3 (1) 2 1 2 3 n n−1 + . . . + n c m (1) n n n−1 n n = c ma + 2 c m an − 2 + 3 c m an − 3 1 2 3 n + ... + nc m n n n = / k c m an − k k k =1 n n n (c) (a + x) n = c m an + c m an − 1 x + c m an − 2 x2 0 1 2 n n−3 3 n n + c ma x + . . . + c mx n 3 Integrating both sides:

Challenge exercise 10 1.

10 10 10 10 10 10 mc m + c mc m + c mc m 0 4 1 3 2 2 10 10 10 10 + c mc m + c mc m 3 1 4 0 20 10 10 10 10 10 10 ` c m = c mc m + c mc m + c mc m 4 0 4 1 3 2 2 10 10 10 10 + c mc m + c mc m 3 1 4 0 10 10 10 10 10 2 = 2 = c mc m + c mc mG + c m 0 4 1 3 2

c

2.

1 n m x

1 n mF x 1 n = < (x + x2) c 1 + m F x = (x + 1 + x2 + x)n = (x2 + 2x + 1)n

(a + x)n + 1

+ C1 n +1 n n RHS = # c m an + c m an − 1 x 0 1 n n + c m an − 2 x2 + . . . + c m xn dx n 2 n n n n − 1 x2 = c ma x + c ma 0 1 2 n n − 2 x3 n xn + 1 + c ma + ... + c m + C2 n n +1 2 3 (a + x) n + 1 n n x2 So + C1 = c m an x + c m an − 1 0 1 n +1 2 n n − 2 x3 n xn + 1 + c ma + ... + c m + C2 n n +1 2 3 n +1 (a + x) n n x2 + C3 = c m an x + c m an − 1 0 1 n +1 2 n n − 2 x3 n xn + 1 + c ma + ... + c m n n +1 2 3 Substitute x = 0: (a + 0) n + 1 n n 02 + C3 = c m an 0 + c m an − 1 0 1 n +1 2 n n − 2 03 n 0n + 1 + c ma + ... + c m n n +1 2 3

(a) LHS = xn (1 + x) n c 1 + = < x(1 + x) c 1 +

LHS = # (a + x)n dx =

20 m. 4 Coefficient of x4 from (1 + x)10 (1 + x)10 is

Coefficient of x4 from (1 + x)20 is c

= 6 (x + 1)2 @ n = (x + 1)2n = RHS

2n m. n n 1 Coefficient of xn in xn (1 + x) n c 1 + m x comes from the terms independent of x in. 1 n (1 + x)n c 1 + m . x n n n n n n n n i.e. c m c m + c m c m + c m c m + . . . + c m c m n n 0 0 1 1 2 2 n 2 n 2 n 2 n 2 2n ` c m = c m + c m + c m + ... + c m n n 0 1 2

(b) Coefficient of xn in (1 + x) 2n is c

3.

Coefficient of xn in (1 + x) n (1 − x) n is n n n n n n c m (−1)n c m + c m (−1)n − 1 c m + . . . + c m (−1) 0 c m . n −1 0 n 1 n 0 2 2 2 n n n i.e. (−1)n c m + (−1) n − 1 c m + . . . + (−1)0 c m n−1 n 0 n n since c m = c m n−k k

ANSWERS

2

/ (−1) k c nk m n

i.e. (1 − x2) n =

6.

k=0 n

k=0

7.

Ck + 1 i.e. Ck + 1 Ck

15 15 − k k 2 m3 k is the coefficient of xk + 1

c

c

=

=

= =

15 m 315 − (k + 1) 2k + 1 k +1 15 c m 315 − (k + 1) 2k + 1 k +1 15 15 − k k 2 m3 k 15! ´2 6 15 − (k + 1)]!(k + 1)!

c

15! ´3 (15 − k)!k! 15! ´ 2 ´ (15 − k)!k!

2n

2n

0 r =0

( 1 + x) n + 1

S(n) = n Ck =

m xk dx

k=0

(1 + x)2n + 1

=

2n xk + 1 = /c m +C k k +1 2n + 1 k=0 Let x = 0: 2n 2n 0k + 1 1 = /c m +C 2n + 1 k = 0 k k + 1 1 =C 2n + 1 2n

n(n − 1) (n − 2) . . . (n − k + 1) 1 ´ 2 ´ 3 ´ ... ´ k

for 1 ≤ k ≤ n

(n − 1) (n − 2) … (n − k + 1) +

1 ´ 2 ´ 3 ´ … ´ (k − 1) (n − 1) (n − 2) . . . (n − k)

1 ´ 2 ´ 3 ´ ... ´ k (n − 1) (n − 2) . . . (n − k + 1)

=1 + 1 ´ 2 ´ 3 ´ . . . ´ (k − 1) (n − 1) (n − 2) . . . (n − k + 1) n = ; E k 1 ´ 2 ´ 3 ´ . . . ´ (k − 1) n(n − 1) (n − 2) . . . (n − k + 1) = 1 ´ 2 ´ 3 ´ . . . ´ (k − 1) ´ k

=

(1 + x)2n + 1

2n 2n xk + 1 1 = /c m + k k + 1 2n + 1 2n + 1 k=0 2n (1 + x ) 2 n + 1 2n xk + 1 1 ` /c m = − k k +1 2n + 1 2n + 1 k=0 (1 + x ) 2 n + 1 − 1 = 2n + 1 x = 2: Let 2n k+1 (1 + 2) 2n + 1 − 1 / c 2kn m 2 = 2n + 1 k +1 k=0 3 2n + 1 − 1 = 2n + 1

`

2

Step 3: Prove S(n) is true. n C1 = n − 1 C0 + n − 1 C1 n−1 =1+ 1 n = 1 ∴ S(n) is true for k = 1 n Ck = n − 1 Ck − 1 + n − 1 Ck

6 15 − (k + 1)]!(k + 1)! ´ 15! ´ 3 2(15 − k)

# (1 + x)2n dx = # / c k

/ c nr m xr dx

2 n

Step 2: Assume S (n − 1) is true. (n − 1) (n − 2) (n − 3) . . . [(n − 1) − k + 1] S (n − 1): n − 1 Ck = 1 ´ 2 ´ 3 ´ ... ´ k (n − 1) (n − 2) (n − 3) . . . (n − k) = 1 ´ 2 ´ 3 ´ ... ´ k n−1 If k = 0: C0 = 1 (by definition) n −1 If k = 1: n − 1 C1 = 1

3(k + 1) 15 (b) C5 = c m 310 25 5 5.

(1 + x) n dx = #

Step 1: Check that S(2) is true by inspection of (1 + x)2 . (1 + x) 2 = 1 + 2x + x2 = 2 C0 + 2 C1 x + 2 C2 x2 2.1 2 2 C0 = 1, 2 C1 = , 2 C2 = =1 1 1.2 ∴ S(n) is true for n = 2.

(a) Ck is the coefficient of xk i.e.

2

n n xr + 1 2 G = =/ c m G n + 1 0 r =0 r r + 1 0 n n n 2r + 1 n 0r + 1 3n + 1 1n + 1 − = /c m − /c m n + 1 n + 1 r =0 r r + 1 r =0 r r + 1 n n 2r + 1 3n + 1 − 1 = /c m n +1 r =0 r r + 1

=

/ (− x2) k

For the coefficient of xn (− x2) k = xn i.e. 2k = n n k= 2 If n is odd, k is not an integer. ` there is no coefficient of xn if n is odd n n 2 ` / (−1)k c m = 0 k k=0 4.

#0

n−k G k

∴ S(n) is proved for all k ≥ 2, and has been proved for k = 1 above.

Chapter 11: Probability

Exercises 11.1 1.

1 30

6.

(a)

2. 4 7

(b)

1 52 3 7

3.

7.

1 6 3 37

4.

8.

1 40 1 12

5.

9.

1 20 000 (a)

11 20

(b)

3 4

651

652

Maths In Focus Mathematics Extension 1 HSC Course

10. (a)

1 6

(b)

1 2

Exercises 11.4

1 3

(c)

1 11. (a) 62

3 (b) 31

1 (c) 2

8 12. (a) 15

7 (b) 15

1 (c) 15

16. (a) 20. 34

7 31

7 31

(b)

21.

1 3

(c)

22. (a)

99 (d) 124

12 31 1 6

23 1 14. , 1 15. 44 2

1 13. 50 1 175

17. 1 3

(b)

18. 8

19.

25 43

5 6

(c)

23. (a) False: outcomes are not equally likely. Each horse and rider has different skills. (b) False: outcomes are not equally likely. Each golfer has different skills. (c) False: outcomes are not dependent on the one before. Each time the coin is tossed, the probability is the same. (d) False: outcomes are not dependent on the one before. Each birth has the same probability of producing a girl or boy. (e) False: outcomes are not equally likely. Each car and driver has different skills.

1.

1 36

6.

(a) 0.0441

9.

(a)

1 4

2.

9 49

15 91

(b)

1.

5 11

14. (a)

1 2400

(b)

1 5 760 000

15. (a)

1 7776

(b)

3125 7776

16. (a)

9 25 000 000

(b)

24 970 009 25 000 000

17. (a)

1 4

9 100

(c)

9 100

18. (a)

1 22

1 11

(c)

7 22

(b)

(b)

6.

98.5%

7 11. 8

3. 22 23

7.

46 12. 49

99.8% 8.

2 13. (a) 15

4.

5 18

0.73

9.

13 (b) 15

5.

0.21

3 10

(b)

3 5

(c)

11 20

15 15. 16

2.

1 (a) 5

3.

(a)

5 26

(b)

9 26

5.

(a)

27 45

(b)

4 9

7.

(a)

21 80

(b)

17 80

(c)

21 40

8.

9.

(a)

7 25

(b)

2 15

(c)

44 75

10. (a)

3 (c) 5

1 (b) 2

3 (d) 5

(c)

(c)

(d)

7 10

(a)

12 13

2 3

19 (e) 50 4.

6.

(a)

1.

(a)

1 4

3.

(a)

5.

(a)

7.

(a) 0.42

(a)

(c) 29 100

3 14

(a)

(b)

(b) 1 10

3 10

13 28

(b)

(b)

2 5

13 20

(c)

(c) 11 20 (c)

4651 7776

(c)

(d)

(c) 1 −

(c)

29 991 25 000 000

15 22

(c) 99.66% 1 2n − 1 = 2n 2n

Exercises 11.5

Exercises 11.3 1.

1 2n

(b)

5 755 201 5 760 000

(c)

38%

10. 91.9%

7 14. 11

1 2n

(c)

(b) 0.34%

1 16 170

12.

35 929 35 937

29 791 35 937

20. (a)

2 9

2.

8 35 937

8. 32.9%

19 99

11.

13. (a)

(b)

25 121

5.

7. 80.4%

3 2075

10.

1 4

4.

(b) 0.6241

19. (a) 61.41%

Exercises 11.2

1 8

3.

9 25

9 28

(c) 3 10

7 20

1 4

(c)

1 900

(b)

1 900

25 169

(b)

80 169

(b)

1 2 (c)

(a)

1 450

1 8

(b)

3 8

4.

(a)

1 25

(c)

(c) 0.49

9.

(a) 0.325

6 11

11. (a)

8.

(a)

(b) 0.0034

2 25

12. (a)

1 25

13. (a)

19 1 249 750

14. (a)

16 75

(b)

38 75

15. (a)

1936 2025

(b)

88 2025

16. (a)

11 20

(b)

3 20

17. (a)

1 1296

(b)

125 324

18. (a)

84 681 1 000 000

1 825

(c)

(b)

(b)

4 27

64 825

(b)

(d)

498 124 975

912 673 1 000 000

441 1000

1 6

152 165

(c)

(b)

(c) 0.997

60 121

(b)

(b)

189 1000

10. (a)

(b)

7 8

6. (a) 27.5% (b) 23.9% (c) 72.5%

(b) 0.09

657 1000

2.

(e)

13 165

1 239 771 1 249 750

(c)

(c)

27 1 000 000

671 1296

ANSWERS

Exercises 11.8

19. (a) 17.6% (b) 11% (c) 21.2% 20. (a)

1488 3025

1 121

(b)

368 (b) 425

22 22. (a) 425

21. (a) 7 (c) 425

1 19

(b)

6 95

(c)

21 190

133 (b) 715

17 23. (a) 65

(d)

17 38

496 (c) 2145

1.

190

2. 56

3.

4.

(a) 1365

(b) 364

6.

(a) 2300

(b)

77 100

9.

(a) 8 145 060

(b)

10.

10 2 760 681

(c) 78

1 216

25 28. (a) 81

5 72

(b)

(c)

40 (b) 81

91 216

27. (a)

1 10

(b)

3 10

(c)

2 5

56 (c) 81

13. (a) 126

343 1331

(b)

336 1331

(c)

988 1331

15. (a) 4368

30. (a)

1 8000

(b)

6859 8000

(c)

1141 8000

16. (a) 350 658 000

(b) 1764

(b)

1 4

(c)

10 000

2.

2 600 000

4.

10 000

5.

10

8.

1 757 600

11. 1000 1 14. 280

9.

6. 1320

12. 90 000 000

3.

248 832 10.

20. (a) 5 852 925

387 420 489 7.

1 523 566

1 1 000 000

2.

(a) 90 (b) 151 200 (c) 30 240 (d) 720 (e) 1 814 400

3.

(a) 120 (b) 60 (c) 40

5.

(a) 64 (b) 16

8.

60

9.

(b)

15 128

(c)

45 1024

2.

(a)

7 128

(b)

7 128

(c)

29 128

3.

(a)

109 375 419 904

(e)

1 015 625 1 679 616

4.

(a)

25 216

5.

(a) 0.254

6.

(a)

7.

(a) 90.1%

(b) 0.4%

(c) 99.6%

8.

(a) 11.4%

(b) 16.4%

(c) 1.2%

9.

(a)

69 984 5 764 801

(b)

(e)

65 536 5 764 801

10.

20

12 25

16 25

(b) (i) 0.25

7. 24 10. (a) 24

11. (a) 40 320 (b) 1152 (c) 10 080 (d) 2880 (e)

14. (a) 120 (b)

3 5

(ii)

4 35

(b) 6 3 28

13. (a) 3 628 500 (b) 362 880

(c) 36

2 15. (a) 362 880 (b) 40 320 (c) 20 160 16. 19 17. (a) 120 (b) 24 (c) 20 160 (d) 2520 (e) 1260 (f) 20 (g) 907 200 (h) 604 800 (i) 277 200 (j) 9 979 200 18. (a) 720

(b) 240 (c) 144 (d) 72 (e) 600 (f) 480

19. (a) 362 880

(c)

12 35

(b)

(b)

875 209 952

5 324

(c)

(d)

(c)

125 324

5 512

(e)

175 419 904

(d)

425 432

11 1024

(d)

(e)

41 1 679 616

7 432

4. (a) 3024 (b) 1680 (c) 672

6. 990

2 7

(b) 6

2 45

105 512

(e) 20 160

(a) 5040 (b) 720

12. (a) 5040 (b) (i)

(c)

(a)

Exercises 11.7 (d) 60

(b) 3 837 240

(b) 140

1.

15. 84

(a) 360 (b) 5040 (c) 9

12. 100 100

Exercises 11.9

1 13. (a) 7776 (b) 7776

1.

84 672 315

5 407 253

17. (a) 10

19. (a) 350 (b) 66

15 49

2 21

Exercises 11.6 1.

(b)

8.

1 86 493 225

12 175

(b)

(a) 735

(c)

(b) 45

14. (a)

29. (a)

18. (a) 560

1 2 036 265

11. (a) 84

(b) 40

5. 1 38 760

7.

24. (a) 0.23 (b) 0.42 (c) 0.995 25. (a) 33% (b) 94% 26. (a)

15

(b) 40 320 20. 19 958 400

144 625

11. (a) (i)

12. (a) 0.23 13.

20

C3 d

(b) 0.01 (b)

(c) 0.0467

243 3125

(ii)

(b) 0.07

(c)

48 625

46 656 823 543 C7 d

(d) 0.198

(d)

(c)

2133 992 (e) 3125 3125

57 591 823 543

3 7 8 13 n d n 11 11

(c) 0.03

3 2 8 6 1 3 5 17 8 n d n n d n ´ C2 d 6 6 11 11

(e) 0.552

(ii) 0.167

(d)

147 456 823 543

653

654

Maths In Focus Mathematics Extension 1 HSC Course

14.

12

C3 d

3 3 1 9 8 5 5 1 3 n d n ´ C5 d n d n 4 4 6 6

16. 7 people

17. (a) 8 sixes

50

(b)

C8 d

18. (a) 25 times (b) 0.0918 19. (a) (i) (b)

101 144

53 112

(c) 0.113 20. (a)

24. (a) (i)

15. 0.000 003 4 1 8 5 42 n d n 6 6 35 72

(ii)

25. (a)

31 72

(iii)

37 72

(a) 80.4%

2.

(a)

728 729

21 50

(b)

3 25

(c)

1

2

3

4

5

6

1

0

1

2

3

4

5

2

1

0

1

2

3

4

3

2

1

0

1

2

3

4

3

2

1

0

1

2

5

4

3

2

1

0

1

6

5

4

3

2

1

0

1 7

4 7

1.

(a)

(b)

2.

3.

34 650

5.

(a)

6.

(a) 1 −

7.

(a) 0.18%

8.

(a) 39 916 800

4.

1 54 145

(ii)

5.

6.

7.

1 6

(iii)

1 2

4 (a) 15

4.

3.

(a) (i)

1 40

(ii)

39 40

(b) k Ck − 3

(a)

1 2 2 5

(b)

(b)

29 100 7 15

(c)

(c)

1 5

11 25

(d)

2 15

8.

(a)

(b)

35 66

9.

(c) 0.25

1 2k

(c)

(c)

9 11

9.

(a) 24

(b) 12

1 10

3! (n − 3)! (n − 1)! 3 ´ 2 ´ 1 ´ (n − 3) (n − 4) . . .

(n − 1) (n − 2) (n − 3) (n − 4) . . . 6 = (n − 1) (n − 2) (d)

(b)

1 7

2 n−1

=

16 25

35 72

(c)

(c) 20 Ck ] 0.73 gk ] 0.27 g20 − k

(b) 113 400

(c) Probability =

(e)

(c) k C9

(b) 967 680

False: the events are independent and there is the 1 same chance next time d n 4 (a)

(b) 0.75

1 2k

(b) 10.7%

11. (a) ] n − 1 g !

1 (b) 10

(b) 1440

33 173 264

(b)

1 2k

(a) 6720

(a) 0.04

10. (a) 5 852 925 1 6 39 (b) 796

23 50

Challenge exercise 11

(b) 1.4% (c) 99.97%

(b) (i)

(b) 0.000 092

(b) 0.127

Test yourself 11 l.

(ii)

64 243

1 56 12. (a)

10. (a) 0.009% (b) 12.9% 11. (a) 60 (b) 75

k! (n − k)! (n − 1)! 4 13

(b)

k! ]n − 1g]n − 2g]n − 3gf ]n − k + 1g

=

25 52

(c)

4 13

13. No—any combination of numbers is equally likely to win. 1 12. (a) 13

14. (a)

3 (b) 13

12 6325

5 (c) 26

(b) 4320 15.

(a)

5 12

(b)

3 10

(ii)

27 160

(iii)

81 200

(c)

11 100

19. (a)

9 40

(b) (i)

18. (a)

1 200

(b)

20. 60 480

22. (a)

80 361

21. (a)

(b)

1 50

40 171

(b)

147 7450

23. (a)

2 9

14. (a) 0 (b)

3 10

1 10

(c)

1 1296

15. (a)

1 7776

(b)

16. (a)

3 10

12 145

(b)

(c)

17. (a)

125 3888 1 144

(b)

5 144

(c)

7 144

(d)

3 144

4 25

(c)

(b)

1 3

2 11

16. (a) 39 916 800 (b) 32 659 200 (c) 17. (a)

640 (b) 15 625

972 13. (a) 3125

1 3

1 15 1 3725

(b)

Practice assessment task set 4

4 5

(d)

1. 3577 3725

− 2016

2.

3360 125 3888

4.

− 448x10

7.

80 8. 660 660 (b) x Z 2.25

5.

3.

20 160

6. 45 697 600 9.

(a) f (2) < 0, f (3) > 0

ANSWERS

10.

1 3

11. (a) 13.3%

12. (a)

1296 2401

864 2401

(b)

(c)

37. f ] 0 g = − 3 < 0 f ]1 g = 1 > 0

(b) 14.1% 1105 2401

So the root lies between 0 and 1. 38.

13. (a)

1 36

(b)

1 6

14. (a)

7 50

(b)

11 20

(c)

11 36

5 36

(d)

(e)

5 12

3 764 768 56 953 125

15. (a)

(b)

3π 8

39. (a) θ = 76° 52l (b) 0.92 cm2

40. $18 399.86 170 859 374 170 859 375

45. (d)

41. (b) 46. (d)

49. (a), (c)

42. (a) 47. (c)

43. (d)

44. (b)

48. (a)

50. (d)

16. 1 400 000 17.

2 540y3 1215y4 1 18y 135y − 5 + − + 4 6 x x x x3 x2

Sample examination papers

5

−

1458y + 729y6 x

1 18. 110

Mathematics—Paper 1 1.

19. a − 6a b + 12ab − 8b 3

2

20. (a) f (0) > 0, f (1) < 0

2

5103 2

24. (a) 26. −

5 33

80 x4

(+ADB straight +) (+ in semicircle) (similarly)

35 66

25. (a)

27.

108 864 390 625

1 12

(b) 101 606 400

(b)

1 4

28. 81a4 − 216a3 b + 216a2 b2 − 96ab3 + 16b4 30.

1 40000

/ c 1k0 m xk 10

k=0

Substitute x = 2

2.

10 ] 1 + 2 g = / c m 2k k k=0 10 10 10 k 3 = /2 c m k k=0 10

10

34.

35. f − 1 (x) = 3 5 36

5 12

(c)

AB BC 1 = = AD DE 2 3.4 1 ` = DE 2 DE = 2 ´ 3.4 = 6.8 cm

(iii)

x−1 2

(b)

(a) (ii) Since AB = BD, AB: AD = 1:2 +ABC = +ADE (corresponding +s, BC < DE) (similarly) +ACB = +AED +A is common ` ∆ ABC 0 (minimum point) At (−1, 7), y m = 6(−1) − 6