Free Essay

In: Science

Submitted By huajui

Words 6354

Pages 26

Words 6354

Pages 26

Focus on Exam 9

1

∫

3 + sin x dx = cos2 x

∫cos

3

2

x

+ sin2 x dx cos x

= ∫[3 sec2 x + (cos x)−2 sin x] dx = ∫[3 sec2 x − (cos x)−2 (−sin x)] dx (cos x)−1 = 3 tan x − +c −1 = 3 tan x + 1 + c cos x 2 ∫sin2 x cos2 x dx = ∫(sin x cos x)2 dx = = = = 1 ∫2 sin 2x

2

dx

sin 2x = 2 sin x cos x 1 sin x cos x = sin 2x 2

∫ 4 sin

1 4

1

2

2x dx

∫

1 − cos 4x dx 2

cos 4x = 1 − 2 sin2 2x sin2 2x 1 − cos 4x = 2

∫ 1 1 = x − sin 4x + c 8 4

3

1 (1 − cos 4x) dx 8

∫

du x 2 dx = 1 − u2 1 − x4

∫

= = =

1 du 1 − u2 2

∫

u = x2 du = 2x dx du x dx = 2

1 1 1+u × ln +c 2 2 1−u

1 1 + x2 ln +c 1 − x2 4

∫ a − x = − 2a dx 1

2 2

ln

a − x + c

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

a+x

2

ACE AHEAD Mathematics (T) Second Term

4 Let x = sin θ dx = cos θ dθ dx = cos θ dθ

∫x

2

dx cos θ dθ = 2 2 1−x sin θ 1 − sin2 θ

∫

= =

∫ sin θ ∫ sin dθ 2

cos θ dθ 2 cos2 θ

θ q 1

x

= ∫cosec2 θ dθ = − cot θ + c 1 − x2 =− x +c 5 Let u = ln x du 1 = dx x dx = du x dx 1 dx ∫ x ln x = ∫ x ln x =

1 − x2

[Shown]

∫u

du

= ln |u| + c = ln |ln x| + c 6 d (tan3 x) = 3 tan2 x sec2 x dx = 3(sec2 x − 1) sec2 x = (3 sec2 x − 3) sec2 x = 3 sec4 x − 3 sec2 x

4 2 3

[Shown]

∫(3 sec x − 3 sec x) dx = tan x + c ∫3 sec4 x dx − ∫3 sec2 x dx = tan3 x + c ∫3 sec4 x dx = ∫3 sec2 x dx + tan3 x + c = 3 tan x + tan3 x + c 1 c ∫ sec4 x dx = tan x + tan3 x + 3 3 1 ∫sec4 x dx = tan x + tan3 x + c′ 3

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

3

7 ∫x sin x cos x dx = x =

1 ∫ 2 sin 2x dx 1

∫ 2 x sin 2x dx 1 1 1 1 = ∫ cos 2x x − ∫ − cos 2x dx 2 2 2 2

1 1 = − x cos 2x + cos 2x dx 4 4

∫

1 1 = − x cos 2x + sin 2x + c 8 4 8

∫

x 1 dx = 2 2 0 1 + x 1 = 2 1 = 2 1 = 2 1 = 2

1 2 0

∫

2x dx 2 0 1 + x

1

[ln |1 + x2|]1 0 [ln (1 + 12) − ln (1 + 02)] (ln 2 − ln 1) ln 2

9

∫

x2 dx 16 − x2

Let x = 4 sin θ dx = 4 cos θ dθ For the lower limit, when x = 0, 0 = 4 sin θ θ=0 For the upper limit, when x = 2, 2 = 4 sin θ 1 sin θ = 2 π θ= 6

∫

π 6

0

16 sin2 θ (4 cos θ dθ) = 16 − 16 sin2 θ = =

∫

π 6

0

16 sin2 θ (4 cos θ dθ) 4 1 − sin2 θ 16 sin2 θ dθ 16 π 6

∫

π 6 π 6

0

∫

0

1 − cos 2θ dθ 2

=8

∫

0

(1 − cos 2θ) dθ

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

4

ACE AHEAD Mathematics (T) Second Term

1 = 8 θ − sin 2θ 2

3

4

π 6

0

=8 =

π − 1 sin π − 0 6 2 3

3 4 π−4 2 3 4 = π− 2 3 3

[Shown]

10

Let t = tan

x 2

dt 1 x = sec2 dx 2 2 dt 1 x = 1 + tan2 dx 2 2 dt 1 = (1 + t2) dx 2 2 dt dx = 1 + t2

For the lower limit, when x = 0, t = tan 0 =0 π For the upper limit, when x = , 2 t = tan =1

π 4

1 + t2 x 1 − t2

2t

∫

π 2

0

5 dx =5 3 sin x + 4 cos x =5 =5

∫

1

3

∫ ∫

0 1 0 1 0

2t 1 − t2 2 + 4 1+t 1 + t2 2 dt 6t + 4 − 4t2

2 dt 1 + t2

dt 3t + 2 − 2t2

1 0 1 0

= −5 = −5

∫ ∫

dt 2t2 − 3t − 2 dt (2t + 1)(t − 2)

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

5

Let

A 1 B = + (2t + 1)(t − 2) 2t + 1 t − 2 1 ≡ A(t − 2) + B(2t + 1)

Letting t = 2, 1 = B(5) 1 B= 5 1 5 Letting t = − , 1 = A − 2 2 A=− \ −5 2 5

−2 1 ∫ 35(2t + 1) + 5(t − 2)4 dt

1 0

∫

dt = −5 0 (2t + 1)(t − 2)

1

= −5 −

3

1 1 ln 2t + 1 + ln t − 2 5 5

1

4

1 0

= 3ln 2t + 1 − ln t − 240 = ln

3

2t + 1 t−2

4

1 0

= ln −3 − ln − = ln

1 2

3 1 2 [Shown]

= ln 6 11

∫

dx 0 1 − sin x x Let t = tan 2 2 tan x 2 x 2

2π 3

tan x = =

1 − tan2

1 + t2 x 1 − t2

2t

2t 1 − t2 x t = tan 2 dt 1 x = sec2 dx 2 2 1 x = 1 + tan2 2 2 1 = (1 + t2) 2

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

6

ACE AHEAD Mathematics (T) Second Term

2 dt = (1 + t2) dx dx = 2 dt 1 + t2 =0 When x = 0 2

When x = 0, t = tan

2π π , t = tan 3 3 = 3 2 dt 1 + t2 2t 1− 1 + t2

∫

2π 3

0

dx = 1 − sin x =

∫ ∫

3

0

3

0

2 dt 1 + t2 − 2t

3

=2 = = = = = = =

∫

0

dt (t − 1)2

3

2(t − 1)−1 3 −1 0

4

3t −214 −

3

0

−2 −2 − −1 3−1 −2 −2 3−1

−2 − 2 3 − 1 3−1 −2 3 3−1 2 3 1− 3 [Shown]

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

7

12

Let t = tan

dt 1 x = sec2 dx 2 2 dt 1 x = 1 + tan2 dx 2 2 dt 1 = (1 + t2) dx 2 2 dt dx = 1 + t2

x 2

For the lower limit, when x = 0, t = tan 0 =0 For the upper limit, when x = 2π π , t = tan 3 2 = 3

2

1 + t2 x 1 − t2

2t

12+dtt dx 3 ∫ 5 + 4 cos x dx = ∫ 5 + 4 1 − t 1 + t

2 π 3

3

3

0

0

2

2

=

∫

3

0

6 dt 5 + 5t + 4 − 4t2

2

=6

∫

3

0

dt 9 + t2

=6×

1 t 3 tan−1 3 3 0

3

4

∫a + x dx 2

2

=

1 x tan−1 + c a a

= 2 tan−1 =2 =

3 − tan−1 0 3

π − 0 6

[Shown]

π 3

13 sin x ≡ A(3 sin x + 4 cos x) + B(3 cos x − 4 sin x) sin x ≡ (3A − 4B)sin x + (4A + 3B)cos x Equating the coefficients of sin x, … 1 = 3A − 4B Equating the coefficients of cos x, … 0 = 4A + 3B Solving and simultaneously, we have 3 4 A= and B = − 25 25

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

8

ACE AHEAD Mathematics (T) Second Term

∫

π 2

0

sin x dx = 3 sin x + 4 cos x = = =

∫

π 2

0

3 4 (3 sin x + 4 cos x) − (3 cos x − 4 sin x) 25 25 dx 3 sin x + 4 cos x 3 cos x − 4 sin x

3 4 ∫ 3 25 − 25 3 sin x + 4 cos x4 dx

0

π 2

3

3 4 x− ln |3 sin x + 4 cos x| 25 25

4

π 2

0

3 π 4 [ln −0 − |3 + 0| − ln |0 + 4|] 25 2 25

= 0.235 14 A B 17 + x ≡ + (4 − 3x)(1 + 2x) 4 − 3x 1 + 2x 17 + x ≡ A(1 + 2x) + B(4 − 3x) 1 1 1 Letting x = − , 16 = B 4 − 3 − 2 2 2 1 1 16 = B 5 2 2 B=3 4 Letting x = , 3 1 4 =A 1+2 3 3 1 2 18 = A 3 3 3 A=5 5 3 17 + x \ = + (4 − 3x)(1 + 2x) 4 − 3x 1 + 2x 18

1 2 1 3

3 4 3 4

∫

17 + x dx = (4 − 3x)(1 + 2x) =

5 3 ∫ 4 − 3x + 1 + 2x dx 5 −3

1 2 1 3

∫

1 2 1 3

−3 3 dx + 4 − 3x 2

1

∫

1 2 1 3

2 dx 1 + 2x

1

3 5 = − [ln |4 − 3x|]2 + [ln |1 + 2x|] 2 1 1 2 3 3 3 =− 3 3 3 2 2 5 ln 4 − − ln 4 − + ln 1 + − ln 1 + 2 3 2 2 3 3 5 5 3 5 = − ln − ln 3 + ln 2 − ln 3 2 2 3

3

4

3

4

5 2 5 = − ln 3 3

2 3 ln 2 5 3 5 5 3 6 = − ln + ln 3 6 2 5 + = 0.577

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

[Shown]

Fully Worked Solution

9

15

A 6x − 6 Bx + C ≡ + (x + 3)(x2 + 3) x + 3 x2 + 3 6x − 6 ≡ A(x2 + 3) + (Bx + C)(x + 3) Letting x = −3, −18 − 6 = 12A −24 = 12A A = −2 Letting x = 0, − 6 = 3A + 3C −6 = −6 + 3C C=0 0 = 4A + (B + C)(4) 0 = −8 + 4B B=2

Letting x = 1,

\

−2 6x − 6 2x ≡ + (x + 3)(x2 + 3) x + 3 x2 + 3 −2 2x ∫ x + 3 + x + 3 dx

2 1 2

∫

6x − 6 dx = 2 1 (x + 3)(x + 3)

2

= −2 [ln |x + 3|]2 + [ln |x2 + 3|]2 1 1 = −2 (ln 5 − ln 4) + ln 7 − ln 4 = −2 ln 5 7 + ln 4 4 7 4 5 4

= ln

3 4

2

= ln 16

28 25

[Shown]

A B C 13 − 11x + 6x2 ≡ + + (x + 3)(x − 2)2 x + 3 (x − 3)2 x − 2 13 − 11x + 6x2 ≡ A(x − 2)2 + B(x + 3) + C(x + 3)(x − 2) Letting x = 2, Letting x = −3, Letting x = 0, 15 = 5B B=3 100 = 25A A=4 13 = 4A + 3B − 6C 13 = 4(4) + 3(3) − 6C C=2

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

10

ACE AHEAD Mathematics (T) Second Term

\

4 13 − 11x + 6x2 3 2 = + + (x + 3)(x2 + 3) x + 3 (x − 2)2 x − 2

∫

13 − 11x + 6x2 = 2 3 (x + 3)(x + 3)

4

∫ x + 3 dx + ∫

3

4

4

3 dx + 2 3 (x − 2)

4

∫

4 3

2 dx x−2

4 = 4 [ln |x + 3|]3 + 3 4 = 4 [ln |x + 3|]3 −

∫

4 3

4 (x − 2)−2 dx + 2[ln |x − 2|]3

3

3 x−2

4

4 3

4 + 2[ln |x − 2|]3

= 4 (ln 7 − ln 6) − = 4 ln

7 3 + + 2 ln 2 6 2 = 3.50 [Shown] 17 1 x + 2x − 15 x2 + 2x − 14 x2 + 2x − 15 1 x2 + 2x − 14 1 =1+ 2 x2 + 2x − 15 x + 2x − 15 1 =1+ (x − 3)(x + 5) A 1 B Let ≡ + (x − 3)(x + 5) x − 3 x + 5

2

3 − 3 + 2(ln 2 − ln 1) 2 1

1 ≡ A(x + 5) + B(x − 3) Letting x = −5, 1 = −8B 1 B=− 8 Letting x = 3, 1 = 8A 1 A= 8 1 x2 + 2x − 14 1 \ 2 =1+ − 8(x − 3) 8(x + 5) x + 2x − 15

∫

x2 + 2x − 14 dx = 2 4 x + 2x − 15

5

1 1 ∫ 1 + 8(x − 3) − 8(x + 5) dx

5 4

= x+ =5+

3

1 1 ln |x − 3| − ln |x + 5| 8 8 ln 2 −

4

5 4

1 8 1 =5+ 8 1 =1+ 8 1 =1+ 8

1 1 1 ln 10 − 4 + ln 1 − ln 9 8 8 8 1 1 ln 2 − ln 10 − 4 + ln 9 8 8 ln 2 × 9 10 9 ln [Shown] 5

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

11

18

∫

x 0

x cos2 x dx = = =

1 + cos 2x ∫ x 2 dx x 0

cos 2x = 2 cos2 x − 1 cos2 x =

∫

1 (x + x cos 2x) dx 0 2 x x

1 + cos 2x 2

∫ 2 x dx + ∫ 2 x cos 2x dx 1 x 1 1 = 3 4 + 3 sin 2x x4 − ∫ 2 2 2 2

0 0 2 x x 0 0

1

x

1

x 0

1 1 sin 2x dx 2 2

Integrating by parts.

=

3 4 3 x2 4 +

x

x 1 1 = + x sin 2x + cos 2x 4 0 4 8 0 0 π2 1 1 1 = − 0 + π sin 2π − 0 + cos 2π − cos 0 4 4 8 8 2 π 1 1 = + − 4 8 8 π2 = [Shown] 4 19 y = 4x2 − 5x ⋅⋅⋅ y = 5x − 6x2 ⋅⋅⋅ Substituting into , 4x2 − 5x = 5x − 6x2 10x2 − 10x = 0 10x(x − 1) = 0 x = 0 or 1 From : When x = 0, y = 0. When x = 1, y = −1. Hence, the points of intersection are (0, 0) and (1, −1). y = 4x2 − 5x = x(4x − 5) The curve cuts the x-axis at the dy 1 = 8x − 5 points (0, 0) and 1 , 0 . dx 4 Since a > 0, the curve has a minimum point. dy At minimum point, =0 dx 8x − 5 = 0 5 x= 8 2 5 5 5 When x = , y = 4 −5 8 8 8 9 = −1 16

3 4 3

0 2 x

1 x sin 2x − 4 0 x 4 ∫ 4 3

x

1 sin 2x dx 0 4 x 4

x

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

12

ACE AHEAD Mathematics (T) Second Term

Hence, the minimum point of the curve is y = 5x − 6x2 = x(5 − 6x) dy = 5 − 12x dx dy =0 dx 5 − 12x = 0 5 x= 12 5 5 5 When x = , y = 5 −6 12 12 12 1 =1 24 At maximum point,

9 5, −1 16. 8

The curve cuts the x-axis at the points 5 (0, 0) and , 0 . 6

Since a < 0, the curve has a maximum point.

2

Hence, the minimum point of the curve is

5 1 12 , 1 24.

The graphs of y = 4x2 − 5x and y = 5x − 6x2 are as shown in the following diagram. y 2 5 , 1 1 12 24 1 A1 O −1 y = 5x − 6x

2

y = 4x 2 − 5x

x 1 A2 (1, −1) y = −x 2

−2

5, 9 −1 8 16

The equation of the chord joining the points of intersection (0, 0) and (1, −1) is y = −x. A1 = =

∫ ∫

1 0 1 0

35x − 6x2 − (−x)4 dx

(6x − 6x2) dx

1

= 33x2 − 2x34 0 = 3(1)2 − 2(1)3 − 0 = 1 unit2

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

13

A2 =

∫ =∫

1 0 1 0

[−x − (4x2 − 5x)]dx (4x − 4x2) dx

x3 1 3 0 1 = 2(1)2 − 4 −0 3 2 = units2 3 2 A1 : A2 = 1 : = 3: 2 3 = 2x2 − 4

3

4

[Shown]

20 The graph of y = 3 ln (x − 2) is as shown in the following diagram. y y = 3 ln (x − 2)

O

2

3

4

x

Required area = =

∫

1

3 1

y dx

To be integrated.

∫

3

3 ln |x − 2| dx

To be kept. Copy back

= [3x ln |x − 2|]3 −

4

∫

4 3

3x

1 dx x−2

Differentiate

= 33x ln |x − 2|4 3 − 3 = 33x ln |x − 2|4 3 − 3

4 3 4

4

∫

4 3

x dx x−2

2 ∫ 1 + x − 2 dx

4 3 4

= 33x ln |x − 2|4 − 33x + 2 ln |x − 2|4 3 = 12 ln 2 − 12 − 6 ln 2 − 9(0) + 9 + 6(0) = 6 ln 2 − 3 [Shown]

1 x − 2 x x−2 2 x 2 \ =1+ x−2 x−2

= 3(4) ln 2 − 3(4 + 2 ln 2) −[3(3) ln 1 − 3(3 + 2 ln 1)]

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

14

ACE AHEAD Mathematics (T) Second Term

21 y = −x2 y = x2 − x3

… …

Substituting into , −x2 = x2 − x3 x3 − 2x2 = 0 x3(x − 2) = 0 x = 0 or 2 From : When x = 0, y = 0 When x = 2, y = −22 = −4 Hence, the points of intersection of the curves are (0, 0) and (2, −4). y = x2 − x3 = x2(1 − x)

The curve intersects the x-axis at the points (0, 0) and (1, 0).

dy = 2x − 3x2 dx d2y = 2 − 6x dx2 At turning points, dy = 0. dx 2x − 3x2 = 0 x = 0 or When x = 0, y = 0 and 2 3

x(2 − 3x) = 0

d2y = 2 − 6(0) dx2 = 2 (x > 0)

Thus, (0, 0) is a minimum point. 2 22 23 When x = , y = − 3 3 3 4 = 27 and d2y 2 =2−6 dx2 3 = −2 (x < 0)

Thus,

4 2, 27 is a maximum point. 3

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

15

The graphs of y = − x2 and y = x2 − x3 are as shown in the following diagram. y y2 = x 2 − x 3 2, 4 3 27 O x

y1 = −x 2 (2, −4)

Required area =

∫ =∫

2

0 2

( y2 − y1) dx 3(x2 − x3 − (−x2)4 dx

2 3

0 2

∫ (2x − x ) dx 2x x =3 − 4 3 4

=

0 3 4 2 0

2 16 = (8) − −0 3 4 1 = 1 units2 [Shown] 3 22 y = 4 x 2 y = 4(x − 1) … …

Substituting into : 42 = 4(x − 1) x 16 = 4(x − 1) x2 4 =x−1 x2 4 = x3 − x2 3 2 x −x − 4=0 By inspection, x = 2 satisfies the equation.

\ (x − 2)(x2 + x + 2) = 0 x−2=0 x=2 or x2 + x + 2 = 0 No real roots because b2 − 4ac = 12 − 4(1)(2) = −7 (< 0)

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

16

ACE AHEAD Mathematics (T) Second Term

When x = 2, y =

4 2 =2

Hence, the point of intersection of the curves is (2, 2). 4 The graphs of the curves y = and y2 = 4 (x − 1) is as shown in the following diagram. x y 3 2 A2 A1 O 1 (2, 2)

y 2 = 4(x − 1)

4 y =x x

y 2 = 4(x − 1)

Required area = A1 + A2 = =

∫ 4 +1 dy + ∫

2 0

y2

3 2

4 dy y

3

2 y3 +y 12 0 8 = +2− 12 8 = + 4 ln 3

3

4

+ [4 ln y]2

3 units 2

0 + (4 ln 3 − 4 ln 2)

2

[Shown]

23 y = x(x + 2)(x − 3) = x3 − x2 − 6x … y = x(x − 3) = x2 − 3x … Substituting into , x3 − x2 − 6x = x2 − 3x 3 x − 2x2 − 3x = 0 x(x2 − 2x − 3) = 0 x(x + 1)(x − 3) = 0 x = 0, −1 or 3 From : When x = −1, y = (−1)2 − 3(−1) =4 When x = 0, y = 0 When x = 3, y = 32 − 3(3) =0 Hence, the points of intersection are (−1, 4), (0, 0) and (3, 0). y = x(x + 2)(x − 3) = x3 − x2 − 6x

The curve cuts the x-axis at the points (−2, 0), (0, 0) and (3, 0).

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

17

dy = 3x2 − 2x − 6 dx d2y = 6x − 2 dx2 dy At turning points, =0 dx 3x2 − 2x − 6 = 0 2 ± (−2)2 − 4(3)(−6) x= 2(3) x = 1.79 or −1.12 When x = 1.79, y = 1.79(1.79 + 2)(1.79 − 3) = −8.21 and d2y = 6(1.79) − 2 dx2 = 8.74 (> 0) \ (1.79, −8.21) is a minimum point. When x = −1.12, y = −1.12(−1.12 + 2)(−1.12 − 3) = 4.06 and d2y = 6(−1.12) − 2 dx2 = −8.72 (< 0) \ (−1.12, 4.06) is a maximum point. y = 1.5(1.5 − 3) 0+3 y = x(x − 3) x= 2 = x2 − 3x The curve cuts the x-axis at the points (0, 0) and (3, 0). Its minimum point is (1.5, −2.25). The graphs of y = x(x + 2)(x − 3) and y = x(x − 3) is as shown in the following diagram. y 8 6 4 A1 2 y = x(x − 3) y = x(x + 2)(x − 3)

(−1.12, 4.06)

−2 −1 O −2 −4 −6 −8

1 A2

2

3

x

(1.5, −2.25)

(1.79, −8.21)

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

18

ACE AHEAD Mathematics (T) Second Term

Required area = Area A1 + Area A2

∫ =∫

= =

0 −1 0 −1

3x3 − x2 − 6x − (x2 − 3x)4 dx + (x3 − 2x2 − 3x) dx +

3 0 3

∫

2

3 0

3x2 − 3x − (x3 − x2 − 6x)4 dx

∫ (−x + 2x + 3x) dx x 4 2x3 3x2 + + 4 3 2

2

3

x 4 2x3 3x2 − − 4 3 2

4 3

−1 3

0

+ −

4

3 0

=0 − =−

3

(−1) 2(−1) 3(−1) 3 2(3)3 3(3)2 − − + − + + 4 3 2 4 3 2

4

4 3

4

4

3 0

−0

1 + 2 − 3 + − 81 + 18 + 27 4 3 2 4 2

[Shown]

5 = 11 units2 6 24 y = e x When x = 0, y = e0 =1 When x → + ∞, y → + ∞

When x → − ∞, y → 0 y = 2 + 3e−x 3 =2+ x e 3 When x = 0, y = 2 + 0 = 5 e 3 When x → + ∞, x → 0 and thus y → 2 e When x → − ∞, y → + ∞ y y = 2 + 3e−x y = ex

5

2 1 O In 3 x

y = ex y = 2 + 3e−x

… …

Substituting into , we have: e x = 2 + 3e−x 3 ex = 2 + x e (e x)2 = 2e x + 3

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

19

(e x)2 − 2e x − 3 = 0 (e x − 3)(e x + 1) = 0 ex = 3 or e x = −1 x = ln 3 (No solution) Hence, the x-coordinate of the point of intersection of the curves y = e x and y = 2 + 3e−x is ln 3. Area of the shaded region =

∫ 32 + 3e − e 4 dx 1 = 32x + 3 e − e 4 −1 ln 3 0 −x x −x x ln 3 0

3 − ex ex 0 3 3 = 2 ln 3 − ln3 − e ln 3 − 2(0) − 0 − e0 e e 3 = 2 ln 3 − − 3 − 0 + 3 + 1 3 = 2.20 units2 = 2x − ln 3

3

4

25 (a) y 2 = x(x − 4)2 y = ± x (x − 4) Hence, the axis of symmetry is the x-axis. (b) Since y 2 ≥ 0, then x(x − 4)2 ≥ 0. Because (x − 4)2 ≥ 0, x(x − 4)2 ≥ 0 if and only if x ≥ 0. Hence, the curve exists only for x ≥ 0. (c) y 2 = x(x − 4)2 = x(x2 − 8x + 16) = x3 − 8x2 + 16x dy 2y = 3x2 − 16x + 16 dx dy 3x2 − 16x + 16 = dx 2y dy At turning points, = 0. dx 3x2 − 16x + 16 =0 2y 3x2 − 16x + 16 = 0 (3x − 4)(x − 4) = 0 x = 4 is not accepted because when x = 4, y = 0 4 dy 0 x= = (undefined). and = 3 dx 0 2 4 4 4 When x = , y 2 = −4 3 3 3 13 =9 27 y = ±3.08 1 1 Hence, the turning points are 1 , 3.08 and 1 , −3.08 . 3 3

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

20

ACE AHEAD Mathematics (T) Second Term

(d) The curve y 2 = x(x − 4)2 is as shown in the following diagram. y 1 1 , 3.08 3

O

4

x

1 1 , −3.08 3

(e) Volume generated = π

∫ y dx = π ∫ x(x − 4) dx = π ∫ (x − 8x + 16x) dx x 8x = π3 − + 8x 4 4 3

2 0 4 0 2 4 0 3 2 4 3 4 0 2

4

= π 64 −

3

8 (64) + 128 3

4

1 = 21 π units3 3 26 y 2 = 6x y = −2x + 6 ... ...

Substituting into , (−2x + 6)2 = 6x 4x2 − 24x + 36 = 6x 4x2 − 30x + 36 = 0 2x2 − 15x + 18 = 0 (2x − 3)(x − 6) = 0 3 x = or 6 2 3 3 From : When x = , y = −2 +6 2 2 =3

When x = 6, y = −2(6) + 6 = −6 Hence, the points of intersection of the curve y 2 = 6x and the straight line y = −2x + 6 are 3, 3 2 and (6, − 6).

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

21

The graphs of y 2 = 6x and y = −2x + 6 are as shown in the following diagram. y 6

y2 = −2x + 6 y12 = 6x V1

3

3 O 3 2 V2

6

x

(6, − 6)

V1 = π =π =π

∫

3 0

y2 2 dy + π 6

∫

6 3 6 3

6−y 2

2

dy

∫

3

0

y4 dy + 1 π 36 4

3 0

∫

(36 − 12y + y 2) dy

3 4

y5 180

y3 + 1 π 36y − 6y 2 + 3 4

3

4

6 3

= π 243 + 1 π 36(6) − 6(6)2 + 216 − 36(3) − 6(3)2 + 27 180 4 3 3 = 27 π + 1 π(72 − 63) 20 4 = 18 π units3 5 V2 = π

3

4

∫

6

0

6x dx − π

∫

6

3

(−2x + 6)2 dx

6 = π [3x2]0 − π

∫

6

3

(4x2 − 24x + 36) dx

= 3π (36 − 0) − π

34x 3

3

− 12x2 + 36x

4

6 3

4 4 = 108π − π (6)3 − 12(6)2 + 36(6) − (3)3 − 12(3)2 + 36(3) 3 3 = 108π − π [72 − 36] = 72π units3 18 π V1 : V2 = 5 72π 1 = 20 = 1:20 [Shown] 27 y = x(4 − x) = 4x − x2 4 y = −1 x … …

3

4

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

22

ACE AHEAD Mathematics (T) Second Term

Substituting into , 4 4x − x2 = −1 x 4x2 − x3 = 4 − x x3 − 4x2 − x + 4 = 0 By inspection, x = 1 satisfies the equation. (x − 1)(x2 − 3x − 4) = 0 (x − 1)(x + 1)(x − 4) = 0 x = 1, −1 or 4 x = −1 is not accepted \ x = 1 or 4 4 From : When x = 1, y = − 1 1 =3 When x = 4, y = 4 −1 4 =0 x2 − 3x − 4 x − 1 x3 − 4x2 − x + 4 x3 − x2 −3x2 − x −3x2 + 3x − 4x + 4 − 4x + 4 0

Hence, the points of intersection of the curves are (1, 3) and (4, 0) for x > 0. 4 The graphs of y = x(4 − x) and y = − 1 for x ≥ 0 are as shown in the following diagram. x y y2 = 4 − 1 x (2, 4)

(1, 3)

y1 = x(4 − x)

O −1

4

x

Volume generated = π =π

∫ ∫ ∫

4

1 4

y22 dx − π

∫

4

1

y12 dx

1 4

x2 (4 − x)2 dx − π

∫

4 1

4 − 1 dx x 16 8 ∫ x − x + 1 dx

4 1 2

2

= π x2(16 − 8x + x2) dx − π

1

∫

=π =π

4

1

(16x2 − 8x3 + x4) dx − π

16 8 ∫ x − x + 1 dx

4 1 2

3

16x3 x5 − 2x4 + 3 5

4

4 1

−π −

3

16 − 8 ln x + x x

4

4 1

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

23

=π

3

16(4)3 45 16 1 − 2(4)4 + − −2+ 3 5 3 5

4

4

−π−

3

16 − 8 ln 4 + 4 − (−16 − 8 ln 1 + 1) 4

3 = 30 π − π(15 − 8 ln 4) 5 3 = 15 π + 8π ln 22 5 3 = 15 π + 16π ln 2 5 3 = 15 + 16 ln 2 π [Shown] 5

28 y = e x When x = 0, y = e0 = 1 When x → + ∞, y → + ∞ When x → − ∞, y → 0 y = 2 + 3e−x 3 y=2+ x e 3 When x = 0, y = 2 + 0 = 5 e 3 When x → + ∞, x → 0 and thus y → 2 e When x → − ∞, y → + ∞ y y = ex 5 y = 2 + 3e−x 2 1 O In 3 x

y = ex … y = 2 + 3e−x … Substituting into , we have: e x = 2 + 3e−x 3 ex = 2 + x e (e x)2 = 2e x + 3 x 2 x (e ) − 2e − 3 = 0 (e x − 3)(e x + 1) = 0 e x = 3 or e x = −1 x = ln 3 (Not possible) Hence, the x-coordinate of the point of intersection of the curves y = e x and y = 2 + 3e−x is ln 3.

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

24

ACE AHEAD Mathematics (T) Second Term

Area of the shaded region =

∫ 32 + 3e − e 4 dx ln 3 0 −x x

= 2x + 3 = 2x −

3 3

1 −1 e

−x

− ex

4

ln 3 0

3 − ex ex

4

ln 3 0

= 2 ln 3 − = 2 ln 3 −

3 3 − e ln 3 − 2(0) − 0 − e0 eln3 e 3 − 3 −0 + 3 + 1 3

= 2.20 units2 29 2x + 1 Ax + B C ≡ + (x2 + 1)(2 − x) x2 + 1 2 − x 2x + 1 ≡ (Ax + B)(2 − x) + C(x2 + 1) Letting x = 2, Letting x = 0, 5 = C(5) C=1 1 = 2B + C 1 = 2B + 1 B=0 3 = (A + B) + 2C 3 = (A + 0) + 2(1) A=1

Letting x = 1,

\

2x + 1 1 = 2x + (x + 1)(2 − x) x + 1 2 − x

2

∫

1 0

2x + 1 dx = (x + 1)(2 − x)

2

∫

1 2

1 0

x dx + x +1

2 1 0 2

∫

1 0

1 dx 2−x

1 0

= =

∫

2x dx − x +1

2

∫

−1 dx 2−x

32 ln (x

1

+ 1)

4

1 0

− [ln (2 − x)]1 0

= 1 (ln 2 − ln 1) − (ln 1 − ln 2) 2 = 1.04 30 (a) 1 x + 2 x2 + x + 2 x2 +2 x x2 + x + 2 x \ 2 =1+ 2 x +2 x +2

2

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

25

∫

x2 + x + 2 dx = x2 + 2 =

∫ 1 + x

2

x dx +2

2

∫ 31 + 2 x

1

2x +2

4 dx

=x+ (b)

1 ln |x2 + 2| + c 2

∫e

x x+1 dx = xe− (x + 1) dx 1 − (x + 1) e x − −e− (x + 1) 1 dx −1 x = − x + 1 + e− (x + 1) dx e x 1 − (x + 1) = − x+1 + e +c e −1 =

∫

∫

∫

=− =− 31 (a) dy 3x − 5 = 2 x dx y= y=

x e x+1 −

1 e x+1 +c

x+1 +c ex + 1

3x − 5 ∫ 2 x dx 3 ∫ 2 x

1 2

5 − − x 2 dx 2

1

1

3 x2 5 y= − 2 3 2 2

3 1

3

x2 1 2

+c

y = x 2 − 5x 2 + c Since the curve passes through the point (1, − 4), then − 4 = (1)2 − 5(1)2 + c −4 = 1 − 5 + c c=0 Hence, the equation of the curve is y = x2 − 5x2 = x2 (x − 5) = x (x − 5) (b) At the x-axis, y = 0 x (x − 5) = 0 x = 0 or 5 x = 0 is ignored because it is given that x > 0. Therefore, x = 5.

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

1 3 1 3 1

26

ACE AHEAD Mathematics (T) Second Term

At a turning point,

dy = 0. dx

3x − 5 =0 2 x 3x − 5 = 0 x= 5 3

5 5 5 When x = , y = −5 3 3 3 = −4.30 dy 3x − 5 = dx 2 x

3 5 − = x2 − x 2 2 2 d2y 3 − 1 5 − 3 = x 2+ x 2 4 dx2 4 = 31 + 53 4x 2 4x 2 5 d2y When x = , 2 = 3 dx 3 45 3

1

1

1 2

+

5 45 3

3 2

(> 0)

2 Hence, 1 , −4.30 is a minimum point. 3 Then curve of y = x (x − 5) is as shown below. y

O −2 −4

x 1 2 3 4 5

2 1 , −4.30 3

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

27

(c) Area of the region bounded by the curve and the x-axis =

∫ y dx = ∫ x − 3x dx

5 0 5 0

3 2

1 2

= =

3

5

2x 2 − 5 2x 2 5 3

5 2

4

3

0

5

2 (5) − 10 (5) − 0 5 3 2 10 = ( 5) − ( 5) 5 3

3 2

5 3

=

2 (25 5

5) −

10 (5 5 ) 3

= 10 5 − = − = 32 20 5 3

50 5 3

20 5 units2 3

∫

3 2

(x − 2)2 dx = x2 =

x ∫

3 2 3 2

2

− 4x + 4 dx x2

4 ∫ 1 − x + 4x dx

−2

−1 = x − 4 ln |x| + 4 x −1

3 3

4

3 2

3 2

= x − 4 ln |x| − = 3 − 4 ln 3 − = = = 33 y = 6 − e x On the x-axis, y = 0. 6 − ex = 0 ex = 6 x = ln 6

4 x

4

4 4 − 2 − 4 ln 2 − 3 2

5 + 4 ln 2 − 4 ln 3 3 5 + 4 (ln 2 − ln 3) 3 5 2 + 4 ln 3 3

[Shown]

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

28

ACE AHEAD Mathematics (T) Second Term

Thus, the curve y = 6 − e x intersects the x-axis at (ln 6, 0). On the y-axis, x = 0. y = 6 − e0 y=5 Thus, the curve y = 6 − e x intersects the y-axis at (0, 5). As x → ∞, y → − ∞ As x → − ∞, y → 6 y 6

5 y = 6 − ex

y = 5e−x (In 5, 1) O In 6 x

y = 5e−x On the y-axis, x = 0. y = 5(e 0) y=5 Therefore, the curve y = 5e−x intersects the y-axis at (0, 5). As x → ∞, y → 0. As x → −∞, y → ∞ The curves y = 6 − e x and y = 5e−x are as shown. y = 6 − ex … y = 5e−x … Substituting into , 6 − e x = 5e−x x 6e − (e x)2 = 5 Letting e x = p, 6p − p2 = 5 p2 − 6p + 5 = 0 (p − 1)(p − 5) = 0 p = 1 or 5 When p = 1, ex = 1 x = ln 1 x=0

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

29

When x = 0, y = 6 − e0 = 5 When p = 5, ex = 5 x = ln 5 When x = ln 5, y = 6 − e ln 5 = 6 − 5 = 1 Hence, the points of intersection are (0, 5) and (ln 5, 1). Area of the shaded region =

∫ 36 − e − 5e 4 dx ln 5 0 x −x

= 6x − e x − = 6x − e x +

3 3

5 −x e (−1) 5 ex

4

ln 5 0

4

ln 5 0

= 6 ln 5 − eln 5 = 6 ln 5 − 5 +

5 5 − 0 − e0 + 0 eln 5 e

5 − (−1 + 5) 5 = 6 ln 5 − 5 + 1 + 1 − 5 = (6 ln 5 − 8) units 2 Volume of the solid generated = π

∫ 36 − e − 5e 4 dx = π∫ 336 − 12e + e − 25e 4 dx ln 5 0 x −x ln 5 0 x 2x −2x

2

2

1 25 −2x = π 36x − 12e x + e2x − e 2 (−2) 1 25 = π 36x − 12e x + e2x − 2x 2 2e

3 3 3 3

4

ln 5 0

4

ln 5 0

1 25 1 25 = π 36 ln 5 − 12eln 5 + e 2 ln 5 + 2 ln 5 − 0 − 12e0 + e0 + 0 2 2e 2 2e 1 25 1 25 = π 36 ln 5 − 12(5) + (25) + − −12 + + 2 2 2 2(25) = π (36 ln 5 − 48) = 12(3 ln 5 − 4)π units3 34 Let u = 1 − x du = − 1 dx dx = − du When x = 0, u = 1. When x = 1, u = 0.

4

4

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

30

ACE AHEAD Mathematics (T) Second Term

∫

1 0

x2 (1 − x) 3 dx = = = = =

1

∫ ∫ ∫

0 1 0 1 0 1 0 1

(1 − u)2 u 3 (−du) − u 3 (1 − u)2 du − u 3 (1 − 2u + u2) du

1 3 1 1

1

∫ − u

+ 2u 3 − u 3 du

7 3 10 3

4

7

3

3

−

u 2u u + − 4 7 10 3 3 3

4 3 7 3

4 3

4

7 3

0

1

3 6 3 = − u + u − u 4 7 10

10 3

4

0

1

3 6 3 = 0 − − (1) + (1) − (1) 3 4 7 10 3 6 3 − + 4 7 10 27 = 140 = 35 (a) y 3

4 3

10

4

−1 −2 O 2 R −2 −3 −4

y2 = x − 2 x

y1 = x 2 − 4

(b) y = x − 2 y = x2 − 4

… …

Substituting into , x2 − 4 = x − 2 x2 − x − 2 = 0 (x − 2)(x + 1) = 0 x = 2 or −1 When x = 2, y = 2 − 2 =0 When x = −1, y = −1 − 2 = −3 Hence, the coordinates of the points of intersection are (2, 0) and (−1, −3).

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

Fully Worked Solution

31

(c) Area of R =

∫ (y − y ) dx = ∫ 3(x − 2) − (x − 4)4 dx = ∫ (− x + x + 2) dx x x = 3− + + 2x4 3 2

−1 2 2 1 2 −1 2 2 −1 3 2 2 −1

2

(−1)3 (−1)2 23 22 + + 2(2) − − + + 2(−1) 3 2 3 2 10 7 = − − 3 6 9 = units2 2 =−

3

4

(d) Volume generated = π =π =π

∫ ∫

2

−1 2

(y12 − y22) dx 3(x2 − 4)2 − (x − 2)24 dx

4 2 2

−1 2

∫ 3(x − 8x + 16) − (x − 4x + 4)4 dx = π ∫ 3(x − 9x + 4x + 12) dx x = π 3 − 3x + 2x + 12x4 5

−1 2 4 2 −1 5 2 3 2 −1

=π =π =

32 − 3(2) 5

5

3

+ 2(2)2 + 12(2) −

(−1)5 − 3(−1)3 + 2(−1)2 + 12(−1) 5

4

372 − − 364 5 5

108 π units3 5

© Oxford Fajar Sdn. Bhd. (008974-T) 2012

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...Chaos and Fractals What is a Fractal A fractal is a never ending pattern. Fractals are never ending complex patterns that are self-similar across different scales. They are created by by being repeated over and over in a feedback loop. Fractals are everywhere. They can be found in nature, algebra, and math. Nature Algebra Geometry Mandelbrot Set This was created in 1980 shortly after the first personal computer in order to calculate numbers thousands and sometimes millions of times. Equation (old)Z=(new)Z2+C We start by plugging a value for the variable C into the simple equation below. Each complex number is actually a point in a 2-dimensional plane. The equation gives an answer, Znew . We plug this back into the equation, as Zold and calculate it again. We are interested in what happens for different starting values of C. When you square a number it gets bigger. Then you square the answer and it get even bigger.. Eventually, it goes to infinity. This is the fate of most starting values of C. Some values of C do not get bigger, but get smaller, or alternate between a set of fixed values. These are the points inside the Mandelbrot Set, which we color black. Outside the Set, all the values of C cause the equation to go to infinity, and the colors are proportional to the speed at which they expand. Ju Fractals in Nature Natural fractals can be found anywhere in nature. Even in our body. It can be the blood vessels in our arms, or...

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...Chaos in the Caribbean Strayer University Forensic Accounting and Fraud Examination ACC 571 Dr. Timothy Brown ------------------------------------------------- 1. Evaluate Avey’s role as an expert witness for the Jamaican government. Avey and his firm were hired by the Jamaican government starting in the early in the 90s to investigate accusations of fraud and mismanagement and prepare reports outlining his findings starting with the Blaise Merchant Bank and Trust Co which spread to two similar but larger cases involving Century National Bank and its related financial entities and Eagle Merchant Bank. As an expert witness, Avey was hired by the Jamaican government to provide forensic investigation and audit support. He utilized specialized investigative skills in carrying out an inquiry conducted in such a manner that the outcome would be applicable to a court of law. In addition he examined evidence regarding assertions to determine its correspondence to established criteria carried out in a manner suitable to the court. Avey conducted his investigation grounded in sound forensic accounting principles where he quickly discovered self dealing in the Blaise Case where money was lent from one Blaise financial entity to companies controlled by its principal shareholders. In the Century Case, dishonesty was the main cause of the problems where the use of depositors’ funds to acquire such assets as real estate (which had also been the case with Blaise) for the...

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