1 Algebra: Topic 1 – Revision of the basics = –1 Add 7 to both sides of the equation. Adding 7...

Topic

1 Worked solutions

1 Algebra: Topic 1 – Revision of the basics

Progress checks

1 (a) 3(2x + 5) = 6x + 15

(b) –2(x + 6) = –2x – 12

(c) 8(x2 + 3x + 4) = 8x2 + 24x + 32

(d) 6(3x – 7x + 9) = 18x – 42x + 54 = –24x + 54

(e) –3(5x – 9) = –15x + 27

(f) –4(x2 + x + 2) = –4x2 – 4x – 8

(g) –(3x + 5) = –3x – 5

(h) –(6x + 8) = –6x – 8

(i) –(x2 – 4x + 8) = –x2 + 4x – 8

(j) –4(5 – x) = 4x – 20

(k) –7(2 – 4x + x2) =– 7x2 + 28x –14

2 (a) 2(3x – 4) + 2(x – 3) = 6x – 8 + 2x – 6 = 8x – 14

(b) 5(4 – 2x) – (x – 4) = 20 – 10x – x + 4 = 24 – 11x

(c) 12(x – 8) – 4(x – 8) = 12x – 96 – 4x + 32 = 8x – 64

(d) 4(2x – 3y) + 6(x + 2y) = 8x – 12y + 6x + 12y = 14x

(e) 2x(x + 4) – x(x + 6) = 2x2 + 8x – x2 – 6x = x2 + 2x

(f) 4x(3x + 1) – 3x(x – 4) = 12x2 + 4x – 3x2 + 12x = 9x2 + 16x

(g) 2x2(x – 6) + x3 + 2x2 = 2x3 – 12x2 + x3 + 2x2 = 3x3 – 10x2

(h) 4x(x + y) – y(x + y) = 4x2 + 4xy – xy – y2 = 4x2 + 3xy – y2

3 (a) (x + 5)(x + 2) = x2 + 2x + 5x + 10 = x2 + 7x + 10

(b) (x – 7)(x + 1) = x2 + x – 7x – 7 = x2 – 6x – 7

(c) (x + 4)(x – 4) = x2 – 4x + 4x – 16 = x2 – 16

(d) (3x + 1)(5x + 3) = 15x2 + 9x + 5x + 3 = 15x2 + 14x + 3

(e) (4x – 1)(2x – 5) = 8x2 – 20x – 2x + 5 = 8x2 – 22x + 5

(f) (5x – 1)(5x + 1) = 25x2 + 5x – 5x – 1 = 25x2 – 1

(g) (2x – 8)(x + 4) = 2x2 + 8x – 8x – 32 = 2x2 – 32

1

(h) (3a + b)(4a + 2b) = 12a2 + 6ab + 4ab + 2b2 = 12a2 + 10ab + 2b2

(i) (5x – y)(4x + y) = 20x2 + 5xy – 4xy – y2 = 20x2 + xy – y2

(j) (6x – 5y)(x – 3y) = 6x2 – 18xy – 5xy + 15y2 = 6x2 – 23xy + 15y2

4 (a) (x + 2)2 + (y + 1)2 = 0

x2 + 4x + 4 + y2 + 2y + 1 = 0

x2 + y2 + 4x + 2y + 5 = 0

(b) (x + 5)2 + (y + 3)2 = 0

x2 + 10x + 25 + y2 + 6y + 9 = 0

x2 + y2 + 10x + 6y + 34 = 0

(c) (x – 7)2 + (y + 2)2 = 0

x2 – 14x + 49 + y2 + 4y + 4 = 0

x2 + y2 – 14x + 4y + 53 = 0

(d) (x – 4)2 + (y + 6)2 = 0

x2 – 8x + 16 + y2 + 12y + 36 = 0

x2 + y2 – 8x + 12y + 52 = 0

(e) (x – 6)2 + (y + 7)2 = 12

x2 – 12x + 36 + y2 + 14y + 49 = 12

x2 + y2 – 12x + 14y + 73 = 0

(f) (x + 2)2 + (y – 5)2 + 7 = 0

x2 + 4x + 4 + y2 – 10y + 25 + 7 = 0

x2 + y2 + 4x – 10y + 36 = 0

(g) (x – 1)2 + (y – 9)2 – 10 = 0

x2 – 2x + 1 + y2 – 18y + 81 – 10 = 0

x2 + y2 – 2x – 18y + 72 = 0

(h) (x + 3)2 + (y – 8)2 – 27 = 0

x2 + 6x + 9 + y2 – 16y + 64 – 27 = 0

x2 + y2 + 6x – 16y + 46 = 0

(i) (x + 1)2 + (y – 1)2 = 18

x2 + 2x + 1 + y2 – 2y + 1 – 18 = 0

x2 + y2 + 2x – 2y – 16 = 0


2

5 (a) 4x2yx y

= 4x

(b) 12x2y3

4x y = 3xy2

(c) 16a3b2c24a2 b

= 23 abc

(d) 24x2y4

8x y5 = 3xy

6 (a) 15abc

5c = 3ab

(b) 4x3

4x y =

x2

y

(c) 6pq3r3pqr

= 2q2

(d) 10x4y3

2x 2y = 5x2y2

(e) 45a3bc2

9ab2c = 5a2c

b

(f) (x + 7)(x + 5)(x + 5)(x + 3) =

(x + 7)(x + 3)

(g) (x + 3)(x + 5)(x + 5)(x + 3) = 1

7 (a) 12x2y + 8xy2 = 4xy(3x + 2y)

(b) 4a2b + 2ab = 2ab(2a + 1)

(c) 24x2y + 6x = 6x(4xy + 1)

(d) 25a3b2c5 + 5a2b3 = 5a2b2(5ac5 + b)

8 (a) x2(x – 1)x(x – 1) = x

(b) x y3

x y = y2

(c) 15x3y3

5x3y2 = 3y

(d) 5(x – 4)

10(x – 2) = x – 4

2(x – 2)

(e) (x + 1)(x – 2)(x – 5)(x + 1) =

x – 2x – 5

(f) x – 3

(x – 3)(x – 1) = 1

(x – 1)

Divide top and bottom by xy.

Divide top and bottom by 4xy.

Divide top and bottom by 8a2b.

Divide top and bottom by 8xy4.

Divide top and bottom by 5c.

Divide top and bottom by 4x.

Divide top and bottom by 3pqr.

Divide top and bottom by 2x2y.

Divide top and bottom by 9abc.

Divide top and bottom by (x + 5).

Divide top and bottom by (x + 3)(x + 5).

Divide top and bottom by x(x – 1).

Divide top and bottom by xy.

Divide top and bottom by 5x3y2.

Divide top and bottom by 5.


Divide top and bottom by (x – 3).

Worked solutions

3

9 x – 7 = 7

x = 14

10 x + 7 = –3

x = –10

11 4x5 = 12

4x = 60

x = 15

12 x5 – 1 = 7

x5 = 8

x = 40

13 2x3 – 1 = 5

2x3 = 6

2x = 18

x = 9

14 2(2x + 1) = 18

4x + 2 = 18

4x = 16

x = 4

15 –6 + x

2 = –1

–6 + x = –2

x = 4

16 (a) 2 – x = 4 + x

2 = 4 + 2x

–2 = 2x

–1 = x

x = –1

Add 7 to both sides of the equation. Adding 7 will eliminate the –7 on the left-hand side.

Subtract 7 from both sides of the equation. Subtracting 7 will eliminate the +7 on the left-hand side of the equation.

Remove the denominator by multiplying both sides by 5.

Remove the multiplier of x by dividing both sides by 4.

Add 1 to both sides.

Multiply both sides by 5.



Divide both sides by 2.

Multiply out the bracket.

Subtract 2 from both sides.


Multiply both sides by 2 to remove the 2 in the denominator. Add 6 to both sides.


4

(b) 4(x – 7) = 3(2x – 10)

4x – 28 = 6x – 30

– 28 = 2x – 30

2 = 2x

x = 1

(c) 5(6x – 3) = 6(2x – 1)

30x – 15 = 12x – 6

18x – 15 = –6

18x = 9

x = 12

(d) 13 (x – 1) = 2x + 4

x – 1 = 6x + 12

– 1 = 5x + 12

–13 = 5x

x = – 135

x = –2.6

17 (a) x – 5

4 = 4x

x – 5 = 16x

– 5 = 15x

x = – 5

15

x = – 13

(b) 6x – 1 = 3(x – 4) + 7

6x – 1 = 3x – 12 + 7

3x – 1 = –5

3x = –4

x = – 43

TAKE NOTEBe careful here. Students often see the numbers 18 and 9 and automatically think that x is 2 which is incorrect.

Worked solutions

5

18 (a) x4 +

x2 = 15

2x + 4x = 120

6x = 120

x = 20

(b) x3 +

x4 = 49

4x + 3x = 588

7x = 588

x = 84

(c) x9 +

2x3 = 42

x + 6x = 378

7x = 378

x = 54

19 x – 3

2 + x + 1

3 = 3

6(x – 3)

2 + 6(x + 1)

3 = 18

3(x – 3) + 2(x + 1) = 18

3x – 9 + 2x + 2 = 18

5x – 7 = 18

5x = 25

x = 5

Multiply both sides by 8 because the denominators 4 and 2 divide exactly into 8.


Multiply both sides by 12 because the denominators 3 and 4 divide exactly into 12. Collect the terms in x.


Multiply both sides by 9 because the denominators 9 and 3 divide exactly into 9. Collect the terms in x together.


2 and 3 are factors of 6 so we multiply both sides by 6.


6

20 x2 –

x5 = 3

10x

2 – 10x

5 = 30

5x – 2x = 30

3x = 30

x = 10

21 14 (x – 1) =

13 (x – 2)

3(x – 1) = 4(x – 2)

3x – 3 = 4x – 8

–3 = x – 8

x = 5

22 2x3 –

x4 = 5

24x

3 – 12x

4 = 60

8x – 3x = 60

5x = 60

x = 12

23 (a) A = πr2

Aπ

= r2

r = A π

(b) A = 4πr2

A

4π = r2

r = A

4π


This step is usually not written down.



Multiply out the brackets.

Subtract 3x from both sides.



This step is usually not written down.


Worked solutions

7

(c) V = 43 πr3

3V = 4πr3

3V4π

= r3

r = 3 3V 4π

24 (a) v = u + at (u)

u = v – at

(b) v = u + at (a)

v – u = at

a = v – u

t

(c) v2 = 2as (s)

s = v 2

2a

(d) v2 = u2 + 2as (a)

v2 – u2 = 2as

a = v 2– u 2

2s

(e) v2 = u2 + 2as (u)

v2 – 2as = u2

u = √ v2 − 2as

(f) s = ut + 12 at2 (a)

s – ut = 12 at2

2(s – ut) = at2

a = 2(s – ut)

t2

(g) y = mx + c (c)

y – mx = c

c = y – mx

Subtract at from both sides.

Subtract u from both sides.

Divide both sides by t.

Divide both sides by 2a.

Subtract u2 from both sides.

Divide both sides by 2s.

Subtract 2as from both sides.

Square root both sides.

Subtract ut from both sides.


Divide both sides by t2.

Subtract mx from both sides.

Always put the subject of the equation on the left-hand side of the equation.


8

(h) y = mx + c (m)

y – c = mx

m = y – c

x

(i) s = 12 (u + v)t (t)

2s = (u + v)t

t = 2s

u + v

(j) E = 12 mv2 (v)

2E = mv2

2Em

= v2

v = 2E m

(k) V = πr2l (l)

l = V

πr2

(l) V = πr2l (r)

Vπl

= r2

r = V πl

Subtract c from both sides.

Divide both sides by x.


Divide both sides by (u + v).


Divide both sides by m.


Divide both sides by πr2.

Divide both sides by πl.


Worked solutions

9

25 (a) x + y = 5 ......................... (1)

5x + 2y = 11 .................... (2)

Multiplying equation (1) by 2 we obtain

2x + 2y = 10

Subtracting the above equation from equation (2) we obtain

5x + 2y = 11

Subtracting 2x + 2y = 10

3x = 1

x = 13

Substituting x = 13 into equation (1) we have

13 + y = 5

Hence, y = 4 23

Substituting x = 13 and y = 4 23 into LHS of equation (2) we obtain

5(13) + 2(42

3) = 11

53 + 91

3 = 11

11 = RHS

Hence the solutions are x = 13 and y = 4 23

(b) 2x – 3y = –5 (1)

5x + 2y = 16 (2)

Multiplying equation (1) by 2 and equation (2) by 3 we obtain

4x – 6y = –10

15x + 6y = 48

Adding 19x = 38

x = 2

Substituting x = 2 into equation (1) we have

2(2) – 3y = –5

4 – 3y = –5

At GCSE level the simultaneous equations you solved usually resulted in whole number (i.e. integer) answers. At this level you can often get fractions so do not automatically assume you have done something wrong if you get fractions.

Both sides of the equation are equal so the values of x and y satisfy the second equation.

>>> TIPAlways specifically say what your answers are. The examiner should not have to wade through your working to find what your answers are.

>>> TIPNotice that the terms in y have opposite sign. It is easier to make these terms the same in value but opposite in sign so that the two equations can be added together in order to eliminate the term in y. It is easier to add the equations so this is why we have chosen to eliminate y rather than x.


10

3y = 9

y = 3

Substituting x = 2 and y = 3 into LHS of equation (2) we obtain

5(2) + 2(3) = 16

16 = RHS

Hence the solutions are x = 2 and y = 3

26 (a) 3x – 5 = x – 1

2x – 5 = –1

2x = 4

x = 2

Substituting x = 2 into the equation y = 3x – 5 we obtain

y = 3(2) – 5

= 1

Checking by substituting x = 2 and y = 1 into y = x – 1 we obtain

1 = 2 – 1

1 = 1

Hence solutions are x = 2 and y = 1.

(b) From the equation 2x + 3y = 8 we have 3y = –2x + 8

Notice that 3y appears in both equations so it is best to substitute the value of 3y into the second equation in order to eliminate y.

5x + 3y = 11

Hence, 5x + (–2x + 8) = 11

3x + 8 = 11

3x = 3

x = 1

Substituting x = 1 into the equation 2x + 3y = 8 we obtain

2(1) + 3y = 8

3y = 6

y = 2


The y-values are equated and the resulting equation solved.

Both sides of the equation are equal, showing that the values of x and y satisfy the second equation.

The 3y is replaced by –2x + 8.

Worked solutions

11

Checking by substituting x = 1 and y = 2 into 5x + 3y = 11 we obtain

5(1) + 3(2) = 11

11 = 11

Hence solutions are x = 1 and y = 2.

27 √ 45 + √ 80 + √ 125 = √ 9 × 5 + √ 16 × 5 + √ 25 × 5 = 3√ 5 + 4√ 5 + 5√ 5 = 12√ 5

28 3√ 3 – √ 2

√ 3 – √ 2 = (3√ 3 – √ 2)(√ 3 + √ 2)

(√ 3 – √ 2)(√ 3 + √ 2) =

9 + 3√ 6 – √ 6 – 23 + √ 6 – √ 6 – 2

= 7 + 2√ 6

1 = 7 + 2√ 6

29 3

√ 3 + √ 75 + (√ 2 × √ 6 ) = √ 3 × 3

√ 3 + √ 25 × 3 + (√ 2 × √ 2 × 3 )

= √ 3 + 5√ 3 + 2√ 3 = 8√ 3

Test yourself 1 (a) 2x + 11 = 25

2x = 14

x = 7

(b) 3x – 5 = 10

3x = 15

x = 5

(c) 15x = 60

x = 4

(d) x

4 = 8

x = 32

(e) 4x

5 = 20

4x = 100

x = 25











12

(f) 2x

3 = –6

2x = –18

x = –9

(g) 5 – x = 7

5 = 7 + x

–2 = x

x = –2

(h) x

7 – 9 = 3

x

7 = 12

x = 84

2 (a) 35x3y2

7xy2 = 5x2

(b) 15ab3c

3ab = 5b2c

(c) (x – 4)(x – 7)(x – 1)(x – 4)

= (x – 7)(x – 1)

(d) (x + 3)2

(x – 6)(x + 3) = (x + 3)

(x – 6)

3 (a) 4(2x – 3) + 5(2x + 1) = 8x – 12 + 10x + 5 = 18x – 7

(b) –2(x + 4) = –2x – 8

(c) –(x – 5) = –x + 5

(d) 4(2x – 6) – (5x – 4) = 8x – 24 – 5x + 4 = 3x – 20

(e) 3(5x – 9) – 4(2x – 6) = 15x – 27 – 8x + 24 = 7x – 3

(f) 4(2x – 7) + 5x – 9 = 8x – 28 + 5x – 9 = 13x – 37

(g) –(3x2 + 4x – 2) = –3x2– 4x + 2

(h) x(x2 – 4x + 8) = x3 – 4x2 + 8x

(i) 3a(a + b) + 2b(a + b) = 3a2 + 3ab + 2ab + 2b2 = 3a2 + 5ab + 2b2

(j) 4a(2a – b) – 3b(2a – b) = 8a2 – 4ab – 6ab + 3b2 = 8a2 – 10ab + 3b2

(k) 5x2(x – 3) + 3x(x + 4) = 5x3 – 15x2 + 3x2 + 12x = 5x3 – 12x2 + 12x

(l) 2x(x – 1) – (x2 – 3x) = 2x2 – 2x – x2 + 3x = x2 + x


Add x to both sides to make x positive.




Divide top and bottom by 7xy2.

Divide top and bottom by 3ab.

Divide top and bottom by (x – 4).


Worked solutions

13

4 (a) (x – 5)(x + 3) = x2 + 3x – 5x – 15 = x2 – 2x – 15

(b) (4x – 1)(x – 5) = 4x2 – 20x – x + 5 = 4x2 – 21x + 5

(c) (2x – 7)(3x + 5) = 6x2 + 10x – 21x – 35 = 6x2 – 11x – 35

(d) (9x – 1)(9x + 1) = 81x2 + 9x – 9x – 1 = 81x2 – 1

(e) (4a – b)(2a + 4b) = 8a2 + 16ab – 2ab – 4b2 = 8a2 + 14ab – 4b2

(f) (5y – 1)(2y + 5) = 10y2 + 25y – 2y – 5 = 10y2 + 23y – 5

5 (a) V = nRT

p

(b) n = pV

RT

(c) T = pV

nR

(d) p = nRT

V

6 f = Eh

7 m = y – c

x

8 (a) λ = cf

(b) V = nc

(c) T = Q

mc

(d) V = 1000n

c

(e) c = 1000n

V

(f) h = Ef

9 (a) √ 48 + 12

√ 3 – √ 27 = 4√ 3 +

12√ 3

√ 3√ 3 – 3√ 3 = 4√ 3 + 4√ 3 – 3√ 3 = 5√ 3

(b) 2 + √ 53 + √ 5

= (2 + √ 5 )(3 – √ 5)(3 + √ 5)(3 – √ 5) =

6 + √ 5 – 59 – 5

= 1 + √ 5

4


14

10 (a) 5

√ 2 =

5

√ 2 × √ 2

√ 2 = 5√ 2

2

(b) 1

3 + √ 5 =

13 + √ 5

× 3 – √ 53 – √ 5

= 3 – √ 59 – 5

= 3 – √ 54

(c) √ 32 + 3√ 2 = √ 16 × 2 + 3√ 2 = 4√ 2 + 3√ 2 = 7√ 2

(d) (2√ 5 )2 = 4 × 5 = 20

11 7x + 4y = 2 .......................... (1)

3x – y = 9 .......................... (2)

Multiplying equation (2) by 4 gives

12x – 4y = 36

7x + 4y = 2

Adding these two equations to eliminate y, we obtain

19x = 38

x = 2

Substituting x = 2 into equation (1) we obtain

7x + 4y = 2

7(2) + 4y = 2

14 + 4y = 2

4y = –12

y = –3

Checking by substituting x = 2 and y = –3 into equation (2) we obtain

3x – y = 9

3(2) – (–3) = 9

6 + 3 = 9

9 = 9

Hence the solutions are x = 2 and y = –3.

Both sides of the equation are equal showing that the values of x and y satisfy the second equation.

Worked solutions

15

Topic

2 Worked solutions

1 Algebra: Topic 2 – Manipulation of algebraic expressions

Progress check

1 (a) (x + 1)2 = x2 + 2x + 1

(b) (x + 11)2 = x2 + 22x + 121

(c) (x + 13)2 = x2 + 26x + 169

(d) (x – 6)2 = x2 – 12x + 36

(e) (x – 11)2 = x2 – 22x + 121

(f) (x + 7)2 = x2 + 14x + 49

2 (a) (x + 3)2 = x2 + 6x + 9

(b) (x + 4)2 = x2 + 8x + 16

(c) (x + 1)2 = x2 + 2x + 1

(d) (x + 6)2 = x2 + 12x + 36

(e) (x + 8)2 = x2 + 16x + 64

(f) (x + 5)2 = x2 + 10x + 25

(g) (x – 4)2 = x2 – 8x + 16

(h) (x – 5)2 = x2 – 10x + 25

(i) (x – 9)2 = x2 – 18x + 81

(j) (x – 7)2 = x2 – 14x + 49

(k) (x – 10)2 = x2 – 20x + 100

(l) (x + 12)2 = x2 + 24x + 144

3 (a) x2 + 4x + 8 = (x + 2)2 – 4 + 8 = (x + 2)2 + 4

(b) x2 + 2x + 6 = (x + 1)2 – 1 + 6 = (x + 1)2 + 5

(c) x2 – 6x + 4 = (x – 3)2 – 9 + 4 = (x – 3)2 – 5

(d) x2 – 2x – 10 = (x – 1)2 – 1 – 10 = (x – 1)2 – 11

(e) x2 – 10x – 2 = (x – 5)2 – 25 – 2 = (x – 5)2 – 27

(f) x2 – 8x + 4 = (x – 4)2 – 16 + 4 = (x – 4)2 – 12

(g) x2 – 6x + 12 = (x – 3)2 – 9 + 12 = (x – 3)2 + 3

16

4 Factorise each of the following expressions.

(a) x2 + 3x + 2 = (x + 1)(x + 2)

(b) x2 + 6x + 8 = (x + 4)(x + 2)

(c) x2 + 10x + 21 = (x + 3)(x + 7)

(d) x2 + 3x – 4 = (x + 4)(x – 1)

(e) x2 – 2x – 3 = (x – 3)(x + 1)

(f) x2 – 3x + 2 = (x – 1)(x – 2)

(g) x2 – 4x – 5 = (x – 5)(x + 1)

(h) x2 + 5x – 14 = (x + 7)(x – 2)

(i) x2 – 5x + 4 = (x – 1)(x – 4)

(j) x2 + 3x – 10 = (x + 5)(x – 2)

5 Factorise each of the following expressions.

(a) 2x2 – x – 3 = (2x – 3)(x + 1)

(b) 2x2 + 9x + 4 = (2x + 1)(x + 4)

(c) 3x2 + 4x + 1 = (3x + 1)(x + 1)

(d) 5x2 + 19x – 4 = (5x – 1)(x + 4)

(e) 5x2 – 7x + 2 = (5x – 2)(x – 1)

(f) 4x2 – 3x – 1 = (4x + 1)(x – 1)

(g) 3x2 + 8x + 5 = (3x + 5)(x + 1)

(h) 2x2 + 3x – 14 = (2x + 7)(x – 2)

(i) 4x2 – 21x + 20 = (4x – 5)(x – 4)

(j) x2 – 3x – 10 = (x – 5)(x + 2)

6 (a) x2 + 3x + 2 = (x + 1)(x + 2)

(b) x2 + 6x + 5 = (x + 5)(x + 1)

(c) x2 + 11x + 24 = (x + 3)(x + 8)

(d) x2 + 10x + 9 = (x + 1)(x + 9)

(e) x2 + 8x + 15 = (x + 5)(x + 3)

(f) x2 – 2x + 1 = (x – 1)(x – 1)

(g) x2 + 5x – 6 = (x + 6)(x – 1)

(h) x2 + 4x – 21 = (x + 7)(x – 3)

(i) x2 – 5x + 6 = (x – 3)(x – 2)

(j) x2 + 7x – 30 = (x + 10)(x – 3)

(k) x2 + 2x – 15 = (x – 3)(x + 5)

7 (a) 2x2 + x – 1 = (2x – 1)(x + 1)

(b) 2x2 + 13x + 6 = (2x + 1)(x + 6)

(c) 4x2 – 3x – 1 = (4x + 1)(x – 1)

(d) 3x2 + 19x– 14 = (3x – 2)(x + 7)

(e) 5x2 + 18x – 8 = (5x – 2)(x + 4)

(f) 8x2 + 30x – 27 = (4x – 3)(2x + 9)

(g) 12x2 + 28x – 5 = (6x – 1)(2x + 5)

(h) 12x2 – 7x + 1 = (4x – 1)(3x – 1)

8 (a) x2 – 1 = (x + 1)(x – 1)

(b) 4x2 – 25 = (2x + 5)(2x – 5)

(c) 4c2 – b2 = (2c + b)(2c – b)

(d) 16x2– 49 = (4x + 7)(4x – 7)

(e) p2 – q2 = (p + q)(p – q)

(f) 25x2 – y2 = (5x + y)(5x – y)

(g) x2 – y2 = (x + y)(x – y)

(h) y2 – 100 = (y + 10)(y – 10)

(i) 4a2 – 1 = (2a + 1)(2a – 1)

(k) c2 – 25 = (c + 5)(c – 5)

9 (a) x = –1 or –2

(b) x = –4 or –2

(c) x = –3 or –7

(d) x = –4 or 1

(e) x = 3 or –1

(f) x = 1 or 2

(g) x = 5 or –1

(h) x = –7 or 2

(i) x = 1 or 4

(j) x = –5 or 2

Worked solutions

17

10 (a) x2 – 10x + 21 = (x – 7)(x – 3) = 0 so x = 7 or 3

(b) a2 – a – 42 = (a + 6)(a – 7) = 0 so a = –6 or 7

(c) 3x2 + 11x – 4 = (3x – 1)(x + 4) = 0 so x = 13 or –4

11 (a) x2 + 4x + 1 = 0

(x + 2)2 – 4 + 1 = 0

(x + 2)2 = 3

x + 2 = ±√ 3

x = √ 3 – 2 or – √ 3 – 2

x = –0.27 or –3.73 (2 d.p.)

(b) 2x2 + 4x – 5 = 0

2(x2 + 2x – 52) = 0

Dividing both sides by 2 we obtain.

x2 + 2x – 52 = 0

(x + 1)2 – 1 – 52 = 0

(x + 1)2 = 3.5

x + 1 = ±√ 3.5

x = √ 3.5 –1 or –√ 3.5 – 1

x = 0.87 or –2.87 (2 d.p.)

12 x2 + 8x – 12 = 0

Completing the square we obtain

(x + 4)2 – 16 – 12 = 0

(x + 4)2 – 28 = 0

(x + 4)2 = 28

x + 4 = ±√ 28

x = ±√ 28 – 4

x = √ 28 – 4 or –√ 28 – 4

x = 1.29 or –9.29 (2 d.p.)


18

13 (a) Comparing the equation 3x2 – 4x + 6 = 0, with ax2 + bx + c = 0 gives a = 3, b = –4 and c = 6.

b2 – 4ac = (–4)2 – 4(3)(6)

= 16 – 72

= –56

As b2 – 4ac < 0, there are no real roots.

(b) Comparing the equation 3x2 + 6x + 2 = 0, with ax2 + bx + c = 0 gives a = 3, b = 6 and c = 2.

Substituting these values into the quadratic equation formula gives:

x = –6 ± √ (6)2 – 4(3)(2)2(3)

= –6 ± √ 36 – 246 = –6 ± √ 12

6 = –6 + √ 126 or –6 – √ 12

6

= –0.42 or –1.58 (2 d.p.).

>>> TIPBe careful here. It is easy to substitute the values for part (a) into this equation.

Worked solutions

19

Test yourself

1 x2 – 3x + 14 = (x –

32)2

– 94 +

14 = (x –

32)2

– 2

2 (a) x2 – 8x + 12 = (x – 4)2 – 16 + 12 = (x – 4)2 – 4

Hence a = –4 and b = –4.

(b) x2 – 8x +12 = 0

(x – 4)2 – 4 = 0

(x – 4)2 = 4

Taking the square root of both sides gives

(x – 4) = ±2

Hence, x – 4 = 2 or x – 4 = –2

x = 2 + 4 or x = –2 + 4

x = 6 or x = 2

3 (a) x2 + 4x + 12 = (x + 2)2 – 4 + 12 = (x + 2)2 + 8

Hence a = 2 and b = 8.

(b) The least value of (x + 2)2 + 8 is 8 and this occurs when x = –2.

4 (i) x2 + 8x – 9 = (x + 4)2 – 16 – 9 = (x + 4)2 – 25

Hence a = 4 and b = –25.

(ii) The minimum point on the curve y = x2 + 8x – 9 will be at (–4, –25)

Hence the least value is –25 and this occurs when x = –4.

5 (a) x2 + 2x + 1 = (x + 1)(x + 1)

(b) x2 + 5x + 6 = (x + 3)(x + 2)

(c) 2x2 + 3x + 1 = (2x + 1)(x + 1)

(d) 3x2 + 10x + 3 = (3x + 1)(x + 3)

(e) x2 – x – 2 = (x – 2)(x + 1)

(f) x2 + 3x – 4 = (x + 4)(x – 1)

(g) x2 + x – 12 = (x – 3)(x + 4)

(h) x2 – 6x + 5 = (x – 1)(x – 5)

(i) x2 – 2x – 35 = (x – 7)(x + 5)

One way to do this is to consider the graph of the function. The minimum point would occur at (–2, 8). The least value is the y-value (i.e. 8) and the value of x for which this occurs is the x-coordinate(i.e. –2). Another way is to look at the function and spot that the bracket squared will only be zero or positive no matter what the value of x is. Whatever the value of the bracket squared is, it will be added to the 8. The smallest value of the whole function would be if the bracket squared was zero and this would occur if x = –2. In this case nothing would be added to the 8 so the least value of x2 + 4x + 12 would be 8.


20

6 (a) 3x2 + 5x + 2 = (3x + 2)(x + 1)

(b) 4x2 + 5x + 1 = (4x + 1)(x + 1)

(c) 5x2 + 21x + 4 = (5x + 1)(x + 4)

(d) 20x2 + 17x + 3 = (4x + 1)(5x + 3)

(e) 3x2 + 11x – 4 = (3x – 1)(x + 4)

(f) 5x2 – 34x – 7 = (5x + 1)(x – 7)

(g) 7x2 – 31x + 12 = (x – 4)(7x – 3)

(h) 6x2 – 5x + 1 = (2x – 1)(3x – 1)

(i) 4x2 + 19x – 30 = (4x – 5)(x + 6)

(j) 8x2 – 49x + 6 = (8x – 1)(x – 6)

(k) 12x2 – 31x + 7 = (3x – 7)(4x – 1)

(l) 9x2 + 89x – 10 = (9x – 1)(x + 10)

7 5x2 – 20x + 10 = 5(x2 – 4x + 2)

= 5[(x – 2)2 – 4 + 2]

= 5(x – 2)2 – 10

Giving a = 5, b = –2 and c = –10.

8 y = x + 4 and y = x2 – 7x + 20

x + 4 = x2 – 7x + 20

x2 – 8x + 16 = 0

(x – 4)(x – 4) = 0

(x – 4)2 = 0

There is just one solution to this equation which proves that the straight line and curve touch.

Solving gives x = 4

When x = 4, y = 4 + 4 = 8

Hence the coordinates of the point of contact are (4, 8).

Remember that the coefficient of x2 needs to be 1 before you complete the square. Here it is necessary to take a 5 out of the square bracket as a factor.

If the line and the curve touch then the resulting equation will have a repeated root.

There is only one solution to the quadratic which means the straight line and curve touch at only one point. The x-coordinate is substituted in the equation of

the line to find the corresponding y-coordinate.

Worked solutions

21

Worked solutions

Progress checks

1 f(x) = x3 – 7x – 6

f(3) = (3)3 – 7(3) – 6 = 27 – 21 – 6 = 0

As f(3) = 0 then (x – 3) is a factor of the function.

2 g(x) = 2x3 – 7x2 + 3x + 1

g(1) = 2(1)3 – 7(1)2 + 3(1) + 1

= 2 – 7 + 3 + 1

= –1

Hence, remainder = –1.

3 f(x) = x3 – 4x2 + x + 8

f(–2) = (–2)3 – 4(–2)2 + (–2) + 8

= –8 – 16 – 2 + 8

= –18

Hence, remainder = –18.

4 f(x) = x3 – 2x2 + 6

f(2) = (2)3 – 2(2)2 + 6

= 8 – 8 + 6

= 6

Hence, remainder = 6.

5 Let f(x) = x3 + 4x2 + x – 6

If (x – 1) is a factor of x3 + 4x2 + x – 6 then f(1) = 0.

Now, f(1) = (1)3 + 4(1)2 + (1) – 6

= 1 + 4 + 1 – 6

= 0

As f(1) = 0, (x – 1) is a factor of x3 + 4x2 + x – 6.

Topic

3

1 Algebra: Topic 3 – The remainder and factor theorems and solving cubic equations

22

6 (a) (x + 2)(x2 + x +1) = x3 + x2 + x + 2x2 + 2x + 2 = x3 + 3x2 + 3x + 2

(b) (x – 4)(x2 – 3x +1) = x3 – 3x2 + x – 4x2 + 12x – 4 = x3 – 7x2 + 13x – 4

(c) (2x – 1)(x2 + x +1) = 2x3 + 2x2 + 2x – x2 – x – 1= 2x3 + x2 + x – 1

(d) (x + 1)(x + 4)(x + 5) = (x + 1)(x2 + 9x + 20)

= x3 + 9x2 + 20x + x2 + 9x + 20

= x3 + 10x2 + 29x + 20

(e) (x – 1)(x + 2)(x + 1) = (x – 1)(x2 + 3x + 2)

= x3 + 3x2 + 2x – x2 – 3x –2 = x3 + 2x2 – x – 2

(f) (x + 3)2(x – 2) = (x + 3)(x + 3)(x – 2) = (x + 3)(x2 + x – 6)

= x3 + x2 – 6x + 3x2 + 3x – 18

= x3 + 4x2 – 3x – 18

(g) (x – 1)2(x – 3) = (x – 1)(x – 1)(x – 3) = (x – 1)(x2 – 4x + 3)

= x3 – 4x2 + 3x – x2 + 4x – 3 = x3 – 5x2 + 7x – 3

7 f(2) = 2(2)3 – 4(2)2 + 2(2) + 1

= 16 – 16 + 4 + 1

= 5

Hence, remainder = 5.

8 Let f(x) = x3 – 2x2 + ax + 6

f(–1) = (–1)3 – 2(–1)2 + a(–1) + 6

= –1 – 2 – a + 6

= – a + 3

Now as (x – 1) is a factor, f(–1) = 0

Hence –a + 3 = 0

Solving, we have a = 3.

9 (i) f(2) = (2)3 – 5(2)2 + 7(2) – 2

= 8 – 20 + 14 – 2

= 0

(ii) x – 2 is a factor of the function.

Worked solutions

23

Test yourself1 Let f(x) = 4x3 + 3x2 – 3x + 1

f(–1) = 4(–1)3 + 3(–1)2 –3(–1) + 1 = 3

Remainder = 3

2 (a) Let f(x) = x3 + 6x2 + ax + 6.

f(–2) = (–2)3 + 6(–2)2 + a(–2) + 6 = 22 – 2a

As x + 2 is a factor, f(–2) = 0

Hence, 22 – 2a = 0

So a = 11

(b) x3 + 6x2 + 11x + 6 = (x + 2)(ax2 + bx + c)

Equating coefficients of x3 gives a = 1

Equating coefficients of x2 gives b + 2a = 6 and since a = 1 this gives b = 4

Equating constant terms gives 2c = 6 so c = 3.

x3 + 6x2 + 11x + 6 = (x + 2)(x2 + 4x + 3)

Now (x + 2)(x2 + 4x + 3) = 0

So (x + 2)(x + 1)(x + 3) = 0

Solving gives x = –2, –1 or –3.

3 (a) (i) f(–2) = (–2)3 – (–2)2 – 4(–2) + 4 = 0

(ii) As there is no remainder, (x + 2) is a factor of x2 – 4x + 4.

(b) x3 – x2 – 4x + 4 = (x + 2)(ax2 + bx + c)

Equating coefficients of x3 gives a = 1.

Equating coefficients of x2 gives –1 = b + 2a so – 1 = b + 2

Hence b = –3.

Equating constant terms gives 4 = 2c so c = 2.

Substituting these values in for a, b and c gives

x3 – x2 – 4x + 4 = (x + 2)(x2 – 3x + 2)

= (x + 2)(x – 2)(x – 1)

Now f(x) = 0 so (x + 2)(x – 2)(x – 1) = 0

Solving gives x = –2, 2 or 1

1 Algebra: Topic 3 – The remainder and factor theorems and solving cubic equations

24

Topic

4 Worked solutions

1 Algebra: Topic 4 – Problem solving and inequalities

Progress check

1 Let Amy’s age = x years so her mother’s age = 3x years.

In 12 years’ time Amy will be x + 12 years and her mother will be 3x + 12 years.

Then Amy will be half the age of her mother so x + 12 = 3x + 12

2

Multiplying both sides by 2 we obtain

2x + 24 = 3x + 12

Solving gives x = 12 years.

Amy’s age is 12 years and her mother’s age is 36.

2 The difference of the two numbers is 1 (with x being the larger number) so we can write this as x – y = 1.

The product of the numbers is 72 and this can be written as

xy = 72

From the first equation we can write x = ( y + 1) and substituting this into the second equation for x we obtain

y( y + 1) = 72

y2 + y – 72 = 0

Factorising, we obtain ( y + 9)( y – 8) = 0.

Solving, we obtain y = –9 or 8.

As the questions says that x and y are both positive, y = 8.

Now xy = 72 so 8x = 72 and hence x = 9.

Hence x = 9 and y = 8.

From the question her mother is three times older.

Both will have aged 12 years.

25

3 Let the integers be x, x + 1, x + 2.

The square of the largest integer minus the square of the smallest integer is equal to 64 can be written as (x + 2)2 – x2 = 64

So we have x2 + 4x + 4 – x2 = 64

4x + 4 = 64

4x = 60

x = 15

Hence the integers are 15, 16, 17

4 (a) x ≥ 5

(b) x ≤ 10

(c) x > –1

(d) x > 4

(e) x ≤ 50

5 (a) 1 ≤ x ≤ 8

(b) –2 ≤ x ≤ 5

(c) 1 < x ≤ 8

(d) –4 < x < 4

(e) 4 ≤ x < 10

6 (a) 3, 4, 5, 6, 7, 8, 9 10

(b) –4, –3, –2, –1, 0

(c) 1, 2, 3, 4, 5, 6, 7

(d) 16, 17, 18

(e) –2, –1, 0, 1, 2

(f) 16, 17, 18, 19

(g) –4, –3, –2, –1, 0, 1, 2, 3, 4

(h) 3, 4, 5, 6, 7, 8, 9

(i) 0, 1, 2, 3, 4, 5

(j) 12, 13, 14, 15, 16, 17, 18

7 (a) x ≤ 3 and x ≥ 8

(b) –3 < x < 7

(c) 4 < x < 14

(d) –4 ≤ x < 5

(e) x < 5 and x ≥ 8

(f) x ≤ –1 and x > 2

8 14 ≤ x ≤ 18

9 (a) 4x – 2 > 3 – x

5x – 2 > 3

5x > 5

x > 1

(b) 2(x + 1) > 8 – x

2x + 2 > 8 – x

3x > 6

x > 2

(c) 2(5x – 3) ≤ 4(x – 3)

10x – 6 ≤ 4x – 12

6x ≤ –6

x ≤ –1

(d) 4 – x < 3x + 7

4 – 4x < 7

– 4x < 3

x > – 34

(e) 9 – 5x ≤ 4 – x

9 – 4x ≤ 4

– 4x ≤ –5

x ≥ 54

(f) 5 – x < 3(x – 2)

5 – x < 3x – 6

5 – 4x < –6

–4x < –11

x > 114

26

10 x2 – 6x + 8 > 0

As the curve has a positive coefficient, it will be -shaped, cutting the x-axis at x = 4 and x = 2.

Sketching the curve for y = x2 – 6x + 8 gives the following:

We want the part of the graph which is above the x-axis.

The range of values for which this occurs are x < 2 and x > 4.

11 1 – 3x < x + 7

– 3x < x + 6

– 4x < 6

x > – 64

x ≥ – 32

y

x2 4

y = x2 – 6x + 8

The inequality sign is reversed because both sides have been divided by a negative quantity (i.e. –4).>>> TIP

If you do not cancel fractions you may lose marks.

Worked solutions

27

12 5x2 + 7x – 6 ≤ 0

Considering the case where 5x2 + 7x – 6 = 0

Factorising gives (5x – 3)(x + 2) = 0

Giving x = 35 or –2 (these are the intercepts on the x-axis)

As the curve y = 5x2 + 7x – 6 has a positive coefficient of x2 the curve will be -shaped.

Sketching the curve for y = 5x2 + 7x – 6 gives the following:

We want the part of the graph which is below or on the x-axis.

Meaning that x lies between –2 and 35 inclusive, which can be written

mathematically as –2 ≤ x ≤ 35 .

y

x

O–2 35

y = 5x2 + 7x – 6


28

13 y = x + 4 and y = x2 – 7x + 20

x + 4 = x2 – 7x + 20

x2 – 8x + 16 = 0

(x – 4)(x – 4) = 0

(x – 4)2 = 0

There is just one solution to this equation which proves that the straight line and curve touch.

Solving gives x = 4

Putting x = 4 into the equation of the straight line

y = 4 + 4 = 8

Hence, the coordinates of the point of contact are (4, 8).

An alternative method for proving that the curve and straight line touch at one point is to find the discriminant and show that it equals zero.

For example, the equation x2 – 8x + 16 = 0 has discriminant b2 – 4ac = (–8)2 – 4(1)(16) = 64 – 64 = 0. This shows there are two real equal roots showing the curve and straight line touch at a single point.

14 Substituting y = 5x + 13 into x2 + y2 = 13 for y gives

x2 + (5x + 13)2 = 13

x2 + 25x2 + 65x + 65x +169 = 13

26x2 + 130x + 156 = 0

Dividing through by 13 we obtain

2x2 + 10x + 12 = 0

(2x + 4)(x + 3) = 0

Hence, x = –2 or –3

Substituting these two values into the linear equation y = 5x + 13 to find the corresponding y-coordinates we obtain

When x = –2, y = 5(–2) + 13 = 3

When x = –3, y = 5(–3) + 13 = –2

Hence, the two points are (–2, 3) and (–3, –2).

If the line and the curve touch then the resulting equation will have a repeated root.

There is only one solution to the quadratic which means the straight line and curve touch at only one point.

Worked solutions

29

15 Solving the two equations y = 3x + 6 and y = x2 – 2x + 1 simultaneously by equating the y-values, we obtain

3x + 6 = x2 – 2x + 1

0 = x2 – 5x – 5

Comparing the equation above, with ax2 + bx + c = 0 gives a = 1, b = –5 and c = –5.

Substituting these values into the quadratic equation formula

x = –b ± √ b2 – 4ac2a

gives:

x = 5 ± √ (–5)2 – 4(1)(–5)2(1)

x = 5 ± √ 25 + 202

= 5 ± √ 452

= 5 + √ 452

or = 5 – √ 452

= 5.85 or –0.85 (2 d.p.)

Test yourself1 (a) 3x – 2 > 7

3x > 9

x > 3

(b) 3(x – 2) > 9

3x – 6 > 9

3x > 15

x > 5

(c) x – 5

7 ≤ –3

x – 5 ≤ –21

x ≤ –16

(d) 3x – 4 < 4x + 6

–x – 4 < 6

–x < 10

x > –10

Note there is no point in trying to factorise this as the question asks that the answer be given to two decimal places. You have to solve this quadratic equation by either completing the square or using the formula. Here we will use the formula.









Add 4 to both sides.Divide both sides by minus 1 and remember to reverse the inequality.


30

2 (a) 2x – 4 > x + 6

x – 4 > 6

x > 10

(b) 4 + x < 6 – 4x

4 + 5x < 6

5x < 2

x < 25

x < 0.4

(c) 2x + 9 ≥ 5(x – 3)

2x + 9 ≥ 5x – 15

–3x + 9 ≥ –15

–3x ≥ –24

x ≤ 8

3 Rearranging the inequality we have x2 – 3x – 18 > 0

Considering the case where x2 – 3x – 18 = 0

Factorising gives (x + 3)(x – 6) = 0

Giving x = –3 or 6 (these are the intercepts on the x-axis)

As the curve y = x2 – 3x – 18 has a positive coefficient of the curve will be -shaped.

Sketching the curve for y = x2 – 3x – 18 gives the following:

Hence, x < –3 and x > 6.

Subtract x from both sides.


Add 4x to both sides.






Divide both sides by –3 remembering to reverse the inequality in the process.

y

x–3 6O

y = x2 – 3x – 18

Note we need the section of the curve which is above (and not on) the x-axis.

Worked solutions

31

4 Factorising x2 – 2x – 15 = 0 gives

(x – 5)(x + 3) = 0

Hence x = 5 or –3

As the coefficient of x2 is positive, the graph of x2 – 2x – 15 is -shaped.

Now x2 – 2x – 15 ≤ 0. This is the region below the x-axis (i.e. where y ≤ 0).

Hence – 3 ≤ x ≤ 5

5 (a) 5 < 2x – 1 ≤ 13

6 < 2x ≤ 14

3 < x ≤ 7

(b) –7 < 3x – 5 < 4

–2 < 3x < 9

– 23 < x < 3

(c) 4(x – 3) ≤ 3(x – 2)

4x – 12 ≤ 3x – 6

x – 12 ≤ – 6

x ≤ 6

TAKE NOTENote with practice you may find you do not need to draw the curve, which will save you a bit of time.

Add 1 to each side.

Divide each side by 2.

Add 5 to each side.

Divide each side by 3.





32

Topic

5 Worked solutions

1 Algebra: Topic 5 – The binomial expansion and probability

Progress check

1 (a) (85) =

n!r!(n – r)! =

8!5!3! =

8 × 7 × 63 × 2 = 56

(b) (73) =

n!r!(n – r)! =

7!3!4! =

7 × 6 × 53 × 2 = 35

(c) (126 ) =

n!r!(n – r)! =

12!6!6! =

12 × 11 × 10 × 9 × 8 × 76 × 5 × 4 × 3 × 2 × 1 = 924

(d) 7C2 = n!

r!(n – r)! = 7!

2!5! = 7 × 62 × 1 = 21

(e) 12C5 =

n!r!(n – r)! =

12!5!7! =

12 × 11 × 10 × 9 × 85 × 4 × 3 × 2 × 1 = 792

(f) 4C2 =

n!r!(n – r)! =

4!2!2! =

4 × 32 × 1 = 6

2 (a) (10) = 1

(b) (21) = 2

(c) (32) = 3

(d) (103 ) = 120

3 (a) (50) = 1

(b) (51) = 5

(c) (85) = 56

(d) (105 ) = 252

4 The formula is as follows:

(a + b)n = an + nan – 1b + n(n – 1)

2! an – 2b2 + n(n – 1)(n – 2)

3! an – 3b3 + …

TAKE NOTEYou have to use the binomial theorem here as it is specified in the question. If you found the answer by multiplying out the brackets you would not gain any marks.

33

Here n = 3, a = 3 and b = 2x.

(3 + 2x)3 = 33 + 3(3)2(2x) + (3)(2)

2! 31(2x)2 + (3)(2)(1)

3! 30(2x)3

= 27 + 54x + 36x2 + 8x3

5 The formula is as follows:

(a + b)n = an + (n1)an – 1b + (n

2)an – 2b2 + … + (nr)an – rbr + … + bn

Here n = 6, a = x and b = 3x

.

Substituting in the values for a, b and n we obtain

(x + 3x)6

= x6 + (61) x5 (3

x) + (62) x4 (3

x)2 + (6

3) x3 (3x)3

+ …

Term in x2 = (62) x4 (3

x)2

To find the coefficients we will expand Pascal’s triangle.

1

1 1

1 2 1

1 3 3 1

1 4 6 4 1

1 5 10 10 5 1

1 6 15 20 15 6 1

As (62) = 15, we have term in x2 = 15x4(3

x)2 = 135x2

6 Obtaining the formula and following the pattern in the terms gives:

(a + b)n = an + (n1)an – 1b + (n

2)an – 2b2 + (n3)an – 3b3 + …

(a + b)5 = a5 + (51)a4b + (5

2)a3b2 + (53)a2b3 + (5

4)ab4 + (55)b5

Putting a = x, b = 2x

, and n = 5 into the equation, gives:

(x + 2x)5

= x5 + 5x4(2x) + 10x3(2

x)2 + 10x2(2

x)3 + 5x(2x)4

+ (2x)5

= x5 + 10x3 + 40x + 80x

+ 80x3 +

32x5

Looking at the above it can be seen that the term in x2 is the third term in the expansion.

The last line of Pascal’s triangle shows the line we need as we need the second number in the line to be a 6 which is the power to which the bracket is to be raised.

The coefficients of this expansion can be found using Pascal’s triangle or worked out using a calculator.


34

7 (a) (a + b)6 = a6 + 6a5b + 15a4b2 + 20a3b3 +15a2b4 + 6ab5 + b6

(1 + x)6 = 1 + 6x + 15x2 + 20x3 +15x4 + 6x5 + x6

(b) (1.02)6 = (1 + 0.02)6

Hence we substitute a = 1 and b = 0.02 into the expansion from part (a).

(1 + 0.02)6

= 16 +6(1)5(0.02) + 15(1)4 (0.02)2 + 20(1)3 (0.02)3 + 15(1)2 (0.02)4 + 6 (1)(0.02)5 + (0.02)6

= 1.1262 (4 d.p.)

8 (i) P(X = r) = (nr)pr(1 – p)n – r

p = 0.4, n = 10 and r = 5.

P(X = 5) = (105 )0.45(1 – 0.4)5

P(X = 5) = (105 )0.45(0.6)5

= 0.2007

= 0.201 (3 s.f. )

(ii) Less than 2 female turtles means 0 or 1 turtle.

P(X = 0) = (100 )0.40(1 – 0.4)10 – 0

= (100 )0.40(0.6)10

= 0.006 047

= 0.006 05 (3 s.f.)

P(X = 1) = (101 )0.41(1 – 0.4)10 – 1

= (101 )0.41(0.6)9

= 0.040 31

= 0.0403 (3 s.f.)

Now, P(X = 0 or 1) = P(X = 0) + P(X = 1)

= 0.006 047 + 0.040 31

= 0.046 357

Probability of less than 2 female turtles

= 0.0464 (3 s.f.)

Note we have used Pascal’s triangle here to determine the coefficients in the expansion which are 1 6 15 20 15 6 1. You could have alternatively used the formula to find these.

We need to find the probability of each of the above and then add the two probabilities together.

Worked solutions

35

9 (a) P(X = 3) = (16)3

= 1

216

= 0.004 630

(b) P (X = 0) = (56)3

= 125216 = 0.5787

(c) Note that at least 2 sixes means 2 or 3 sixes.

We first find the probability of 2 sixes and then add this to the probability of obtaining 3 sixes already found in part (a).

P(X = 2) = (32)(1

6)2(1 – 16)3

– 2

= (32)(1

6)2(56)1

= 0.069 44

Hence, probability at least 2 sixes = P(2 sixes) + P(3 sixes)

= 0.069 44 + 0.004 630

= 0.07407

= 0.0741 (3 s.f.)

Test yourself

1 (a + b)n = an + (n1)an – 1b + (n

2)an – 2b2 + …

(nr) =

n!r!(n – r)!

The term in x2 is given by:

(n2)an – 2b2

Here a = 2, b = 3x and n = 5

So the term in x2 is 5!

2!(5 – 2)! (2)3(3x)2 = 10 × 8 × 9x2 = 720x2

Hence, the coefficient of x2 is 720.

You could use the binomial formula but here we can use the AND law and the fact that the probability of throwing one six is 16. Note the probability of not

getting a six = 1 – 16 = 56.


36

2 (1 + x)n = 1 + nx + n(n – 1)x2

2! + n(n – 1)(n – 2)x3

3! +…

In this case we substitute 3x for x and 6 for n.

Hence (1 + 3x)6 = 1 + (6)(3x) + (6)(5)(3x)2

2 × 1 + (6)(5)(4)(3x)3

3 × 2 × 1

= 1 + 18x + 135x2 + 540x3

3 P(X = r) = (nr)pr(1 – p)n – r

p = 0.25 and n = 20.

P(X = 4) = (204 )0.254(1 – 0.25)20 – 4

= (204 )0.254(0.75)16

= 0.1897 (4 s.f.)

4 P(X = 15) = (2015)0.815(1 – 0.8)5

= 0.1746 (4 s.f.)

5 P(X = r) = (nr)pr(1 – p)n – r

Now p = 0.12, n = 10, r = 1, so we have

P(X = 1) = (101 )0.121(1 – 0.12)10 – 1

P(X = 1) = (101 )0.121(0.88)9

= 0.3798 (4 s.f.)

Worked solutions

37

Topic

6 Worked solutions

Progress checks

1 (a) Negative

(b) Zero

(c) Positive

(d) Negative

(e) Positive

(f) Infinite

(g) Negative

(h) Zero

2 (a) Gradient = y2−y1

x2−x1 = 9−34−2 =

62=3

(b) Gradient = y2−y1

x2−x1 = 12−09−1 =

128 =

32

(c) Gradient = y2−y1

x2−x1 =

0−3−5−(−2) =

−3−3 = 1

(d) Gradient = y2−y1

x2−x1 =

1−10−5−(−1) =

−9−4 =

94

(e) Gradient = y2−y1

x2−x1 = 6−(−3)1−0 =

91 = 9

(f) Gradient = y2−y1

x2−x1 = 2−(−2)4−1 =

43

(g) Gradient = y2−y1

x2−x1 = −5−(−4)−1−10 =

−1−11 =

111

3 (a) (x1 + x2

2 , y1 + y2

2 ) = (1+32 , 2 + 8

2 )=(2,5)(b) (x1 + x2

2 , y1 + y2

2 ) = (0+42 , 2 + 1

2 ) = (2, 32)

(c) (x1 + x2

2 , y1 + y2

2 ) = (−2+02 , 5+(−5)

2 )=(−1,0)(d) (x1 + x2

2 , y1 + y2

2 ) = (−8+(−2)2 , 4+(−6)

2 )=(−5,−1)

2 Coordinate geometry: Topic 6 – Coordinate geometry of straight lines

38

(e) (x1 + x2

2 , y1 + y2

2 ) = (10+(−3)2 , 12+0

2 ) = (72 , 6)

(f) (x1 + x2

2 , y1 + y2

2 ) = (−3+(−4)2 , −4+6

2 ) = (−72 , 1)

(g) (x1 + x2

2 , y1 + y2

2 ) = (8+(−5)2 , −1+7

2 ) = (32 ,3)4 (a) √(x2−x1)2 + (y2−y1)2 = √(5−1)2+(9−5)2 = √32=5.66

(b) √(x2−x1)2 + (y2−y1)2 = √(6−3)2+(9−4)2 = √34=5.83

(c) √(x2−x1)2 + (y2−y1)2 = √(6−1)2+(12−0)2 = √169=13.0

(d) √(x2−x1)2 + (y2−y1)2 = √(2−(−3))2+(6−2)2 = √41=6.40

(e) √(x2−x1)2 + (y2−y1)2 = √(0−(−5))2+(4−0)2 = √41=6.40

(f) √(x2−x1)2 + (y2−y1)2 = √(0−(−12))2+(10−5)2 = √169=13.0

(g) √(x2−x1)2 + (y2−y1)2 = √(−7−(−6))2+(2−(−3))2 = √26=5.10

5 (a) A to B = ( 3−1) C to D = ( 3−1) Hence AB = CD

(b) A to B = (51) C to D = (51) Hence AB = CD

(c) A to B = ( 6−1)

C to D = ( 6−1)

Hence AB = CD

(d) A to B = ( 6−1)

C to D = (61)

Hence AB = CD

Worked solutions

39

(e) A to B = (−24 ) C to D = (−4−2) Hence AB = CD

(f) A to B = (61)

C to D = (16)

Hence AB = CD

(g) A to B = (12)

C to D = (21)

Hence AB = CD

6 (a) √(x2−x1)2 + (y2−y1)2 = √(6−1)2+(7−4)2 = √34=5.83

(b) √(x2−x1)2 + (y2−y1)2 = √(5−0)2+(17−5)2 = √169=13.0

(c) √(x2−x1)2 + (y2−y1)2 = √(−3−[−1])2+(0−[−5])2 = √29=5.39

(d) √(x2−x1)2 + (y2−y1)2 = √(2−4)2+(4−[−1])2 = √29=5.39

7 (a) m=3,c = 2

(b) m = 2, c=3

(c) m = 23, c = 1

(d) m=−43, c=3

(e) m = 12, c=−

32

(f) m = 12, c=−

43

8 The mid-point of a line joining the points (x1, y1) and (x2, y2) is given by:

(x1 + x2

2 , y1 + y2

2 ) Mid-point of AB = (−5+12 ,

12 + 42 )=(−2,8)


40

9 (a) Mid-point of PQ, M = (−4+42 , 3+5

2 )=(0,4)(b) P to Q = (8

2) PQ2 = 82 + 22

PQ2 = 64 + 4

PQ2 = 68

PQ2 = 4 × 17

PQ2 = 2 √17

10 (a) Length = √(x2−x1)2 + (y2−y1)2 = √(6−1)2+(1−(−2))2

= √25+9 = √34=5.83

(b) Length = √(x2−x1)2 + (y2−y1)2 = √(0−[−4])2+(−3−0)2

= √16 + 9 = √25=5

(c) Length = √(x2−x1)2 + (y2−y1)2 = √(4−0)2+(7−8)2

= √16 +1 = √17=4.12

11 Let the coordinates of B be (x, y).

The x-coordinate of the mid-point = x + 2

2

The x-coordinate is 4, so 4 = x + 2

2

Solving gives x = 6

The y-coordinate of the mid-point = y+3

2

The y-coordinate is 4, so 4 = y+3

2 Solving gives y=5

Hence, coordinates of Bare(6,5)

12 Gradient of line y2−y1

x2−x1 =

k−25−0 =

k−25

Gradient of AB = 45, so

k−25 =

45

Remember that the mid-point of the line joining the points (x1, y1) and (x2, y2) is

(x1 + x2

2, y1 + y2

2 ).

Worked solutions

41

Solving, gives k = 6

13 (a) Gradient = 4

(b) Gradient = 32

(c) Gradient = 14

(d) Gradient = 23

14 (a) As point Bliesontheline,itscoordinateswillsatisfytheequationoftheline.

Hence,5(2)−2(k) = 2

Solving gives k = 4

(b) Mid-point = (x1 + x2

2 , y1 + y2

2 ) = (4 + 22 ,

9 + 42 )=(3,6.5)

15 4x−3y = c

Substitutingthecoordinates(−3,2)intothisequationgives

4(−3)−3(2)=c

Solving gives c=−18.

Hence required equation of line is 4x−3y=−18

16 (a) Gradient of AB = 3–13–1 = 1

Gradient of BC = 0–36–3=–1

(b) Productofthegradients=(1)(−1)=−1.

As m1m2=−1,AB and BCareperpendiculartoeachother.

17 (a) Gradient of PQ = 0–64–0=–

32

For equation of line PQ,

y –y1 = m(x –x1) where m =–32 and (x1, y1)=(0,6)

y –6=–32 (x –0)

2y–12=–3x

2y=–3x + 12

Both gradients are found using the formula:

Gradient = y2 − y1

x2 − x1


42

(b) Mid-point of PQ = (x1 + x2

2 , y1 + y2

2 ) = (0+42 , 6+0

2 )=(2,3) Gradient of line perpendicular to PQ is given by

m × (–32)=−1 m =

23

Equation of line perpendicular to PQ,

y –y1 = m(x – x1) where m = 23 and (x1, y1)=(2,3)

y –3 = 23 (x –2)

3y –9=2x –4

3y = 2x +5

18 (a) y = 6x−5

y=6(1)−5

y = 1 which is the y-valueofthepointsothepointliesontheline.

(b) y = 6x−5

y=6(0)−5

y=−5whichisthey-valueofthepointsothepointliesontheline.

(c) y = 6x−5

y=6(2)−5

y=12−5

y = 7 as this is not the y-coordinateofthepoint,thepointdoesnotlieontheline.

(d) y = 6x−5

y = 6(12)−5

y=3−5

y=−2

This is the y-valueofthepoint,sothepointliesontheline.

(e) y = 6x−5

y=6(−1)−5

y=−11

This is not the y-valueofthepoint,sothepointdoesnotlieontheline.

Worked solutions

43

19 (a) y = 4x−1

y=4(0)−1

y=−1

(b) y=3x+5

y=3(0)+5

y=5

(c) 4x−2y=0

4(0)−2y=0

−2y=0

y=0

(d) 5y−x = 2

5y−0=2

y = 25

(e) 2x + y−1=0

2(0)+y−1=0

0+y−1=0

y = 1

(f) y−x−3=0

y−0−3=0

y =3

20 (a) y = 4x + 2

0=4x + 2

−2=4x

x =–12

(b) y=−3x+15

0=−3x+15

−15=−3x

x=5

(c) 3x + 2y = 12

3x+2(0)=12

x = 4

(d) 5y−x = 9

5(0)−x = 9

x=−9

(e) 5x−7y=25

5x −7(0)=25

5x=25

x=5

(f) x−y+7=0

x−0+7=0

x=−7

(g) 5x−3y−10=0

5x−3(0)−10=0

x = 2


44

21 7x + 2y = 19 .................................... (1)

x –y = 4 .......................................... (2)

Multiplying equation (2) by 2 gives

2x –2y = 8

7x + 2y = 19

Adding the above two equations gives

9x = 27

x=3

Substituting x=3intoequation(1)weobtain

7(3)+2y = 19

21 + 2y = 19

2y=−2

y=−1

Checking by substituting x =3andy=−1intoequation(2)

LHS=3−(−1)=4

RHS = 4

So LHS = RHS

Pointofintersectionis(3,–1).

22 7y =5x–27.......................................... (1)

4y =3x–16........................................... (2)

Multiplyingequation(1)by3andequation(2)by5givesthefollowing

21y =15x–81

20y =15x–80

Subtracting these two equations gives

y=−1

Substituting y=−1intoequation(1)gives

−7=5x−27

20=5x

x = 4

Notethat−81−(−80)=−1. Be very careful when subtracting negative numbers.

Worked solutions

45

Checking in equation (2)

LHS=4(−1)=–4

RHS=3(4)−16=–4

So LHS = RHS

Pointofintersectionis(4,–1).

23 Rearranging each equation so that it is the form y = mx + c.

y−3x+4=0soy=3x−4.Thislinehasagradientof3.

4y−12x +1=0soy =3x−14.Thislinealsohasagradientof3.

Both lines have the same gradient and are therefore parallel so they will notintersect.

24 4x + y = 1

3x + 2y = 7

Multiplyingthefirstequationby2gives

8x + 2y = 2 .................................................... (1)

3x + 2y = 7 .................................................... (2)

(1)−(2)gives

5x=−5

Hence, x=−1

Substituting x=−1intoequation(1)gives

−8+2y = 2

2y=10

y=5

Checking by substituting x=−1andy=5intoequation(2)gives

LHS=3(−1)+2(5)=−3+10=7

RHS = 7

So LHS = RHS

Coordinates of Aare(−1,5)


46

Test yourself

1 (a) Gradient of AB = y2−y1

x2−x1 = 1–04–1 =

13

Gradient of CD = y2−y1

x2−x1 =

4–32–(–1) =

13

As the gradients of AB and CDarethesamethetwolinesareparallel.

(b) Gradient of AB = 13 and AB passes through point A(1,0)soequationofAB is:

y –y1 = m(x –x1)

y –0=13 (x–1)

3y = x –1

Rearranging this equation so that it is in the form asked for by the question gives:

x –3y –1 =0

2 (a) Gradient of AB = y2−y1

x2−x1 =

–1–4k–(–7) =

–5k + 7

But gradient of AB=–12 so

–5

k + 7=–12

–5×2=–1(k + 7)

–10=–k–7

Giving k=3.

(b) Theproductofthegradientsofperpendicularlinesis–1.Hence,

m(–1 2)=–1 Hence gradient of BC = 2

Equation of BC is:

y –y1 = m(x –x1) where m = 2 and (x1, y1)=(3,–1).

y –(–1)=2(x–3)

y + 1 = 2x–6

2x –y –7 =0

The equation is multiplied through by the common denominator, 2(k +7).

Worked solutions

47

3 (a)

Gradient of AB = y2−y1

x2−x1 =

6–21–(–3) =

44 = 1

Gradient of BC = y2−y1

x2−x1 = 1–66–1 =

–55 =–1

Productofgradients=(1)(–1)=–1provingthatthetwolinesareperpendiculartoeachother.

(b) Length = √(x2−x1)2 + (y2−y1)2

Putting the coordinates A(–3,2)and B (1, 6) into the formula gives

AB = √(1−[–3])2+(6−2)2 = √16 + 16 = √32 units

Using the coordinates B (1, 6) and C (6, 1) in the formula gives

BC = √(6−1)2+(1−6)2 = √25+25 = √50 units

(c) Tan AĈB = ABBC

= √32√50

= √ 16 × 2

√25×2 = 4√ 25√ 2

= 45

4 Theequationofthelineparallelto3x + 2y=5,willbeoftheform

3x + 2y = c.

Substitutingthecoordinates(3,4)intotheequation,weobtain

3(3)+2(4)=c

Giving c=17.

Hence, the equation of the line is

3x + 2y = 17

–3 –2 –1 1 2 3 4 5 6 7

B (1, 6)

C (6, 1)

(–3, 2) A

y

x0

6

5

4

3

2

1

TAKE NOTEThis formula needs toberemembered.Ifyou forget it you can plot the two points on a sketch graph and form a triangle and use Pythagoras' theorem tofindthelengthofthehypotenuse.


48

5 (a) Gradient of PQ = y2−y1

x2−x1 =

2–00–(–4) =

24 =

12

Equation of line PQ with gradient 12 and passing through the point

Q(0,2)is

y –y1 = m(x –x1)

y –2=12 (x –0)

2y –4=x

Hence, equation of line is 2y –x = 4

(b) Mid-point of PQ = (x1 + x2

2 , y1 + y2

2 ) = (–4+02 , 0+2

2 ) =(–2,1) Gradient of PQ =

12

Gradient of line perpendicular to PQ, m is given by

(m)(12)=–1(i.e.theproductsofthegradientsoftwo

linesare–1iftheyareperpendicular).

Hence, m=–2

Equation of required line through mid-point is:

y –y1 = m(x –x1)

y –1=–2(x –[–2])

y –1=–2x –4

y =–2x –3

6 (a)

Gradient of PQ = y2−y1

x2−x1 = 5–03–1 =

52

>>> TIPFor questions involving coordinate geometry, it is always worthwhile spending a little time sketching a graph showing the positions of thecoordinates.Itistheneasier to see the shape formed when certain points are joined up withlines.Youcanalsocheck the signs of any gradients you have found numerically.

–4 –3 –2 –1 1 2 3 4

R (–2, 6)

Q (3, 5)

P (1, 0)

y

x0

6

5

4

3

2

1

-1

-2

Worked solutions

49

(b) Equation of straight line PQwhichpassesthrough(1,0)andhas

gradient 52 is given by:

y –y1 = m(x –x1) where m = 52 and (x1, y1)=(1,0)

y –0=52 (x –1)

y = 52 (x –1)

2y =5(x –1)

2y =5x –5

2y –5x =–5

(c)

As lines RS and PQ are opposite sides of a parallelogram, they are parallelandthereforehavethesamegradient.

Hence, gradient of RS = 52.

Equation of straight line RSwhichpassesthrough(–2,6)andhas

gradient 52 is given by:

y –y1 = m(x –x1) where m = 52 and (x1, y1)=(–2,6)

y –6=52 (x –[–2])

y –6=52 (x + 2)

2y –12=5(x + 2)

2y –12 =5x +10

2y –5x = 22

Multiply both sides by 2 to remove the denominator in thefraction.

–4 –3 –2 –1 1 2 3 4

R (–2, 6)

Q (3, 5)

S

y

x0

6

5

4

3

2

1

–1

–2

P (1, 0)

Youcannowformtheparallelogram and it is possible to see from the sketch the rough position of point S.Ifyougiveitsomethought you can get the coordinates of point S using thediagram.


50

(d) TofindthecoordinatesofthepointS, the equations for lines RS and SParesolvedsimultaneously.

2y–5x = 22 ...................................................... (1)

5y + x = 1 ............................................................ (2)

Multiplyingequation(2)by5gives

25y+5x =5

2y–5x = 22

Adding the two above equations gives

27y = 27

So, y = 1

Substituting y = 1 into equation (2) and solving gives

5+x = 1

So, x=–4

Hence the coordinates of point Sare(–4,1)

7 (a) Because there are lots of points in this question, it is worth spending alittletimedoingasketchshowingtheirpositions.

1 2 3 4 5 6 7 8 9

D (5, –2)

C (9, –4)

y

x

3

2

1

0

–1

–2

–3

–4

–5

–6

–7B (6, –7)

(2, –5) A

Worked solutions

51

Gradient of AB = y2−y1

x2−x1 = –7–(–5)6–2 =

–24 =–

12

(b) Gradient of DC = y2−y1

x2−x1 = –4–(–2)9–5 =

–24 =–

12

Gradient of AB = gradient of DC, so lines are parallel

(c) The vector to go from A to B is ( 4–2).

The vector to go from D to C is ( 4–2).

Thesevectorsarethesamesothelinesarethesamelength.

(d) The vector to go from A to D is (33). The vector to go from B to C is (33). These vectors are the same so the lines are both parallel and

thesamelength.

Hence, this and the answer to parts (b) and (c) prove that ABCDisaparallelogram.

8 (a)

Mid-point, M = (x1 + x2

2 , y1 + y2

2 ) = (4 + 12 ,

2+52 ) = (52 ,

72)

(b) Gradient of AB = y2−y1

x2−x1 = 5–21–4 =

3(–3)=–1

(c) Gradient of MC = y2−y1

x2−x1 = 5–724–52

= (32)(32) = 1

Product of gradients of lines AB and MC=(–1)(1)=–1 Astheproductoftwoperpendicularlinesis–1soAB and MC areperpendicular.

The two lines at right angles which have the vector as the hypotenuse are the same for both triangles so the lengths of the hypotenusewillbethesame.

1 2 3 4 5 6

B (1, 5) C (4, 5)

M

y

x

6

5

4

3

2

1

0

A (4, 2)


52

(d) Equation of line MC with gradient 1 and passing through the point C(4,5)is

y –y1 = m(x –x1)

y –5=1(x –4)

y –5=x –4

Hence, equation of MC is y = x + 1

9 (a) 2x+3y=5

3y=–2x+5

y=– 23 x +

53

Gradientofline=– 23

(b) Equationoflinewithgradient– 23 and passing through the

point R(3,3)is

y –y1 = m(x –x1)

y –3=–23(x –3)

3y–9=–2x + 6

Hence,equationoflineis3y=–2x +15

Now, the equation of the y-axis is x=0.

Solving this simultaneously with the equation of the line, we have

3y=0 +15soy=5

Hence, Sisthepoint(0,5)

10 (a) 4x+5y=10

5y=–4x+10

y=–45 x + 2

Hence, gradient of AB=–45

Worked solutions

53

(b) Substituting the coordinates of C into the equation of the line we obtain

LHS=4(–5)+5(6)

=–20+30

=10

RHS=10

As LHS = RHS, point Cliesontheline.

(c) As lines are perpendicular, m ×–45=–1,som =

54

Gradient of line at right-angles to AB = 54

Equation of line with gradient 54 and passing through the

point C(–5,6)is

y –y1 = m(x –x1)

y –6=54 (x –[–5])

4y –24=5(x +5)

4y –24=5x +25

Hence, equation of line is 4y =5x + 49


54

Topic

1 Worked solutionsTopic

7 Worked solutions

2 Coordinate geometry: Topic 7 – Coordinate geometry of the circle

Progress check

1 (a) x2 + y2 = 1

This equation is in the form x2 + y2 = r2

Hence r = √ 1 = 1

(b) x2 + y2 = 9


Hence r = √ 9=3

(c) x2 + y2=25


Hence r = √25=5 (d) x2 + y2–4=0

x2 + y2 = 4


Hence r = √ 4 = 2

(e) x2 + y2–49=0

x2 + y2 = 49


Hence r = √ 49 = 7

(f) 4x2 + 4y2 = 16

Dividing both sides by 4 we obtain

x2 + y2 = 4


Hence r = √ 4 = 2

Note that the radius of a circle cannot be zero asitisalength.

55

(g) 8x2 + 8y2 = 72

Dividing both sides by 8 we obtain

x2 + y2 = 9


Hence r = √ 9=3

(h) 3(x2 + y2)–27=0

Dividingbothsidesby3weobtain

x2 + y2–9=0

x2 + y2 = 9


Hence r = √ 9=3

(i) y2=16–x2

x2 + y2 = 16


Hence r = √ 16 = 4

(j) x2 + y2–5=0

x2 + y2=5


Hence r = √5

(k) x2 + y–50=0

x2 + y2=50


Hence r = √50 = √25×2=5√ 2

2 (a) x2 + y2 = 9

(b) x2 + y2 = 16

(c) x2 + y2 = 81

(d) x2 + y2 = 6

(e) x2 + y2 = 12

(f) x2 + y2=45

(g) x2 + y2 = 18

3 (a) (x –3)2 + (y + 1)2 = 9

x2–6x + 9 + y2 + 2y + 1 = 9

x2 + y2–6x + 2y+1=0

(b) (x–2)2 + (y + 4)2 = 16

x2–4x + 4 + y2 + 8y + 16 = 16

x2 + y2–4x + 8y+4=0

(c) (x–1)2 + (y–3)2 = 1

x2–2x + 1 + y2–6y + 9 = 1

x2 + y2–2x–6y+9=0

(d) (x + 4)2 + (y–5)2=25

x2 + 8x + 16 + y2–10y+25=25

x2 + y2 + 8x–10y+16=0

(e) (x+5)2 + (y–1)2 = 9

x2+10x+25+y2–2y + 1 = 9

x2 + y2+10x–2y+17=0

(f) (x–6)2 + (y + 7)2 = 49

x2–12x+36+y2 + 14y + 49 = 49

x2 + y2–12x + 14y+36=0

(g) (x–5)2 + (y –4)2 = 16

x2–10x+25+y2–8y + 16 = 16

x2 + y2–10x –8y +25 =0

(h) x2 + (y–1)2 = 4

x2 + y2–2y + 1 = 4

x2 + y2–2y – 3=0


56

Test yourself1 (a) Comparing the equation x2 + y2–8x–6y=0withtheequation

x2 + y2 + 2gx + 2fy + c=0wecanseeg=–4,f=–3andc =0.

Centre Ahascoordinates(–g,– f)=(4,3)

Radius = √ g2 + f 2–c = √(–4)2+(–3)2–0 = √25=5

2 (a) Comparing the equation x2 + y2–4x + 6y–3=0withtheequation

x2 + y2 + 2gx + 2fy + c=0wecanseeg=–2,f=3andc =–3.

Centre Ahascoordinates(–g,– f)=(2,–3)

Radius = √ g2 + f 2–c = √(–2)2+(3)2–(–3) = √ 16 = 4

(b) If point P (2, 1) lies on the circle its coordinates will satisfy the equationofthecircle.

x2 + y2–4x + 6x–3=0

x2 + y2–4x + 6y–3=(2)2 + (1)2–4(2)+6(1)–3=4+1–8+6–3=0

Both sides of the equation equal zero so P(2,1)liesonthecircle.

3 (a) Equation of the circle is:

(x–a)2 + (y–b)2 = r2

(x–2)2 + (y–3)2=25

x2–4x + 4 + y2–6y+9=25

x2 + y2–4x–6y–12=0

(b) Gradient of line joining the centre A(2,3)toP(5,7)

= 7–35–2 =

43

Gradientoftangent=–34

Equation of tangent is

y–7=–34 (x –5)

4y–28=–3x+15

4y+3x–43=0

AP is a radius of the circle andwillmakeanangleof90°to the tangent at point P.

Worked solutions

57

4 (a) x2 + y2–4x + 8y+4=0

Completing the squares for x and y gives

(x–2)2 + (y + 4)2–4–16+4+0

(x–2)2 + (y + 4)2–16=0

(x–2)2 + (y + 4)2 = 16

Centreofcircleisat(2,–4)

(b) If the point P(6,–4)liesonthecircle,thecoordinateswillsatisfytheequationofthecircle.

Hence, (x–2)2 + (y + 4)2=(6–2)2+(–4+4)2

= 42+0

= 16

This is the same as the RHS of the equation so the point lies onthecircle.

5 (a) Centre of circle is at the mid-point of the diameter AB.

Mid-point of line joining A(1,–4)andB(9,10)is

(1 + 92 ,

–4+102 ) =(5,3)

(b) Distancebetweenthepoints(1,–4)and(5,3)isgivenby

r = √ (x2−x1)2 + (y2−y1)2

r = √(5−1)2+(3−[−4])2

r = √ 42 + 72

r = √65

(c) The equation of a circle having centre (a, b) and radius r is given by

(x–a)2 + (y–b)2 = r2

Forthiscircle,centreis(5,3)andradiusis√65. (x–5)2 + (y–3)2=65

Multiplying out the brackets we obtain

x2–10x+25+y2–6y+9=65

x2 + y2–10x–6y–31=0

The distance from the mid-point to the circumference is theradiusofthecircle.

Make sure you give the equation of the circle in the format asked for in the question(i.e.x2 + y2 + ax + by + c=0inthiscase).


58

6 Radius of the circle = √40=6.32 The length of a straight line joining the two points (x1, y1) and (x2, y2) is

given by:

√(x2−x1)2 + (y2−y1)2

Distanceofthepoint(4,3)fromthecentreofthecircle(0,0) = √(x2−x1)2 + (y2−y1)2

= √(4–0)2+(3–0)2 = √ 16 + 9=25=5

This distance is less than the radius of the circle, so the point lies inside thecircle.

7 (a) Themid-pointofthediameterofthecircleisthecentreofthecircle.

Mid-point of AB = (0+42 , –3+1

2 ) =(2,–1) Socentreofcircleisat(2,–1).

(b) Radius, r, is distance from B (4,1)tocentreofcircle(2,–1).

r2 = (x–a)2 + (y –b)2

=(4–2)2+(1–[–1])2

= 22 + 22

= 8

r = √ 8 = √ 4 × 2 = 2√ 2

Worked solutions

59

Topic


8 Worked solutions

2 Coordinate geometry: Topic 8 – Inequalities and linear programming

Progress check

1 (a) x≤3

(b) y≥2

(c) y≤x

(d) y ≥–2x + 8

(e) y ≥12 x+3

2 (a) Region 2

(b) Region5

(c) Region 1

(d) Region3

(e) Region 4

3 Youfirsthavetoobtainsomepointstobeabletodraweachline.Notethatas the inequalities all have an equals component in the question, all the lineswillbesolidlines.

y=0(Thislineisthex-axis)

x = 2 (This is a vertical line at x = 2)

y = 13 x ( When x=0,y=0andchoosingasuitablepointonthex-axis such as x = 12 so y = 13[12]=4)

2y=12–x (When x=0,y = 6 and when y=0,x =12)

Adding the above lines to the graph, we obtain the following:

x

y

121110

9876543210

1 2 3 4 5 6 7 8 9 10 11 12

2y = 12 – x

x = 2

y = x13

60

We now add the shadings to each line to show the sides of the line that do notsatisfytheinequality.

For y≥0weshadebelowthex-axis.

For x≥2weshadetotheleftofthelinex=2.

For y≤13 x we shade above the line y =

13 x.

For 2y≤12–x we shade above the line 2y=12–x.

Adding the shadings to the lines we now have the following:

Be careful to select the correct region as the region whichfitsalltheinequalities(i.e.theallowableregion).Youarelookingfortheregionwhichiscompletely enclosed by lines where the shading is ontheoppositesidetotheregion.

We now shade in all the regions on the graph not in thefeasibleregion.

Remember we are shading theareanotrequired.

x 1 2 3 4 5 6 7 8 9 10 11 12

2y = 12 – xFEASIBLE REGION

y = x13

y

121110

9876543210

x = 2

x 1 2 3 4 5 6 7 8 9 10 11 12

2y = 12 – xFEASIBLE REGION

y = x13

y

121110

9876543210

x = 2

Worked solutions

61

4 5 (a) 3x + 4y≤100

(b) 20x+30y≤1500

(c) y<3x

6 (a) x≤500

y≤500

x + y ≤800

x≥13 y

(b) Amount of money per week=0.8x+0.9y

Note than no more than meanslessthanorequalto.

Note that the cost of each cookie must be converted into pounds, so we divide each cost in penceby100.

7 Regarding the constraints on the purchase costs, we have

20000y+15000x≤500000.

Note that you could simplify this by dividing both sides of the inequality by5000togive4y +3x≤100.

Regarding the constraints made by the popularity of the vans, we have

y≥3x

Regarding the constraints of the running costs, we have

40x+50y≤1200

Notethatthisinequalitycouldbesimplifiedbydividingbothsides by10togive

4x+5y≤120

x

y

151413121110

9876543210

1 2 3 4 5 6 7 8 9 10 11 12 13 14

y = –1.5x + 15

2y = 10 – x

y = x

Upto100meansless than or equal to100.


62

Test yourself1 (a) x≤150

y≤120

x + y≤200

(b) For x + y=200,whenx=0,y=200andwheny=0,x=200.

x=0andy=0arethey- and x-axesrespectively.

x=150isaverticallineandy=120isahorizontalline.

Adding these lines to the graph and shading the regions that are excludedweobtain.

(c) Profit=2x+5y

Now,oneoftheverticesinthefeasible(i.e.unshaded)regionwillmaximisetheprofit.

Therearefourvertices;(150,0),(0,120),(150,50)and(80,120).

ThesecaneachbesubstitutedintotheProfitequationinturnuntilthemaximumvalueisfound.Youcanseethatsomeofthesecoordinatescanbediscounted.Forexample(150,0)and(150,50)have the same x-coordinate but the one with the higher y-coordinate willbetheonlyonetoconsider.

When x=0andy=120theprofitis2(0)+5(120)=£600.

When x=80andy=120theprofitis2(80)+5(120)=£760andthisisthehighestvalue.

Hence,80ofbagAand120ofbagBshouldbeproduced.

x

y

200

175

150

125

100

75

50

25

0 25 50 75 100 125 150 175 200

x + y = 200

y = 120

x = 150

The values of x and y are substituted into Profit=2x+5y.

Worked solutions

63

2 (a) For 4x + y = 24, when x=0,y = 24 and when y=0,x=6.

Adding this line and the other lines to the graph we obtain:

(b) For line 2x + y=20,whenx=0,y=20andwheny=0,x=10.

Adding this line to the graph we have:

x

y

24

22

20

18

16

14

12

10

8

6

4

2

0 1 2 3 4 5 6 7 8 9 10

4x + y = 24

y = 10

x = 2

x

y

24

22

20

18

16

14

12

10

8

6

4

2

0 1 2 3 4 5 6 7 8 9 10

4x + y = 24

2x + y = 20

y = 10

x = 2


64

The smallest value of 2x + y will lie in the feasible region and be parallel to the line 2x + y=20.Thepointwilllieonorneartooneoftheverticesoftheunshadedregion.Thepoint(3,0)givesthesmallest value of 2x + y which is 6, and this occurs when x=3and y=0.

3 (a) The total number of tents cannot be more than 18, so x + y≤18.

(b) There are 72 children, so 8x+3y≥72.

(c) For x + y = 18, when x=0,y = 18 and when y=0,x=18.

For 8x+3y = 72, when x=0,y = 24 and when y=0,x=9.

x =0andy=0arethey- and x-axesrespectively.

(d) Total cost of hiring tents, is 80x+20y.

x

y

24

22

20

18

16

14

12

10

8

6

4

2

0 2 4 6 8 10 12 14 16 18

x + y = 18

8x + 3y = 72

Worked solutions

65

(e) Plottingtheline80x +20y=400,whenx=0,y=20andwhen y=0,x=5.

The minimum cost is given by the line that is parallel to the objective functiondrawninthefeasibleregionthatisfurthesttotheleft.Thepoint P does not have integer coordinates so we need to look for a point near Pwithintegercoordinates.

The point Q is near to Pandhascoordinates(4,14).Ifthesecoordinates are substituted into the expression for the cost (i.e.80x+20y),thecostis80(4)+20(14)=600.

It is worth just checking other points near to Pinthefeasibleregion.

e.g.point(5,13)givesacostof80(5)+20(13)=660 point(5,12)givesacostof80(5)+20(12)=640

Note that we require a minimum cost and both these points give a highercost.

Hence the point Q(4, 14) gives the values of x and y that give the smallestcost. Hence4largetentsand14smalltentsneedtobeused.

Notewehavechosen400asthecosthere.Wehavechosen this number to make themathseasieras80and20bothdivideinto400exactly.Alwaysmakesurethat the number you choose can be shown on the graph youhavedrawn.

x

y

24

22

20

18

16

14

12

10

8

6

4

2

0 2 4 6 8 10 12 14 16 18

x + y = 18

8x + 3y = 72

P Q

80x + 20y = 400


66

Topic


9 Worked solutions

Progress checks

1 (a) cos 30° = 15x

x = 15

cos 30° = 17.32 cm (2 d.p.)

(b) cos 40° = x

12 Hence, x = 12 cos 40°

= 9.19 cm (2 d.p.)

2 (a) tan θ = 103

θ = tan–1(103 )

= 73.3° (nearest 0.1°)

(b) sin θ = 1013

θ = sin –1(1013)

= 50.3° (nearest 0.1°)

3 sin θ = bc

(i.e. sin θ = opposite

hypotenuse) and cos θ = ac

(i.e. cos θ = adjacent

hypotenuse) Hence

sin θcos θ =

|bc|a c

= bc

× ca

= ba

Now, ba

= oppositeadjacent = tan θ

Hence, sin θcos θ = tan θ

x is the hypotenuse and the 15 cm side is the adjacent so we use cos θ = adjacent

hypotenuse .

x is the adjacent and the 12 cm side is the hypotenuse so we use cos θ = adjacent

hypotenuse .

Remember that when you divide by fractions you turn the bottom fraction upside down and replace the division by a multiplication.

3 Trigonometry: Topic 9 – Trigonometric ratios and the graphs of sine, cosine and tangent

67

4

Cos θ is positive in the first and fourth quadrants. The solution in the first quadrant is found by entering cos–1(1

2) into the calculator giving the answer θ = 60°. By symmetry, the other value is found by subtracting 60° from 360° giving the other solution θ = 300°.

Hence solutions are 60° and 300°.

5 The first angle is found by entering sin–1(√ 32 ) into the calculator.

This gives θ = 60°.

By the symmetry of the graph, the other angle is 180° – 60° = 120°.

Hence the values of θ in the required range are 60° and 120°.

6 The solution using trigonometric graphs is used here. The alternative CAST method could also have been used.

x = tan–1(–12) = –26.6°

Drawing the graph of y = tan x you can see where this angle is on the graph.

By the symmetry of the graph you can see that the line y = –1

2 cuts the tan graph in two places in the required range and that these values of x are at 180 – 26.6 = 153.4° and 360 – 26.6 = 333.4°.

Hence x = 153.4° or 333.4°.

90°

180° 0°

270°

S A

T

θ2

C

θpV

1

–1

90˚ 180˚ 270˚ 360˚ �

y

√ 32

90°–26.6°

0 180° 270° 360°

y y = tan x

x

As a check find the tan of both of these angles. You should obtain –1

2 in each case.


68

7 2sinθ = 1

sinθ = 12

θ = sin–1(12)

θ = 30°

θ = 180 – 30° = 150°

Hence θ = 30° or 150°

8 (a) θ = sin–1 (–0.23) = –13.3°

Sketching a sine graph to look for symmetry we have.

Notice the –13.3° solution. The line y = –0.23 cuts in two places in the required range. By symmetry these two solutions will be

θ = 180 + 13.3 = 193.3° or θ = 360 – 13.3 = 346.7°

Hence θ = 193° or 347° (nearest degree)

1

0.5

0 30° 150° 180° 360° �

y

1

–1

–0.23°

–13.3°

0 270°180° 360° �

y

y = sin �

180° + 13.3° 360° – 13.3°

Note you could also have used the CAST method here.

The above is not in the required range.

Worked solutions

69

(b) cos θ = – 0.72

θ = cos–1(–0.72)

= 136.1°

Drawing a sketch graph to find other solutions.

Other solution, θ = 360 – 136.1 = 223.9°

Solutions are θ = 136° or 224° (nearest degree)

(c) Tan θ = 2.45

θ = tan–1 (2.45)

θ = 67.8°

Sketching a graph to find other solutions.

Other solution, θ = 180 + 67.8 = 247.8°

Solutions are θ = 68° or 248° (nearest degree)

136.1° 360 – 136.1°

1

0

–0.72

90° 180° 360°270°

y

�

90°0 180°

2.45

67.8°270° 360°

y y = tan x

x

180 + 67.8°


70

9 The smallest angle of any triangle is always opposite the smallest side.

Drawing a sketch of the triangle and calling the smallest angle θ we have

Applying the cosine rule to this triangle we obtain

52 = 72 + 102 – 2 × 7 × 10 cos θ

25 = 49 + 100 – 140 cos θ

cos θ = 0.8857

θ = cos–1 (0.8857)

= 27.7° (nearest 0.1°)

10 (a) Using the cosine rule we obtain

132 = 112 + 92 – 2 × 11 × 9 cos θ

169 = 121 + 81 – 198 cos θ

Solving gives θ = 80.4° (nearest 0.1°)

(b) Area = 12 × 9 × 11 sin 80.4°

= 48.81 cm2

= 48.8 cm2 (1 d.p.)

11 (a) Area = 12 × 25 × 30 ×

35

= 225 cm2

�

7 cm5 cm

10 cm

The formula Area = 12 bc sin A is used here.

The formula Area = 12 bc sin A is used here. Sin θ = 35 is substituted into this equation.

Worked solutions

71

(b)

Adjacent = 4

cos θ = adjacent

hypotenuse = 45

(c) Using the cosine rule we obtain

AC2 = 252 + 302 – 2 × 25 × 30 × 45

AC2 = 325

AC = √ 325

AC = 18.0 cm (3 s.f.)

Test yourself1 (a)

(b) tan θ = 1, so θ = tan–1(1) = 45° or 180 + 45 = 225°

A right-angled triangle is drawn with the opposite of length 3 and the hypotenuse of length 5.

Ѳ

53

4

The length of the adjacent can be found by either using Pythagoras’ theorem or remembering that the sides are in the ratio of 3 : 4 : 5.

90°0 180° 270° 360°

yy = tan θ

θ

The first value of θ is found using a calculator (or the knowledge that tan 45° = 1) and the second value is found using the symmetry of the graph. The same section of the graph from 0° to 45° is repeated starting at 180° so we need to add 45° to 180° to determine the second angle (i.e. 225°).


72

2 (a) As the multiple of the angle is 2 we need to look at the range from 0 to 720°.

It is best to do a quick sketch of the sine graph like this. You can see there are two places where sin x = 1.

2x = sin–1(1)

2x = 90° or 450°

Hence, x = 45° or 225°

(b) tan x = 2

x = tan–1(2)

= 63.4°

Sketching the graph to find the other angles in the range.

The other solution is x = 180 + 63.4 = 243.4°.

Hence, x = 63.4° or 243.4°

0x

–1

180° 360° 540° 720°

y

1

To find x, these two angles are divided by 2.

90°0 180°

2

270° 360°

yy = tan x

x

Worked solutions

73

3 (a) Area = 12 bc sin A =

12 × 12 × 8 × sin 150° = 24 cm2

(b) Using the cosine rule

a2 = b2 + c2 – 2bc cos A

= 122 + 82 – 2 × 12 × 8 cos 150°

= 374.2769

a = 19.3462

a = 19.3 cm (1 d.p.)

4 (a) (0°, 0), (180°, 0), (360°, 0), (540°, 0), (720°, 0)

(b) (90°, 1), (270°, –1), (450°, 1), (630°, –1)

5 (a) Using the sine rule we obtain

x

sin 84° = 12

sin 40°

x = 12 sin 84°

sin 40°

x = 18.5664

= 18.6 cm (3 s.f.)

(b) Angle BAC = 180 – (84 + 40) = 56°

Area = 12 bc sin A =

12 × 18.6 × 12 × sin 56°

= 92.521

= 92.5 cm2 (3 s.f.)

Remember not to round off answers to the required number of decimal places until the final answer.

Look for pairs of angles and opposite sides. If you have two pairs that include one unknown (either side or angle) then use the sine rule.


74

Topic


10 Worked solutions

Progress check

1 cos x = –2 sin x

Dividing both sides by cos x, we obtain cos xcos x =

–2 sin xcos x

Now sin xcos x = tan x so 1 = –2 tan x

giving tan x = – 12

x = tan–1(– 12) = –26.6°

We can see the other solutions in the required range by sketching a graph of y = tan x.

You can see from the graph that there are two angles in the range where

tan x = – 12 .

Using the symmetry of the graph you can see that the two values of x are 180 – 26.6 = 153.4° and 360 – 26.6 = 333.4°.

Hence the solutions are 153.4° or 333.4°.

Note that this is outside the range specified in the question.

360°270°90°–90°–26.6°

180°

yy = tan x

x12

–

3 Trigonometry: Topic 10 – Trigonometric identities and solving trigonometric equations

75

2 cos2 θ + sin2 θ = 1 so sin2 θ = 1 – cos2 θ

Substituting sin2 θ = 1 – cos2 θ into the original identity gives

cos2 θ – sin2 θ = cos2 θ – (1 – cos2 θ)

= cos2 θ – 1 + cos2 θ

= 2cos2 θ – 1

3 (a)

By Pythagoras we have

(√ 5)2= x2 + 22

5 = x2 + 4

Hence x = 1.

sin θ = opposite

hypotenuse = 1

√ 5

(b) tan θ = sin θcos θ =

| 1√5|2√5

= 1

√ 5 × √ 5

2 = 12

4 6cos2 θ + 5cos θ – 6 = 0

Factorising, we obtain (3cos θ – 2)(2cos θ + 3) = 0

Hence, cos θ = 23 or cos θ = –

32 (which is ignored as smallest value is –1)

cos θ = 23 so θ = cos–1(2

3) giving θ = 48.2° or 311.8° (nearest 0.1°).

Ѳ

x

2

√ 5

This is a quadratic in cos θ. This needs to be factorised before solving.

Note that the maximum and minimum values of cos θ or sin θ are 1 and –1.

The cos θ graph or the CAST method is used to find the other value of θ.


76

5 (a) 5sin θ = 3cos θ

Dividing both sides by cos θ we obtain

5 sin θcos θ = 3

5tan θ = 3

tan θ = 35

(b) θ = tan–1(35)

θ = 31° (nearest degree)

6 12cos2 x – 5sin x – 10 = 0

12(1 – sin2 x) – 5sin x – 10 = 0

12sin2 x + 5sin x – 2 = 0

(4sin x – 1) (3sin x + 2) = 0

sin x = 14 or sin x = –

23

When sin x = 14, x = 14.5° or 165.5°

When sin x = –23, x = 221.8° or 318.2°

Hence x = 14.5°, 165.5°, 221.8° or 318.2°

7 Distance travelled by John in 3 h = 6 × 3 = 18 km.

Distance travelled by Amy in 3 h = 15 km.

Using the cosine rule

Drawing a diagram to show the distances and angles we have:

a2 = b2 + c2 – 2bc cos A

x2 = 182 + 152 – 2 × 18 × 15 cos 50°

x = 14.2090 km

x = 14.2 km (1 d.p.)

Note that sin θcos θ = tan θ.

x

50°15 km

18 km

A

N

Worked solutions

77

8 (a) By Pythagoras’ theorem we obtain

PR2 = 92 + 92

= 162

= 81 × 2

Taking the square root of both sides gives

PR = √ 81 × 2

= 9√ 2

(b)

By Pythagoras’ theorem we obtain

VP2 = 82 + (4.5√ 2)2

Solving, gives VP = 10.2 cm (3 s.f.)

(c) Let angle OPV = θ

Tan θ = 8

4.5√ 2

Hence, θ =51.5° (nearest 0.1°)

9 (a) By Pythagoras’ theorem

AC2 = 92 + 62

Giving AC = 10.8167 cm

= 10.8 cm (3 s.f.)

(b) By Pythagoras’ theorem

AF2 = 42 + 10.81672

Giving AF = 11.5326 cm

= 11.5 cm (3 s.f.)

8 cm

OP

V

cm9√ 22

TAKE NOTEIt is always a good idea to simplify things by drawing the triangle you are working on separately.


78

(c)

Let angle CAF = θ

Cos θ = 10.816711.5326

θ = 20.3° (to the nearest 0.1°)

(d) In order to obtain the greatest angle down the slope, think of the path a ball would take if it was released at the top of the slope.

It would follow a line parallel to the lines AE or BF.

We can use triangle BCF to find this angle.

Let angle CBF = α so tan α = 46 giving α = 33.7° (nearest 0.1°)

10 (a)

(b) Notice that in the above diagram we have two sides of a triangle and we need to find the other side, representing the path back to the harbour. If we find the included angle between the two original paths we can use the cosine rule to find the length of the other side of the triangle.

Ѳ

11.5326 cm

F

CA 10.8167 cm

N

N

8 km

10 km

10°

H

40°

Worked solutions

79

We can draw some additional angles on the original diagram like this.

You can now see that the included angle is 140 + 10 = 150°.

Let the length of the path back to the harbour = x.

Using the cosine rule we have

x2 = 102 + 82 – 2 × 10 × 8cos 150°

Solving gives x = 17.4 km.

(c) To find the bearing, the angle marked θ in the diagram needs to be found.

Using the sine rule, we have

17.4

sin 150° = 8

sin θ

Solving gives θ = 13.3°

Now, bearing = 180 + 10 + 13 = 203°

N

N

8 km

10 km

10°

H

40°

40°

140°

Always extend the north line down past the point and look for any alternate angles. In this case you can spot an alternate angle of 40°.

Remember bearings are always three figures so 13.3° is rounded to the nearest whole number which is 13°.

N

N

8 km

10 km

10°

H

40°

40°

140°

Ѳ


80

Test yourself1 Using the identity cos2 θ + sin2 θ = 1 we have sin2 θ = 1 – cos2 θ.

Substituting sin2 θ = 1 – cos2 θ into the equation given, we obtain

3(1 – cos2 θ) + 5cos θ – 5 = 0

3 – 3cos2 θ + 5cos θ – 5 = 0

Rearranging and simplifying we obtain

3cos2 θ – 5cos θ + 2 = 0

Factorising gives

(3cos θ – 2)(cos θ – 1) = 0

Hence 3cos θ – 2= 0 or cos θ – 1 = 0

So cos θ = 23 or cos θ = 1

θ = cos–1(23) or θ = cos–1(1)

θ = 48.2°, 311.8° or θ = 0°, 360°

Hence θ = 0°, 48.2°, 311.8° or 360°

2 sin 2x = √ 3 cos 2x

Dividing both sides by sin 2x we obtain

tan 2x = √ 3

2x = tan–1(√ 3) Using the CAST method to find the angles, we have

This is a quadratic in cos θ.

The above angles are found using either the trigonometric graph of cos θ or the CAST method.

Note that because the multiple of the angle is 2 (i.e. tan 2x) we need to consider twice the range. Hence we need to consider the angles from 0° to 720°, including the values themselves. 90°

180° 0°

270°

S A

T C

Ѳ2

ѲpV

Worked solutions

81

Using the diagram on page 81 θpV = tan–1(√ 3) = 60°

θ 2 = 180 + 60 = 240°

θ 3 = 360 + 60 = 420°

θ4 = 360+ 240 = 600°

2x = 60°, 240°, 420°, 600°

Hence, x = 30°, 120°, 210°, 300°

3 (a) First we copy the diagram, marking the lighthouse on it and some of the angles we know.

Using speed = distancetime , we have distance = speed × time

Hence AB = 24 × 2 = 48 nautical miles

(b) Angle ABL = 90 + 30 = 120°

Using the cosine rule

a2 = b2 + c2 – 2bc cos A

AL2 = 482 + 52 – 2 × 48 × 5cos 120°

= 2569

AL = √ 2569 = 50.6853

AL = 50.7 nautical miles (to one decimal place)

Distance from port A to the lighthouse is 50.7 nautical miles (1 d.p.)

Note that an angle of 360° takes you back to the start again and you now add the angles you have found to 360°

N

N

LighthouseLB

A

30°

30°

Notice that we know the two sides and the included angle and want to find the missing side. We therefore need to use the cosine rule.


82

(c)

Using the sine rule to find angle θ we have

48

sin θ = 50.6853sin 120

Rearranging and solving gives θ = 55.1°

Bearing = 270 – 55 = 215°

4 sin θ = –3cos θ

Dividing both sides by cos θ we obtain

Hence, tan θ = –3

Here we will use the CAST method but the trigonometric graphs could also have been used to find the angles.

Tan θ is negative in the second and fourth quadrants.

Now tan–1(–3) = –71.57°

= –71.6° (nearest 0.1°)

–71.6° is an angle measured from 0° in a clockwise direction. This would be 360 – 71.6 = 288.4° in the normal direction (i.e θpV). The other value θ would be 180 – 71.6 = 108.4°.

Hence the two values of θ satisfying the equation are 108.4° and 288.4°.

We use the identity sin θcos θ = tan θ.

The diagram needs some further information adding to it. Adding a north line through point L and calling the angle ALB = θ and marking all the known lengths, and the bearing we are asked to find, the following diagram is obtained.

N

N N

Lighthouse

Bearing

LB

A

30°

48

5

50.6853

30° Ѳ

Note that the bearing has to be a whole number of degrees.

90°

71.6°

71.6°

180° 0°

270°

S A

T C

Ѳ

ѲpV

>>> TIPIt is easy to make a mistake when working out angles so always check your answers by finding the sin, cos or tan of the angles to check that you get back to the original value, i.e. –3 for tan θ in this case.

Worked solutions

83

5 (a)

By Pythagoras’ theorem we have

202 = VR2 + 162

400 = VR2 + 256

VR2 = 144

VR = 12 cm

(b) By Pythagoras’ theorem we have

202 = VS2 + 122

400 = VS2 + 144

VS2 = 256

VS = 16 cm

By Pythagoras’ theorem we have

162 = RS2 + 122

256 = RS2 + 144

RS2 = 112

RS = 10.6 cm ( 3 s.f.)

20 cm 20 cm

12 cm

V

SPQ

16 cm 16 cm

12 cm

R

SQ P

20 cm

16 cm

V

R Q


84

(c)

Cos θ = 10.616

θ = cos–1(10.616 )

= 48.5° (3 s.f.)

16 cm

10.6 cm

V

R SѲ

Worked solutions

85

Topic


11 Worked solutions

Progress check

1 (a) y = 4x3 + 6x2 – 3x + 1 dydx

= 12x2 + 12x – 3

(b) y = 6x5 + 8x4 – 3x3 + 1 dydx

= 30x4 + 32x3 – 9x2

(c) y = 7x4 + 8x3 – 9x2 + 1 dydx

= 28x3 + 24x2 – 18x

(d) y = 10x3 – 7x2 – 9x + 11 dydx

= 30x2 – 14x – 9

(e) y = 20x2 – 7x – 21 dydx

= 40x – 7

(f) y = 5x2 – 7x – 5 dydx

= 10x – 7

2 (a) y = (x + 2)(x + 1) = x2 + 3x + 2 dydx

= 2x + 3

(b) y = (x + 4)(x + 2) = x2 + 6x + 8 dydx

= 2x + 6

(c) y = (x – 3)(x + 2) = x2 – x – 6 dydx

= 2x – 1

(d) y = (x – 4)(x – 2) = x2 – 6x + 8 dydx

= 2x – 6

(e) y = (x – 4)2 = x2 – 8x + 16 dydx

= 2x – 8

(f) y = (x + 2)(x – 2) = x2 – 4 dydx

= 2x

(g) y = (x + 5)2 = x2 + 10x + 25 dydx

= 2x + 10

(h) y = x(x2 + 2x + 1) = x3 + 2x2 + x dydx

= 3x2 + 4x + 1

(i) y = x(3x2 + 6x + 9) = 3x3 + 6x2 + 9x dydx

= 9x2 + 12x + 9

(j) y = x2(x2 + 6x + 9) = x4 + 6x3 + 9x2 dydx

= 4x3 + 18x2 + 18x

(k) y = (3x + 2)(2x – 5) = 6x2 – 11x – 10 dydx

= 12x – 11

4 Calculus: Topic 11 – Differentiation

86

(l) y = (5x – 1)(4x – 7) = 20x2 – 39x + 7 dydx

= 40x – 39

(m) y = (2x + 3)2 = 4x2 + 12x + 9 dydx

= 8x + 12

(n) y = (3x – 2)2 = 9x2 – 12x + 4 dydx

= 18x – 12

3 (a) y = (x + 1)(x2 + 3x + 1) = x3 +3x2 + x + x2 + 3x + 1 = x3 + 4x2 + 4x + 1 dydx

= 3x2 + 8x + 4

(b) y = (x – 1)(x2 + 5x – 1) = x3 + 5x2 – x – x2 – 5x + 1 = x3 + 4x2 – 6x + 1 dydx

= 3x2 + 8x – 6

(c) y = (x – 5)(x2 – 2x + 4) = x3 – 2x2 + 4x – 5x2 + 10x – 20 = x3 – 7x2 + 14x – 20 dydx

= 3x2 – 14x + 14

(d) y = (2x + 1)(x2 – 2x + 4) = 2x3 – 4x2 + 8x + x2 – 2x + 4 = 2x3 – 3x2 + 6x + 4 dydx

= 6x2 – 6x + 6

(e) y = (4x – 3)(x2 + 5x – 3) = 4x3 + 20x2 – 12x – 3x2 – 15x + 9 = 4x3 + 17x2 – 27x + 9 dydx

= 12x2 + 34x – 27

(f) y = (5x – 1)(x2 + 5x – 3) = 5x3 + 25x2 – 15x – x2 – 5x + 3 = 5x3 + 24x2 – 20x + 3 dydx

= 15x2 + 48x – 20

(g) y = (x + 2)(x + 3)(x + 5) = (x + 2)(x2 + 8x + 15) = x3 + 8x2 + 15x + 2x2 + 16x + 30

= x3 + 10x2 + 31x + 30

dydx

= 3x2 +20x + 31

(h) y = (x + 6)(x + 5)(x + 7) = (x + 6)(x2 +12x + 35) = x3 + 12x2 + 35x + 6x2 + 72x + 210

= x3 + 18x2 + 107x + 210

dydx

= 3x2 + 36x + 107

(i) y = (x – 3)(x – 2)(x + 1) = (x – 3)(x2 – x – 2) = x3 – 4x2 + x + 6 dydx

= 3x2 – 8x + 1

(j) y = (x + 1)2(x + 2) = (x + 1)(x + 1)(x + 2) = (x + 1)(x2 + 3x + 2)

= x3 + 4x2 + 5x + 2

dydx

= 3x2 + 8x + 5

Worked solutions

87

(k) y = (2x – 1)2(x – 3) = (2x – 1)(2x – 1)(x – 3) = (2x – 1)(2x2 – 7x + 3)

= 4x3 – 16x2 + 13x – 3

dydx

= 12x2 – 32x + 13

(l) y = (3x + 4)2(x – 1) = (3x + 4)(3x + 4)(x – 1) = (3x + 4)(3x2 + x – 4)

= 9x3 + 15x2 – 8x – 16

dydx

= 27x2 + 30x – 8

4 y = 3x2 – 2x + 5

dydx

= 6x – 2

When x = 2

dydx

= 6(2) – 2

= 10

Equation of a straight line having gradient m and passing through the point (x1, y1) is given by:

y – y1 = m(x – x1)

In this case m = 10 and (x1, y1) = (2, 13), so

y – 13 = 10(x – 2)

y – 13 = 10x – 20

Equation of the tangent to the curve at (2, 13) is y = 10x – 7


88

5 y = x3 + 4x2 – 8x + 2

dydx

= 3x2 + 8x – 8

When x = 2

dydx

= 3(2)2 + 8(2) – 8

= 20

When x = 2, y = 23 + 4(2)2 – 8(2) + 2 = 10.

Equation of the tangent at (2, 10) is

y – 10 = 20(x – 2)

y – 10 = 20x – 40

y = 20x – 30

To find the gradient of the normal we use m1 m2 = –1

So, (20) m2 = –1 (where m2 is the gradient of the normal)

Giving gradient of the normal, m2 = – 1

20


y – y1 = m(x – x1)

In this case m = – 1

20 and (x1, y1) = (2, 10), so

y – 10 = – 1

20(x – 2)

20y – 200 = –x + 2

Hence, equation of the normal at P is x + 20y = 202

x = 2 is substituted into the equation of the curve to find the corresponding y-coordinate.

Worked solutions

89

6 y = 2x3 – 3x2 – 4x + 2

dydx

= 6x2 – 6x – 4

When x = 2

dydx

= 6(2)2 – 6(2) – 4

= 8


y – y1 = m(x – x1) In this case m = 8 and (x1, y1) = (2, –2), so

y – (–2) = 8(x – 2)

y + 2 = 8x – 16

y = 8x – 18

Equation of the tangent to the curve at (2, –2) is y = 8x – 18

7 (a) y = x2 – 3x + 2

dydx

= 2x – 3

When x = 1

dydx

= 2(1) – 3

= –1

When x = 2

dydx

= 2(2) – 3

= 1


90

(b) When x = 1, y = 12 – 3(1) + 2 = 0.

Equation of tangent having gradient –1 and passing through (1, 0) is

y – 0 = –1(x – 1)

y = –x + 1

When x = 2, y = 22 – 3(2) + 2 = 0.

Equation of tangent having gradient 1 and passing through (2, 0) is

y – 0 = 1(x – 2)

y = x – 2

(c) Solving the equations of the two tangents simultaneously:

y = –x + 1 .............................................................. (1)

y = x – 2 .................................................................... (2)

Adding these two equations we obtain

2y = –1

y = – 12

Substituting y = – 12 into equation (1) we obtain

– 12 = –x + 1

x = 32

Hence the coordinates of the point where the two tangents intersect

are (32 , –

12).

Note that in order to use the formula for the equation of a straight line it is necessary to know the y-coordinate of the point through which the line passes. We therefore substitute the x-coordinate into the equation of the curve to find the y-coordinate.

You can perform a check by substituting the values for x and y into each side of equation (2).

Worked solutions

91

Test yourself1 (a) y = 4x2 – 30x – 3

dydx

= 8x – 30

When x = 4, dydx

= 8(4) – 30 = 2

(b) Gradient of the tangent is 2 so if the gradient of the normal is m then the product of the gradients of the normal and tangent are –1.

Hence m(2) = –1

So m = – 12

When x = 4, y = 4(4)2 – 30(4) – 3 = 64 – 120 – 3 = –59

Equation of the normal having gradient – 12 and passing through

(4, –59) is

y + 59 = – 12 (x – 4)

2y + 118 = – x + 4

2y + x + 114 = 0

2 (a) y = x3 + 3x2 – 9x – 8

dydx

= 3x2 + 6x – 9

Now at the stationary points dydx

= 0, so we have

3x2 + 6x – 9 = 0

x2 + 2x – 3 = 0

(x + 3)(x – 1) = 0

x = –3 or 1

Hence there is a stationary point at x = 1.

(b) d2ydx2 = 6x + 6

When x = 1, d2ydx2 = 6(1) + 6 = 12 > 0 showing there is a minimum point

at x = –1.

Divide both sides by 3 to make the factorising of the quadratic equation easier. Notice that there are two

stationary points but this question is only concerned with one of them.

The first derivative is differentiated again to find the second derivative.

Equation for line of gradient m passing through (x1, y1) is y – y1 = m (x – x1)


92

3 (a) y = x3 – 6x2 + 9x + 1

dydx

= 3x2 – 12x + 9

(b) When x = 3, dydx

= 3(3)2 – 12(3) + 9 = 0

(c) Now at the stationary points dydx

= 0, so we have

3x2 – 12x + 9 = 0

x2 – 4x + 3 = 0

(x – 3)(x – 1) = 0

x = 3 or 1

When x = 3, y = x3 – 6x2 + 9x + 1 = 33 – 6(3)2 + 9(3) + 1 = 1

When x = 1, y = x3 – 6x2 + 9x + 1 = 13 – 6(1)2 + 9(1) + 1 = 5

Hence the two stationary points are (3, 1) and (1, 5).

(d) Finding the second order derivative

d2ydx2 = 6x – 12

When x = 3, d2ydx2 = 18 – 12 = 6.

The positive value shows that (3, 1) is a minimum point.

When x = 1, d2ydx2 = 6 – 12 = –6.

The negative value shows that (1, 5) is a maximum point.

4 y = 23 x3 +

12 x2 – 6x

dydx

= 2x2 + x – 6 = (2x – 3)(x + 2)

At the stationary points dydx

= 0

(2x – 3)(x + 2) = 0

Solving gives x = 32 or –2

Substituting x = 32 into the equation of the curve to find the y-coordinate

gives

y = 23 (3

2)3 +

12 (3

2)2 – 6(3

2) = 94 +

98 – 9 = –55

8

Note that the fact that the gradient is 0 at x = 3 means there is a stationary point at x = 3.

Divide both sides by 3 to make the factorising of the quadratic equation easier.

Notice that we know one of the factors from the answer to part (a). There is a stationary point at x =1, so (x – 1) must be a factor.

TAKE NOTEMake sure you find both the x- and y-coordinates of any stationary points if you are asked.

Worked solutions

93

Putting x = –2 into the equation of the curve to find the y-coordinate gives

y = 23 (– 2)3 +

12 (– 2)2 – 6(– 2) = –

163 + 2 + 12 = 8

23

Finding the second order derivative

d2ydx2 = 4x + 1

When x = 32,

d2ydx2 = 7.

The positive value shows that (32 , –55

8 ) is a minimum point.

When x = –2, d2ydx2 = –7.

The negative value shows that (–2, 823 )is a maximum point.

5 y = x3 – 6x2 + 12x + 1

dydx

= 3x2 – 12x + 12 = 3(x2 – 4x + 4) = 3(x – 2) (x – 2) = 3(x – 2)2

At the stationary points dydx

= 0

3(x – 2)2 = 0

Solving gives x = 2 so there is only one stationary point.

To determine the y-coordinate of the stationary point, we substitute x = 2 into the equation of the curve.

y = 23 – 6(2)2 + 12(2) + 1 = 9

So, the stationary point of curve C is at (2, 9).


94

Topic


12 Worked solutions

Progress check

1 (a) 4x3 dx = 4x4

4 + c = x4 + c

(b) 5x dx = 5x2

2 + c

(c) 3 dx = 3x + c

(d) –6x2 dx = –6x3

3 + c = –2x3 + c

(e) 16x3 dx = 16x4

4 + c = 4x4 + c

(f) 1 dx = x + c

4 Calculus: Topic 12 – Integration

2 (a) (5x2 + 3x – 5) dx = 5x3

3 + 3x2

2 – 5x + c

(b) (7x3 – 4x2 + 8x – 1) dx = 7x4

4 – 4x3

3 + 8x2

2 – x + c = 7x4

4 – 4x3

3 + 4x2 – x + c

(c) (12 x2 + 5x – 3) dx =

x3

6 + 5x2

2 – 3x + c

(d) (x – 3)(x + 2) = x2 – x – 6 (x2 – x – 6) dx = x3

3 – x2

2 – 6x + c

(e) (x – 3)2 = x2 – 6x + 9 (x2 – 6x + 9) dx = x3

3 – 6x2

2 + 9x + c = x3

3 – 3x2 + 9x + c

(f) x(x2 + 6x – 5) = x3 + 6x2 – 5x (x3 + 6x2 – 5x) dx = x4

4 – 6x3

3 – 5x2

2 + c = x4

4 – 2x3 – 5x2

2 + c

3 (a) 4x3 dx = x4 + c

(b) 6x2 dx = 2x3 + c

(c) (2x + 1) dx = x2 + x + c

95

(d) (8x2 + 4x – 1) dx = 8x3

3 + 2x2 – x + c

(e) (5x4 + 4x3 – 6x) dx = x5 + x4 – 3x2 + c

(f) (x2 + 4x – 5) dx = x3

3 + 2x2 – 5x + c

(g) (x + 5)(x + 1) dx = (x2 + 6x + 5) dx = x3

3 + 3x2 + 5x + c

(h) (x – 1)(x + 1) dx = (x2 – 1) dx = x3

3 – x + c

(i) (2x – 5)(x + 1) dx = (2x2 – 3x – 5) dx = 2x3

3 – 3x2

2 – 5x + c

(j) (x + 5)2 dx = (x2 + 10x + 25) dx = x3

3 + 5x2 + 25x + c

(k) x(x + 1) dx = (x2 + x) dx = x3

3 + x2

2 + c

(l) x2(2x + 1) dx = (2x3 + x2) dx = x4

2 + x3

3 + c

(m) (6x5 + 20x4 – 6x2 – x + 9) dx = x6 + 4x5 – 2x3 – x2

2 + 9x + c

4 (x3

4 – x2

2) dx = x4

4 × 4 – x3

2 × 3 + c = x4

16 – x3

6 + c

5 (a) (5x4 – 6x2 + 9) dx = 5x5

5 – 6x3

3 + 9x + c = x5 – 2x3 + 9x + c

(b) (8x3 – 6x2 + 10x + 5) dx = 8x4

4 – 6x3

3 + 10x2

2 + 5x + c = 2x4 – 2x3 + 5x2 + 5x + c

(c) (4 – x)2 dx = ∫(16 – 8x + x2) dx = 16x – 8x2

2 + x3

3 + c = 16x – 4x2 + x3

3 + c

(d) x4

3 dx = x5

3 × 5 + c = x5

15 + c

(e) (x2

2 + x3) dx =

x3

2 × 3 + x2

3 × 2 + c = x3

6 + x2

6 + c

6 (a) 3x2 dx = 3x3

3 + c = x3 + c

(b) 16x3 dx = 16x4

4 + c = 4x4 + c

(c) 13 x2 dx = x3

9 + c

(d) 5 dx = 5x + c


96

7 dydx

= 4x + 5 so y = (4x + 5) dx = 4x2

2 + 5x + c = 2x2 + 5x + c

Now as point (2, 9) lies on the curve, these coordinates will satisfy the equation of the curve. Hence,

9 = 2(2)2 + 5(2) + c

9 = 8 + 10 + c

c = –9

Hence, the equation of the curve is y = 2x2 + 5x – 9.

8 y = (6x2 + 10x + 2) dx

= 6x3

3 + 10x2

2 + 2x + c

= 2x3 + 5x2 + 2x + c

Now when x = 1, y = 1.

1 = 2(1)3 + 5(1)2 + 2(1) + c

1 = 2 + 5 + 2 + c

c = –8

Hence, y = 2x3 + 5x2 + 2x – 8

9 (a) dydx

= 2x – 2

y = (2x – 2) dx

= x2 – 2x + c

The point (3, –5) lies on the curve, so these coordinates satisfy the equation of curve.

Hence y = x2 – 2x + c

–5 = 32 – 2(3) + c

c = –8

Equation of the curve is y = x2 – 2x – 8

(b) When y = 0, x2 – 2x – 8 = 0

Factorising we obtain (x – 4)(x + 2) = 0

Solving gives x = 4 or –2.

Hence, coordinates of the points of intersection of the x-axis are (–2, 0) and (4, 0).

Worked solutions

97

(c) dydx

= 2x – 2

At the minimum point, dydx

= 0, so 2x – 2 = 0. Solving gives x = 1.

When x = 1, y = 12 – 2(1) – 8 = –9

Hence the minimum point is (1, –9).

Test yourself1 y = (4 – 2x – 3x2) dx

= 4x – 2x2

2 – 3x3

3 + c

= 4x – x2 – x3 + c

When x = 0, y = 1 so we have

1 = 4(0) – (0)2 – (0)3 + c

c = 1.

Hence the equation of the curve is

y = 4x – x2 – x3 + 1

2 (a) (4x – 3x3) dx

= 4x2

2 – 3x4

4 + c

= 2x2 – 3x4

4 + c

(b) 2x(3x2 – 5x + 1)


6x3 – 10x2 + 2x

(6x3 – 10x2 + 2x) dx

= 6x4

4 – 10x3

3 + 2x2

2 + c

= 3x4

2 – 10x3

3 + x2 + c

Always cancel fractions if possible.


98

(c) (12 x2 – x +

12) dx

= x3

2 × 3 – x2

2 + x2 + c

= x3

6 – x2

2 + x2 + c

3 (25t4 – 12t3 + 15t2 – 9t + 2) dt

= 25t5

5 – 12t4

4 + 15t3

3 – 9t2

2 + 2t + c

= 5t5 – 3t4 + 5t3 – 9t2

2 + 2t + c

4 2

15 (x2 – 30)


2

15 x2 – 4

( 215 x2 – 4) dx

= 2x3

45 – 4x + c

5 (6t2 – 2t + 1) dt = 6t3

3 – 2t2

2 + t + c

= 2t3 – t2 + t + c

6 dydx

= 15x2 + 8x +1

y = (15x2 + 8x + 1) dx

= 5x3 + 4x2 + x + c

When x = 1, y =3 so 3 = 5(1)3 + 4(1)2 + 1 + c giving c = –7.

Equation of the curve is y = 5x3 + 4x2 + x – 7

7 (a) dydx

= 3 – 2x

y = ∫(3 – 2x) dx

= 3x – x2 + c

Substituting (0, 0) for x and y into the equation we obtain

0 = 0 – 0 + c

Hence c = 0.

Equation of curve is y = 3x – x2

(b) Solving the equation of the x-axis (i.e. y = 0) with the equation of the curve we obtain

3x – x2 = 0

x(3 – x) = 0

Solving gives x = 3.

Hence the point is (3, 0).

Worked solutions

99

Topic


13 Worked solutions

Progress check

1 0

1

(6x2 – 2x + 5) dx = [6x3

3 – 2x2

2 + 5x]0

1

= [2x3 – x2 + 5x]0

1

=[(2 × 13 – 12 + 5(1)) – (0 – 0 + 0)] = [(2 – 1 + 5) – (0)]

= 6

2 2

3

(6x2 – 4x – 5) dx = [6x3

3 – 4x2

2 – 5x]2

3

= [2x3 – 2x2 – 5x]2

3

= [(2(3)3 – 2(3)2 – 5(3)) – (2(2)3 – 2(2)2 – 5(2))] = [(54 – 18 – 15) – (16 – 8 – 10)]

= [21 – (–2)]

= 23

3 0

0.5

y dx = 13

0

0.5 (2x – 3x2) dx

= 13[2x2

2 – 3x3

3 ]0

0.5

= 13[x2 – x3]

0

0.5

= 13[((0.5)2 – (0.5)3) – (0)]

= 0.0417 (3 s.f.)

4 Calculus: Topic 13 – Definite integration

100

4 0

2 (x2 + 4x – 3) dx = [x3

3 + 4x2

2 – 3x]0

2

= [x3

3 + 2x2 – 3x]0

2

= [(83 + 8 – 6) – (0)]

= 423

5 0

3 (x2 + 6x + 4) dx –

0

3 (x2 – 4x) dx

= 0

3 (x2 + 6x + 4 – x2 + 4x) dx

We now collect the terms to produce the following:

0

3 (10x + 4) dx = [10x2

2 + 4x]0

3

= [5x2 + 4x]0

3

= [(5(3)2 + 4(3)) – (0)]

= 57

6 0

1 (x + 5)(x + 6) dx =

0

1 (x2 + 11x + 30) dx

= [x3

3 + 11x2

2 + 30x]0

1

= [(13 +

112 + 30) – (0)]

= 35.8 (3 s.f.)

7 Shaded area = –3

3 (9 – x2) dx

= [9x – x3

3]–3

3

= [(9(3) – 33

3 ) – (9(–3) – (–3)3

3 )] = 36

>>> TIPIt is important to note that you can only combine the integrals if the limits are the same.

Worked solutions

101

8 (a) To find where the curve cuts the x-axis we substitute y = 0 into the equation of the curve.

Hence x2 – 4 = 0

(x – 2)(x + 2) = 0

Solving, gives x = 2 or – 2.

Note as the curve has a positive coefficient of x2, the curve will be -shaped.

When x = 0, y = –4

Sketching the curve we obtain:

(b) 2

3 (x2 – 4) dx = [x3

3 – 4x]2

3

= [(33

3 – 4(3)) – (23

3 – 4(2))] = (9 – 12) – (8

3 – 8) =

73

0

2 (x2 – 4) dx = [x3

3 – 4x]0

2

= [(23

3 – 4(2)) – (0)] = –

163

(c) The positive value represents the area above the x-axis and the negative value represents the area below the x-axis.

y

x

y = x2 – 4

0 2–2

–4


102

9 (a) Solve the equations of the curve and straight line simultaneously to find the coordinates of the points of intersection A and B.

Equating the y-values gives:

9 – x2 = x + 3

x2 + x – 6 = 0

(x + 3)(x – 2) = 0

Solving gives x = –3 or 2

Put both values of x into the equation of the straight line to find the corresponding y-coordinates.

When x = –3, y = (– 3) + 3 = 0

When x = 2, y = 2 + 3 = 5

By looking at the graph A is (–3, 0) and B is (2, 5).

(b) Area under the curve between x = –3 and x = 2 is given by

–3

2 (9 – x2) dx = [(9x –

x3

3)]–3

2

= [(9(2) – (2)3

3 ) – (9(–3) – (–3)3

3 )] = [(18 –

83) – (–27 + 9)]

= 1513 + 18

= 3313

Area of right-angled triangle with side AB as the hypotenuse

= 12 × 5 × 5

= 12.5

Required area = 3313 – 12.5

= 20.8 (3 s.f.)

Form a quadratic equation, then factorise and finally solve it.

Looking at the diagram this is point A.

Worked solutions

103

Test yourself1

1

4 (6x2 – 2) dx = [2x3 – 2x]1

4

= [(2(4)3 – 2(4)) – (2(1)3 – 2(1))] = 128 – 8 – 2 + 2

= 120

2 Area = –3

3 y dx

= –3

3 (9 – x2) dx

= [(9x – x3

3)]–3

3

= [(9(3) – (3)3

3 ) – (9(–3) – (–3)3

3 )] = [(27 – 9) – (–27 + 9)]

= 36

3 (a) 0

2 (x2 – 4x + 2) dx = [x3

3 – 4x2

2 + 2x]0

2

= [x3

3 – 2x2 + 2x]0

2

= [(23

3 – 2(2)2 + 2(2)) – (0)] =

83 – 8 + 4

= –113

(b) The negative sign means that the area is below the x-axis.

4 –1

2 (x – 3)(x + 4) dx =

–1

2 (x2 + x – 12) dx

= [x3

3 + x2

2 – 12x]–1

2

= [(23

3 + 22

2 – 12(2)) – ((–1)3

3 + (–1)2

2 – 12(–1))] = [(8

3 + 2 – 24) – (– 13 +

12 + 12)]

= – 31.5


104

5 Area = 0

1 (x2 + 1) dx

= [x3

3 + x]0

1

= [(13 + 1) – (0)]

= 43 or 11

3

6 (a) 0

2 x(x2 – 6x + 3) dx =

0

2 (x3 – 6x2 + 3x) dx

= [x4

4 – 2x3 + 3x2

2 ]0

2

= [(24

4 – 2(2)3 + 3(2)2

2 ) – (0)] = 4 – 16 + 6

= –6

(b) 0

2 (x – 3)

3 dx = 13

0

2 (x – 3) dx

= 13 [x2

2 – 3x]0

2

= 13 [(2 – 6) – (0)]

= – 43

(c) –2

2 (12x2 – 4x + 1) dx = [4x3 – 2x2 + x]

–2

2

= [(4(2)3 – 2(2)2 + 2) – (4(–2)3 – 2(–2)2 – 2)]

= 32 – 8 + 2 + 32 + 8 + 2

= 68

Worked solutions

105

Topic


14 Worked solutions

4 Calculus: Topic 14 – Application of calculus to kinematics

Progress check

1 (a) u = 0 ms–1, v = ?, a = 0.9 ms–2 , t = 10 s

Using v = u + at

v = 0 + 0.9 × 10

= 9 ms–1

(b) Using s = 12 (u + v)t

s = 12 (0 + 9)10 = 45 m

2 (a) Taking the upward velocity as positive, we have

u = 20 ms–1, v = 0 ms–1, a = g = – 9.8 ms–2

Using v2 = u2 + 2as gives 0 = 202 + 2 × (–9.8) × s

Solving for s, gives s = 20.4 m

(b) Using s = ut + 12 at2

0 = 20t + 12 × (–9.8) × t2

0 = 20t – 4.9t2

0 = t(20 – 4.9t)

t = 0 or 4.1 s

Hence time = 4.1 s

3 (a) Taking the downward direction as positive.

u = 0.8 ms–1, v = ?, a = g = 9.8 ms–2, t = 3.5 s

Using v = u + at we have v = 0.8 + 9.8 × 3.5 = 35.1 ms–1


s = 0.8 × 3.5 + 12 × 9.8 × 3.52 = 62.8 m

The displacement, s, is zero when the stone returns to its point of projection.

t = 0 is ignored as a possible time.

106

4 (a) Taking upwards as the positive direction, we have

u = 10 ms–1, v = 0 ms–1, a = g = –9.8 ms–2

Using v = u + at gives

0 = 10 – 9.8t

Hence, t = 1.02 s

(b) Using v2 = u2 + 2as gives

0 = 102 + 2 × (–9.8)s

Hence s = 5.1 m

5 (a) s = 12t3 + 9

v = dsdt

= 36t2

(b) a = dvdt

= 72t

When t = 2, a = 72 × 2 = 144 ms–2

6 (a) v = 0.64t3 – 0.36t2

a = dvdt

= 1.92t2 – 0.72t

(b) s = v dt

= (0.64t3 – 0.36t2) dt

= 0.64t4

4 – 0.36t3

3 + c

When t = 0, s = 0

Substituting these two values into the expression for s we obtain

0 = 0.64(0)4

4 – 0.36(0)3

3 + c

Solving gives c = 0

Hence s = 0.64t4

4 – 0.36t3

3 = 0.16t4 – 0.12t3

When t = 10, s = 0.16(10)4 – 0.12(10)3 = 1480 m

t = 2 is substituted into the expression for a.

You now need to determine the value of the constant, c. You need a value of t and the corresponding value of s to substitute into the expression for s.

We write the expression with the value of c included.

Worked solutions

107

7 (a) v = a dt

= (3 – 0.1t) dt

= 3t – 0.1t2

2 + c

When t = 0, v = 0 so 0 = 3(0) – 0.1(0)2

2 + c

Solving gives c = 0.

The expression for the velocity is v = 3t – 0.1t2

2 .

(b) When t = 10, v = 3(10) – 0.1(10)2

2 v = 30 – 5

= 25 ms–1

(c) When t = 30 s, a = 3 – 0.1(30) = 0 ms–2

The acceleration is momentarily zero at t = 30 s.

When t = 31 s, a = 3 – 0.1(31) = –0.1 ms–2 (note that this a deceleration which means the lorry does not travel at constant velocity but instead slows down).

Hence at t = 30 s, the lorry momentarily stops accelerating.

(d) s = v dt

= (3t – 0.1t2

2 ) dt

= 3t2

2 – 0.1t3

6 + c

When t = 0, s = 0 so substituting these values into the above expression gives

0 = 3(0)2

2 – 0.1(0)3

6 + c

Solving gives c = 0

s = 3t2

2 – 0.1t3

6

When t = 30, s = 3(30)2

2 – 0.1(30)3

6

= 1350 – 450

= 900 m


108

8 v = 6t + 4

s = v dt

= (6t + 4) dt

= 6t2

2 + 4t + c

= 3t2 + 4t + c

When t = 0, s = 0 so we have 0 = 3(0)2 + 4(0) + c and solving gives c = 0.

Hence s = 3t2 + 4t

When t = 2 s, s = 3(2)2 + 4(2)= 20 m

When t = 5 s, s = 3(5)2 + 4(5)= 95 m

Distance travelled between the times t = 2 s and t = 5 s is 95 – 20 = 75 m

9 (a) (i) v = 6t2 – 2t + 8

a = dvdt

= 12t – 2

(ii) When t = 1, a = 12(1) – 2 = 10 ms–2

(b) s = v dt

= (6t2 – 2t + 8)dt

= 6t3

3 – 2t2

2 + 8t + c

= 2t3 – t2 + 8t + c

When t = 0, s = 0 so 0 = 2(0)3 – (0)2 + 8(0) + c

Solving gives c = 0.

Hence, the expression for the displacement is

s = 2t3 – t2 + 8t

The particle is at the origin at t = 0 so we know s = 0 as the origin is the point from which the distance is measured.

Worked solutions

109

Test yourself1 (a) u = 5 ms–1, a = 10 ms–2, t = 6 s and we need to find v.

Using v = u + at

v = 5 + 10 × 6

= 65 ms–1


s = 5 × 6 + 12 × 10 × 62

= 30 + 180

= 210 m

2 (a)

(b) u = 0 ms–1, v = ?, a = 0.9 ms–2, t = 5 s

Using v = u + at

gives v = 0 + 0.9 × 5 = 4.5 ms–1

(c) Acceleration = Gradient of the graph between t = 25 and t = 33 s

= 0 – 4.533 – 25

= –0.56 ms–2

Hence deceleration = 0.56 ms–2

(d) Distance travelled = area under the velocity–time graph

= 12 (20 + 33) × 4.5

= 119.25 m

v (ms–1)

5 25 33 t (s)0

Note that if you say that this is a deceleration, then you need to remove the minus sign.


110

3 (a) v = 64 – 1

27 t3

When the car comes to rest, v = 0 so we have

0 = 64 – 1

27 t3

Hence 1

27 t3 = 64

t3 = 64 × 27

Taking the cube root of both sides gives t = 12 s

(b) s = v dt

= (64 – 1

27 t3)dt

= 64t – 1

108 t4 + c

When t = 0, s = 0 so c = 0.

When t = 12, s = 64 × 12 − 1

108 × 124 = 768 – 192 = 576 m

4 (a) a = 2t + 3

s = v dt

= (2t + 3) dt

= t2 + 3t + c

When t = 0, v = 10. Substituting these values into the above equation to find c, we have.

v = t2 + 3t + c

10 = 02 + 0 + c

c = 10

Hence v = t2 + 3t + 10

When t = 3, v = 32 + 3(3) + 10

= 28 ms–1

Worked solutions

111

(b) s = v dt

= (t2 + 3t + 10) dt

= t3

3 + 3t2

2 + 10t + c

When t = 0, s = 0. Substituting these values into the above equation to find c, we have

c = 0

Hence s = t3

3 + 3t2

2 + 10t

When t = 3, s = 33

3 + 3(3)2

2 + 10(3)

= 9 + 13.5 + 30

= 52.5 m

5 (a) s = 120 m, u = 20 ms–1, v = 32 ms–1, a = ?

Using v2 = u2 + 2as, we obtain

322 = 202 + 2a × 120

Solving gives a = 2.6 ms–2

(b) Using v = u + at, we obtain

32 = 20 + 2.6t

Solving we obtain t = 4.62 s

(c) u = 20 ms–1, a = 2.6 ms–2, t = 20 s, v = ?

Using v = u + at, we obtain

v = 20 + 2.6 × 20

Solving we obtain v = 72 ms–1

(d) s = ?, u = 20 ms–1, a = 2.6 ms–2, t = 30 s

Using s = ut + 12 at2, we obtain

s = 20 × 30 + 12 × 2.6 × 302

s = 1770 m


112

(e)

Motion under constant acceleration, means the velocity–time graph will be a straight line.

Velocity(ms–1)

20

4.62 Time (s)0

A

B32

If the acceleration varies with time the velocity–time graph for the motion will be a curve. The acceleration is represented by the gradient of the curve. As the gradient of the curve varies with time, calculus is used to find the gradient of the graph at a particular time.

Worked solutions

113

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