1 Algebra: Topic 1 – Revision of the basics
Topic
1
Worked solutions
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
1 Algebra: Topic 1 – Revision of the basics
(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
x2 – 12x + 36 + y2 + 14y + 49 = 12
(e)(x – 6)2 + (y + 7)2 = 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
x2 – 2x + 1 + y2 – 18y + 81 – 10 = 0
(g)(x – 1)2 + (y – 9)2 – 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
x2 + 2x + 1 + y2 – 2y + 1 – 18 = 0
(i)(x + 1)2 + (y – 1)2 = 18
2
x2 + y2 + 2x – 2y – 16 = 0
Worked solutions
4x2y
5(a) x y
= 4x
12x2y3
= 3xy2
(b)
4x y
16a3b2c 2
(c) 2 = abc
24a b 3
24x2y4 3x
=
(d)
8x y5
y
Divide top and bottom by xy.
Divide top and bottom by 4xy.
15abc
= 3ab
5c
4x3 x2
(b) =
4x y y
6pq3r
(c) = 2q2
3pqr
10x4y3
(d) 2 = 5x2y2
2x y
45a3bc2 5a2c
(e) 2 =
9ab c
b
(x + 7)(x + 5) (x + 7)
=
(f)
(x + 5)(x + 3) (x + 3)
(x + 3)(x + 5)
(g)
=1
(x + 5)(x + 3)
6(a)
7
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).
(a) 12x2y + 8xy2 = 4xy(3x + 2y)
(b)4a2b + 2ab = 2ab(2a + 1)
(c)24x2y + 6x = 6x(4xy + 1)
(d)25a3b2c5 + 5a2b3 = 5a2b2(5ac5 + b)
x2(x – 1)
8(a) x(x – 1) = x
(b)
(c)
(d)
(e)
(f)
x y3 2
=y
x y
15x3y3
= 3y
5x3y2
5(x – 4)
x–4
=
10(x – 2) 2(x – 2)
(x + 1)(x – 2) x – 2
=
(x – 5)(x + 1) x – 5
x–3
1
=
(x – 3)(x – 1) (x – 1)
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 + 1).
Divide top and bottom by (x – 3).
3
1 Algebra: Topic 1 – Revision of the basics
Add 7 to both sides of the
equation. Adding 7 will
eliminate the –7 on the lefthand side.
Remove the denominator by
multiplying both sides by 5.
Add 1 to both sides.
Add 1 to both sides.
9
x–7=7
10
x + 7 = –3
x = 14
Subtract 7 from both sides of the equation. Subtracting 7
will eliminate the +7 on the left-hand side of the equation.
x = –10
4x
= 12
5
4x = 60
11
Remove the multiplier of x by dividing both sides by 4.
x = 15
x
–1=7
5
x
= 8
5
x = 40
12
Multiply both sides by 5.
2x
–1=5
3
2x
= 6
3
13
2x = 18
Multiply out the bracket.
Divide both sides by 2.
Multiply both sides by
2 to remove the 2 in the
denominator.
x = 9
142(2x + 1) = 18
4x + 2 = 18
–6 + x = –2
x=4
16
(a) 2 – x = 4 + x
–2 = 2x
x = –1
Divide both sides by 2.
Subtract 2 from both sides.
x=4
–6 + x
= –1
2
15 4
4x = 16
Multiply both sides by 3.
2 = 4 + 2x
–1 = x
Add 6 to both sides.
Worked solutions
(b)4(x – 7) = 3(2x – 10)
4x – 28 = 6x – 30
– 28 = 2x – 30
(c)5(6x – 3) = 6(2x – 1)
2 = 2x
x=1
30x – 15 = 12x – 6
18x – 15 = –6
18x = 9
x=
1
2
1
(d)(x – 1) = 2x + 4
3
17(a)
TAKE NOTE
Be careful here.
Students often see the
numbers 18 and 9 and
automatically think that
x is 2 which is incorrect.
x – 1 = 6x + 12
– 1 = 5x + 12
–13 = 5x
x=–
13
5
x = –2.6
x–5
= 4x
4
x – 5 = 16x
– 5 = 15x
x=–
5
15
1
3
(b)6x – 1 = 3(x – 4) + 7
x=–
6x – 1 = 3x – 12 + 7
3x – 1 = –5
3x = –4
x=–
4
3
5
1 Algebra: Topic 1 – Revision of the basics
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.
Divide both sides by 7.
Multiply both sides by 9
because the denominators 9
and 3 divide exactly into 9.
2 and 3 are factors of 6 so we
multiply both sides by 6.
18
(a)
x x
+ = 15
4 2
2x + 4x = 120
6x = 120
Divide both sides by 6.
x = 20
x x
(b)+ = 49
3 4
4x + 3x = 588
(c)
19
7x = 588 Collect the terms in x.
x = 84
x 2x
+ = 42
9 3
Collect the terms in x together.
x + 6x = 378
Divide both sides by 7.
7x = 378
x = 54
x–3 x+1
+
=3
2
3
6(x – 3) 6(x + 1)
+
= 18
2
3
3(x – 3) + 2(x + 1) = 18
3x – 9 + 2x + 2 = 18
6
5x – 7 = 18
5x = 25
x=5
Worked solutions
x x
– =3
2 5
10x 10x
= 30
–
2
5
20
5x – 2x = 30
3x = 30
x = 10
1
Multiply both sides by 10.
This step is usually not written down.
Divide both sides by 3.
1
(x – 1) = (x – 2)
21
4
3
Multiply both sides by 12.
3(x – 1) = 4(x – 2)
3x – 3 = 4x – 8
Subtract 3x from both sides.
–3 = x – 8
x=5
2x x
– =5
3 4
24x 12x
= 60
–
3
4
22
8x – 3x = 60
Multiply out the brackets.
5x = 60
x = 12
Add 8 to both sides.
Multiply both sides by 12.
This step is usually not written down.
Divide both sides by 5.
23(a)A = πr2
A 2
=r
π
r=
A
π
(b)
A = 4πr2
A
= r2
4π
r=
A
4π
7
1 Algebra: Topic 1 – Revision of the basics
4
(c)
V = πr3
3
3V = 4πr3
3V
= r3
4π
Subtract at from both sides.
Subtract u from both sides.
Divide both sides by 2a.
Subtract u2 from both sides.
Subtract 2as from both sides.
Subtract ut from both sides.
Divide both sides by t2.
Subtract mx from both sides.
8
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
Divide both sides by t.
(c) v 2 = 2as (s)
s=
(d)
v 2 = u2 + 2as (a)
v 2 – u2 = 2as
(e)
v 2
2a
a=
v 2– u 2
2s
v 2 = u2 + 2as (u)
v 2 – 2as = u2
u = √ v2 − 2as
1
s = ut + at2 (a)
2
1
s – ut = at2
2
(f)
2(s – ut) = at2
(g)
a=
Square root both sides.
Multiply both sides by 2.
2(s – ut)
t2
y = mx + c (c)
y – mx = c
Divide both sides by 2s.
c = y – mx
Always put the subject of the equation on
the left-hand side of the equation.
Worked solutions
(h)y = mx + c (m)
y – c = mx
y–c
x
m =
1
(i)
s = (u + v)t (t)
2
2s = (u + v)t
t=
2s
u+v
Subtract c from both sides.
Divide both sides by x.
Multiply both sides by 2.
Divide both sides by (u + v).
1
(j)
E = mv2 (v)
2
Multiply both sides by 2.
2E
= v2
m
Square root both sides.
2E = mv2
v =
2E
m
Divide both sides by m.
(k)V = πr2l (l)
Divide both sides by πr2.
(l)V = πr2l (r)
Divide both sides by πl.
V
πr2
l=
V
= r2
πl
r=
V
πl
Square root both sides.
9
1 Algebra: Topic 1 – Revision of the basics
25(a)
x + y = 5 .......................... (1)
5x + 2y = 11 ..................... (2)
Multiplying equation (1) by 2 we obtain
Subtracting the above equation from equation (2) we obtain
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.
>>>
TIP
Always specifically say
what your answers are.
The examiner should not
have to wade through
your working to find what
your answers are.
>>> TIP
Notice 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
2x + 2y = 10
5x + 2y = 11
Subtracting 2x + 2y = 10
3x = 1
1
3
1
Substituting x = into equation (1) we have
3
1
+y=5
3
x=
Hence, y = 4 23
Substituting x = 13 and y = 4 23 into LHS of equation (2) we obtain
5(13) + 2(423) = 11
5
3
+ 913 = 11
11 = RHS
Both sides of the equation are equal so the
values of x and y satisfy the second equation.
Hence the solutions are x = 13 and y = 4 23
(b)
2x – 3y = –5 (1)
5x + 2y = 16 (2)
4x – 6y = –10
Multiplying equation (1) by 2 and equation (2) by 3 we obtain
15x + 6y = 48
Adding 19x = 38
x=2
Substituting x = 2 into equation (1) we have
2(2) – 3y = –5
4 – 3y = –5
Worked solutions
3y = 9
y=3
Substituting x = 2 and y = 3 into LHS of equation (2) we obtain
5(2) + 2(3) = 16
16 = RHS
Both sides of the equation are equal so the
values of x and y satisfy the second equation.
Hence the solutions are x = 2 and y = 3
26 (a)3x – 5 = x – 1
The y-values are equated and the resulting equation solved.
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
Both sides of the equation are equal, showing that the values
of x and y satisfy the second equation.
Hence solutions are x = 2 and y = 1.
(b) From the equation 2x + 3y = 8 we have 3y = –2x + 8
The 3y is replaced by –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
y=2
3y = 6
11
1 Algebra: Topic 1 – Revision of the basics
Both sides of the equation
are equal so the values of
x and y satisfy the second
equation.
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
3 3 – 2 (3 3 – 2)( 3 + 2) 9 + 3 6 – 6 – 2 7 + 2 6
28 √3 – √2 = ( √3 – √2)( √3 + √2) = 3 + √6 – √6 – 2 = 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
Subtract 11 from both sides.
Add 5 to both sides.
Divide both sides by 15.
1(a)2x + 11 = 25
2x = 14
x=7
(b)
3x – 5 = 10
3x = 15
x=5
Divide both sides by 7.
Divide both sides by 3.
(c)
15x = 60
x=4
x
Multiply both sides by 4.
(d)= 8
4
Multiply both sides by 5.
x = 32
4x
(e)= 20
5
4x = 100
x = 25
12
√
Divide both sides by 4.
Worked solutions
2x
(f)= –6
3
Multiply both sides by 3.
2x = –18
x = –9
Add x to both sides to make x positive.
(g) 5 – x = 7
5 = 7 + x
Subtract 7 from both sides.
–2 = x
x = –2
x
–9=3
(h)
7
x
= 12
7
Add 9 to both sides.
Multiply both sides by 7.
x = 84
35x3y2
(a)
= 5x2
2
2
7xy
15ab3c
= 5b2c
(b)
3ab
3
(d) (x – 4)(x – 7)
(x – 1)(x – 4)
(x + 3)2
(x – 6)(x + 3)
=
Divide top and bottom by 3ab.
(x – 7)
(x – 1)
=
(x + 3)
(x – 6)
Divide top and bottom by (x – 4).
Divide top and bottom by (x + 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
(c) Divide top and bottom by 7xy2.
(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
(l) 2x(x – 1) – (x2 – 3x) = 2x2 – 2x – x2 + 3x = x2 + x
(k) 5x2(x – 3) + 3x(x + 4) = 5x3 – 15x2 + 3x2 + 12x = 5x3 – 12x2 + 12x
13
1 Algebra: Topic 1 – Revision of the basics
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
(d) (9x – 1)(9x + 1) = 81x2 + 9x – 9x – 1 = 81x2 – 1
(c) (2x – 7)(3x + 5) = 6x2 + 10x – 21x – 35 = 6x2 – 11x – 35
(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 =
6
7
(b) n =
(c) T =
(d) p =
f=
nRT
p
pV
RT
pV
nR
nRT
V
E
h
m=
y–c
x
c
8 (a) λ = f
9
14
(b) V =
(c) T =
(d) V =
(e) c =
(f) h =
(a) n
c
Q
mc
1000n
c
1000n
V
E
f
√ 48 +
12
√3
– √ 27 = 4√ 3 +
12√ 3
√ 3√ 3
– 3√ 3 = 4√ 3 + 4√ 3 – 3√ 3 = 5√ 3
2 + √ 5 (2 + √ 5 )(3 – √ 5) 6 + √ 5 – 5 1 + √ 5
(b) =
=
=
3 + √ 5 (3 + √ 5)(3 – √ 5)
4 9 – 5 Worked solutions
10 (a) 5
√2
=
1
5
√2
=
√ 2 = 5√ 2
√ 2 2
×
1
×
3 – √5
(b) (d) 2√ 5 = 4 × 5 = 20
3 + √5
3 + √5
3 – √5
=
3 – √5
9–5
=
3 – √5
4
(c) √ 32 + 3√ 2 = √ 16 × 2 + 3√ 2 = 4√ 2 + 3√ 2 = 7√ 2
(
)
2
117x + 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
Both sides of the equation are equal showing that
the values of x and y satisfy the second equation.
Hence the solutions are x = 2 and y = –3.
15
1 Algebra: Topic 2 – M
anipulation of algebraic
expressions
Topic
2
Worked solutions
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
16
(g) x2 – 6x + 12 = (x – 3)2 – 9 + 12 = (x – 3)2 + 3
Worked solutions
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) x – 2x – 3 = (x – 3)(x + 1)
2
(f) x – 3x + 2 = (x – 1)(x – 2)
2
(g) x – 4x – 5 = (x – 5)(x + 1)
2
(h) x + 5x – 14 = (x + 7)(x – 2)
2
(i) x – 5x + 4 = (x – 1)(x – 4)
2
5
(j) x2 + 3x – 10 = (x + 5)(x – 2)
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
17
1 Algebra: Topic 2 – Manipulation of algebraic expressions
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
1
(c)3x2 + 11x – 4 = (3x – 1)(x + 4) = 0 so x = or –4
3
2
11(a) x + 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 –
)
5
=0
2
Dividing both sides by 2 we obtain.
5
x2 + 2x – = 0
2
5
(x + 1)2 – 1 – = 0
2
12
(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.)
x2 + 8x – 12 = 0
Completing the square we obtain
(x + 4)2 – 16 – 12 = 0
(x + 4)2 = 28
x + 4 = ±√ 28
x = ±√ 28 – 4
x = 1.29 or –9.29 (2 d.p.)
18
(x + 4)2 – 28 = 0
x = √ 28 – 4 or –√ 28 – 4
Worked solutions
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)
>>> TIP
Be careful here. It is
easy to substitute the
values for part (a) into
this equation.
= 16 – 72
= –56
As b2 – 4ac < 0, there are no real roots.
Substituting these values into the quadratic equation formula gives:
(b) Comparing the equation 3x2 + 6x + 2 = 0, with ax2 + bx + c = 0 gives
a = 3, b = 6 and c = 2.
–6 ± √ (6)2 – 4(3)(2)
x=
2(3)
=
–6 ± √ 36 – 24 –6 ± √ 12 –6 + √ 12 –6 – √ 12
=
=
or
6
6
6
6
= –0.42 or –1.58 (2 d.p.).
19
1 Algebra: Topic 2 – Manipulation of algebraic expressions
Test yourself
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.
( )
( )
1
32 9 1
32
2
x
–
3x
+
x
–
–
x
–
–2
=
+
=
1
4
2
4 4
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)
20
Worked solutions
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)
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.
7
(l)9x2 + 89x – 10 = (9x – 1)(x + 10)
5x2 – 20x + 10= 5(x2 – 4x + 2)
= 5[(x – 2)2 – 4 + 2]
= 5(x – 2)2 – 10
8
Giving a = 5, b = –2 and c = –10.
y = x + 4 and y = x2 – 7x + 20
x + 4 = x2 – 7x + 20
x2 – 8x + 16 = 0
If the line and the curve touch then the resulting
equation will have a repeated root.
(x – 4)(x – 4) = 0
There is only one solution to
the quadratic which means
the straight line and curve
touch at only one point.
(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
The x-coordinate is substituted in the equation of
the line to find the corresponding y-coordinate.
Hence the coordinates of the point of contact are (4, 8).
21
1 Algebra: Topic 3 – The remainder and factor
theorems and solving cubic equations
Topic
3
Worked solutions
Progress checks
1f(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.
2g(x) = 2x3 – 7x2 + 3x + 1
g(1) = 2(1)3 – 7(1)2 + 3(1) + 1
=2–7+3+1
= –1
Hence, remainder = –1.
3f(x) = x3 – 4x2 + x + 8
f(–2) = (–2)3 – 4(–2)2 + (–2) + 8
= –8 – 16 – 2 + 8
= –18
Hence, remainder = –18.
4f(x) = x3 – 2x2 + 6
f(2)= (2)3 – 2(2)2 + 6
5
=8–8+6
=6
Hence, remainder = 6.
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
22
=1+4+1–6
=0
As f(1) = 0, (x – 1) is a factor of x3 + 4x2 + x – 6.
Worked solutions
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
= x3 + 3x2 + 2x – x2 – 3x –2 = x3 + 2x2 – x – 2
(e)(x – 1)(x + 2)(x + 1)= (x – 1)(x2 + 3x + 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
7f(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
9
Solving, we have a = 3.
(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.
23
1 Algebra: Topic 3 – The remainder and factor theorems and solving cubic equations
Test yourself
1
2
Let f(x) = 4x3 + 3x2 – 3x + 1
f(–1) = 4(–1)3 + 3(–1)2 –3(–1) + 1 = 3
Remainder = 3
(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 x2 gives b + 2a = 6 and since a = 1
this gives b = 4
x3 + 6x2 + 11x + 6 = (x + 2)(x2 + 4x + 3)
3
Equating coefficients of x3 gives a = 1
Equating constant terms gives 2c = 6 so c = 3.
Now (x + 2)(x2 + 4x + 3) = 0
So (x + 2)(x + 1)(x + 3) = 0
Solving gives x = –2, –1 or –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.
Equating coefficients of x3 gives a = 1.
(b) x3 – x2 – 4x + 4 = (x + 2)(ax2 + bx + c)
Equating coefficients of x2 gives –1 = b + 2a so – 1 = b + 2
Equating constant terms gives 4 = 2c so c = 2.
Hence b = –3.
Substituting these values in for a, b and c gives
x3 – x2 – 4x + 4 = (x + 2)(x2 – 3x + 2)
24
= (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 4 – P
roblem solving and
inequalities
Topic
4
Worked solutions
Progress check
From the question her
mother is three times older.
Both will have aged 12 years.
1
Let Amy’s age = x years so her mother’s age = 3x years.
Then Amy will be half the age of her mother so x + 12 =
In 12 years’ time Amy will be x + 12 years and her mother will be
3x + 12 years.
3x + 12
2
Multiplying both sides by 2 we obtain
Solving gives x= 12 years.
2
The difference of the two numbers is 1 (with x being the larger number) so
we can write this as x – y= 1.
2x + 24 = 3x + 12
Amy’s age is 12 years and her mother’s age is 36.
The product of the numbers is 72 and this can be written as
xy = 72
y( y + 1) = 72
From the first equation we can write x = ( y + 1) and substituting this into
the second equation for x we obtain
y2 + y – 72 = 0
Factorising, we obtain ( y + 9)( y – 8) = 0.
As the questions says that x and y are both positive, y = 8.
Solving, we obtain y = –9 or 8.
Now xy = 72 so 8x = 72 and hence x = 9.
Hence x = 9 and y = 8.
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 x + 4x + 4 – x = 64
2
2
4x + 4 = 64
x = 15
4x = 60
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
(d) –4 < x < 4
6
(c) 1 < x ≤ 8
(e) 4 ≤ x < 10
(a) 3, 4, 5, 6, 7, 8, 9 10
(b) –4, –3, –2, –1, 0
(d) 16, 17, 18
(c) 1, 2, 3, 4, 5, 6, 7
(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
26
(b) –3 < x < 7
(d) –4 ≤ x < 5
(c) 4 < x < 14
(e)
x < 5 and x ≥ 8
(f)
x ≤ –1 and x > 2
8
9
14 ≤ x ≤ 18
(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
3
x > – 4
(e)9 – 5x ≤ 4 – x
9 – 4x ≤ 4
– 4x ≤ –5
5
x ≥
4
(f) 5 – x < 3(x – 2)
5 – x < 3x – 6
5 – 4x < –6
–4x < –11
11
x >
4
Worked solutions
10
x2 – 6x + 8 > 0
Sketching the curve for y = x2 – 6x + 8 gives the following:
As the curve has a positive coefficient, it will be -shaped, cutting the
x-axis at x = 4 and x = 2.
y
y = x2 – 6x + 8
2
>>>
TIP
If you do not cancel
fractions you may lose
marks.
4
x
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
6
x > – 4
3
x ≥ – 2
The inequality sign is
reversed because both sides
have been divided by a
negative quantity (i.e. –4).
27
1 Algebra: Topic 4 – Problem solving and inequalities
125x2 + 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:
y = 5x2 + 7x – 6
y
x
–2
28
O
3
5
We want the part of the graph which is below or on the x-axis.
3
Meaning that x lies between –2 and inclusive, which can be written
5
3
mathematically as –2 ≤ x ≤ .
5
Worked solutions
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.
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
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.
Hence, the coordinates of the point of contact are (4, 8).
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.
14Substituting 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
2x2 + 10x + 12 = 0
Dividing through by 13 we obtain
Hence, (2x + 4)(x + 3) = 0
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).
29
1 Algebra: Topic 4 – Problem solving and inequalities
15
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.
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.
x=
Substituting these values into the quadratic equation formula
x=
x=
–b ± √ b2 – 4ac
gives:
2a
5 ± √ (–5)2 – 4(1)(–5)
2(1)
5 ± √ 25 + 20 5 ± √ 45 5 + √ 45
5 – √ 45
=
=
or =
= 5.85 or –0.85 (2 d.p.)
2
2
2
2
Test yourself
1
Add 2 to both sides.
(a)3x – 2 > 7
3x > 9
Divide both sides by 3.
x>3
Multiply out the brackets.
(b)3(x – 2) > 9
3x – 6 > 9
3x > 15
Add 6 to both sides.
Divide both sides by 3.
x > 5
x–5
(c) ≤ –3
7
Subtract 4x from both sides.
30
Divide both sides by minus
1 and remember to reverse
the inequality.
x – 5 ≤ –21
x ≤ –16
(d)3x – 4 < 4x + 6
–x – 4 < 6
–x < 10
x > –10
Multiply both sides by 7.
Add 5 to both sides.
Add 4 to both sides.
Worked solutions
Subtract x from both sides.
Add 4x to both sides.
Divide both sides by 5.
Multiply out the brackets.
Subtract 9 from both sides.
2
(a)2x – 4 > x + 6
x–4>6
Add 4 to both sides.
x > 10
(b)4 + x < 6 – 4x
4 + 5x < 6
5x < 2
x<
Subtract 4 from both sides.
2
5
x < 0.4
(c)2x + 9 ≥ 5(x – 3)
2x + 9 ≥ 5x – 15
Subtract 5x from both sides.
Divide both sides by –3 remembering to
reverse the inequality in the process.
–3x + 9 ≥ –15
3
–3x ≥ –24
x≤8
Rearranging the inequality we have x2 – 3x – 18 > 0
Considering the case where x2 – 3x – 18 = 0
Factorising gives (x + 3)(x – 6) = 0
y
Giving x = –3 or 6 (these are
the intercepts on the x-axis)
Note we need the section
of the curve which is above
(and not 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:
y = x2 – 3x – 18
–3
O
6
x
Hence, x < –3 and x > 6.
31
1 Algebra: Topic 4 – Problem solving and inequalities
TAKE NOTE
Note with practice you
may find you do not
need to draw the curve,
which will save you a
bit of time.
4Factorising 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.
Hence Now x2 – 2x – 15 ≤ 0. This is the region below the x-axis (i.e. where y ≤ 0).
5
–3≤x≤5
(a)5 < 2x – 1 ≤ 13
6 < 2x ≤ 14
3 < x ≤ 7
(b) –7 < 3x – 5 < 4
–2 < 3x < 9
2
– < x < 3
3
Divide each side by 2.
Add 5 to each side.
Divide each side by 3.
(c)4(x – 3) ≤ 3(x – 2)
Multiply out the brackets.
x – 12 ≤ – 6
Add 12 to both sides.
4x – 12 ≤ 3x – 6
32
Add 1 to each side.
x≤6
Subtract 3x from both sides.
1 Algebra: Topic 5 – T
he binomial expansion and
probability
Topic
5
Worked solutions
Progress check
1(a)(5) = r!(n – r)! = 5!3! =
8
()
n!
8!
8×7×6
= 56
3×2
7
n!
7! 7 × 6 × 5
=
=
= 35
(b) =
3 r!(n – r)! 3!4!
3×2
( )
12
n!
12! 12 × 11 × 10 × 9 × 8 × 7
=
=
= 924
(c) =
6
r!(n – r)! 6!6!
6×5×4×3×2×1
7
C2 =
(d)
(e)
(f)
C5 =
12
C2 =
4
n!
7! 7 × 6
=
=
= 21
r!(n – r)! 2!5! 2 × 1
n!
12! 12 × 11 × 10 × 9 × 8
=
=
= 792
r!(n – r)! 5!7!
5×4×3×2×1
n!
4! 4 × 3
=
=
=6
r!(n – r)! 2!2! 2 × 1
2(a)(0) = 1
TAKE NOTE
You 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.
1
()
()
( )
()
()
()
( )
2
(b) = 2
1
3
(c) = 3
2
10
(d) = 120
3
5
3(a) 0 = 1
5
(b) = 5
1
8
(c) = 56
5
10
(d) = 252
5
4
The formula is as follows:
(a + b)n = an + nan – 1b +
n(n – 1) n – 2 2 n(n – 1)(n – 2) n – 3 3
a b +
a b +…
2!
3!
33
1 Algebra: Topic 5 – The binomial expansion and probability
Here n = 3, a = 3 and b = 2x.
(3 + 2x)3= 33 + 3(3)2(2x) +
= 27 + 54x + 36x2 + 8x3
5
Looking at the above it
can be seen that the term
in x2 is the third term in
the expansion.
(3)(2) 1
(3)(2)(1) 0
3 (2x)2 +
3 (2x)3
2!
3!
The formula is as follows:
(a + b)n = an +
(1n)a
n–1
b+
(2n)a
3
Here n = 6, a = x and b = .
x
b +…+
n–2 2
(nr)a
b + … + bn
n–r r
Substituting in the values for a, b and n we obtain
( )
x+
() () () () () ()
36 6 6 5 3
6 4 32 6 3 33
=x +
x
+
x
+
x
+…
x
1
x
2
x
3
x
Term in x2 =
(62) x (3x)
2
4
To find the coefficients we will expand Pascal’s triangle.
1
11
121
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.
1331
14641
15101051
As
6
1 6152015 6 1
()
Obtaining the formula and following the pattern in the terms gives:
(1n)a b + (2n)a b + (3n)a b + …
5
5
5
5
5
(a + b) = a + ( )a b + ( )a b + ( )a b + ( )ab + ( )b
1
2
3
4
5
(a + b)n = an +
5
The coefficients of this expansion
can be found using Pascal’s triangle
or worked out using a calculator.
34
()
6
32
= 15, we have term in x2 = 15x4
= 135x2
2
x
5
n–1
n–2 2
4
3 2
()
()
2 3
n–3 3
4
2
Putting a = x, b = , and n = 5 into the equation, gives:
x
( )
x+
()
5
() ()
25 5
2
22
23
24 25
= x + 5x4
+ 10x3
+ 10x2
+ 5x
+
x
x
x
x
x
x
= x5 + 10x3 + 40x +
80 80 32
+
+
x x3 x5
Worked solutions
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.
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).
= 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 + 0.02)6
= 1.1262 (4 d.p.)
8(i) P(X = r) = ( r )pr(1 – p)n – r
n
p = 0.4, n = 10 and r = 5.
(105)0.4 (1 – 0.4)
10
P(X = 5) = ( )0.4 (0.6)
5
P(X = 5) =
5
5
5
5
= 0.2007
We need to find the
probability of each of the
above and then add the two
probabilities together.
= 0.201 (3 s.f. )
(ii) Less than 2 female turtles means 0 or 1 turtle.
(100)0.4 (1 – 0.4)
10
= ( )0.4 (0.6)
0
P(X = 0) =
0
10 – 0
10
= 0.006 047
= 0.006 05 (3 s.f.)
(101)0.4 (1 – 0.4)
10
= ( )0.4 (0.6)
1
P(X = 1) =
0
1
1
10 – 1
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.)
35
1 Algebra: Topic 5 – The binomial expansion and probability
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.
1
(16) = 216
3
9
(a)P(X = 3) =
(b) P (X = 0) =
= 0.004 630
= 0.5787
(56) = 125
216
3
Note the probability of not
getting a six = 1 – 16 = 56.
(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).
(32)(16) (1 – 16)
3 1 5
= ( )( ) ( ) 2 6 6
2
P(X = 2)=
2
3–2
1
= 0.069 44
Hence, probability at least 2 sixes = P(2 sixes) + P(3 sixes)
= 0.069 44 + 0.004 630
= 0.0741 (3 s.f.)
= 0.07407
Test yourself
1(a + b)n = an + (1)an – 1b + (2)an – 2b2 + …
n
n
(nr) = r!(nn!– r)!
The term in x2 is given by:
(2n)a
b
n–2 2
Here a = 2, b = 3x and n = 5
36
So the term in x2 is
5!
(2)3(3x)2 = 10 × 8 × 9x2 = 720x2
2!(5 – 2)!
Hence, the coefficient of x2 is 720.
Worked solutions
n(n – 1)x2 n(n – 1)(n – 2)x3
+…
+
2!
3!
2
(1 + x)n = 1 + nx +
Hence (1 + 3x)6 = 1 + (6)(3x) +
In this case we substitute 3x for x and 6 for n.
= 1 + 18x + 135x2 + 540x3
3P(X = r) = ( r )pr(1 – p)n – r n
p = 0.25 and n = 20.
P(X = 4) =
=
(6)(5)(3x)2 (6)(5)(4)(3x)3
+
2×1
3×2×1
(204)0.25 (1 – 0.25)
4
(204)0.25 (0.75)
4
= 0.1897 (4 s.f.)
= 0.1746 (4 s.f.)
20 – 4
16
4P(X = 15) = (15)0.815(1 – 0.8)5
20
5P(X = r) = ( r )pr(1 – p)n – r
n
Now p = 0.12, n = 10, r = 1, so we have
(101)0.12 (1 – 0.12)
10
P(X = 1) = ( )0.12 (0.88)
1
P(X = 1) =
1
1
10 – 1
9
= 0.3798 (4 s.f.)
37
2 Coordinate geometry: Topic 6 – Coordinate
geometry of straight lines
Topic
6
Worked solutions
Progress checks
1(a)Negative
(b)Zero
(c)Positive
(d)Negative
(e)Positive
(f) Infinite
(g)Negative
2
(h)Zero
(a) Gradient =
y2 − y1 9 − 3 6
=
= =3
x2 − x 1 4 − 2 2
(c) Gradient =
y2 − y1
0−3
−3
=
=
=1
x2 − x1 −5 − (− 2) −3
(b) Gradient =
(d) Gradient =
(e) Gradient =
(f) Gradient =
(g) Gradient =
(
(
(
(
3(a)
(b)
(c)
38
(d)
y2 − y1 12 − 0 12 3
=
=
=
x2 − x 1 9 − 1
8 2
y2 − y1
1 − 10
−9 9
=
=
=
x2 − x1 −5 − (− 1) −4 4
y2 − y1 6 − (−3) 9
=
= =9
x2 − x 1
1−0
1
y2 − y1 2 − (−2) 4
=
=
x2 − x 1
4−1
3
y2 − y1 −5 − (−4) −1
1
=
=
=
x2 − x1 −1 −10 −11 11
)
)
)
)
(
(
(
(
)
)( )
)
)
x1 + x2 y1 + y2
1+3 2+8
=
= (2, 5)
,
,
2
2
2
2
x1 + x2 y1 + y2
0+4 2+1
3
=
= 2,
,
,
2
2
2
2
2
x1 + x2 y1 + y2
−2 + 0 5 + (−5)
=
= (−1, 0)
,
,
2
2
2
2
x1 + x2 y1 + y2
−8 + (−2) 4 + (−6)
=
= (−5, −1)
,
,
2
2
2
2
Worked solutions
(e)
(f)
(g)
(
(
(
)(
)(
)(
)( )
)( )
)( )
x1 + x2 y1 + y2
10 + (−3) 12 + 0
7
=
= , 6
,
,
2
2
2
2
2
x1 + x2 y1 + y2
−3 + (−4) −4 + 6
7
=
= − , 1
,
,
2
2
2
2
2
x1 + x2 y1 + y2
8 + (−5) −1 + 7
3
=
= , 3
,
,
2
2
2
2
2
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
(g)
√(x2 − x1)2 + (y2 − y1)2 = √(−7 − (−6))2 + (2 − (−3))2 = √26 = 5.10
C to D =
(f)
√(x2 − x1)2 + (y2 − y1)2 = √(0 − (−12))2 + (10 − 5)2 = √169 = 13.0
5(a)A to B =
( )
( )
3
−1
3
−1
Hence AB = CD
(b) A to B =
C to D =
()
()
5
1
5
1
Hence AB = CD
(c) A to B =
C to D =
( )
( )
6
−1
6
−1
Hence AB = CD
(d) A to B =
C to D =
( )
()
6
−1
6
1
Hence AB = CD
39
2 Coordinate geometry: Topic 6 – Coordinate geometry of straight lines
(e) A to B =
C to D =
( )
( )
−2
4
−4
−2
Hence AB = CD
(f) A to B =
C to D =
()
()
6
1
1
6
Hence AB = CD
(g) A to B =
C to D =
()
()
1
2
2
1
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
2
(c)
m= ,c=1
3
4
(d)
m = − , c = 3
3
1
3
(e)
m = , c = − 2
2
1
4
(f)
m = , c = − 2
3
8 The mid-point of a line joining the points (x1, y1) and (x2, y2) is given by:
x1 + x2 y1 + y2
,
2
2
−5 + 1 12 + 4
Mid-point of AB =
= (−2, 8)
,
2
2
(
40
)
(
)
Worked solutions
9
(a) Mid-point of PQ, M =
(b) P to Q =
()
8
2
PQ2 = 82 + 22
PQ2 = 68
PQ2 = 64 + 4
(
)
−4 + 4 3 + 5
= (0, 4)
,
2
2
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
= √16 +1 = √17 = 4.12
(c) Length = √(x2 − x1)2 + (y2 − y1)2 = √(4 − 0)2 + (7 − 8)2
Remember that the midpoint of the line joining the
points (x1, y1) and (x2, y2) is
(
11
)
x1 + x2 y1 + y2
.
,
2
2
Let the coordinates of B be (x, y).
x+2
2
x+2
The x-coordinate is 4, so 4 =
2
The x-coordinate of the mid-point =
Solving gives x = 6
y+3
2
y+3
The y-coordinate is 4, so 4 =
2
Solving gives y = 5
The y-coordinate of the mid-point =
12
Hence, coordinates of B are (6, 5)
Gradient of line
y2 − y1 k − 2 k − 2
=
=
x2 − x 1 5 − 0
5
4
Gradient of AB = , so
5
k−2 4
=
5
5
41
2 Coordinate geometry: Topic 6 – Coordinate geometry of straight lines
13
Solving, gives k = 6
(a) Gradient = 4
3
(b) Gradient =
2
(c) Gradient =
14
(d) Gradient =
1
4
2
3
(a) As point B lies on the line, its coordinates will satisfy the equation of the line.
Hence, 5(2) − 2(k) = 2
Solving gives k = 4
(b) Mid-point =
154x − 3y = c
(
)(
)
x1 + x2 y1 + y2
4+2 9+4
=
= (3, 6.5)
,
,
2
2
2
2
Substituting the coordinates (−3, 2) into this equation gives
4(−3) − 3(2) = c
Solving gives c = −18.
Hence required equation of line is 4x − 3y = −18
3–1
=1
3–1
0–3
= –1
Gradient of BC =
6–3
(b) Product of the gradients = (1)(−1) = −1.
16
17
(a) Gradient of AB=
As m1m2 = −1, AB and BC are perpendicular to each other.
(a) Gradient of PQ =
0–6
3
= – 4–0
2
For equation of line PQ,
3
y – y1 = m(x – x1) where m = – and (x1, y1) = (0, 6)
2
3
y – 6 = – (x – 0)
2
2y – 12 = –3x
2y = –3x + 12
42
Both gradients are found
using the formula:
Gradient =
y2 − y1
x2 − x 1
Worked solutions
(
)(
)
x1 + x2 y1 + y2
0+4 6+0
=
= (2, 3)
,
,
2
2
2
2
Gradient of line perpendicular to PQ is given by
Equation of line perpendicular to PQ,
(b) Mid-point of PQ =
( )
3
m × – = −1
2
2
m =
3
2
y – y1 = m(x – x1) where m = and (x1, y1) = (2, 3)
3
2
y – 3 = (x – 2)
3
3y – 9 = 2x – 4
3y = 2x + 5
18(a)y = 6x − 5
y = 6(1) − 5
y = 1 which is the y-value of the point so the point lies on the line.
(b) y = 6x − 5
y = 6(0) − 5
y = −5 which is the y-value of the point so the point lies on the line.
(c) y = 6x − 5
y = 6(2) − 5
y = 7 as this is not the y-coordinate of the point, the point does not lie on the line.
y = 12 − 5
(d) y = 6x − 5
y = 6(12) − 5
y=3−5
y = −2
This is the y-value of the point, so the point lies on the line.
(e) y = 6x − 5
y = 6(−1) − 5
This is not the y-value of the point, so the point does not lie on the line.
y = −11
43
2 Coordinate geometry: Topic 6 – Coordinate geometry of straight lines
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
(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
0 = 4x + 2
−2 = 4x
x = –
1
2
(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
2
y =
5
20(a)y = 4x + 2
y=3
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
44
x=2
Worked solutions
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 = 3 into equation (1) we obtain
7(3) + 2y = 19
21 + 2y = 19
2y = −2
y = −1
Checking by substituting x = 3 and y = −1 into equation (2)
So LHS = RHS
22
LHS = 3 − (−1) = 4
RHS = 4
Point of intersection is (3, –1).
7y = 5x – 27 .......................................... (1)
4y = 3x – 16 ........................................... (2)
Multiplying equation (1) by 3 and equation (2) by 5 gives the following
20y = 15x – 80
21y = 15x – 81
Note that −81 − (−80) = −1.
Be very careful when
subtracting negative
numbers.
Subtracting these two equations gives
y = −1
Substituting y = −1 into equation (1) gives
−7 = 5x − 27
20 = 5x
x=4
45
2 Coordinate geometry: Topic 6 – Coordinate geometry of straight lines
Checking in equation (2)
So LHS = RHS
23
LHS = 4(−1) = –4
RHS = 3(4) − 16 = –4
Point of intersection is (4, –1).
Rearranging each equation so that it is the form y = mx + c.
y − 3x + 4 = 0 so y = 3x − 4. This line has a gradient of 3.
1
4y − 12x + 1 = 0 so y = 3x − . This line also has a gradient of 3.
4
Both lines have the same gradient and are therefore parallel so they will
not intersect.
24
4x + y = 1
3x + 2y = 7
Multiplying the first equation by 2 gives
8x + 2y = 2 .................................................... (1)
3x + 2y = 7 .................................................... (2)
(1) − (2) gives
Hence, −8 + 2y = 2
46
2y = 10
y=5
Checking by substituting x = −1 and y = 5 into equation (2) gives
x = −1
Substituting x = −1 into equation (1) gives
5x = −5
LHS = 3(−1) + 2(5) = −3 + 10 = 7
RHS = 7
So LHS = RHS
Coordinates of A are (−1, 5)
Worked solutions
Test yourself
1
(a) Gradient of AB = Gradient of CD = y2 − y1 1 – 0 1
=
=
x2 − x 1 4 – 1 3
y2 − y1
4–3
1
=
= x2 − x1 2 – (–1) 3
As the gradients of AB and CD are the same the two lines are parallel.
(b) Gradient of AB = 13 and AB passes through point A (1, 0) so equation of
AB 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
The equation is multiplied
through by the common
denominator, 2(k + 7).
y2 − y1 –1 – 4
–5
=
=
x2 − x1 k – (–7) k + 7
1
But gradient of AB = – so
2
–5
1
= – k+7
2
2
(a) Gradient of AB = –5 × 2 = –1(k + 7)
–10 = –k – 7
Giving k = 3.
(b) The product of the gradients of perpendicular lines is –1. Hence,
( )
1
m – = –1
2
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
47
2 Coordinate geometry: Topic 6 – Coordinate geometry of straight lines
3
(a)
y
B (1, 6)
6
5
4
3
(–3, 2) A
2
C (6, 1)
1
–3
TAKE NOTE
This formula needs
to be remembered. If
you forget it you can
plot the two points
on a sketch graph and
form a triangle and use
Pythagoras' theorem
to find the length of the
hypotenuse.
–2
Gradient of AB = Gradient of BC = 0
1
2
3
4
5
6
7
y2 − y1 1 – 6 –5
=
= = –1
x2 − x 1 6 – 1 5
Product of gradients = (1)(–1) = –1 proving that the two lines are
perpendicular to each other.
Putting the coordinates A (–3, 2) and B (1, 6) into the formula gives
Using the coordinates B (1, 6) and C (6, 1) in the formula gives
AB = √(1 − [–3])2 + (6 − 2)2 = √16 + 16 = √32 units
BC = √(6 − 1)2 + (1 − 6)2 = √25 + 25 = √50 units
(c)Tan AĈB =
AB √ 32
=
=
BC √ 50
3(3) + 2(4) = c
3x + 2y = c.
Substituting the coordinates (3, 4) into the equation, we obtain
Giving c = 17.
√ 16 × 2 = 4√ 2 = 4
√ 25 × 2 5√ 2 5
The equation of the line parallel to 3x + 2y = 5, will be of the form
x
y2 − y1
6–2
4
=
= =1
x2 − x1 1 – (–3) 4
(b) Length = √(x2 − x1)2 + (y2 − y1)2
4
48
–1
Hence, the equation of the line is
3x + 2y = 17
Worked solutions
5
y2 − y1
2–0
2 1
=
= =
x2 − x1 0 – (–4) 4 2
1
Equation of line PQ with gradient and passing through the point
2
Q (0, 2) is
(a) Gradient of PQ = y – y1 = m(x – x1)
1
y – 2 = (x – 0)
2
2y – 4 = x
Hence, equation of line is 2y – x = 4
(b) Mid-point of PQ =
(
)(
)
x1 + x2 y1 + y2
–4 + 0 0 + 2
=
= (–2, 1)
,
,
2
2
2
2
1
Gradient of PQ = 2
Gradient of line perpendicular to PQ, m is given by
1
= –1 (i.e. the products of the gradients of two
(m)
2
lines are –1 if they are perpendicular).
Hence, ()
m = –2
Equation of required line through mid-point is:
y – y1 = m(x – x1)
>>> TIP
For questions involving
coordinate geometry,
it is always worthwhile
spending a little time
sketching a graph
showing the positions of
the coordinates. It is then
easier to see the shape
formed when certain
points are joined up
with lines. You can also
check the signs of any
gradients you have found
numerically.
y – 1 = –2(x – [–2])
y – 1 = –2x – 4
6
y = –2x – 3
(a)
R (–2, 6)
y
6
Q (3, 5)
5
4
3
2
–4 –3
–2
1
–1 0
-1
-2
Gradient of PQ = 1
P (1, 0)
2
3
4
y2 − y1 5 – 0 5
=
=
x2 − x 1 3 – 1 2
x
49
2 Coordinate geometry: Topic 6 – Coordinate geometry of straight lines
Multiply both sides by 2 to
remove the denominator in
the fraction.
(b) Equation of straight line PQ which passes through (1, 0) and has
5
gradient is given by:
2
5
y – y1 = m(x – x1) where m = and (x1, y1) = (1, 0)
2
5
y – 0 = (x – 1)
2
5
y = (x – 1)
2
2y = 5(x – 1)
2y = 5x – 5
2y – 5x = –5
You can now form the
parallelogram and it is
possible to see from the
sketch the rough position of
point S. If you give it some
thought you can get the
coordinates of point S using
the diagram.
(c)
y
6
R (–2, 6)
Q (3, 5)
5
4
3
2
S
–4 –3
–2
1
–1 0
–1
–2
1
P (1, 0)
2
3
4
x
As lines RS and PQ are opposite sides of a parallelogram, they are
parallel and therefore have the same gradient.
5
Hence, gradient of RS = .
2
Equation of straight line RS which passes through (–2, 6) and has
5
gradient is given by:
2
5
y – y1 = m(x – x1) where m = and (x1, y1) = (–2, 6)
2
5
y – 6 = (x – [–2])
2
5
y – 6 = (x + 2)
2
2y – 12 = 5(x + 2)
2y – 12 = 5x + 10
50
2y – 5x = 22
Worked solutions
(d) To find the coordinates of the point S, the equations for lines RS and
SP are solved simultaneously.
2y – 5x = 22 ...................................................... (1)
5y + x = 1 ............................................................ (2)
25y + 5x = 5
Multiplying equation (2) by 5 gives
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
Hence the coordinates of point S are (–4, 1)
So,x = –4
7
(a) Because there are lots of points in this question, it is worth spending
a little time doing a sketch showing their positions.
y
3
2
1
0
–1
1
2
3
4
5
6
7
–3
–4
–6
–7
9
x
D (5, –2)
–2
–5
8
C (9, –4)
(2, –5) A
B (6, –7)
51
2 Coordinate geometry: Topic 6 – Coordinate geometry of straight lines
y2 − y1 –7 – (–5) –2
1
=
= = – x2 − x 1
6–2
4
2
y − y –4 – (–2) –2
1
= = – (b) Gradient of DC = 2 1 =
x2 − x 1
9–5
4
2
Gradient of AB = Gradient of AB = gradient of DC, so lines are parallel
4
(c) The vector to go from A to B is
.
–2
4
The vector to go from D to C is
.
–2
These vectors are the same so the lines are the same length.
3
.
(d) The vector to go from A to D is
3
3
The vector to go from B to C is
.
3
These vectors are the same so the lines are both parallel and
the same length.
The two lines at right angles
which have the vector as the
hypotenuse are the same for both
triangles so the lengths of the
hypotenuse will be the same.
8(a)
( )
( )
()
()
Hence, this and the answer to parts (b) and (c) prove that
ABCD is a parallelogram.
y
6
5
B (1, 5)
4
C (4, 5)
M
3
2
A (4, 2)
1
0
1
2
(
3
4
5
)(
6
x
) ( )
x1 + x2 y1 + y2
4+1 2+5
5 7
=
= ,
,
,
2
2
2
2
2 2
y −y 5–2
3
(b) Gradient of AB = 2 1 =
=
= –1 x2 − x1 1 – 4 (–3)
Mid-point, M =
y2 − y1 5 – 72 (32)
=
=
=1
(c) Gradient of MC = x2 − x1 4 – 52 (32)
52
Product of gradients of lines AB and MC = (–1)(1) = –1
As the product of two perpendicular lines is –1 so AB and MC
are perpendicular.
Worked solutions
(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 9
Hence, equation of MC is y = x + 1
2
Gradient of line = – 3
(a)2x + 3y = 5
3y = –2x + 5
2
5
y = – x +
3
3
2
(b) Equation of line with gradient – and passing through the
3
point R (3, 3) is
y – y1 = m(x – x1)
2
y – 3 = – (x – 3)
3
3y – 9 = –2x + 6
Now, the equation of the y-axis is x = 0.
10 Hence, equation of line is 3y = –2x + 15
Solving this simultaneously with the equation of the line, we have
3y = 0 + 15 so y = 5
Hence, S is the point (0, 5)
(a)4x + 5y = 10 5y = –4x + 10
4
y = – x + 2
5
4
Hence, gradient of AB = – 5
53
2 Coordinate geometry: Topic 6 – Coordinate geometry of straight lines
(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 C lies on the line.
4
5
(c) As lines are perpendicular, m × – = –1, so m =
5
4
5
Gradient of line at right-angles to AB = 4
5
Equation of line with gradient and passing through the
4
point C (–5, 6) is
y – y1 = m(x – x1)
5
y – 6 = (x – [–5]) 4
4y – 24 = 5(x + 5)
4y – 24 = 5x + 25 54
Hence, equation of line is 4y = 5x + 49
2 Coordinate geometry: Topic 7 – Coordinate
geometry of the circle
Topic
1
7
Worked solutions
Progress check
Note that the radius of
a circle cannot be zero
as it is a length.
1(a)x2 + y2 = 1
This equation is in the form x2 + y2 = r2
Hence r = √ 1 = 1
(b)
x 2 + y2 = 9
This equation is in the form x2 + y2 = r2
Hence r = √ 9 = 3
(c)
x2 + y2 = 25
This equation is in the form x2 + y2 = r2
Hence r = √ 25 = 5
(d)
x 2 + y2 – 4 = 0
x2 + y2 = 4
This equation is in the form x2 + y2 = r2
Hence r = √ 4 = 2
(e)
x2 + y2 – 49 = 0
x2 + y2 = 49
This equation is in the form x2 + y2 = r2
Hence r = √ 49 = 7
(f)4x2 + 4y2 = 16
Dividing both sides by 4 we obtain
This equation is in the form x2 + y2 = r2
x2 + y2 = 4
Hence r = √ 4 = 2
55
2 Coordinate geometry: Topic 7 – Coordinate geometry of the circle
(g)8x2 + 8y2 = 72
3(a)
(x – 3)2 + (y + 1)2 = 9
Dividing both sides by 8 we obtain
x2 – 6x + 9 + y2 + 2y + 1 = 9
This equation is in the form x2 + y2 = r2
x2 + y2 = 9
Hence r = √ 9 = 3
(h) 3(x2 + y2) – 27 = 0
Dividing both sides by 3 we obtain
x2 + y2 – 9 = 0
x2 + y2 = 9
This equation is in the form x2 + y2 = r2
Hence r = √ 9 = 3
(i)
y2 = 16 – x2
x2 + y2 = 16
This equation is in the form x2 + y2 = r2
Hence r = √ 16 = 4
(j)
x 2 + y2 – 5 = 0
x2 + y2 = 5
This equation is in the form x2 + y2 = r2
Hence r = √ 5
(k)
x2 + y – 50 = 0
x2 + y2 = 50
This equation is in the form x2 + y2 = r2
Hence r = √ 50 = √ 25 × 2 = 5√ 2
2(a)x2 + y2 = 9
(b)
x2 + y2 = 16
(c)
x2 + y2 = 81
(d)
x 2 + y2 = 6
(e)
x2 + y2 = 12
(f)
x2 + y2 = 45
(g)
x2 + y2 = 18
56
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
Worked solutions
Test yourself
1
(a) Comparing the equation x2 + y2 – 8x – 6y = 0 with the equation
x2 + y2 + 2gx + 2fy + c = 0 we can see g = –4, f = –3 and c = 0.
Centre A has coordinates (– 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 = 0 with the equation
x2 + y2 + 2gx + 2fy + c = 0 we can see g = –2, f = 3 and c = –3.
Centre A has coordinates (– 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
equation of the circle.
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
3
Both sides of the equation equal zero so P (2, 1) lies on the circle.
(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
AP is a radius of the circle
and will make an angle of 90°
to the tangent at point P.
x2 + y2 – 4x – 6y – 12 = 0
(b) Gradient of line joining the centre A (2, 3) to P (5, 7)
=
7–3 4
=
5–2 3
3
Gradient of tangent = – 4
Equation of tangent is
3
y – 7 = – (x – 5)
4
4y – 28 = –3x + 15
4y + 3x – 43 = 0
57
2 Coordinate geometry: Topic 7 – Coordinate geometry of the circle
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
Centre of circle is at (2, –4)
(b) If the point P (6, –4) lies on the circle, the coordinates will satisfy the
equation of the circle.
Hence, (x – 2)2 + (y + 4)2 = (6 – 2)2 + (–4 + 4)2
5
= 16
This is the same as the RHS of the equation so the point lies
on the circle.
(a) Centre of circle is at the mid-point of the diameter AB.
Mid-point of line joining A(1, –4) and B(9, 10) is
(1 +2 9 , –4 +2 10) = (5, 3)
(b) Distance between the points (1, –4) and (5, 3) is given by
The distance from the midpoint to the circumference is
the radius of the circle.
Make sure you give the equation of the
circle in the format asked for in the
question (i.e. x2 + y2 + ax + by + c = 0 in
this case).
58
= 42 + 0
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
(x – 5)2 + (y – 3)2 = 65
For this circle, centre is (5, 3) and radius is √ 65.
Multiplying out the brackets we obtain
x2 – 10x + 25 + y2 – 6y + 9 = 65
x2 + y2 – 10x – 6y – 31 = 0
Worked solutions
6
Radius of the circle = √ 40 = 6.32
√(x2 − x1)2 + (y2 − y1)2
The length of a straight line joining the two points (x1, y1) and (x2, y2) is
given by:
Distance of the point (4, 3) from the centre of the circle (0, 0)
= √(x2 − x1)2 + (y2 − y1)2
This distance is less than the radius of the circle, so the point lies inside
the circle.
= √ (4 – 0)2 + (3 – 0)2 = √ 16 + 9 = 25 = 5
7
(a) The mid-point of the diameter of the circle is the centre of the circle.
Mid-point of AB =
(0 +2 4 , –32+ 1) = (2, –1)
So centre of circle is at (2, –1).
(b)Radius, r, is distance from B (4, 1) to centre of circle (2, –1).
r2 = (x – a)2 + (y – b)2
= (4 – 2)2 + (1 – [–1])2
= 22 + 22
= 8
r = √ 8 = √ 4 × 2 = 2√ 2
59
2 Coordinate geometry: Topic 8 – Inequalities
and linear programming
Topic
6
8
Worked solutions
Progress check
1(a)x ≤ 3
2
(a) Region 2
(c) Region 1
(b) y ≥ 2
(d)
y ≥ –2x + 8
(c)
y≤x
(e)
y ≥ 12 x + 3
3
(b) Region 5
(d) Region 3
(e) Region 4
You first have to obtain some points to be able to draw each line. Note that
as the inequalities all have an equals component in the question, all the
lines will be solid lines.
y = 0 (This line is the x-axis)
x = 2 (This is a vertical line at x = 2)
2y = 12 – x (When x = 0, y = 6 and when y = 0, x =12)
y = 13 x (When x = 0, y = 0 and choosing a suitable point on the x-axis such
as x = 12 so y = 13 [12] = 4)
Adding the above lines to the graph, we obtain the following:
y
x=2
12
11
10
9
8
7
6
5
4
y = 13 x
3
2
1
0
60
2y = 12 – x
1
2 3
4 5
6
7 8
9 10 11 12
x
Worked solutions
We now add the shadings to each line to show the sides of the line that do
not satisfy the inequality.
For y ≥ 0 we shade below the x-axis.
For x ≥ 2 we shade to the left of the line x = 2.
1
1
For y ≤ x we shade above the line y = x.
3
3
Remember we are shading
the area not required.
For 2y ≤ 12 – x we shade above the line 2y = 12 – x.
Adding the shadings to the lines we now have the following:
y
x=2
12
11
10
9
8
7
6
5
4
3
2
1
0
1
2 3
4 5
x=2
11
9
8
7
6
5
4
3
2
0
1
2 3
7 8
9 10 11 12
x
Be careful to select the correct region as the region
which fits all the inequalities (i.e. the allowable
region). You are looking for the region which is
completely enclosed by lines where the shading is
on the opposite side to the region.
10
1
6
2y = 12 – x
y
12
FEASIBLE
REGION
y = 13 x
4 5
FEASIBLE
REGION
6
7 8
We now shade in all the regions on the graph not in
the feasible region.
y = 13 x
2y = 12 – x
9 10 11 12
x
61
2 Coordinate geometry: Topic 8 – Inequalities and linear programming
4
y
15
14
y=x
y = –1.5x + 15
13
12
9
4
x + y ≤ 800
1
x ≥ y
3
2y = 10 – x
3
2
1
0
1
2 3
4 5
6
7 8
9 10 11 12 13 14
x
(b) Amount of
money per
week = 0.8x + 0.9y
Note that the cost
of each cookie
must be converted
into pounds, so we
divide each cost in
pence by 100.
7
Regarding the constraints on the purchase costs, we have
Note that you could simplify this by dividing both sides of the inequality
by 5000 to give 4y + 3x ≤ 100.
20 000y + 15 000x ≤ 500 000.
Regarding the constraints made by the popularity of the vans, we have
y ≥ 3x
Regarding the constraints of the running costs, we have
Note that this inequality could be simplified by dividing both sides
by 10 to give
62
Note than no more than
means less than or equal to.
y ≤ 500
8
5
(b) 20x + 30y ≤ 1500
Up to 100 means
less than or equal
to 100.
6(a)x ≤ 500
10
6
(a) 3x + 4y ≤ 100
(c)
y < 3x
11
7
5
40x + 50y ≤ 1200
4x + 5y ≤ 120
Worked solutions
Test yourself
1(a)x ≤ 150
y ≤ 120
x + y ≤ 200
(b)For x + y = 200, when x = 0, y = 200 and when y = 0, x = 200.
x = 0 and y = 0 are the y- and x-axes respectively.
x = 150 is a vertical line and y = 120 is a horizontal line.
Adding these lines to the graph and shading the regions that are
excluded we obtain.
x = 150
y
200
175
150
x + y = 200
y = 120
125
100
75
50
25
0
50
(c) Profit = 2x + 5y
The values of x and y
are substituted into
Profit = 2x + 5y.
25
75 100 125 150 175 200
x
Now, one of the vertices in the feasible (i.e. unshaded) region will
maximise the profit.
These can each be substituted into the Profit equation in turn
until the maximum value is found. You can see that some of these
coordinates can be discounted. For example (150, 0) and (150, 50)
have the same x-coordinate but the one with the higher y-coordinate
will be the only one to consider.
There are four vertices; (150, 0), (0, 120), (150, 50) and (80, 120).
When x = 0 and y = 120 the profit is 2(0) + 5(120) = £600.
When x = 80 and y = 120 the profit is 2(80) + 5(120) = £760 and this
is the highest value.
Hence, 80 of bag A and 120 of bag B should be produced.
63
2 Coordinate geometry: Topic 8 – Inequalities and linear programming
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:
y
x=2
24
22
20
18
16
14
4x + y = 24
12
y = 10
10
8
6
4
2
0
1
2
3
4
5
6
7
8
9
10
x
(b) For line 2x + y = 20, when x = 0, y = 20 and when y = 0, x = 10.
Adding this line to the graph we have:
y
x=2
24
22
20
18
16
14
2x + y = 20
12
10
y = 10
8
6
4
2
0
64
4x + y = 24
1
2
3
4
5
6
7
8
9
10
x
Worked solutions
The smallest value of 2x + y will lie in the feasible region and be
parallel to the line 2x + y = 20. The point will lie on or near to one
of the vertices of the unshaded region. The point (3, 0) gives the
smallest value of 2x + y which is 6, and this occurs when x = 3 and
y = 0.
3
(a) The total number of tents cannot be more than 18, so x + y ≤ 18.
(c)For x + y = 18, when x = 0, y = 18 and when y = 0, x = 18.
(b) There are 72 children, so 8x + 3y ≥ 72.
For 8x + 3y = 72, when x = 0, y = 24 and when y = 0, x = 9.
x = 0 and y = 0 are the y- and x-axes respectively.
y
24
22
20
18
8x + 3y = 72
16
14
12
10
x + y = 18
8
6
4
2
0
2
4
6
8
10
12
14
16
18
x
(d) Total cost of hiring tents, is 80x + 20y.
65
2 Coordinate geometry: Topic 8 – Inequalities and linear programming
Note we have chosen 400
as the cost here. We have
chosen this number to make
the maths easier as 80 and
20 both divide into 400
exactly. Always make sure
that the number you choose
can be shown on the graph
you have drawn.
(e) Plotting the line 80x + 20y = 400, when x = 0, y = 20 and when
y = 0, x = 5.
y
24
22
20
8x + 3y = 72
16
P Q
18
14
12
10
x + y = 18
8
6
80x + 20y = 400
4
2
0
66
2
4
6
8
10
12
14
16
18
x
The minimum cost is given by the line that is parallel to the objective
function drawn in the feasible region that is furthest to the left. The
point P does not have integer coordinates so we need to look for a
point near P with integer coordinates.
The point Q is near to P and has coordinates (4, 14). If these
coordinates are substituted into the expression for the cost
(i.e. 80x + 20y), the cost is 80(4) + 20(14) = 600.
It is worth just checking other points near to P in the feasible region.
e.g. point (5, 13) gives a cost of 80(5) + 20(13) = 660
point (5, 12) gives a cost of 80(5) + 20(12) = 640
Note that we require a minimum cost and both these points give a
higher cost.
Hence the point Q(4, 14) gives the values of x and y that give the
smallest cost. Hence 4 large tents and 14 small tents need to be used.
3 Trigonometry: Topic 9 – Trigonometric ratios
and the graphs of sine, cosine and tangent
Topic
1
9
Worked solutions
Progress checks
x is the hypotenuse and the
15 cm side is the adjacent so
adjacent
we use cos θ = hypotenuse
.
15
x
15
x =
= 17.32 cm (2 d.p.)
cos 30°
x
(b) cos 40° =
12
Hence, x = 12 cos 40°
1
(a) cos 30° =
2(a)tan θ =
x is the adjacent and the 12 cm
side is the hypotenuse so we
adjacent
use cos θ = hypotenuse
.
= 9.19 cm (2 d.p.)
10
3
(103)
θ = tan–1
= 73.3° (nearest 0.1°)
(b) sin θ =
10
13
(10
13)
θ = sin –1
= 50.3° (nearest 0.1°)
3sin θ = c (i.e. sin θ = hypotenuse) and cos θ = c (i.e. cos θ = hypotenuse)
b
b
Remember that when you
divide by fractions you
turn the bottom fraction
upside down and replace the
division by a multiplication.
sin θ | c |
Hence
=
cos θ ac
opposite
a
adjacent
b c
= × c a
b
a
b opposite
Now, =
= tan θ
a adjacent
sin θ
= tan θ
Hence,
cos θ
=
67
3 Trigonometry: Topic 9 – Trigonometric ratios and the graphs of sine, cosine and tangent
4
90°
S
180°
T
5
θ2
Cos θ is positive in the first and fourth quadrants. The
solution in the first quadrant is found by entering cos–1(12)
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°.
A
θpV
0°
Hence solutions are 60° and 300°.
C
270°
(√23) into the calculator.
The first angle is found by entering sin–1
This gives θ = 60°.
y
√3
2
6
1
90˚
–1
180˚
270˚
360˚
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°.
�
The solution using trigonometric graphs is used here.
The alternative CAST method could also have been used.
–26.6°
0
x = tan–1(–12) = –26.6°
y = tan x
y
Drawing the graph of y = tan x you can see where this angle
is on the graph.
x
90°
180°
270°
360°
By the symmetry of the graph you can see that the line
y = –12 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°.
As a check find the tan of both of
these angles. You should obtain
–12 in each case.
68
Worked solutions
72sinθ = 1
sinθ =
y
1
Note you could also have
used the CAST method here.
0.5
0
30°
θ = sin–1
θ = 180 – 30° = 150°
θ = 30°
150° 180°
360° �
(12)
The above is not in the
required range.
1
2
Hence θ = 30° or 150°
8(a)θ = sin–1 (–0.23) = –13.3°
y
1
–13.3°
Sketching a sine graph to look for symmetry we have.
180° + 13.3°
0
–0.23°
180°
y = sin �
270°
360° – 13.3°
360°
�
–1
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)
69
3 Trigonometry: Topic 9 – Trigonometric ratios and the graphs of sine, cosine and tangent
(b)cos θ = – 0.72
θ = cos–1(–0.72)
= 136.1°
Drawing a sketch graph to find other solutions.
y
1
0
90°
–0.72
136.1°
180°
360 – 136.1°
270°
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.
y = tan x
y
2.45
0
70
67.8°
90°
x
180°
180 + 67.8°
270°
360°
Other solution, θ = 180 + 67.8 = 247.8°
Solutions are θ = 68° or 248° (nearest degree)
360° �
Worked solutions
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
7 cm
�
5 cm
10 cm
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)
10
= 27.7° (nearest 0.1°)
(a) Using the cosine rule we obtain
132 = 112 + 92 – 2 × 11 × 9 cos θ
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.
169 = 121 + 81 – 198 cos θ
Solving gives θ= 80.4° (nearest 0.1°)
1
(b)
Area = × 9 × 11 sin 80.4°
2
11(a)
= 48.81 cm2
= 48.8 cm2 (1 d.p.)
1
3
Area = × 25 × 30 ×
2
5
= 225 cm2
71
3 Trigonometry: Topic 9 – Trigonometric ratios and the graphs of sine, cosine and tangent
(b)
A right-angled triangle is
drawn with the opposite of
length 3 and the hypotenuse
of length 5.
5
Ѳ
Adjacent = 4
cos θ =
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.
3
4
adjacent
4
=
hypotenuse 5
(c) Using the cosine rule we obtain
AC2 = 252 + 302 – 2 × 25 × 30 ×
AC2 = 325
AC = √ 325
4
5
AC = 18.0 cm (3 s.f.)
Test yourself
1(a)
y
y = tan θ
0
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
90°
180°
270°
360°
θ
(b) tan θ = 1, so θ = tan–1(1) = 45° or 180 + 45 = 225°
Worked solutions
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.
y
1
0
180°
–1
To find x, these two angles
are divided by 2.
360°
540°
720°
x
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.
y
y = tan x
2
0
90°
x
180°
270°
360°
The other solution is x = 180 + 63.4 = 243.4°.
Hence, x = 63.4° or 243.4°
73
3 Trigonometry: Topic 9 – Trigonometric ratios and the graphs of sine, cosine and tangent
3
1
1
(a) Area = bc sin A = × 12 × 8 × sin 150° = 24 cm2
2
2
(b) Using the cosine rule
a2 = b2 + c2 – 2bc cos A
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.
= 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)
5
(a) Using the sine rule we obtain
(b) (90°, 1), (270°, –1), (450°, 1), (630°, –1)
x
12
=
sin 84° sin 40°
x =
12 sin 84°
sin 40°
x = 18.5664
= 18.6 cm (3 s.f.)
(b) Angle BAC = 180 – (84 + 40) = 56°
1
1
Area = bc sin A = × 18.6 × 12 × sin 56°
2
2
74
= 92.521
= 92.5 cm2 (3 s.f.)
3 Trigonometry: Topic 10 – Trigonometric
identities and solving trigonometric equations
Topic
10
1
Worked solutions
Progress check
1cos x = –2 sin x
Dividing both sides by cos x, we obtain
sin x
= tan x so 1 = –2 tan x
cos x
1
giving tan x = – 2
1
x = tan–1 – = –26.6°
2
We can see the other solutions in the required range by sketching a graph
of y = tan x.
Now
Note that this is outside
the range specified in the
question.
cos x –2 sin x
=
cos x
cos x
( )
y
–26.6°
–90°
1
–
2
y = tan x
90°
180°
270°
360°
x
You can see from the graph that there are two angles in the range where
1
tan x = – .
2
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°.
75
3 Trigonometry: Topic 10 – Trigonometric identities and solving trigonometric equations
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)
√5
Ѳ
x
2
By Pythagoras we have
( )
2
√ 5 = x2 + 22
5 = x2 + 4
Hence x = 1.
sin θ =
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
4
1
opposite
=
hypotenuse √ 5
1
sin θ |√5| 1 √ 5 1
(b)tan θ =
=
=
=
×
cos θ 2 √ 5 2 2
√5
6cos2 θ + 5cos θ – 6 = 0
Factorising, we obtain (3cos θ – 2)(2cos θ + 3) = 0
2
3
Hence, cos θ = or cos θ = – (which is ignored as smallest value is –1)
3
2
2
2
giving θ = 48.2° or 311.8° (nearest 0.1°).
cos θ = so θ = cos–1
3
3
()
Worked solutions
5(a)5sin θ = 3cos θ
sin θ
Note that cos
θ = tan θ.
Dividing both sides by cos θ we obtain
sin θ
=3
5 cos θ
5tan θ = 3
tan θ =
3
5
(35 )
(b)
θ = tan–1
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
1
2
or sin x = –
4
3
1
When sin x = , x = 14.5° or 165.5°
4
2
When sin x = – , x = 221.8° or 318.2°
3
sin x =
Hence x = 14.5°, 165.5°, 221.8° or 318.2°
7
Distance travelled by John in 3 h = 6 × 3 = 18 km.
Using the cosine rule
Distance travelled by Amy in 3 h = 15 km.
Drawing a diagram to show the distances and
angles we have:
N
x
18 km
A
50°
15 km
a2 = b2 + c2 – 2bc cos A
x2 = 182 + 152 – 2 × 18 × 15 cos 50°
x = 14.2 km (1 d.p.)
x = 14.2090 km
77
3 Trigonometry: Topic 10 – Trigonometric identities and solving trigonometric equations
8
(a) By Pythagoras’ theorem we obtain
PR2 = 92 + 92
= 162
= 81 × 2
Taking the square root of both sides gives
TAKE NOTE
It is always a good idea
to simplify things by
drawing the triangle
you are working on
separately.
PR = √ 81 × 2
= 9√ 2
(b)
V
8 cm
P
9√ 2 cm
2
O
By Pythagoras’ theorem we obtain
Solving, gives VP = 10.2 cm (3 s.f.)
VP2 = 82 + (4.5√ 2)2
(c) Let angle OPV = θ
9
8
4.5√ 2
Hence, θ =51.5° (nearest 0.1°)
(a) By Pythagoras’ theorem
Tan θ =
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
78
= 11.5 cm (3 s.f.)
Worked solutions
(c)
F
11.5326 cm
A
Ѳ
C
10.8167 cm
Let angle CAF = θ
Cos θ =
10.8167
11.5326
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.
4
Let angle CBF = α so tan α = giving α = 33.7° (nearest 0.1°)
6
10(a)
N
N
40°
10 km
10°
8 km
H
(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.
79
3 Trigonometry: Topic 10 – Trigonometric identities and solving trigonometric equations
We can draw some additional angles on the original diagram like this.
N
N
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°.
40°
140°
8 km
10 km
10°
40°
H
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°
(c)
Solving gives x = 17.4 km.
Ѳ
N
Remember bearings are
always three figures so 13.3°
is rounded to the nearest
whole number which is 13°.
80
N
40°
H
140°
8 km
40°
10 km
10°
o find the bearing, the angle marked T
θ in the diagram needs to be found.
Using the sine rule, we have
17.4
8
=
sin 150° sin θ
Solving gives θ = 13.3°
Now, bearing = 180 + 10 + 13 = 203°
Worked solutions
Test yourself
1
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
This is a quadratic in cos θ.
Rearranging and simplifying we obtain
Factorising gives
3cos2 θ – 5cos θ + 2 = 0
(3cos θ – 2)(cos θ – 1) = 0
Hence 3cos θ – 2= 0 or cos θ – 1 = 0
2
So cos θ = or cos θ = 1
3
2
or θ = cos–1(1)
θ = cos–1
3
()
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.
θ = 48.2°, 311.8° or θ = 0°, 360°
2
Hence θ = 0°, 48.2°, 311.8° or 360°
sin 2x = √ 3 cos 2x
Dividing both sides by sin 2x we obtain
tan 2x = √ 3 (√ 3)
2x = tan–1
Using the CAST method to find the angles, we have
S
180°
T
90°
Ѳ2
A
0°
ѲpV
270°
C
81
3 Trigonometry: Topic 10 – Trigonometric identities and solving trigonometric equations
Note that an angle of 360°
takes you back to the start
again and you now add the
angles you have found to
360°
(√ 3) = 60°
Using the diagram on page 81 θpV = tan–1
θ 2 = 180 + 60 = 240°
θ4 = 360+ 240 = 600°
θ 3 = 360 + 60 = 420°
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.
N
B
N
A
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.
30°
Lighthouse
L
30°
Using speed = distance
time , 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)
82
Distance from port A to the lighthouse is 50.7 nautical miles (1 d.p.)
Worked solutions
(c)
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.
Note that the bearing has
to be a whole number of
degrees.
We use the identity
sin θ
cos θ = tan θ.
It 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.
30°
50.6853
L
Lighthouse
Bearing
Using the sine rule to find angle θ we have
48 50.6853
=
sin θ sin 120
Rearranging and solving gives θ = 55.1°
Bearing = 270 – 55 = 215°
4sin θ = –3cos θ
Dividing both sides by cos θ we obtain
Here we will use the CAST method but the trigonometric graphs could also
have been used to find the angles.
>>> TIP
48
30°
A
5
Ѳ
B
N
N
N
Hence, tan θ = –3
Tan θ is negative in the second and fourth quadrants.
Now tan–1(–3) = –71.57°
= –71.6° (nearest 0.1°)
S
180°
T
90°
71.6°
ѲpV
Ѳ
A
0°
71.6°
270°
C
–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°.
83
3 Trigonometry: Topic 10 – Trigonometric identities and solving trigonometric equations
5(a)
V
20 cm
R
16 cm
Q
By Pythagoras’ theorem we have
202 = VR2 + 162
400 = VR2 + 256
VR2 = 144
V
VR = 12 cm
(b)By Pythagoras’ theorem we have
202 = VS2 + 122
400 = VS2 + 144
20 cm
Q
12 cm
20 cm
VS2 = 256
VS = 16 cm
P
S
By Pythagoras’ theorem we have
162 = RS2 + 122
RS = 10.6 cm (3 s.f.)
84
16 cm
16 cm
256 = RS2 + 144
RS2 = 112
R
Q
12 cm
S
P
Worked solutions
(c)V
16 cm
R
10.6 cm
Cos θ =
10.6
16
Ѳ
S
(10.6
16 )
θ = cos–1
= 48.5° (3 s.f.)
85
4 Calculus: Topic 11 – Differentiation
Topic
11
6
Worked solutions
Progress check
dy
= 12x2 + 12x – 3
dx
dy
= 30x4 + 32x3 – 9x2
(b) y = 6x5 + 8x4 – 3x3 + 1 dx
dy
(c) y = 7x4 + 8x3 – 9x2 + 1 = 28x3 + 24x2 – 18x
dx
dy
= 30x2 – 14x – 9
(d) y = 10x3 – 7x2 – 9x + 11 dx
dy
= 40x – 7
(e) y = 20x2 – 7x – 21 dx
dy
= 10x – 7
(f) y = 5x2 – 7x – 5 dx
dy
2(a)y = (x + 2)(x + 1) = x2 + 3x + 2 dx = 2x + 3
dy
= 2x + 6
(b) y = (x + 4)(x + 2) = x2 + 6x + 8 dx
dy
= 2x – 1
(c) y = (x – 3)(x + 2) = x2 – x – 6 dx
dy
= 2x – 6
(d) y = (x – 4)(x – 2) = x2 – 6x + 8 dx
dy
= 2x – 8
(e) y = (x – 4)2 = x2 – 8x + 16 dx
dy
= 2x
(f) y = (x + 2)(x – 2) = x2 – 4 dx
dy
= 2x + 10
(g) y = (x + 5)2 = x2 + 10x + 25 dx
dy
= 3x2 + 4x + 1
(h) y = x(x2 + 2x + 1) = x3 + 2x2 + x dx
dy
= 9x2 + 12x + 9
(i) y = x(3x2 + 6x + 9) = 3x3 + 6x2 + 9x dx
dy
= 4x3 + 18x2 + 18x
(j) y = x2(x2 + 6x + 9) = x4 + 6x3 + 9x2 dx
dy
= 12x – 11
(k) y = (3x + 2)(2x – 5) = 6x2 – 11x – 10 dx
1(a)y = 4x3 + 6x2 – 3x + 1 86
Worked solutions
(l) y = (5x – 1)(4x – 7) = 20x2 – 39x + 7 dy
= 40x – 39
dx
dy
= 8x + 12
dx
dy
= 18x – 12
(n) y = (3x – 2)2 = 9x2 – 12x + 4 dx
(m) y = (2x + 3)2 = 4x2 + 12x + 9 dy
= 3x2 + 8x + 4
dx
dy
= 3x2 + 8x – 6
y = (x – 1)(x2 + 5x – 1) = x3 + 5x2 – x – x2 – 5x + 1 = x3 + 4x2 – 6x + 1 dx
dy
= 3x2 – 14x + 14
y = (x – 5)(x2 – 2x + 4) = x3 – 2x2 + 4x – 5x2 + 10x – 20 = x3 – 7x2 + 14x – 20 dx
dy
= 6x2 – 6x + 6
y = (2x + 1)(x2 – 2x + 4) = 2x3 – 4x2 + 8x + x2 – 2x + 4 = 2x3 – 3x2 + 6x + 4 dx
dy
= 12x2 + 34x – 27
y = (4x – 3)(x2 + 5x – 3) = 4x3 + 20x2 – 12x – 3x2 – 15x + 9 = 4x3 + 17x2 – 27x + 9 dx
dy
= 15x2 + 48x – 20
y = (5x – 1)(x2 + 5x – 3) = 5x3 + 25x2 – 15x – x2 – 5x + 3 = 5x3 + 24x2 – 20x + 3 dx
3(a)y = (x + 1)(x2 + 3x + 1) = x3 +3x2 + x + x2 + 3x + 1 = x3 + 4x2 + 4x + 1 (b)
(c)
(d)
(e)
(f)
(g) y = (x + 2)(x + 3)(x + 5) = (x + 2)(x2 + 8x + 15) = x3 + 8x2 + 15x + 2x2 + 16x + 30
= x3 + 10x2 + 31x + 30
dy
= 3x2 +20x + 31
dx
(h) y = (x + 6)(x + 5)(x + 7) = (x + 6)(x2 +12x + 35) = x3 + 12x2 + 35x + 6x2 + 72x + 210
= x3 + 18x2 + 107x + 210
dy
= 3x2 + 36x + 107
dx
(i) y = (x – 3)(x – 2)(x + 1) = (x – 3)(x2 – x – 2) = x3 – 4x2 + x + 6 (j) y = (x + 1)2(x + 2) = (x + 1)(x + 1)(x + 2) = (x + 1)(x2 + 3x + 2)
dy
= 3x2 – 8x + 1
dx
= x3 + 4x2 + 5x + 2
dy
= 3x2 + 8x + 5
dx
87
4 Calculus: Topic 11 – Differentiation
(k) y = (2x – 1)2(x – 3) = (2x – 1)(2x – 1)(x – 3) = (2x – 1)(2x2 – 7x + 3)
= 4x3 – 16x2 + 13x – 3
dy
= 12x2 – 32x + 13
dx
(l) y = (3x + 4)2(x – 1) = (3x + 4)(3x + 4)(x – 1) = (3x + 4)(3x2 + x – 4)
4
= 9x3 + 15x2 – 8x – 16
dy
= 27x2 + 30x – 8
dx
y = 3x2 – 2x + 5
dy
= 6x – 2
dx
When x = 2
dy
= 6(2) – 2
dx
= 10
Equation of a straight line having gradient m and passing through the
point (x1, y1) is given by:
In this case m = 10 and (x1, y1) = (2, 13), so
y – 13 = 10x – 20
y – y1 = m(x – x1)
88
y – 13 = 10(x – 2)
Equation of the tangent to the curve at (2, 13) is y = 10x – 7
Worked solutions
5
y = x3 + 4x2 – 8x + 2
dy
= 3x2 + 8x – 8
dx
When x = 2
dy
= 3(2)2 + 8(2) – 8
dx
x = 2 is substituted into
the equation of the curve
to find the corresponding
y-coordinate.
= 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)
1
Giving gradient of the normal, m2 = – 20
Equation of a straight line having gradient m and passing through the
point (x1, y1) is given by:
y – y1 = m(x – x1)
1
In this case m = – and (x1, y1) = (2, 10), so
20
1
y – 10= – (x – 2)
20
20y – 200 = –x + 2
Hence, equation of the normal at P is x + 20y = 202
89
4 Calculus: Topic 11 – Differentiation
6
y = 2x3 – 3x2 – 4x + 2
dy
= 6x2 – 6x – 4
dx
When x = 2
dy
= 6(2)2 – 6(2) – 4
dx
=8
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 = 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
dy
= 2x – 3
dx
dy
= 2(1) – 3
dx
When x = 1
= –1
When x = 2
90
dy
= 2(2) – 3
dx
=1
Worked solutions
(b) When x = 1, y = 12 – 3(1) + 2 = 0.
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.
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
1
y = – 2
Substituting y = – 12 into equation (1) we obtain
You can perform a check
by substituting the values
for x and y into each side of
equation (2).
1
– = –x + 1
2
3
x =
2
Hence the coordinates of the point where the two tangents intersect
are
(32, – 12).
91
4 Calculus: Topic 11 – Differentiation
Test yourself
1(a)y = 4x2 – 30x – 3
dy
= 8x – 30
dx
When x = 4,
dy
= 8(4) – 30 = 2
dx
(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
1
So m = – 2
Equation for line of gradient
m passing through (x1, y1) is
y – y1 = m (x – x1)
When x = 4, y = 4(4)2 – 30(4) – 3 = 64 – 120 – 3 = –59
1
Equation of the normal having gradient – and passing through
2
(4, –59) is
1
y + 59 = – (x – 4)
2
2y + 118 = – x + 4
2y + x + 114 = 0
2(a)y = x3 + 3x2 – 9x – 8
Divide both sides by 3 to
make the factorising of the
quadratic equation easier.
dy
= 3x2 + 6x – 9
dx
Now at the stationary points
3x2 + 6x – 9 = 0
x2 + 2x – 3 = 0
(x + 3)(x – 1) = 0
The first derivative is
differentiated again to find
the second derivative.
92
dy
= 0, so we have
dx
(b)
x = –3 or 1
Hence there is a stationary point at x = 1.
d2y
= 6x + 6
dx2
Notice that there are two
stationary points but this
question is only concerned
with one of them.
d2y
When x = 1, 2 = 6(1) + 6 = 12 > 0 showing there is a minimum point
dx
at x = –1.
Worked solutions
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.
3(a)y = x3 – 6x2 + 9x + 1
dy
= 3x2 – 12x + 9
dx
dy
(b)When x = 3, = 3(3)2 – 12(3) + 9 = 0
dx
dy
(c) Now at the stationary points = 0, so we have
dx
2
3x – 12x + 9 = 0
x2 – 4x + 3 = 0
(x – 3)(x – 1) = 0
x = 3 or 1
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.
When x = 3, y = x3 – 6x2 + 9x + 1 = 33 – 6(3)2 + 9(3) + 1 = 1
TAKE NOTE
When x = 1, y = x3 – 6x2 + 9x + 1 = 13 – 6(1)2 + 9(1) + 1 = 5
Make sure you find
both the x- and
y-coordinates of any
stationary points if
you are asked.
Hence the two stationary points are (3, 1) and (1, 5).
(d) Finding the second order derivative
d2y
= 6x – 12
dx2
d2y
When x = 3, 2 = 18 – 12 = 6.
dx
The positive value shows that (3, 1) is a minimum point.
d2y
When x = 1, 2 = 6 – 12 = –6.
dx
The negative value shows that (1, 5) is a maximum point.
4
2
1
y = x3 + x2 – 6x
3
2
dy
= 2x2 + x – 6 = (2x – 3)(x + 2)
dx
dy
At the stationary points = 0
dx
(2x – 3)(x + 2) = 0
3
Solving gives x = or –2
2
3
Substituting x = into the equation of the curve to find the y-coordinate
2
gives
() () ()
2 33 1 32
3
9 9
+ –6
= + – 9 = –558
y= 3 2
2 2
2
4 8
93
4 Calculus: Topic 11 – Differentiation
Putting x = –2 into the equation of the curve to find the y-coordinate gives
2
1
16
y = (– 2)3 + (– 2)2 – 6(– 2) = – + 2 + 12 = 8 23
3
2
3
Finding the second order derivative
d2y
= 4x + 1
dx2
3 d2y
When x = , 2 = 7.
2 dx
The positive value shows that (32 , –558 ) is a minimum point.
The negative value shows that (–2, 823 )is a maximum point.
When x = –2,
5
y = x3 – 6x2 + 12x + 1
dy
= 3x2 – 12x + 12 = 3(x2 – 4x + 4) = 3(x – 2) (x – 2) = 3(x – 2)2
dx
dy
At the stationary points = 0
dx
d2y
= –7.
dx2
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
4 Calculus: Topic 12 – Integration
Topic
12
1
Worked solutions
Progress check
1(a)
(b)
(c)
(d)
(e)
(f)
4x4
+ c = x4 + c
4
5x2
5x dx =
+ c
2
4x3 dx =
3 dx = 3x + c
–6x3
+ c = –2x3 + c
3
16x4
16x3 dx =
+ c = 4x4 + c
4
–6x2 dx =
1 dx = x + c
5x3 3x2
+
– 5x + c
3
2
7x4 4x3 8x2
7x4 4x3
3
2
–
+
–x+c=
–
+ 4x2 – x + c
(b) (7x – 4x + 8x – 1) dx =
4
3
2
4
3
x3 5x2
– 3x + c
(c) (12 x2 + 5x – 3) dx = +
6 2
x3 x2
2
2
(x – x – 6) dx = – – 6x + c
(d)(x – 3)(x + 2) = x – x – 6
3 2
x3 6x2
x3
(x2 – 6x + 9) dx = –
+ 9x + c = – 3x2 + 9x + c
(e)(x – 3)2 = x2 – 6x + 9
3 2
3
4
3
2
x 6x 5x
x4
5x2
2
3
2
3
2
3
(x + 6x – 5x) dx = –
–
+ c = – 2x –
+c
(f) x(x + 6x – 5) = x + 6x – 5x
4
3
2
4
2
2(a) (5x2 + 3x – 5) dx =
3(a)
4x3 dx = x4 + c
6x2 dx = 2x3 + c
(b)
(2x + 1) dx = x2 + x + c
(c)
95
4 Calculus: Topic 12 – Integration
(d)
(e)
(f)
(8x2 + 4x – 1) dx =
8x3
+ 2x2 – x + c
3
(5x4 + 4x3 – 6x) dx = x5 + x4 – 3x2 + c
(x2 + 4x – 5) dx =
x3
+ 2x2 – 5x + c
3
x3
(g) (x + 5)(x + 1) dx = (x + 6x + 5) dx = + 3x2 + 5x + c
3
x3
(h) (x – 1)(x + 1) dx = (x2 – 1) dx = – x + c
3
2x3 3x2
2
–
– 5x + c
(i) (2x – 5)(x + 1) dx = (2x – 3x – 5) dx =
3
2
x3
(j) (x + 5)2 dx = (x2 + 10x + 25) dx = + 5x2 + 25x + c
3
3
2
x x
(k) x(x + 1) dx = (x2 + x) dx = + + c
3 2
x4 x3
(l) x2(2x + 1) dx = (2x3 + x2) dx = + + c
2 3
x2
5
4
2
6
5
3
(m) (6x + 20x – 6x – x + 9) dx = x + 4x – 2x – + 9x + c
2
x3 x2
x4
x3
x4 x3
dx
=
–
–
+
c
=
– +c
4 4 2
4×4 2×3
16 6
5x5 6x3
4
2
5(a) (5x – 6x + 9) dx = 5 – 3 + 9x + c = x5 – 2x3 + 9x + c
8x4 6x3 10x2
–
+
+ 5x + c = 2x4 – 2x3 + 5x2 + 5x + c
(b) (8x3 – 6x2 + 10x + 5) dx =
4
3
2
8x2 x3
x3
2
2
2
+ + c = 16x – 4x + + c
(c) (4 – x) dx = ∫(16 – 8x + x ) dx = 16x –
2 3
3
x4
x5
x5
dx =
+c=
+c
(d)
3
3×5
15
x2 x
x3
x2
x3 x2
+ dx =
+
+c= + +c
(e)
2 3
2×3 3×2
6 6
3x3
2
(a)
3x
dx
=
+ c = x3 + c
6
3
16x4
3
+ c = 4x4 + c
(b) 16x dx =
4
1 2
x3
x dx = + c
(c)
3
9
(
)
(
(d)
96
)
5 dx = 5x + c
2
Worked solutions
7
dy
4x2
= 4x + 5 so y = (4x + 5) dx =
+ 5x + c = 2x2 + 5x + c
dx
2
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
8
Hence, the equation of the curve is y = 2x2 + 5x – 9.
y = (6x2 + 10x + 2) dx
6x3 10x2
=
+
+ 2x + c
3
2
= 2x3 + 5x2 + 2x + c
Now when x = 1, y = 1.
1=2+5+2+c
1 = 2(1)3 + 5(1)2 + 2(1) + c
c = –8
Hence, y = 2x3 + 5x2 + 2x – 8
9
(a) dy
= 2x – 2
dx
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
Hence, coordinates of the points of intersection of the x-axis are
(–2, 0) and (4, 0).
Solving gives x = 4 or –2.
97
4 Calculus: Topic 12 – Integration
(c)
dy
= 2x – 2
dx
dy
= 0, so 2x – 2 = 0. Solving gives x = 1.
dx
When x = 1, y = 12 – 2(1) – 8 = –9
At the minimum point,
Hence the minimum point is (1, –9).
Test yourself
1
y = (4 – 2x – 3x2) dx
= 4x –
2x2 3x3
–
+c
2
3
= 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)
4x2 3x4
–
+c
2
4
3x4
+c
= 2x2 –
4
(b)2x(3x2 – 5x + 1)
=
Always cancel fractions if
possible.
(4x – 3x3) dx
Multiplying out the brackets we obtain
6x3 – 10x2 + 2x
(6x3 – 10x2 + 2x) dx
6x4 10x3 2x2
–
+
+c
4
3
2
3x4 10x3 2
–
+x +c
=
2
3
=
98
Worked solutions
(
)
1
1
(c) x2 – x + dx
2
2
3
2
x
x x
– + +c
=
2×3 2 2
x3 x2 x
= – + + c
6 2 2
3
(25t4 – 12t3 + 15t2 – 9t + 2) dt
25t5 12t4 15t3 9t2
–
+
–
+ 2t + c
=
5
4
3
2
9t2
5
4
3
+ 2t + c
= 5t – 3t + 5t –
2
2
4 15 (x2 – 30)
5
6
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.
)
= 3x – x2 + c
Substituting (0, 0) for x and y into the
equation we obtain
0=0–0+c
Hence c = 0.
= 2t3 – t2 + t + c
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
6t3 2t2
(6t – 2t + 1)dt =
+t+c
–
3
2
dy
y = ∫(3 – 2x) dx
2 2
x – 4
15
2 2
x – 4 dx
15
2x3
– 4x + c
=
45
2
Equation of the curve is y = 5x3 + 4x2 + x – 7
7(a)dx= 3 – 2x
Multiplying out the brackets we obtain
(
dy
= 15x2 + 8x +1
dx
3x – x2 = 0
x(3 – x) = 0
Solving gives x = 3.
Hence the point is (3, 0).
99
4 Calculus: Topic 13 – Definite integration
Topic
13
6
Worked solutions
Progress check
1
1
0
(6x2 – 2x + 5) dx=
[
[
= 2x3 – x2 + 5x
]
1
2
0
=[(2 × 13 – 12 + 5(1)) – (0 – 0 + 0)]
= [(2 – 1 + 5) – (0)]
=6
[
[
= 2x3 – 2x2 – 5x
]
]
3
2
3
2
= [(2(3) – 2(3) – 5(3)) – (2(2)3 – 2(2)2 – 5(2))]
= [21 – (–2)]
3
[ ]
[ ]
1 2x2 3x3
=
–
3 2
3
0.5
0.5
0
1 2 3
x –x
3
0
1
= [((0.5)2 – (0.5)3) – (0)]
3
=
= 0.0417 (3 s.f.)
2
= [(54 – 18 – 15) – (16 – 8 – 10)]
= 23
0.5
0.5
1
3 0 y dx= 3 0 (2x – 3x2) dx
100
1
0
6x3 4x2
(6x
–
4x
–
5)
dx=
–
– 5x
2 2
3
2
3
]
6x3 2x2
–
+ 5x
3
2
Worked solutions
2
[
[
[(
x3 4x2
(x
+
4x
–
3)
dx
=
+
– 3x
4 0
3 2
2
x3
=
+ 2x2 – 3x
3
=
>>>
TIP
It is important to note
that you can only
combine the integrals if
the limits are the same.
5
3
0
=
(x + 6x + 4) dx –
0
3
0
0
(x2 – 4x) dx
(x2 + 6x + 4 – x2 + 4x) dx
We now collect the terms to produce the following:
3
0
[
[
(10x + 4) dx
= 57
0
(x + 5)(x + 6) dx =
=
7 Shaded area=
1
0
[
[(
2
0
(x2 + 11x + 30) dx
]
3
–3
= 35.8 (3 s.f.)
(9 – x2) dx
[ ]
[( ) (
= 9(3) –
= 36
1
0
) ]
1 11
+
+ 30 – (0)
3 2
x3
= 9x –
3
0
x3 11x2
+
+ 30x
=
3
2
]
3
3
= [(5(3) + 4(3)) – (0)]
1
]
10x2
=
+ 4x
2
= 5x2 + 4x
6
2
) ]
0
8
+ 8 – 6 – (0)
3
= 423
2
3
]
]
2
3
–3
)]
33
(–3)3
– 9(–3) –
3
3
101
4 Calculus: Topic 13 – Definite integration
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.
Sketching the curve we obtain:
When x = 0, y = –4
y
–2
3
y = x2 – 4
0
–4
2
[ ]
[( ) (
( )
x3
(b)(x – 4) dx =
– 4x
3
2
2
=
=
2
0
x
3
2
= (9 – 12) –
(x2 – 4) dx=
)]
33
23
– 4(3) –
– 4(2)
3
3
7
3
8
–8
3
[ ]
[( ) ]
x3
– 4x
3
2
0
23
– 4(2) – (0)
3
16
= – 3
(c) The positive value represents the area above the x-axis and the
negative value represents the area below the x-axis.
102
=
Worked solutions
9 (a)
Form a quadratic equation,
then factorise and finally
solve it.
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
Looking at the diagram this
is point A.
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
2
[(
[(
[(
)]
x3
(9 – x ) dx= 9x –
3
–3
2
= 9(2) –
= 18 –
2
–3
)(
)]
(2)3
(–3)3
– 9(–3) –
3
3
)
]
8
– (–27 + 9)
3
1
= 15 + 18
3
1
= 33
3
Area of right-angled triangle with side AB as the hypotenuse
1
= ×5×5
2
= 12.5
= 20.8 (3 s.f.)
1
Required area = 33 – 12.5
3
103
4 Calculus: Topic 13 – Definite integration
Test yourself
4
1
1
[
(6x – 2) dx= 2x – 2x
2
3
]
4
1
= [(2(4)3 – 2(4)) – (2(1)3 – 2(1))]
= 128 – 8 – 2 + 2
= 120
2 Area=
=
3
–3
3
–3
[(
[(
y dx
(9 – x2) dx
)]
x3
= 9x –
3
= 9(3) –
3
–3
)(
)]
(3)3
(–3)3
– 9(–3) –
3
3
= [(27 – 9) – (–27 + 9)]
= 36
2
[
[
[(
]
]
x3 4x2
(a)
(x
–
4x
+
2)
dx=
–
+ 2x
3
3 2
0
2
x3
– 2x2 + 2x
=
3
=
4
2
0
) ]
(b) The negative sign means that the area is below the x-axis.
2
–1
(x – 3)(x + 4) dx=
2
–1
[
[(
[(
(x2 + x – 12) dx
]
x3 x2
=
+ – 12x
3 2
=
= – 31.5
=
2
–1
)(
)]
2 2
(–1)3 (–1)2
+ – 12(2) –
+
– 12(–1)
3 2
3
2
3
104
0
23
– 2(2)2 + 2(2) – (0)
3
8
= –8+4
3
= –113
2
2
)(
)]
8
1 1
+ 2 – 24 – – + + 12
3
3 2
Worked solutions
5
1
Area =
0
(x2 + 1) dx
[ ]
[( ) ]
x3
=
+x
3
1
0
1
=
+ 1 – (0)
3
4
= or 113
3
2
6(a) 0 x(x
2
– 6x + 3) dx =
2
(x – 3)
dx
(b)
3
0
(c)
2
–2
2
0
[
[(
(x3 – 6x2 + 3x) dx
]
x4
3x2
=
– 2x3 +
4
2
=
0
= –6
1
= (x – 3) dx
3 0
[
1 x2
= – 3x
3 2
]
2
0
1
= [(2 – 6) – (0)]
3
4
= – 3
[
(12x – 4x + 1) dx= 4x – 2x + x
2
) ]
2
3(2)2
– (0)
– 2(2)3 +
4
2
4
= 4 – 16 + 6
2
2
3
2
]
2
–2
= [(4(2)3 – 2(2)2 + 2) – (4(–2)3 – 2(–2)2 – 2)]
= 68
= 32 – 8 + 2 + 32 + 8 + 2
105
4 Calculus: Topic 14 – Application of calculus to
kinematics
Topic
14
6
Worked solutions
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
1
(b) Using s = (u + v)t
2
1
s = (0 + 9)10 = 45 m
2
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
The displacement, s, is zero
when the stone returns to its
point of projection.
t = 0 is ignored as a possible
time.
1
s = ut + at2
2
1
0 = 20t + × (–9.8) × t2
2
(b) Using 0 = 20t – 4.9t2
0 = t(20 – 4.9t)
t = 0 or 4.1 s
3
Hence time = 4.1 s
(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
106
1
(b) Usings = ut + at2
2
1
s = 0.8 × 3.5 + × 9.8 × 3.52 = 62.8 m
2
Worked solutions
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
t = 2 is substituted into the
expression for a.
v =
ds
dt
(b)
a=
dv
dt
= 36t2
= 72t
When t = 2, a = 72 × 2 = 144 ms–2
6(a)v = 0.64t3 – 0.36t2
a=
dv
= 1.92t2 – 0.72t
dt
(b)
s= v dt
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.
= (0.64t3 – 0.36t2) dt
0.64t4 0.36t3
=
–
+c
4
3
When t = 0, s = 0
Substituting these two values into the expression for s we obtain
0 =
0.64(0)4 0.36(0)3
–
+c
4
3
Solving gives c = 0
Hence s =
0.64t4 0.36t3
–
= 0.16t4 – 0.12t3
4
3
When t = 10, s = 0.16(10)4 – 0.12(10)3 = 1480 m
107
4 Calculus: Topic 14 – Application of calculus to kinematics
7(a)v =
a dt
= (3 – 0.1t) dt
= 3t –
0.1t2
+c
2
When t = 0, v = 0 so 0 = 3(0) –
Solving gives c = 0.
The expression for the velocity is v = 3t –
(b) When t = 10, v = 3(10) –
v = 30 – 5
0.1(0)2
+c
2
= 25 ms–1
0.1(10)2
2
0.1t2
.
2
(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
dt
2
3t2 0.1t3
–
+c
2
6
When t = 0, s = 0 so substituting these values into the above
expression gives
3(0)2 0.1(0)3
–
+c
0
=
2
6
Solving gives c = 0
3t2 0.1t3
–
s=
2
6
When t = 30, s =
108
3(30)2 0.1(30)3
–
2
6
= 1350 – 450
= 900 m
Worked solutions
8
v = 6t + 4
s= v dt
= (6t + 4) dt
=
6t2
+ 4t + c
2
= 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 =
dv
dt
= 12t – 2
(ii)
When t = 1, a = 12(1) – 2 = 10 ms–2
(b) s = v dt
= (6t2 – 2t + 8)dt
=
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.
6t3 2t2
–
+ 8t + c
3
2
= 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
109
4 Calculus: Topic 14 – Application of calculus to kinematics
Test yourself
1(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
1
(b) Using s = ut + at2
2
1
s = 5 × 6 + × 10 × 62
2
= 30 + 180
= 210 m
2(a)
v (ms–1)
0
5
25
(b) u = 0 ms–1, v = ?, a = 0.9 ms–2, t = 5 s
33
t (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
Note that if you say that this
is a deceleration, then you
need to remove the minus
sign.
0 – 4.5
33 – 25
= –0.56 ms–2
Hence deceleration = 0.56 ms–2
(d) Distance travelled= area under the velocity–time graph
110
=
1
= (20 + 33) × 4.5
2
= 119.25 m
Worked solutions
3(a)
v = 64 –
1 3
t
27
0 = 64 –
1 3
t
27
When the car comes to rest, v = 0 so we have
Hence
1 3
t = 64
27
t3 = 64 × 27
Taking the cube root of both sides gives t = 12 s
s = v dt
= 64t –
(b)
=
(64 – 271 t )dt
3
1 4
t + c
108
When t = 0, s = 0 so c = 0.
When t = 12, s = 64 × 12 −
4(a)a = 2t + 3
1
× 124 = 768 – 192 = 576 m
108
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
111
4 Calculus: Topic 14 – Application of calculus to kinematics
(b) s= v dt
= (t2 + 3t + 10) dt
=
t3 3t2
+
+ 10t + c
3 2
When t = 0, s = 0. Substituting these values into the above equation to
find c, we have
c=0
Hence s =
t3 3t2
+
+ 10t
3 2
When t = 3, s =
33 3(3)2
+
+ 10(3)
3
2
= 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
Using v = u + at, we obtain
Solving we obtain v = 72 ms–1
(c) u = 20 ms–1, a = 2.6 ms–2, t = 20 s, v = ?
v = 20 + 2.6 × 20
(d) s = ?, u = 20 ms–1, a = 2.6 ms–2, t = 30 s
1
Using s = ut + at2, we obtain
2
1
s = 20 × 30 + × 2.6 × 302
2
s = 1770 m
112
Worked solutions
(e)
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.
Velocity
(ms–1)
32
20
0
B
A
4.62
Time (s)
Motion under constant acceleration, means the velocity–time graph
will be a straight line.
113