Cambridge International AS and A Level Mathematics Pure Mathematics 1

Chapter

(^4)
P1^
f: x  1016 x (approx.)
f−^1 : x  10 −^16 x (approx.)
(iv) (a) Function but no inverse
function since fares are
banded.
Activity 4.1 (Page 117)
(i)
f(x) = x^2 ; f−^1 (x) = x
(ii)
f(x) = 2 x; f−^1 (x) = 12 x
(iii)
f(x) = x + 2; f−^1 (x) = x − 2
(iv)
f(x) = x^3 + 2; f−^1 (x) = 3 x– 2
y = f(x) and y = f−^1 (x) appear to
be reflections of each other in
y = x.
Exercise 4B (Page 120)
  1 (i) 8 x^3
(ii) 2 x^3
(iii) (x + 2)^3
(iv) x^3 + 2
(v) 8(x + 2)^3
(vi) 2(x^3 + 2)
(vii) 4 x
(viii) [(x + 2)^3 + 2]^3
(ix) x + 4
  2 (i) f−^1 (x) = x–7 2
(ii) f−^1 (x) = 4 − x
(iii) f−^1 (x) = 24 xx–
(iv) f−^1 (x) = x+3, x  − 3
  3 (i), (ii)
  4 (i) fg
(ii) g^2
(iii) fg^2
(iv) gf
  5 (i) x
(ii) (^1) x
(iii) (^1) x
(iv) (^1) x
  6 (i) a = 3
(ii)
(iii) f(x)  3
(iv) Function f is not one-to-one
when domain is .
Inverse exists for function
with domain x  −2.
  7 f−^1 : x  3 x– 43 , x ∈ .
The graphs are reflections of
each other in the line y = x.
  8 (i) a = 2, b = − 5
(ii) Translation –

• 
  

2

5









(iii) y  − 5

(iv) c = − 2

(v)

x

y y = f(x)

y = f–1(x)

O

x

y

O

y = f(x)

y = f–1(x)

x

y

2

2

y = f(x)

y = f–1(x)

O

x

y

y = f(x)

y = f–1(x)

O

2

2

y

O x

y = f(x)

y = f–1(x)

y = x

(3, 2)

(2, 3)

y

x = –2 O

(–2, 3)

7

x

x

y

(–2, –5)

O

y = g(x)

x

y

O

y = g(x)

y = x

y = g–1(x)

(–2, –5)

(–5, –2)