**Frequently Asked Questions in SPM for Functions**

We will discuss some pass year SPM exam questions based on the four concepts :

**Relations – SPM 2010**

**Functions – SPM 2007**

**Composite Functions – SPM 2007**

**Composite & Inverse Functions – SPM 2010**

**Relations**** – SPM 2010 Paper1 Question 1**

Diagram 1 shows the relation between set X and set Y in the graph form.

State

(a) The relation in the form of ordered pairs.

(b) The type of the relation

(c) The range of the relation

(3 marks)

Solution:

a) Analysis

- The question required you to state the relation in the form of ordered pairs,
**eg: {(a,b),(c,d)}**, - Hence we can state it as
**{(1,p),(2,r),(3,s),(4,p)}**. - Remember that the element of set X come first, follow by the element from set Y, eg: (element from set X, element from set Y).

b) Analysis

- We need to know the objects and images
- Objects – 1,2,3,4 (from set X)
- Images – p,r,s (from set Y)
- 1 & 4 could be mapped to p, so we can state the relation as
**many to one relation****.**

c) Analysis

- The range is all the images being mapped, in this case the
**range = {p,r,s}** - Take note that q is not included as it does not belong to any object

**Functions –**** SPM 2007 Paper1 Question 2**

Given that f : x –> | x – 3 |, find the values of x such that f(x) =5. [*2marks*]

Solution:

Analysis

Given | x – 3 | and f(x) = 5, meant that you have both positive (5) and negative (-5) image value, so,

x – 3 = 5 or x – 3 = -5

x = 8 or x = -2

Thus, **x = 8 or x = -2**

**Composite Functions**** – SPM 2007 Paper1 Question 3**

The following information is about the function h and the composite function h^{2}.

h : x –> ax + b, where a and b are constants, and a > 0h^{2} : x –> 36x – 35 |

Find the value of a and b [*3marks*]

Solution:

Given h(x) = ax + b

h^{2}(x) = hh(x)

= h( ax + b )

= a*( ax + b )* + b **[ insert ax + b into the function h(x) again ]**

= a^{2}x + ab + b

If we compare both composite functions

h

^{2}(x) = 36x – 35h

^{2}(x) = a^{2 }x + ab + b

you will see that a^{2} = 36,

**a = 6 [ Since a > 0, so we do not accept -6 here ]**

ab + b = -35

6b + b = -35

7b = -35

** b = -5**

**So a = 6, b = -5**

**Composite Functions**** & Inverse Functions – SPM 2010 Paper1 Question 2**

Given the functions g:x –> 2*x* + 1 and h:x* –> *3*x* + 6, find

(a) g^{-1}(x)

(b) hg^{-1}(9)

(3 marks)

Solution:

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