 ## 1.7.3 Function, SPM Practice (Long Question)

Question 5: In diagram below, the function g maps set P to set Q and the function h maps set Q to set R. Find (a) in terms of x, the function (i) which maps set Q to set P, (ii) h(x). (b) the value of x such that gh(x) = 8x + 1. Solution: … Read more

## 1.7.2 Function, SPM Practice (Long Question)

Question 3: Given that f : x → hx + k and f2 : x → 4x + 15.   (a) Find the value of h and of k. (b) Take the value of h > 0, find the values of x for which f (x2 ) = 7x Solution: (a) Given f (x) = hx … Read more

## 1.7.1 Function, SPM Practice (Long Question)

Question 1: The function f and g is defined by f:x↦2x−3 g:x↦ 2 x ;x≠0 Find the expression for each of the following functions (a) ff, (b) gf, (c) f-1 , Calculate the value of x such that ff(x) = gf(x). Solution: (a) ff( x )=f[ f( x ) ]          =f( 2x−3 )          =2( 2x−3 … Read more

## 1.6.2 Function, SPM Practice (Short Question)

Question 5: It is given the functions g(x) = 3x and h(x) = m – nx, where m and n are constants. Express m in terms of n such that hg(1) = 4. Solution: hg( x )=h( 3x )  =m−n( 3x )  =m−3nx hg( 1 )=4 m−3n( 1 )=4 m−3n=4 m=4+3n Question 6: Given the … Read more

## 1.6.1 Function, SPM Practice (Short Question)

Question 1:Diagram below shows the relation between set M and set N in the graph form.State(a) the range of the relation,(b) the type of the relation between set M and set N.Solution: (a) Range of the relation = {p, r, s}.(b) Type of the relation between set M and set N is many to one relation. … Read more

## 1.5.7 Inverse Function Example 7

Example 7 (Comparison Method)Given that f : x → 2 h x − 3 k , x≠3k , where h and k are constants and f − 1 : x → 14 + 24 x x , x≠0, find the value of h and of k. Solution:

## 1.5.6 Inverse Function Example 6 (Comparison Method)

Example 6 (Comparison Method)If f : x ↦ m x − n x − 2 , x ≠ 2   and f − 1 : x ↦ 5 − 2 x 2 − x , x ≠ 2. . Find the value of m and of n,

## 1.5.5 Inverse Function Example 5

Example 5 If g : x ↦ m − x x − 3 , x ≠ 3   and g − 1 ( 5 ) = 14  .  Find the value of m. Correction:m – 14 = 55m = 55 + 14m = 69

## 1.5.4 Inverse Function Example 4

Example 4:(a) If f : x → x – 2, find f -1 (5), (b) if f:x→ x+9 x−5 , x≠5, find  f −1 (3). Solution:   (a) f (x) = x– 2 Let y = f -1 (5) f (y) = 5 y – 2 = 5 y = 7 therefore, f -1 (5) = 7 (b) f(x)= … Read more

## 1.5.3 Inverse Function Example 3

Example 3Find the inverse function of f ( x ) = 3 x + 2 5 x + 3