1.5 Function, SPM Practice (Short Question)


Question 4:
It is given the functions g(x) = 3x and h(x) = mnx, where m and n are constants.
Express m in terms of n such that hg(1) = 4.

Solution:
hg( x )=h( 3x )          =mn( 3x )          =m3nx hg( 1 )=4 m3n( 1 )=4 m3n=4 m=4+3n


Question 5:
Given the function g : x → 3x – 2, find  
(a) the value of x when g(x) maps onto itself,
(b) the value of k such that g(2 – k) = 4k.

Solution:
(a)
  g( x )=x 3x2=x 3xx=2      2x=2         x=1

(b)
      g( x )=3x2 g( 2k )=4k 3( 2k )2=4k 63k2=4k        7k=4              k= 4 7


Question 6:
Given the functions f : xpx + 1, g : x → 3x – 5 and fg(x) = 3px + q.  
Express p in terms of q.

Solution:
f( x )=px+1, g( x )=3x5 fg( x )=p( 3x5 )+1          =3px5p+1 Given fg( x )=3px+q 3px5p+1=3px+q       5p+1=q            5p=q1               5p=1q                 p= 1q 5


Question 7:
Given the functions h : x → 3x + 1, and gh : x → 9x2 + 6x – 4, find  
(a) h-1 (x),
(b) g(x).

Solution:
(a)
Let  h 1 ( x )=y, thus  h( y )=x        3y+1=x             3y=x1               y= x1 3    h 1 ( x )= x1 3 h 1 :x x1 3

(b)
g[ h( x ) ]=9 x 2 +6x4 g( 3x+1 )=9 x 2 +6x4 Let y=3x+1 thus     x= y1 3      g( y )=9 ( y1 3 ) 2 +6( y1 3 )4             = 9 ( y1 ) 2 9 +2( y1 )4             = y 2 2y+1+2y24             = y 2 5  g( x )= x 2 5

Leave a Reply

Your email address will not be published. Required fields are marked *