1.6 Function, SPM Practice (Long Question)


Question 3:
Given that f : xhx + 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 + k
f2 (x) = ff (x) = f (hx + k)
= h (hx + k) + k
= h2 x + hk + k

f2 (x) = 4x + 15
h2 x + hk + k = 4x + 15
h2 = 4
h = ± 2
when, h = 2
hk + k = 15
2k + k = 15
k = 5

When, h = –2
hk + k = 15
–2k + k = 15
k = –15

(b)
h > 0, h = 2, k = 5
Given f (x) = hx + k
f (x) = 2x + 5 f (x2 ) = 7x
2 (x2 ) + 5 = 7x
2x2 – 7x + 5 = 0
(2x – 5)(x –1) = 0
2x – 5 = 0    or    x –1= 0
x = 5/2                     x = 1


Question 4:
The function f is denoted by f:x 1+x 1x ,x1.  Find  f 2 , f 3 , f 4  and hence write down the functions  f 51  and  f 52 .

Solution:
f( x )= 1+x 1x ,x1 f 2 ( x )=f[ f( x ) ]=f( 1+x 1x )          = 1+( 1+x 1x ) 1( 1+x 1x ) = 1x+1+x 1x 1x1x 1x          = 2 2x = 1 x f 3 ( x )=f[ f 2 ( x ) ]=f( 1 x )          = 1+( 1 x ) 1( 1 x ) = x1 x x+1 x          = x1 x+1 f 4 ( x )=f[ f 3 ( x ) ]=f( x1 x+1 )           = 1+( x1 x+1 ) 1( x1 x+1 ) = x+1+x1 x+1 x+1x+1 x+1           = 2x 2 =x f 5 ( x )=f[ f 4 ( x ) ]=f( x )= 1+x 1x ( recurring ) f 51 ( x )= f 3 [ f 48 ( x ) ]= f 3 ( x )              = x1 x+1 f 52 ( x )= f 4 [ f 48 ( x ) ]= f 4 ( x )=x



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