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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−3nxhg(1)=4m−3n(1)=4m−3n=4m=4+3nhg(x)=h(3x) =m−n(3x) =m−3nxhg(1)=4m−3n(1)=4m−3n=4m=4+3n

Question 6:
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)=x3x−2=x3x−x=2 2x=2x=1  g(x)=x3x−2=x3x−x=2 2x=2x=1

(b)
  g(x)=3x−2g(2−k)=4k3(2−k)−2=4k6−3k−2=4k   −7k=−4 k=47  g(x)=3x−2g(2−k)=4k3(2−k)−2=4k6−3k−2=4k   −7k=−4 k=47

Question 7:
Given the functions f : x → px + 1, g : x → 3x – 5 and fg(x) = 3px + q.  
Express p in terms of q.

Solution:
f(x)=px+1, g(x)=3x−5fg(x)=p(3x−5)+1 =3px−5p+1Given fg(x)=3px+q3px−5p+1=3px+q  −5p+1=q   −5p=q−1  5p=1−qp=1−q5f(x)=px+1, g(x)=3x−5fg(x)=p(3x−5)+1 =3px−5p+1Given fg(x)=3px+q3px−5p+1=3px+q  −5p+1=q   −5p=q−1  5p=1−qp=1−q5

Question 8:
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=x3y=x−1  y=x−13∴  h−1(x)=x−13h−1:x↦x−13Let h−1(x)=y,thus  h(y)=x   3y+1=x3y=x−1  y=x−13∴  h−1(x)=x−13h−1:x↦x−13


(b)
g[h(x)]=9x2+6x−4g(3x+1)=9x2+6x−4Let y=3x+1thus  x=y−13 g(y)=9(y−13)2+6(y−13)−4=9(y−1)29+2(y−1)−4=y2−2y+1+2y−2−4=y2−5∴ g(x)=x2−5

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