Question 6 (8 marks):
It is given that g : x → 2x – 3 and h : x → 1 – 3x.
(a) Find
(i) h (5)
(ii) the value of k if g(k+2)=17h(5),
(iii) hg(x).
(b) Hence, sketch the graph of y = | hg(x) | for –1 ≤ x ≤ 3.
State the range of y.
Solution:
(a)(i)
h(x)=1−3xh(5)=1−3(5) =−14
(a)(ii)
g(x)=2x−3g(k+2)=17h(5)2(k+2)−3=17(−14)2k+4−3=−22k=−3k=−32
(a)(iii)
g(x)=2x−3, h(x)=1−3xhg(x)=h(2x−3) =1−3(2x−3) =1−6x+9 =10−6x
(b)
y = |hg(x)|,
y = |10 – 6x|
Range of y : 0 ≤ y ≤ 16
It is given that g : x → 2x – 3 and h : x → 1 – 3x.
(a) Find
(i) h (5)
(ii) the value of k if g(k+2)=17h(5),
(iii) hg(x).
(b) Hence, sketch the graph of y = | hg(x) | for –1 ≤ x ≤ 3.
State the range of y.
Solution:
(a)(i)
h(x)=1−3xh(5)=1−3(5) =−14
(a)(ii)
g(x)=2x−3g(k+2)=17h(5)2(k+2)−3=17(−14)2k+4−3=−22k=−3k=−32
(a)(iii)
g(x)=2x−3, h(x)=1−3xhg(x)=h(2x−3) =1−3(2x−3) =1−6x+9 =10−6x
(b)
y = |hg(x)|,
y = |10 – 6x|
Range of y : 0 ≤ y ≤ 16