Date | May 2008 | Marks available | 6 | Reference code | 08M.1.hl.TZ2.4 |
Level | HL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 4 | Adapted from | N/A |
Question
Let f(x)=4x+2, x≠−2 and g(x)=x−1.
If h=g∘f , find
(a) h(x) ;
(b) h−1(x) , where h−1 is the inverse of h.
Markscheme
(a) h(x)=g(4x+2) (M1)
=4x+2−1(=2−x2+x) A1
(b) METHOD 1
x=4y+2−1(interchanging x and y) M1
Attempting to solve for y M1
(y+2)(x+1)=4(y+2=4x+1) (A1)
h−1(x)=4x+1−2(x≠−1) A1 N1
METHOD 2
x=2−y2+y(interchanging x and y) M1
Attempting to solve for y M1
xy+y=2−2x(y(x+1)=2(1−x)) (A1)
h−1(x)=2(1−x)x+1(x≠−1) A1 N1
Note: In either METHOD 1 or METHOD 2 rearranging first and interchanging afterwards is equally acceptable.
[6 marks]
Examiners report
This question was generally well done, with very few candidates calculating f∘g rather than g∘f.