| Date | May 2009 | Marks available | 3 | Reference code | 09M.1.sl.TZ2.1 |
| Level | SL only | Paper | 1 | Time zone | TZ2 |
| Command term | Find | Question number | 1 | Adapted from | N/A |
Question
Let f(x)=x2f(x)=x2 and g(x)=2x−3g(x)=2x−3 .
Find g−1(x)g−1(x) .
Find (f∘g)(4)(f∘g)(4) .
Markscheme
for interchanging x and y (may be done later) (M1)
e.g. x=2y−3x=2y−3
g−1(x)=x+32g−1(x)=x+32 (accept y=x+32,x+32y=x+32,x+32 ) A1 N2
[2 marks]
METHOD 1
g(4)=5g(4)=5 (A1)
evidence of composition of functions (M1)
f(5)=25f(5)=25 A1 N3
METHOD 2
f∘g(x)=(2x−3)2f∘g(x)=(2x−3)2 (M1)
f∘g(4)=(2×4−3)2f∘g(4)=(2×4−3)2 (A1)
= 25 A1 N3
[3 marks]
Examiners report
Many candidates performed successfully in finding the inverse function, as well as the composite at a specified value of x.
Many candidates performed successfully in finding the inverse function, as well as the composite at a specified value of x. Some candidates made arithmetical errors especially if they expanded the binomial before substituting x=4x=4 .
