| Date | May 2011 | Marks available | 2 | Reference code | 11M.1.sl.TZ1.1 |
| Level | SL only | Paper | 1 | Time zone | TZ1 |
| Command term | Find | Question number | 1 | Adapted from | N/A |
Question
Let f(x)=7−2xf(x)=7−2x and g(x)=x+3g(x)=x+3 .
Find (g∘f)(x)(g∘f)(x) .
Write down g−1(x)g−1(x) .
Find (f∘g−1)(5)(f∘g−1)(5) .
Markscheme
attempt to form composite (M1)
e.g. g(7−2x)g(7−2x) , 7−2x+37−2x+3
(g∘f)(x)=10−2x(g∘f)(x)=10−2x A1 N2
[2 marks]
g−1(x)=x−3g−1(x)=x−3 A1 N1
[1 mark]
METHOD 1
valid approach (M1)
e.g. g−1(5)g−1(5) , 22 , f(5)f(5)
f(2)=3f(2)=3 A1 N2
METHOD 2
attempt to form composite of f and g−1g−1 (M1)
e.g. (f∘g−1)(x)=7−2(x−3)(f∘g−1)(x)=7−2(x−3) , 13−2x13−2x
(f∘g−1)(5)=3(f∘g−1)(5)=3 A1 N2
[2 marks]
Examiners report
A majority of candidates found success in the opening question. Common errors in (a) were to give f∘gf∘g or to multiply f by g.
For (b) some gave the inverse as the reciprocal function 1x+31x+3 , or wrote x=y+3x=y+3 .
Most candidates chose to find a composite in (c), sometimes making simple errors when working with brackets and a negative sign. Only a handful used the more efficient f(2)=3f(2)=3 . Additionally, it was not uncommon for candidates to give a correct substitution but not complete the result. Simple expressions such as (7−2x)+3(7−2x)+3 should be finished as 10−2x10−2x .
