Date | November 2014 | Marks available | 2 | Reference code | 14N.2.sl.TZ0.1 |
Level | SL only | Paper | 2 | Time zone | TZ0 |
Command term | Find | Question number | 1 | Adapted from | N/A |
Question
Let \(f(x) = 2x + 3\) and \(g(x) = {x^3}\).
Find \((f \circ g)(x)\).
Solve the equation \((f \circ g)(x) = 0\).
Markscheme
attempt to form composite (in any order) (M1)
eg\(\;\;\;f({x^3}),{\text{ }}{(2x + 3)^3}\)
\((f \circ g)(x) = 2{x^3} + 3\) A1 N2
[2 marks]
evidence of appropriate approach (M1)
eg\(\;\;\;2{x^3} = - 3\), sketch
correct working (A1)
eg\(\;\;\;{x^3} = \frac{{ - 3}}{2}\), sketch
\( - 1.14471\)
\(x = \sqrt[3]{{\frac{{ - 3}}{2}}}\;\;\;{\text{(exact), }} - 1.14{\text{ }}[ - 1.15,{\text{ }} - 1.14]\) A1 N3
[3 marks]
Total [5 marks]
Examiners report
Generally well done, though there were some careless errors with the substitution into \(f\) in part (ai) and rearranging the equation in part (b). Although candidates understood that they were supposed to solve the equation \(2{x^3} + 3 = 0\), many wrote \(2{x^3} = 3\) or \(x = \sqrt {\frac{3}{2}} \). The majority of the candidates chose an algebraic method instead of using their GDC.
Generally well done, though there were some careless errors with the substitution into \(f\) in part (ai) and rearranging the equation in part (b). Although candidates understood that they were supposed to solve the equation \(2{x^3} + 3 = 0\), many wrote \(2{x^3} = 3\) or \(x = \sqrt {\frac{3}{2}} \). The majority of the candidates chose an algebraic method instead of using their GDC.