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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 \((f \circ g)(x)\).

[2]
a.

Solve the equation \((f \circ g)(x) = 0\).

[3]
b.

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]

a.

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]

b.

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.

a.

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.

b.

Syllabus sections

Topic 2 - Functions and equations » 2.1 » Composite functions.
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