IB Maths AA SL 2.10 Solving equations: practice questions

(a(i)) Mark scheme

• A1 \(f(0) = -2\);

A model answer: \(f(0) = 6\ln 1 - 0 - 2 = -2\)

(a(ii)) Mark scheme

• A1 \(f(30) = -11.4\) (\(6\ln 31 - 32 = -11.3960...\));

A model answer: \(f(30) = 6\ln 31 - 32 = -11.4\) (\(-11.3960...\))

(b) Mark scheme

• (M1) attempt to find at least one zero (graph, table or equation solver);
• A1 \(x = 0.523\) and \(x = 14.4\) (0.522628..., 14.4101...);

A model answer: Graphing \(f\) on the GDC, the zeros are \(x = 0.523\) (0.522628...) and \(x = 14.4\) (14.4101...).

(c) Mark scheme

• A1 endpoints at about \((0, -2)\) and \((30, -11.4)\);
• A1 \(x\)-intercepts at about \(x = 0.5\) and \(x = 14.4\);
• A1 a single maximum at about \((5, 3.75)\), with the curve concave down;

A model answer: A smooth curve that starts at \((0, -2)\), rises to cross the \(x\)-axis at \(x \approx 0.5\), reaches a maximum at about \((5, 3.75)\), then falls, crossing the \(x\)-axis again at \(x \approx 14.4\), and ends at \((30, -11.4)\).