(i)     Sketch the graphs of \(f\) and \(g\) on the same axes.

(ii)     Find the area of \(R\) .

Find the area of \(R\) .

Consider all values of \(m\) such that the graphs of \(f\) and \(g\) intersect. Find the value of \(m\) that gives the greatest value for the area of \(R\) .

 (i)

   A1A1     N2

Notes: Award A1 for the graph of \(f\) positive, increasing and concave up.

    Award A1 for graph of \(g\) increasing and linear with \(y\)-intercept of \(0\).

    Penalize one mark if domain is not [\( - 5\), \(5\)] and/or if \(f\) and \(g\) do not intersect in the first quadrant.

[2 marks]

(ii)
attempt to find intersection of the graphs of \(f\) and \(g\)     (M1)

eg   \({{\rm{e}}^{\frac{x}{4}}} = x\)

\(x = 1.42961 \ldots \)     A1

valid attempt to find area of \(R\)     (M1)

eg   \(\int {(x - {{\rm{e}}^{\frac{x}{4}}}} ){\rm{d}}x\) ,  \(\int_0^1 {(g - f)} \) , \(\int {(f - g)} \)

area \( = 0.697\)     A2     N3


[5 marks]

attempt to find intersection of the graphs of \(f\) and \(g\)     (M1)

eg   \({{\rm{e}}^{\frac{x}{4}}} = x\)

\(x = 1.42961 \ldots \)     A1

valid attempt to find area of \(R\)     (M1)

eg   \(\int {(x - {{\rm{e}}^{\frac{x}{4}}}} ){\rm{d}}x\) ,  \(\int_0^1 {(g - f)} \) , \(\int {(f - g)} \)

area \( = 0.697\)     A2     N3

[5 marks]

recognize that area of \(R\) is a maximum at point of tangency     (R1)

eg   \(m = f'(x)\)

equating functions     (M1)

eg   \(f(x) = g(x)\) , \({{\rm{e}}^{\frac{x}{4}}} = mx\)

\(f'(x) = \frac{1}{4}{{\rm{e}}^{\frac{x}{4}}}\)     (A1)

equating gradients     (A1)

eg   \(f'(x) = g'(x)\) , \(\frac{1}{4}{{\rm{e}}^{\frac{x}{4}}} = m\)

attempt to solve system of two equations for \(x\)     (M1)

eg   \(\frac{1}{4}{{\rm{e}}^{\frac{x}{4}}} \times x = {{\rm{e}}^{\frac{x}{4}}}\)

\(x = 4\)     (A1)

attempt to find \(m\)     (M1)

eg   \(f'(4)\) , \(\frac{1}{4}{{\rm{e}}^{\frac{4}{4}}}\)

\(m = \frac{1}{4}e\) (exact), \(0.680\)     A1     N3

[8 marks]

There was a flaw with the domain noted in this question. While not an error in itself, it meant that part (b) no longer assessed what was intended. The markscheme included a variety of solutions based on candidate work seen, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected, and these were looked at during the grade award meeting.

While some candidates sketched accurate graphs on the given domain, the majority did not. Besides the common domain error, some exponential curves were graphed with several concavity changes.

There was a flaw with the domain noted in this question. While not an error in itself, it meant that part (b) no longer assessed what was intended. The markscheme included a variety of solutions based on candidate work seen, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected, and these were looked at during the grade award meeting.

In part (a)(ii), most candidates found the intersection correctly. Those who used their GDC to evaluate the integral numerically were usually successful, unlike those who attempted to solve with antiderivatives. A common error was to find the area of the region enclosed by \(f\) and \(g\) (although it involved a point of intersection outside of the given domain), rather than the area of the region enclosed by \(f\) and \(g\) and the \(y\)-axis.

There was a flaw with the domain noted in this question. While not an error in itself, it meant that part (b) no longer assessed what was intended. The markscheme included a variety of solutions based on candidate work seen, and examiners were instructed to notify the IB assessment centre of any candidates adversely affected, and these were looked at during the grade award meeting.

While some candidates were able to show some good reasoning in part (b), fewer were able to find the value of \(m\) which maximized the area of the region. In addition to the answer obtained from the restricted domain, full marks were awarded for the answer obtained by using the point of tangency.