Solve \(f(x) = g(x)\).

Find the area of the region enclosed by the graphs of \(f\) and \(g\).

valid approach     (M1)

eg sketch

0, 1.73843

\(x = 0,{\text{ }}x = 1.74{\text{ }}\left( {{\text{accept }}(0,{\text{ }}0){\text{ and }}(1.74,{\text{ }}3.02)} \right)\)     A1A1     N3

[3 marks]

integrating and subtracting functions (in any order)     (M1)

eg\(\,\,\,\,\,\)\(\int {g - f} \)

correct substitution of their limits or function (accept missing \({\text{d}}x\))

(A1)

eg\(\,\,\,\,\,\)\(\int_0^{1.74} {g - f,{\text{ }}\int {3\ln (x + 1) - {x^2}} } \)

Note:     Do not award A1 if there is an error in the substitution.

1.30940

1.31     A1     N3

[3 marks]

Candidates often did not make the connection between parts (a) and (b). The extraordinary number of failed analytical approaches in part (a) and correct use of the GDC to find the limits in part (b) suggests that candidates are equating the command term “solve” to mean use an algebraic approach to solve equations or inequalities, instead of their GDC. Many candidates appeared to interpret part (a) as something they should do by hand and often did not recognize that their answer to part (a) were the limits in part (b). Quite a few candidates failed to interpret a GDC solution of \(x = 5 \times {10^{ - 14}}\) correctly as \(x = 0\) and others found the solution \(x = 1.74\) as the only solution, ignoring the second intersection point until part (b).

Candidates often did not make the connection between parts (a) and (b). The extraordinary number of failed analytical approaches in part (a) and correct use of the GDC to find the limits in part (b) suggests that candidates are equating the command term “solve” to mean use an algebraic approach to solve equations or inequalities, instead of their GDC. Many candidates appeared to interpret part (a) as something they should do by hand and often did not recognize that their answer to part (a) were the limits in part (b). Quite a few candidates failed to interpret a GDC solution of \(x = 5 \times {10^{ - 14}}\) correctly as \(x = 0\) and others found the solution \(x = 1.74\) as the only solution, ignoring the second intersection point until part (b).