Sketch the graph of \(f\) on the following grid.

Solve \(f(x) = 0\).

The region enclosed by the graph of \(f\) and the \(x\)-axis is rotated \(360°\) about the \(x\)-axis.

Find the volume of the solid formed.

     A1A1A1     N3

Note:     Award A1 for both endpoints in circles,

A1 for approximately correct shape (concave up to concave down).

Only if this A1 for shape is awarded, award A1 for maximum point in circle.

\(x = 1\;\;\;x = 1.83928\)

\(x = 1{\text{ (exact)}}\;\;\;x = 1.84{\text{ }}[1.83,{\text{ }}1.84]\)     A1A1     N2

[2 marks]

attempt to substitute either (FT ) limits or function into formula with \({f^2}\)     (M1)

eg\(\;\;\;\)\(V = \pi \int_1^{1.84} {{f^2},{\text{ }}\int {{{( - {x^4} + 2{x^3} - 1)}^2}{\text{d}}x} } \)

\(0.636581\)

\(V = 0.637{\text{ }}[0.636,{\text{ }}0.637]\)     A2     N3

[3 marks]

Total [8 marks]

Despite being a straightforward question, and although most candidates had a roughly correct shape for their graph, their sketches were either out of scale or missed one of the endpoints. In part (b), a few did not give both answers despite going on to use 1.84 in part (c).

Part (c) proved difficult for most candidates, as only a small number could write the correct expression for the volume: some included the correct limits but did not square the function, whilst others squared the function but did not write the correct limits in the integral. Many did not find a volume, or found an incorrect volume. The latter included finding the integral from 0 to 2, or dividing the region into three parts, showing a lack of understanding of “enclosed”.

Despite being a straightforward question, and although most candidates had a roughly correct shape for their graph, their sketches were either out of scale or missed one of the endpoints. In part (b), a few did not give both answers despite going on to use 1.84 in part (c).

Part (c) proved difficult for most candidates, as only a small number could write the correct expression for the volume: some included the correct limits but did not square the function, whilst others squared the function but did not write the correct limits in the integral. Many did not find a volume, or found an incorrect volume. The latter included finding the integral from 0 to 2, or dividing the region into three parts, showing a lack of understanding of “enclosed”.

Despite being a straightforward question, and although most candidates had a roughly correct shape for their graph, their sketches were either out of scale or missed one of the endpoints. In part (b), a few did not give both answers despite going on to use 1.84 in part (c).

Part (c) proved difficult for most candidates, as only a small number could write the correct expression for the volume: some included the correct limits but did not square the function, whilst others squared the function but did not write the correct limits in the integral. Many did not find a volume, or found an incorrect volume. The latter included finding the integral from 0 to 2, or dividing the region into three parts, showing a lack of understanding of “enclosed”.