The function f is defined by

\[f(x) = {({x^3} + 6{x^2} + 3x - 10)^{\frac{1}{2}}},{\text{ for }}x \in D,\]

where \(D \subseteq \mathbb{R}\) is the greatest possible domain of f.

(a)     Find the roots of \(f(x) = 0\).

(b)     Hence specify the set D.

(c)     Find the coordinates of the local maximum on the graph \(y = f(x)\).

(d)     Solve the equation \(f(x) = 3\).

(e)     Sketch the graph of \(\left| y \right| = f(x),{\text{ for }}x \in D\).

(f)     Find the area of the region completely enclosed by the graph of \(\left| y \right| = f(x)\)

(a)     solving to obtain one root: 1, – 2 or – 5     A1

obtain other roots     A1

[2 marks]

(b)     \(D = x \in [ - 5,{\text{ }} - 2] \cup [1,{\text{ }}\infty {\text{)}}\) (or equivalent)     M1A1

Note: M1 is for 1 finite and 1 infinite interval.

[2 marks]

(c)     coordinates of local maximum \( - 3.73 - 2 - \sqrt 3 ,{\text{ }}3.22\sqrt {6\sqrt 3 } \)     A1A1

[2 marks]

(d)     use GDC to obtain one root: 1.41, – 3.18 or – 4.23     A1

obtain other roots     A1

[2 marks]

(e)

    A1A1A1

Note: Award A1 for shape, A1 for max and for min clearly in correct places, A1 for all intercepts.

Award A1A0A0 if only the complete top half is shown.

[3 marks]

(f)     required area is twice that of \(y = f(x)\) between – 5 and – 2     M1A1

answer 14.9     A1     N3

Note: Award M1A0A0 for \(\int_{ - 5}^{ - 2} {f(x){\text{d}}x = 7.47 \ldots } \) or N1 for 7.47.

[3 marks]

Total [14 marks]

This was a multi-part question that was well answered by many candidates. The main difficulty was sketching the graph and this meant that the last part was not well answered.