A function \(f\) is defined in the interval \(\left] { - k,{\text{ }}k} \right[\), where \(k > 0\). The gradient function \({f'}\) exists at each point of the domain of \(f\).

The following diagram shows the graph of \(y = f(x)\), its asymptotes and its vertical symmetry axis.

(a)     Sketch the graph of \(y = f'(x)\).

Let \(p(x) = a + bx + c{x^2} + d{x^3} +  \ldots \) be the Maclaurin expansion of \(f(x)\).

(b)     (i)     Justify that \(a > 0\).

(ii)     Write down a condition for the largest set of possible values for each of the parameters \(b\), \(c\) and \(d\).

(c)     State, with a reason, an upper bound for the radius of convergence.

(a)

A1 for shape, A1 for passing through origin     A1A1

Note: Asymptotes not required.

[2 marks]

(b)     \(p(x) = \underbrace {f(0)}_a + \underbrace {f'(0)}_bx + \underbrace {\frac{{f''(0)}}{{2!}}}_c{x^2} + \underbrace {\frac{{{f^{(3)}}(0)}}{{3!}}}_d{x^3} +  \ldots \)

(i)     because the y-intercept of \(f\) is positive     R1

(ii)     \(b = 0\)     A1

\(c \geqslant 0\)     A1A1

Note: A1 for \( > \) and A1 for \( = \).

\(d = 0\)     A1

[5 marks]

(c)     as the graph has vertical asymptotes \(x =  \pm k,{\text{ }}k > 0\),     R1

the radius of convergence has an upper bound of \(k\)     A1

Note: Accept \(r < k\).

[2 marks]

Overall candidates made good attempts to parts (a) and most candidates realized that the graph contained the origin; however many candidates had difficulty rendering the correct shape of the graph of \(f'\). Part b(i) was also well answered although some candidates where not very clear and digressed a lot. Part (ii) was less successful with most candidates scoring just part of the marks. A small number of candidates answered part (c) correctly with a valid reason.