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.