A family of cubic functions is defined as ${f_k}(x) = {k^2}{x^3} - k{x^2} + x,{\text{ }}k \in {\mathbb{Z}^ + }$ .

(a)     Express in terms of k

(i)     ${{f'}_k}(x){\text{ and }}{{f''}_k}(x)$ ;

(ii)     the coordinates of the points of inflexion ${P_k}$ on the graphs of ${f_k}$ .

(b)     Show that all ${P_k}$ lie on a straight line and state its equation.

(c)     Show that for all values of k, the tangents to the graphs of ${f_k}$ at ${P_k}$ are parallel, and find the equation of the tangent lines.

(a)     (i)     ${{f'}_k}(x) = 3{k^2}{x^2} - 2kx + 1$     A1

${{f''}_k}(x) = 6{k^2}x - 2k$     A1

(ii)     Setting $f''(x) = 0$     M1

$\Rightarrow 6{k^2}x - 2k = 0 \Rightarrow x = \frac{1}{{3k}}$     A1

$f\left( {\frac{1}{{3k}}} \right) = {k^2}{\left( {\frac{1}{{3k}}} \right)^3} - k{\left( {\frac{1}{{3k}}} \right)^2} + \left( {\frac{1}{{3k}}} \right)$     M1

$= \frac{7}{{27k}}$     A1

Hence, ${P_k}{\text{ is }}\left( {\frac{1}{{3k}},\frac{7}{{27k}}} \right)$

[6 marks]

(b)     Equation of the straight line is $y = \frac{7}{9}x$     A1

As this equation is independent of k, all ${P_k}$ lie on this straight line     R1

[2 marks]

(c)     Gradient of tangent at ${P_k}$ :

$f'({P_k}) = f'\left( {\frac{1}{{3k}}} \right) = 3{k^2}{\left( {\frac{1}{{3k}}} \right)^2} - 2k\left( {\frac{1}{{3k}}} \right) + 1 = \frac{2}{3}$     M1A1

As the gradient is independent of k, the tangents are parallel.     R1

$\frac{7}{{27k}} = \frac{2}{3} \times \frac{1}{{3k}} + c \Rightarrow c = \frac{1}{{27k}}$     (A1)

The equation is $y = \frac{2}{3}x + \frac{1}{{27k}}$     A1

[5 marks]

Total [13 marks]

Many candidates scored the full 6 marks for part (a). The main mistake evidenced was to treat k as a variable, and hence use the product rule to differentiate. Of the many candidates who attempted parts (b) and (c), few scored the R1 marks in either part, but did manage to get the equations of the straight lines.