The function \(f\) is defined as \(f(x) = a{x^2} + bx + c\) where \(a,{\text{ }}b,{\text{ }}c \in \mathbb{R}\).

Hayley conjectures that \(\frac{{f({x_2}) - f({x_1})}}{{{x_2} - {x_1}}} = \frac{{f'({x_2}) + f'({x_1})}}{2},{\text{ }}x1 \ne x2\).

Show that Hayley’s conjecture is correct.

\(\frac{{f({x_2}) - f({x_1})}}{{{x_2} - {x_1}}} = \frac{{ax_2^2 + b{x_2} + c - (ax_1^2 + b{x_1} + c)}}{{{x_2} - {x_1}}}\)     (M1)

\( = \frac{{a(x_2^2 - x_1^2) + b({x_2} - {x_1})}}{{{x_2} - {x_1}}}\)     A1

\( = \frac{{a({x_2} - {x_1})({x_2} + {x_1}) + b({x_2} - {x_1})}}{{{x_2} - {x_1}}}\)     (A1)

\( = a({x_2} + {x_1}) + b\,\,\,\,\,({x_1} \ne {x_2})\)     A1

\(\frac{{f'({x_2}) + f'({x_1})}}{2} = \frac{{(2a{x_2} + b) + (2a{x_1} + b)}}{2}\)     M1

\( = \frac{{2a({x_2} + {x_1}) + 2b}}{2}\)

\( = a({x_2} + {x_1}) + b\)     A1

so Hayley’s conjecture is correct     AG

[6 marks]

This was generally answered very well. A small minority attempted to ‘prove’ the result by substituting specific values into the identity and thus gained little or no credit. Some started by assuming the result to be correct, then manipulated both sides until they derived an obvious identity. Reluctantly, they gained credit for this, though such an approach should be discouraged.