\(\mathrm{f}(x) = 5 + 8x - 2x^2\) Express \(\mathrm{f}(x)\) in the form \(A - B(x - C)^2\), where \(A\), \(B\) and \(C\) are positive constants to be found....

Here \(\mathrm{f}(x) = 5 + 8x - 2x^2\) has a negative coefficient of \(x^2\), so the graph is a downward parabola with a maximum, not a minimum. The requested form \(A - B(x - C)^2\) is built for exactly that shape: a positive constant \(A\) with a non-negative quantity \(B(x - C)^2\) subtracted from it. Every part of the question then follows from that single completed square, which is why the later parts all say "hence".

(a) Express \(\mathrm{f}(x)\) in the form \(A - B(x - C)^2\) [4]

First reorder into descending powers and take out the factor \(-2\) from the \(x^2\) and \(x\) terms only:

\[5 + 8x - 2x^2 = -2\left(x^2 - 4x\right) + 5 \qquad \textbf{[M1]}\]

The [M1] is for extracting \(-2\) correctly. Dividing \(8x\) by \(-2\) gives \(-4x\), and that sign change is where most marks are lost: \(-2(x^2 + 4x) + 5\) is a different function.

\[x^2 - 4x = (x - 2)^2 - 4 \qquad \textbf{[M1]}\]

The second [M1] is for completing the square inside the bracket: half of \(-4\) is \(-2\), and \(2^2 = 4\) is subtracted back.

\[\mathrm{f}(x) = -2\left[(x - 2)^2 - 4\right] + 5 = -2(x - 2)^2 + 8 + 5 = 13 - 2(x - 2)^2 \qquad \textbf{[A1]}\] \[A = 13, \quad B = 2, \quad C = 2 \qquad \textbf{[A1]}\]

The first [A1] is for the correct completed square, the second for identifying the three constants against the requested form. Multiplying \(-2\) by \(-4\) gives \(+8\), so the constant becomes \(8 + 5 = 13\); a candidate who writes \(-8 + 5 = -3\) loses both accuracy marks while keeping both method marks. The question says \(A\), \(B\) and \(C\) are positive, which is a built-in check: if any of your three values comes out negative, the sign handling has gone wrong somewhere.

(b) The maximum value and where it occurs [2]

Since \((x - 2)^2 \ge 0\) for all real \(x\), the term \(2(x - 2)^2\) is never negative, so subtracting it can only reduce \(13\). Therefore \(\mathrm{f}(x) \le 13\), with equality exactly when \((x - 2)^2 = 0\).

The maximum value of \(\mathrm{f}(x)\) is \(13\) [B1], occurring at \(x = 2\) [B1]. These are two independent B marks, so one can be earned without the other. No differentiation is required, and none is expected; the completed square already contains the answer.

(c) Solve \(\mathrm{f}(x) = 0\) in simplest surd form [3]

"Hence" and "exact values in simplest surd form" together rule out the quadratic formula as the intended route and rule out a decimal answer entirely. Use the completed square:

\[13 - 2(x - 2)^2 = 0 \quad\Rightarrow\quad (x - 2)^2 = \frac{13}{2} \qquad \textbf{[M1]}\] \[x - 2 = \pm\sqrt{\frac{13}{2}} = \pm\frac{\sqrt{13}}{\sqrt{2}} = \pm\frac{\sqrt{26}}{2} \qquad \textbf{[M1]}\] \[x = 2 \pm \frac{\sqrt{26}}{2} = \frac{4 + \sqrt{26}}{2} \quad\text{or}\quad \frac{4 - \sqrt{26}}{2} \qquad \textbf{[A1]}\]

The first [M1] is for isolating the squared bracket, the second for taking the square root with \(\pm\) attached. That \(\pm\) is the trap this part is built to catch: taking only the positive root gives one solution instead of two and the accuracy mark is lost. The rationalising step \(\sqrt{13/2} = \sqrt{26}/2\) is what "simplest surd form" is asking for; a surd left in a denominator is not simplest form.

(d) Hence solve \(\mathrm{f}(x) \gt 0\) [2]

This is where the shape of the graph does the work. The coefficient of \(x^2\) is negative, so the parabola opens downwards, is above the axis between its two roots and below the axis outside them [M1]. Hence

\[\frac{4 - \sqrt{26}}{2} \lt x \lt \frac{4 + \sqrt{26}}{2} \qquad \textbf{[A1]}\]

The [M1] is for the correct region ("between the roots"), and it is a follow-through mark: a candidate whose roots in (c) were wrong can still earn it by placing their own roots correctly. The [A1] needs the exact roots from (c) in a single correct inequality. The classic error is to give \(x \lt \frac{4 - \sqrt{26}}{2}\) or \(x \gt \frac{4 + \sqrt{26}}{2}\), which is the answer for a positive coefficient of \(x^2\), and would be right for \(\mathrm{f}(x) \lt 0\) here.

Check: \(\sqrt{26} \approx 5.10\), so the roots are about \(-0.55\) and \(4.55\). Test the midpoint \(x = 2\): \(\mathrm{f}(2) = 5 + 16 - 8 = 13 \gt 0\), which sits inside the stated interval, and \(\mathrm{f}(5) = 5 + 40 - 50 = -5\), which is outside it and negative. Both agree with the answer.