2026-08-31T11:35:05.318389 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL Figure 1 shows a sketch of part of the curve \(C\...

(a) Find \(\dfrac{dy}{dx}\) [2]

\(y = 2x^{3} - 9x^{2} + 12x\) is a polynomial, so differentiate term by term with the rule "multiply by the power, then reduce the power by one".

\[\frac{dy}{dx} = 6x^{2} - 18x + 12 \quad \textbf{[M1][A1]}\]

The M mark is for a correct differentiation attempt, meaning at least one term reduced in power correctly, so it survives a slip in a coefficient. The A mark needs all three terms right. The constant-free linear term \(12x\) differentiates to \(12\), not to \(12x\) and not to \(0\).

(b) Coordinates of \(A\) and \(B\) [4]

Stationary points are where the gradient is zero, so solve \(\dfrac{dy}{dx} = 0\). Taking out the common factor \(6\) first makes the quadratic trivially factorisable, which is quicker and safer than the formula here.

\[6x^{2} - 18x + 12 = 0 \ \Rightarrow \ 6\left(x^{2} - 3x + 2\right) = 0 \ \Rightarrow \ 6(x - 1)(x - 2) = 0 \quad \textbf{[M1][A1]}\]

so \(x = 1\) or \(x = 2\). The M mark is for equating the derivative to zero and attempting a solution; the A mark is for both roots.

Now substitute into the equation of the curve, not the derivative:

• \(x = 1\): \(y = 2 - 9 + 12 = 5\), so \(A(1,\, 5)\). [A1]
• \(x = 2\): \(y = 2(8) - 9(4) + 12(2) = 16 - 36 + 24 = 4\), so \(B(2,\, 4)\). [A1]

Each accuracy mark is independent. Read the labelling off the figure the way the question sets it up: the stationary point at the smaller \(x\) is \(A\).

(c) \(\dfrac{d^{2}y}{dx^{2}}\) and the nature of each point [3]

Differentiate again:

\[\frac{d^{2}y}{dx^{2}} = 12x - 18 \quad \textbf{[B1]}\]

This is a B mark, awarded on sight for the correct expression, independent of anything else in the question. The test is: a negative second derivative means the gradient is decreasing, so the curve is bending downwards and the point is a maximum; a positive second derivative means a minimum.

• At \(x = 1\): \(12(1) - 18 = -6\), which is negative, so \(A(1,\, 5)\) is a maximum point. [A1]
• At \(x = 2\): \(12(2) - 18 = 6\), which is positive, so \(B(2,\, 4)\) is a minimum point. [A1]

Quote the numerical value and its sign, then the conclusion. Writing only "maximum" without the supporting value forfeits the accuracy mark.

(d) Range of values of \(x\) for which \(y\) is decreasing [2]

A function is decreasing exactly where its gradient is negative, so the condition is

\[\frac{dy}{dx} \lt 0 \quad \Rightarrow \quad 6(x - 1)(x - 2) \lt 0 \quad \textbf{[M1]}\]

The quadratic \(6(x-1)(x-2)\) opens upwards, so it is below the axis strictly between its roots:

\[1 \lt x \lt 2 \quad \textbf{[A1]}\]

The M mark is for recognising that "decreasing" means a negative first derivative and using the roots found in part (b); the A mark is for the correct interval.

Common wrong turns on this question. The one this part is built to catch is writing the answer as two separate outside intervals, \(x \lt 1\) or \(x \gt 2\). That is where the quadratic is positive, so it describes where \(y\) is increasing. Sketch or recall the upward parabola, and the "between the roots" region is unmistakable. A second slip is to use \(\dfrac{d^{2}y}{dx^{2}} \lt 0\), which describes concavity, not decrease.

Check. The answer must be consistent with parts (b) and (c): the curve rises to the maximum at \(x = 1\), falls between \(x = 1\) and \(x = 2\), then rises again after the minimum at \(x = 2\). The interval of decrease therefore runs exactly from one stationary point to the other, and the values \(y = 5\) then \(y = 4\) confirm the fall.