2026-08-31T11:34:13.055263 image/svg+xml Matplotlib v3.10.9, https://matplotlib.org/ © EAGLE BEACON GLOBAL Figure 5 shows the curve \(C\) with equation \(y ...

The idea underlying the whole question is that a drawn graph is a solving machine. Where a curve crosses the \(x\)-axis, the equation "curve \(= 0\)" is satisfied; where a curve meets a line, the equation "curve \(=\) line" is satisfied. So a single drawn cubic can solve a whole family of related equations, provided each one is first rearranged into the form "the plotted expression \(=\) something easy to draw". Because the answers are read off a printed graph, they are only as accurate as the reading allows, which is why every part specifies one decimal place.

(a) The three values of \(x\) for which \(x^3 - 3x^2 + 2 = 0\) [3]

Setting \(y = 0\) means looking for the points where the curve \(C\) shown in Figure 5 crosses the \(x\)-axis, and it crosses three times [M1]. Reading each crossing to one decimal place:

\[x = -0.7 \qquad \textbf{[A1]}\] \[x = 1.0 \quad\text{and}\quad x = 2.7 \qquad \textbf{[A1]}\]

The [M1] is for recognising that the solutions are the \(x\)-axis intercepts, and the two accuracy marks cover the three readings. Note that these values are exact to the accuracy demanded: the curve factorises as \(\left(x - 1\right)\left(x^2 - 2x - 2\right)\), so the true roots are \(1\) and \(1 \pm \sqrt{3}\), that is \(-0.732\ldots\) and \(2.732\ldots\), which round to \(-0.7\) and \(2.7\). A candidate should not attempt that factorisation here, because the question says "use the graph", but it confirms the readings.

(b) Show the intersections with \(y = 2x - 4\) satisfy \(x^3 - 3x^2 - 2x + 6 = 0\) [2]

At any point where the line meets the curve the two \(y\)-values are equal:

\[x^3 - 3x^2 + 2 = 2x - 4 \qquad \textbf{[M1]}\] \[x^3 - 3x^2 - 2x + 2 + 4 = 0 \quad\Rightarrow\quad x^3 - 3x^2 - 2x + 6 = 0 \qquad \textbf{[A1]}\]

The [M1] is for equating the two expressions, the [A1] for the printed cubic. Subtracting \(2x\) and adding \(4\) to both sides gives \(-2x\) and \(+6\); the sign of the constant is where this goes wrong, since \(2 - (-4) = 6\), not \(-2\). Because the target is printed, the rearrangement is the whole answer. This part is also the key to part (c): it tells you which line to draw in order to solve the new cubic.

(c) Draw \(y = 2x - 4\) on Figure 5 and find the three solutions [4]

To draw a straight line, plot two points and join them with a ruler. Convenient choices within the range of the axes are \(x = 0\), giving \(y = -4\), and \(x = 3\), giving \(y = 2\), so the line passes through \((0,\, -4)\) and \((3,\, 2)\) [M1]. A third point such as \((1,\, -2)\) is worth plotting as a check that the three are collinear.

The line cuts the curve three times, and the solutions of \(x^3 - 3x^2 - 2x + 6 = 0\) are the \(x\)-coordinates of those crossings [M1]. Reading them to one decimal place:

\[x = -1.4 \qquad \textbf{[A1]}\] \[x = 1.4 \quad\text{and}\quad x = 3.0 \qquad \textbf{[A1]}\]

The first [M1] is for a correct line, evidenced by two correct points; the second is for identifying the intersections as the solutions rather than, say, reading off \(y\)-values. The two accuracy marks cover the three readings, and each depends on the line actually being drawn, so an answer with no line on the figure cannot score them.

Exactly, the roots are \(x = 3\) and \(x = \pm\sqrt{2}\), since the cubic factorises as \(\left(x - 3\right)\left(x^2 - 2\right)\), and \(\sqrt{2} = 1.414\ldots\) rounds to \(1.4\). That is a useful confirmation, but the marks here are for the graphical reading. Two errors recur: reading the \(y\)-coordinate of an intersection instead of the \(x\)-coordinate, and drawing \(y = 2x + 4\) by misreading the sign of the intercept, which shifts the line up by eight units and gives only one intersection.

(d) The set of values of \(k\) for which \(x^3 - 3x^2 + 2 = k\) has three distinct real roots [2]

The equation says the curve equals a constant, so the relevant line is the horizontal line \(y = k\), and the number of real roots is the number of times that horizontal line cuts \(C\). Reading the turning points off Figure 5, the local maximum is at \((0,\, 2)\) and the local minimum is at \((2,\, -2)\), so a horizontal line cuts the curve three times only when its height lies strictly between those two values [M1]:

\[-2 \lt k \lt 2 \qquad \textbf{[A1]}\]

The [M1] is for identifying the two stationary values as the boundaries, the [A1] for the correct strict inequality. The strictness is the point of the question. At \(k = 2\) the line passes exactly through the maximum, so two of the three roots coincide and there are only two distinct roots; the same happens at \(k = -2\) through the minimum. Writing \(-2 \le k \le 2\) therefore loses the accuracy mark. Above \(k = 2\) or below \(k = -2\) the horizontal line cuts the curve only once.

Check: part (a) is a special case of part (d) with \(k = 0\), and \(0\) does lie strictly between \(-2\) and \(2\), which is consistent with the three roots found there. The turning-point values can also be verified from the equation: differentiating gives \(3x^2 - 6x = 3x(x - 2)\), which is zero at \(x = 0\) and \(x = 2\), and substituting back gives \(y = 2\) and \(y = -2\), exactly the values read from the graph.