Further Pure Mathematics 4PM1 | Paper 1 Mock 01 | Written Paper 1

Two quadratic inequalities, 8 marks in total. The method for both is the same: get everything on one side, factorise, find the critical values, and then decide the sign of the expression in each region. A sketch of the parabola or a sign table is what makes the last step reliable. The second part carries the extra trap of a variable denominator.

(a) Solve \(2x^2 - 5x - 3 \gt 0\) [3 marks]

Factorise first: \[2x^2 - 5x - 3 = (2x + 1)(x - 3) \quad \textbf{M1}\] The M1 is for an attempt to factorise or to solve the quadratic; the formula or completing the square would earn it equally. The critical values are where the expression is zero: \[x = -\tfrac{1}{2} \quad \text{and} \quad x = 3 \quad \textbf{A1}\] Now the sign. The coefficient of \(x^2\) is positive, so the parabola opens upwards and lies above the axis outside its two roots and below between them. Since we want the expression to be positive, \[x \lt -\tfrac{1}{2} \quad \text{or} \quad x \gt 3 \quad \textbf{A1}\] The final A1 needs both regions and the word "or". Writing \(-\tfrac{1}{2} \gt x \gt 3\) is meaningless, because no number is simultaneously less than \(-\tfrac{1}{2}\) and greater than 3.

(b) Solve \(\dfrac{x + 4}{x - 2} \le 3\) [5 marks]

The one thing you must not do is multiply both sides by \((x - 2)\), because that expression changes sign at \(x = 2\) and multiplying an inequality by a negative quantity reverses it. The safe manoeuvre is to multiply by \((x - 2)^2\), which is positive for every \(x \ne 2\) and so leaves the inequality sign alone. Record the excluded value first: \(x \ne 2\), since the original expression is undefined there.

\[(x + 4)(x - 2) \le 3(x - 2)^2 \quad \textbf{M1}\] \[x^2 + 2x - 8 \le 3x^2 - 12x + 12 \quad \textbf{A1}\] \[0 \le 2x^2 - 14x + 20 \ \Rightarrow\ x^2 - 7x + 10 \ge 0 \quad \textbf{A1}\] \[(x - 2)(x - 5) \ge 0 \ \Rightarrow\ x \le 2 \ \text{or} \ x \ge 5 \quad \textbf{M1}\]

Finally impose the exclusion. The value \(x = 2\) satisfies \((x-2)(x-5) \ge 0\) but is not in the domain of the original inequality, so it must be removed: \[x \lt 2 \quad \text{or} \quad x \ge 5 \quad \textbf{A1}\] The second M1 is for solving the new quadratic inequality correctly in structure; the last A1 is reserved for the fully correct final set, including the strict inequality at \(x = 2\) and the inclusive one at \(x = 5\). The A marks are not follow-through, so an expansion slip earlier costs them.

An alternative route that also earns full marks is to subtract 3 and combine into a single fraction: \[\frac{x+4}{x-2} - 3 = \frac{x + 4 - 3(x-2)}{x-2} = \frac{10 - 2x}{x - 2} \le 0,\] then use a sign table on the critical values \(x = 2\) and \(x = 5\). Both routes must end with \(x = 2\) excluded.

Common wrong turns on this question

• Multiplying by \((x - 2)\) in part (b) and obtaining only \(x \ge 5\). This is the error the question is built to catch, and it silently discards the whole region to the left of 2.
• Forgetting to exclude \(x = 2\), giving \(x \le 2\) and so claiming a solution where the expression is undefined.
• Cancelling the factor \((x - 2)\) from \((x+4)(x-2) \le 3(x-2)^2\). Dividing by a quantity that may be negative or zero is not permitted.
• Getting the direction wrong in part (a) by assuming "greater than zero" means "between the roots". Sketch the parabola, or test one convenient value.

How to check

Test one value from each region. In part (a), \(x = 0\) gives \(-3\), which is not positive, correctly excluding the middle region; \(x = 4\) gives \(9 \gt 0\). In part (b), \(x = 0\) gives \(\dfrac{4}{-2} = -2 \le 3\), so \(x \lt 2\) belongs; \(x = 3\) gives \(\dfrac{7}{1} = 7\), which is not at most 3, so the middle region is correctly excluded; \(x = 5\) gives exactly \(\dfrac{9}{3} = 3\), so the boundary is correctly inclusive. Three substitutions settle a whole inequality.

Eleven marks on a product of two binomial expansions, ending with a numerical estimate. The strategy is to expand each bracket separately as far as \(x^2\), then multiply the two series keeping only the terms up to \(x^2\). Part (d) then exploits the fact that a truncated series is an excellent approximation when \(x\) is small.

(a) The first three terms of \((1 + 2x)^6\) [3 marks]

\[(1 + 2x)^6 = 1 + \binom{6}{1}(2x) + \binom{6}{2}(2x)^2 + \ldots \quad \textbf{B1}\] \[= 1 + 6(2x) = 1 + 12x \quad \textbf{B1}\] \[+ \ 15(4x^2) = +\ 60x^2 \quad \textbf{B1}\]

So \((1 + 2x)^6 = 1 + 12x + 60x^2 + \ldots\). Three separate B marks, one for the structure and one for each of the \(x\) and \(x^2\) terms, so a slip in one does not cost the others. The point to watch is \((2x)^2 = 4x^2\): the 2 is squared as well, which is why the coefficient is \(15 \times 4 = 60\) and not 15.

(b) The first three terms of \((1 - x)^4\) [2 marks]

\[(1 - x)^4 = 1 + \binom{4}{1}(-x) + \binom{4}{2}(-x)^2 + \ldots = 1 - 4x \quad \textbf{B1}\] \[+ \ 6x^2 \quad \textbf{B1}\]

The alternating signs come from the powers of \(-x\): odd powers are negative and even powers positive, so \((-x)^2 = +x^2\) and the \(x^2\) coefficient is \(+6\), not \(-6\).

(c) Show the expansion of \(\mathrm{f}(x)\) begins \(1 + 8x + 18x^2\) [3 marks]

The result is printed, so the marks are for the multiplication. Collect the products whose indices add to each required power: \[\mathrm{f}(x) = \bigl(1 + 12x + 60x^2 + \ldots\bigr)\bigl(1 - 4x + 6x^2 + \ldots\bigr)\] \[\text{Constant: } 1 \times 1 = 1 \quad \textbf{B1}\] \[\text{Term in } x: \ 1(-4) + 12(1) = -4 + 12 = 8 \quad \textbf{B1}\] \[\text{Term in } x^2: \ 1(6) + 12(-4) + 60(1) = 6 - 48 + 60 = 18 \quad \textbf{B1}\] There are exactly three ways to produce an \(x^2\) term, and all three must appear: constant times \(x^2\), \(x\) times \(x\), and \(x^2\) times constant. Terms in \(x^3\) and beyond are ignored because only the first three are wanted, so there is no need to expand either bracket further. Note the near-cancellation in the \(x^2\) coefficient: \(6 - 48 + 60\) is only 18, so dropping any one contribution changes the answer dramatically.

(d) An estimate for \((1.02)^6 \times (0.99)^4\) [3 marks]

Match the numerical bases to the algebraic ones. With \(x = 0.01\), \[1 + 2x = 1 + 0.02 = 1.02 \quad \text{and} \quad 1 - x = 1 - 0.01 = 0.99 \quad \textbf{B1}\] so \(\mathrm{f}(0.01) = (1.02)^6 \times (0.99)^4\) exactly. Substituting into the truncated expansion: \[\text{Estimate} = 1 + 8(0.01) + 18(0.01)^2 \quad \textbf{M1}\] \[= 1 + 0.08 + 0.0018 = 1.0818 \ \text{(4 d.p.)} \quad \textbf{A1}\] The B1 is for identifying that \(x = 0.01\) reproduces both bases, which is the insight the part is testing. Note \((0.01)^2 = 0.0001\), so the \(x^2\) term contributes \(0.0018\); the successive terms shrink by roughly a factor of 100, which is exactly why three terms suffice for four decimal places.

Common wrong turns on this question

• Writing \((2x)^2\) as \(2x^2\) in part (a), giving \(30x^2\) instead of \(60x^2\).
• Making the \(x^2\) coefficient in part (b) negative. Squaring \(-x\) gives a positive result.
• Finding only two of the three \(x^2\) contributions in part (c), most often omitting \(12 \times (-4)\).
• Choosing \(x = 0.02\) in part (d) because \(1.02\) is visible. The algebraic form is \(1 + 2x\), so \(2x = 0.02\) and \(x = 0.01\); the same \(x\) must also satisfy \(1 - x = 0.99\), which confirms the choice.
• Writing \((0.01)^2\) as \(0.001\) and reporting \(1.098\).
• Rounding to fewer than four decimal places.

How to check

Evaluate the target directly: \((1.02)^6 = 1.126162\ldots\) and \((0.99)^4 = 0.960596\ldots\), whose product is \(1.081787\ldots\), which rounds to \(1.0818\). The three-term estimate therefore agrees to all four decimal places, confirming both the expansion and the substitution. A structural check on part (c) is also worth doing: the constant term of \(\mathrm{f}(x)\) must be \(1^6 \times 1^4 = 1\), and the \(x\) coefficient must be \(6(2) + 4(-1) = 8\), matching the answer.

The area of the finite region \(R\) bounded by \(C\) and \(l\) shown shaded is found by the standard "top curve minus bottom curve" principle. Between the two intersection points the line lies above the parabola, so the area is the area under the line minus the area under the curve over the same interval. That is why the question first makes you locate the intersections in part (b) and then evaluate the two definite integrals separately in parts (c) and (d): the limits of integration are the \(x\)-coordinates of the intersections, and nothing can be integrated until they are known.

(a) Complete the square and write down the minimum point [3]

Half of \(-4\) is \(-2\), so \((x - 2)^2 = x^2 - 4x + 4\), and four must be given back:

\[x^2 - 4x + 7 = (x - 2)^2 - 4 + 7 = (x - 2)^2 + 3 \qquad \textbf{[M1 A1]}\]

Matching against the requested form \((x + a)^2 + b\) gives \(a = -2\) and \(b = 3\). The [M1] is for the completing-the-square structure, the [A1] for both constants. The form has a plus sign inside the bracket, so \(a = -2\); quoting \(a = 2\) is the standard way to lose the accuracy mark here.

Since \((x - 2)^2 \ge 0\) with equality only at \(x = 2\), the least value of \(y\) is \(3\), so the minimum point of \(C\) is \((2,\, 3)\) [B1]. This is written down from the completed square, not obtained by differentiation.

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

At an intersection the two \(y\)-values agree, so equate the equations:

\[x^2 - 4x + 7 = x + 3 \qquad \textbf{[M1]}\] \[x^2 - 5x + 4 = 0 \qquad \textbf{[A1]}\] \[(x - 1)(x - 4) = 0 \quad\Rightarrow\quad x = 1 \text{ or } x = 4 \qquad \textbf{[M1]}\]

Substituting into the simpler equation, the line \(y = x + 3\), gives the points \(A(1,\, 4)\) and \(B(4,\, 7)\) [A1]. Always use the line for the \(y\)-values: it is one step instead of three and the arithmetic is safer. The final accuracy mark needs full coordinates, since the question asks for points; a pair of \(x\)-values alone is an incomplete answer.

(c) Show that \(\displaystyle\int_{1}^{4}\left(x^2 - 4x + 7\right)\mathrm{d}x = 12\) [4]

Integrate term by term, raising each index by one and dividing by the new index:

\[\int \left(x^2 - 4x + 7\right)\mathrm{d}x = \frac{x^3}{3} - 2x^2 + 7x \qquad \textbf{[M1 A1]}\]

The [M1] is for an attempt at integration, shown by at least one term correct; the [A1] is for all three terms. Because this is a definite integral, no constant of integration is required and none should be written; on an indefinite integral its omission would cost a mark, so the distinction is worth holding in mind. Note \(-4x\) integrates to \(-2x^2\), not \(-4x^2/2\) left unsimplified, and the constant \(7\) integrates to \(7x\).

Now substitute the upper limit and subtract the value at the lower limit:

\[= \left(\frac{64}{3} - 32 + 28\right) - \left(\frac{1}{3} - 2 + 7\right) \qquad \textbf{[M1]}\] \[= \left(\frac{64}{3} - 4\right) - \left(\frac{1}{3} + 5\right) = \frac{64}{3} - \frac{1}{3} - 9 = 21 - 9 = 12 \qquad \textbf{[A1]}\]

The [M1] is for correct substitution of both limits in the right order, and it survives an arithmetic slip; the [A1] is for the printed value \(12\). Since the answer is given, the working must be complete: writing \(12\) with no evaluation earns nothing. The frequent error is subtracting in the wrong order or losing the bracket around the lower-limit value, which flips the signs of its terms.

(d) Hence show that the area of \(R\) is \(\dfrac{9}{2}\) [6]

The region \(R\) lies between the line above and the curve below, over \(1 \le x \le 4\). Deal with the line first:

\[\int_{1}^{4}(x + 3)\,\mathrm{d}x = \left[\frac{x^2}{2} + 3x\right]_{1}^{4} \qquad \textbf{[M1 A1]}\] \[= (8 + 12) - \left(\frac{1}{2} + 3\right) = 20 - \frac{7}{2} = \frac{33}{2} \qquad \textbf{[M1 A1]}\]

The first [M1 A1] pair is for the integration, the second for the evaluation. The limits are the same \(1\) and \(4\) found in part (b), because that is the horizontal extent of \(R\).

Now the subtraction principle. The area of \(R\) is the area under the line minus the area under the curve, both taken between the same limits [M1], and part (c) has already supplied the second of these:

\[\text{Area of } R = \frac{33}{2} - 12 = \frac{33}{2} - \frac{24}{2} = \frac{9}{2} \qquad \textbf{[A1]}\]

The [M1] is for the correct principle, so it is earned by a candidate who subtracts the right way round even with a numerical slip; the [A1] is for the printed value. Subtracting the wrong way gives \(-\tfrac{9}{2}\), and an area cannot be negative, which is the built-in signal that the order has been reversed. The way to decide the order without guessing is to test a value between the limits: at \(x = 2\), the line gives \(y = 5\) and the curve gives \(y = 3\), so the line is on top.

An equivalent single-integral method scores the same marks and is less error-prone:

\[\int_{1}^{4}\left[(x + 3) - \left(x^2 - 4x + 7\right)\right]\mathrm{d}x = \int_{1}^{4}\left(-x^2 + 5x - 4\right)\mathrm{d}x = \left[-\frac{x^3}{3} + \frac{5x^2}{2} - 4x\right]_{1}^{4}\] \[= \left(-\frac{64}{3} + 40 - 16\right) - \left(-\frac{1}{3} + \frac{5}{2} - 4\right) = \frac{8}{3} - \left(-\frac{11}{6}\right) = \frac{16}{6} + \frac{11}{6} = \frac{27}{6} = \frac{9}{2}\]

which confirms the answer. Note that the integrand \(-x^2 + 5x - 4\) is the negative of the quadratic from part (b), which is no accident: it vanishes exactly at the intersection points, so it is positive throughout the interior of \(R\), another confirmation that the line is the upper boundary there.

Check: a rough magnitude test. The region is about \(3\) units wide and, at its widest point \(x = 2.5\), the vertical gap between line and curve is \(5.5 - 3.25 = 2.25\), so an area of \(4.5\) is entirely plausible for a lens-shaped region of those dimensions.

Part (a) is a direct application of the remainder theorem: the remainder on dividing \(\mathrm{g}(x)\) by \((x - a)\) is simply \(\mathrm{g}(a)\), so no division is needed. Part (b) extends the same idea to a quadratic divisor, and the extension rests on one structural fact: the remainder on dividing by a quadratic is of lower degree than the divisor, so it is at most linear, which is exactly why the question tells you it has the form \(ax + b\).

(a) Remainder when \(\mathrm{g}(x)\) is divided by \((x - 3)\) [2]

By the remainder theorem the remainder is \(\mathrm{g}(3)\) [M1]. Substituting into \(\mathrm{g}(x) = x^4 - 2x^3 + 3x - 5\):

\[\mathrm{g}(3) = 81 - 2(27) + 9 - 5 = 81 - 54 + 9 - 5 = 31 \qquad \textbf{[A1]}\]

The [M1] is for identifying \(x = 3\) as the value to substitute, so it survives an arithmetic slip; the [A1] is for the value \(31\). Note that \(\mathrm{g}(x)\) has no \(x^2\) term, so nothing is contributed there. The sign trap is the divisor: \((x - 3)\) means \(x = +3\). Long division would also reach \(31\) but wastes several minutes for the same two marks.

(b) Find \(a\) and \(b\), where the remainder on dividing by \(\left(x^2 - x - 6\right)\) is \(ax + b\) [4]

Factorise the divisor first, because that is what makes substitution possible:

\[x^2 - x - 6 = (x - 3)(x + 2)\]

The divisor is quadratic, so the remainder is at most linear, and the division statement is

\[\mathrm{g}(x) = (x - 3)(x + 2)\,\mathrm{Q}(x) + ax + b \qquad \textbf{[M1]}\]

where \(\mathrm{Q}(x)\) is the quotient. The [M1] is for setting up this identity. Its power is that substituting either root of the divisor kills the \(\mathrm{Q}(x)\) term entirely, whatever \(\mathrm{Q}(x)\) happens to be, so the quotient never has to be found.

Put \(x = 3\). The bracket \((x - 3)\) vanishes, so

\[3a + b = \mathrm{g}(3) = 31 \qquad \textbf{[A1]}\]

which reuses part (a) and is why the two parts sit together. Put \(x = -2\), which makes \((x + 2)\) vanish:

\[\mathrm{g}(-2) = 16 - 2(-8) + 3(-2) - 5 = 16 + 16 - 6 - 5 = 21, \quad\text{so } -2a + b = 21 \qquad \textbf{[A1]}\]

The evaluation of \(\mathrm{g}(-2)\) is where marks go astray: \((-2)^4 = +16\) and \((-2)^3 = -8\), so the term \(-2x^3\) becomes \(-2 \times (-8) = +16\), a double negative that turns positive. Getting it as \(-16\) gives \(\mathrm{g}(-2) = -11\) and wrecks the rest.

Now solve the pair. Subtracting the second equation from the first eliminates \(b\):

\[(3a + b) - (-2a + b) = 31 - 21 \quad\Rightarrow\quad 5a = 10 \quad\Rightarrow\quad a = 2\] \[b = 31 - 3a = 31 - 6 = 25 \qquad \textbf{[A1]}\]

So \(a = 2\) and \(b = 25\), and the remainder is \(2x + 25\). The final [A1] requires both constants. Each of the two accuracy marks before it is independent, one per substitution, so a slip in \(\mathrm{g}(-2)\) costs only that mark and the last one, while the method mark and the \(x = 3\) mark both stand.

The wrong turn this part is designed to catch is assuming the remainder is a constant, as it is for a linear divisor. Dividing by a quadratic can leave a linear remainder, which is why two unknowns and therefore two substitutions are needed. A second, subtler error is substituting \(x = 3\) and \(x = 2\), reading the factors from the coefficients rather than from \((x - 3)(x + 2)\).

Check: the remainder should reproduce both function values. At \(x = 3\), \(2(3) + 25 = 31 = \mathrm{g}(3)\), and at \(x = -2\), \(2(-2) + 25 = 21 = \mathrm{g}(-2)\). Both agree, which confirms \(a\) and \(b\) without carrying out the division. If more assurance is wanted, the full long division gives quotient \(x^2 - x + 5\), so \(\mathrm{g}(x) = \left(x^2 - x - 6\right)\left(x^2 - x + 5\right) + 2x + 25\), and multiplying out returns \(x^4 - 2x^3 + 3x - 5\) exactly.

This question runs one idea through six parts: the number of times a line meets a curve is the number of real roots of the quadratic you get by eliminating \(y\), and that number is controlled by the discriminant. A tangent touches once, so the discriminant is zero; two distinct crossings need it positive. Once part (c) has produced the quadratic in \(x\), parts (d), (e) and (f) are three different questions about the same discriminant, which is why they are worth so many marks between them.

(a) Express \(x^2 - 6x + 10\) in the form \((x - p)^2 + q\) [2]

Half of \(-6\) is \(-3\), so \((x - 3)^2 = x^2 - 6x + 9\) and nine must be given back:

\[x^2 - 6x + 10 = (x - 3)^2 - 9 + 10 = (x - 3)^2 + 1 \qquad \textbf{[M1]}\] \[p = 3, \quad q = 1 \qquad \textbf{[A1]}\]

The [M1] is for the completing-the-square structure \((x - 3)^2 - 9 + 10\); the [A1] is for both constants. The requested form has a minus sign inside the bracket, so \(p = 3\) here, positive; watch the sign convention in each question because it changes between papers.

(b) Coordinates of the minimum point of \(C\) [1]

\((x - 3)^2 \ge 0\) always, and is zero only at \(x = 3\), so the smallest value of \(y\) is \(1\), reached at \(x = 3\). The minimum point is \((3,\, 1)\) [B1]. This is written down from the completed square, not found by differentiation, which is what "hence" signals. A very common slip is to give \((-3,\, 1)\): the vertex sits where the bracket vanishes, so the \(x\)-coordinate is \(+3\).

(c) Show that the intersections satisfy \(x^2 - 8x + 10 - k = 0\) [2]

The target is printed, so all the credit is in the elimination. At any common point of \(C\) and \(l\) the two \(y\)-values are equal:

\[x^2 - 6x + 10 = 2x + k \qquad \textbf{[M1]}\] \[x^2 - 6x - 2x + 10 - k = 0 \quad\Rightarrow\quad x^2 - 8x + 10 - k = 0 \qquad \textbf{[A1]}\]

The [M1] is for equating the two expressions, the [A1] for the tidy printed form. Subtracting \(2x\) gives \(-8x\), not \(-4x\), and the constant term is \(10 - k\), a single quantity that must be carried as such into the discriminant later.

(d) Find \(k\) and the coordinates of \(P\) [5]

The line is a tangent, so it meets \(C\) exactly once, so the quadratic from (c) has a repeated root, so its discriminant is zero:

\[b^2 - 4ac = 0 \qquad \textbf{[M1]}\] \[(-8)^2 - 4 \times 1 \times (10 - k) = 0 \qquad \textbf{[M1]}\] \[64 - 40 + 4k = 0 \quad\Rightarrow\quad 4k = -24 \quad\Rightarrow\quad k = -6 \qquad \textbf{[A1]}\]

The first [M1] is for knowing that tangency means zero discriminant; the second is for substituting \(a = 1\), \(b = -8\), \(c = 10 - k\) correctly. Expanding \(-4(10 - k)\) as \(-40 - 4k\) is the standard error and loses the [A1], giving \(k = 6\).

For the point of contact, put \(k = -6\) back into the quadratic:

\[x^2 - 8x + 16 = 0 \quad\Rightarrow\quad (x - 4)^2 = 0 \quad\Rightarrow\quad x = 4 \qquad \textbf{[M1]}\] \[y = 2(4) + (-6) = 2, \quad\text{so } P(4,\, 2) \qquad \textbf{[A1]}\]

The [M1] here is follow-through: a candidate with the wrong \(k\) can still earn it by solving their own quadratic and finding \(y\) from the line. The [A1] needs the correct pair. The repeated-root shortcut \(x = -\dfrac{b}{2a} = 4\) is quicker and avoids a factorising slip.

(e) Set of values of \(k\) for two distinct points [3]

Two distinct intersections means two distinct real roots, so the same discriminant must now be strictly positive:

\[64 - 4(10 - k) \gt 0 \qquad \textbf{[M1]}\] \[64 - 40 + 4k \gt 0 \quad\Rightarrow\quad 24 + 4k \gt 0 \qquad \textbf{[M1]}\] \[k \gt -6 \qquad \textbf{[A1]}\]

The first [M1] is for using \(b^2 - 4ac \gt 0\), the second for correct expansion, the [A1] for the final set. Two traps: writing \(\ge\) instead of \(\gt\) admits the tangent case, where the two points coincide, so it is wrong; and dividing an inequality by a negative number without reversing it. Notice the structural sense of the answer: \(k = -6\) is the tangent found in (d), and any larger \(k\) lifts the line clear of the touching position so that it cuts twice.

(f) Given \(k = 10\), find the two intersection points [3]

Substituting \(k = 10\) into the quadratic from (c) makes the constant term vanish:

\[x^2 - 8x + 10 - 10 = 0 \quad\Rightarrow\quad x^2 - 8x = 0 \qquad \textbf{[M1]}\] \[x(x - 8) = 0 \quad\Rightarrow\quad x = 0 \text{ or } x = 8 \qquad \textbf{[A1]}\]

Using \(y = 2x + 10\) gives the points \((0,\, 10)\) and \((8,\, 26)\) [A1].

The wrong turn this part is designed to catch is dividing through by \(x\) to get \(x = 8\) only. Dividing by a quantity that may be zero destroys the root \(x = 0\), and half the answer with it; always factorise instead. The second [A1] also requires the \(y\)-values, since the question asks for coordinates.

Check: both points must lie on \(C\) as well as on \(l\). At \(x = 0\), \(y = 0 - 0 + 10 = 10\); at \(x = 8\), \(y = 64 - 48 + 10 = 26\). Both agree. Consistency with (e) is a further check: \(k = 10 \gt -6\), so two distinct points were expected, and two were found.

Four marks on a trigonometric equation. The equation \(2\sin^2\theta + 3\cos\theta = 3\) mixes \(\sin\) and \(\cos\), which cannot be solved directly. The standard move is to use the Pythagorean identity \(\sin^2\theta + \cos^2\theta = 1\) to eliminate one function in favour of the other. Here the \(\sin\) appears only as \(\sin^2\theta\), so replacing it with \(1 - \cos^2\theta\) leaves an equation in \(\cos\theta\) alone. Going the other way, replacing \(\cos\theta\) by \(\pm\sqrt{1 - \sin^2\theta}\), would introduce a square root and a sign ambiguity, which is why this direction is the right one.

(a) Show the equation can be written as \(2\cos^2\theta - 3\cos\theta + 1 = 0\) [2 marks]

The target is given, so all the credit is in the derivation. Substituting \(\sin^2\theta = 1 - \cos^2\theta\): \[2\bigl(1 - \cos^2\theta\bigr) + 3\cos\theta = 3 \quad \textbf{M1}\] Expanding and collecting everything on the left: \[2 - 2\cos^2\theta + 3\cos\theta - 3 = 0 \ \Rightarrow\ -2\cos^2\theta + 3\cos\theta - 1 = 0\] Multiplying through by \(-1\) to give a positive leading coefficient: \[2\cos^2\theta - 3\cos\theta + 1 = 0 \quad \textbf{A1}\] The M1 is for the correct use of the identity; the A1 is for reaching the printed form, including the sign change. Every intermediate line must be shown, since a candidate who writes down the given result unsupported earns nothing.

(b) Solve for \(0 \le \theta \lt 360^\circ\) [2 marks]

Treat the result as a quadratic in \(\cos\theta\). It factorises: \[\bigl(2\cos\theta - 1\bigr)\bigl(\cos\theta - 1\bigr) = 0 \ \Rightarrow\ \cos\theta = \tfrac{1}{2} \ \text{ or } \ \cos\theta = 1 \quad \textbf{M1}\] Now solve each. Cosine is positive in the first and fourth quadrants, so \(\cos\theta = \tfrac{1}{2}\) gives a principal value of \(60^\circ\) and a second solution at \(360^\circ - 60^\circ = 300^\circ\). The equation \(\cos\theta = 1\) has its only solution in the interval at \(\theta = 0^\circ\), the start of the range: \[\theta = 0^\circ, \ 60^\circ, \ 300^\circ \quad \textbf{A1}\] The M1 is for solving the quadratic to reach values of \(\cos\theta\), so it survives if only some of the angles are then found; the A1 requires all three angles and no extras. The interval notation is worth reading carefully: \(0^\circ\) is included because of the \(\le\), while \(360^\circ\) is excluded because of the strict inequality, so \(360^\circ\) must not be offered as a fourth solution.

Where the marks are

Only one of the four marks is a method mark, which makes this question unforgiving in part (a): the accuracy mark demands the exact printed rearrangement. In part (b), the danger is losing the A1 for an incomplete solution set rather than for a wrong one.

Common wrong turns on this question

• Using \(\sin^2\theta = 1 + \cos^2\theta\) or misremembering the identity, which produces the wrong quadratic.
• Dividing through by \(\cos\theta\) at some stage, which is invalid where \(\cos\theta = 0\) and destroys solutions.
• Missing \(\theta = 0^\circ\). Because \(\cos\theta = 1\) has a single solution in the range and it sits right on the boundary, it is the answer most often dropped. This is the omission the question is designed to catch.
• Offering \(\theta = 360^\circ\) as well. The range is \(\theta \lt 360^\circ\).
• Giving \(\theta = 240^\circ\) for \(\cos\theta = \tfrac{1}{2}\). That is the second solution for \(\cos\theta = -\tfrac{1}{2}\); for positive cosine the second solution is \(360^\circ - 60^\circ\).
• Working in radians. The interval is stated in degrees.

How to check

Substitute each angle into the original equation, not the rearranged one, so that an error in part (a) would be exposed. At \(\theta = 0^\circ\): \(2(0) + 3(1) = 3\), correct. At \(\theta = 60^\circ\): \(\sin 60^\circ = \dfrac{\sqrt{3}}{2}\), so \(2 \times \dfrac{3}{4} + 3 \times \dfrac{1}{2} = 1.5 + 1.5 = 3\), correct. At \(\theta = 300^\circ\): \(\sin 300^\circ = -\dfrac{\sqrt{3}}{2}\), and squaring removes the sign, so the value is again \(1.5 + 1.5 = 3\), correct. A final count check: a quadratic in \(\cos\theta\) with two distinct roots in \([-1, 1]\) usually yields up to four angles in a full revolution, but \(\cos\theta = 1\) is a boundary value contributing only one, which is why three solutions is the right total here.

Twelve marks of vector geometry in a parallelogram. Figure 2 shows \(OACB\) with \(\overrightarrow{OA} = \mathbf{a}\) and \(\overrightarrow{OB} = \mathbf{b}\), \(M\) the midpoint of \(BC\), and \(X\) the point where \(OM\) crosses the diagonal \(AB\). The whole question turns on one powerful idea: if \(\mathbf{a}\) and \(\mathbf{b}\) are not parallel, then any point of the plane has exactly one expression as \(\lambda\mathbf{a} + \mu\mathbf{b}\). Writing the same point two ways therefore forces the coefficients to match, and that gives simultaneous equations.

(a) \(\overrightarrow{OC}\) [1 mark]

In the parallelogram \(OACB\), the side \(AC\) is equal and parallel to \(OB\), so travelling \(O\) to \(A\) to \(C\) is \(\mathbf{a}\) followed by \(\mathbf{b}\): \[\overrightarrow{OC} = \mathbf{a} + \mathbf{b} \quad \textbf{B1}\]

(b) Show that \(\overrightarrow{OM} = \tfrac{1}{2}\mathbf{a} + \mathbf{b}\) [2 marks]

Since the result is given, the marks are for the route. First find the side \(BC\): \[\overrightarrow{BC} = \overrightarrow{OC} - \overrightarrow{OB} = (\mathbf{a} + \mathbf{b}) - \mathbf{b} = \mathbf{a}, \ \text{so} \ \overrightarrow{BM} = \tfrac{1}{2}\mathbf{a} \quad \textbf{M1}\] Then travel \(O\) to \(B\) to \(M\): \[\overrightarrow{OM} = \overrightarrow{OB} + \overrightarrow{BM} = \mathbf{b} + \tfrac{1}{2}\mathbf{a} = \tfrac{1}{2}\mathbf{a} + \mathbf{b} \quad \textbf{A1}\]

(c) Two expressions for \(\overrightarrow{OX}\) [3 marks]

Along \(OM\), \(X\) is a fraction \(t\) of the way: \[\overrightarrow{OX} = t\left(\tfrac{1}{2}\mathbf{a} + \mathbf{b}\right) = \tfrac{1}{2}t\,\mathbf{a} + t\,\mathbf{b} \quad \textbf{B1}\] Along \(AB\), start at \(A\) and move a fraction \(s\) of the way towards \(B\). The direction is \[\overrightarrow{AB} = \overrightarrow{OB} - \overrightarrow{OA} = \mathbf{b} - \mathbf{a} \quad \textbf{B1}\] so \[\overrightarrow{OX} = \overrightarrow{OA} + s\,\overrightarrow{AB} = \mathbf{a} + s(\mathbf{b} - \mathbf{a}) = (1 - s)\mathbf{a} + s\,\mathbf{b} \quad \textbf{B1}\] Three independent B marks. Both expressions must be written purely in terms of \(\mathbf{a}\), \(\mathbf{b}\) and the single parameter, with the brackets expanded ready for comparison.

(d) The values of \(s\) and \(t\) [3 marks]

The two expressions describe the same point, and \(\mathbf{a}\) and \(\mathbf{b}\) are non-parallel, so the coefficients may be equated: \[\text{Coefficients may be compared because } \mathbf{a} \ \text{and} \ \mathbf{b} \ \text{are not parallel.} \quad \textbf{M1}\] \[\mathbf{a}: \ \tfrac{1}{2}t = 1 - s; \qquad \mathbf{b}: \ t = s \quad \textbf{A1}\] Substituting \(t = s\) into the first equation: \[\tfrac{1}{2}s = 1 - s \ \Rightarrow\ \tfrac{3}{2}s = 1 \ \Rightarrow\ s = \tfrac{2}{3}, \qquad t = \tfrac{2}{3} \quad \textbf{A1}\] The M1 rewards the justification as well as the act of comparing; it is the mathematical content of the part. The final A1 needs both values.

(e) The ratio \(AX : XB\) [1 mark]

Since \(\overrightarrow{AX} = s\,\overrightarrow{AB} = \tfrac{2}{3}\overrightarrow{AB}\), the point \(X\) is two thirds of the way from \(A\) to \(B\), leaving one third: \[AX : XB = 2 : 1 \quad \textbf{B1}\] Read from \(s\) directly as \(s : (1 - s)\). Quoting \(1 : 2\) reverses the segments.

(f) \(\left|\overrightarrow{OX}\right|\) with \(\mathbf{a} = 4\mathbf{i} + \mathbf{j}\), \(\mathbf{b} = -2\mathbf{i} + 5\mathbf{j}\) [2 marks]

Use either expression with \(s = t = \tfrac{2}{3}\). Taking the second gives \(\overrightarrow{OX} = \tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b}\): \[\overrightarrow{OX} = \tfrac{1}{3}(4\mathbf{i} + \mathbf{j}) + \tfrac{2}{3}(-2\mathbf{i} + 5\mathbf{j}) = \left(\tfrac{4}{3} - \tfrac{4}{3}\right)\mathbf{i} + \left(\tfrac{1}{3} + \tfrac{10}{3}\right)\mathbf{j} = \tfrac{11}{3}\mathbf{j} \quad \textbf{M1}\] The \(\mathbf{i}\) components cancel exactly, so \(\overrightarrow{OX}\) points straight up the \(\mathbf{j}\) direction and its modulus is simply the size of that one component: \[\left|\overrightarrow{OX}\right| = \tfrac{11}{3} \quad \textbf{A1}\] Using the first expression is an equally good check: \(\tfrac{2}{3}\left(\tfrac{1}{2}\mathbf{a} + \mathbf{b}\right) = \tfrac{1}{3}\mathbf{a} + \tfrac{2}{3}\mathbf{b}\), the same vector.

Common wrong turns on this question

• Taking \(M\) as the midpoint of \(OC\) or of \(AC\). Read the figure: \(M\) is the midpoint of \(BC\), which is why \(\overrightarrow{OM}\) carries a full \(\mathbf{b}\) and only half of \(\mathbf{a}\).
• Writing \(\overrightarrow{AB} = \mathbf{a} - \mathbf{b}\). "To minus from" gives \(\mathbf{b} - \mathbf{a}\); the reversed version leads to \(s = \tfrac{1}{3}\) and a wrong ratio.
• Equating coefficients without stating that \(\mathbf{a}\) and \(\mathbf{b}\) are non-parallel. That justification is what the M1 in part (d) is for; the step is invalid for parallel vectors.
• Assuming \(X\) is the midpoint of \(AB\) because \(OM\) looks central in the figure. The diagonal is cut in the ratio \(2 : 1\), not \(1 : 1\).
• Being surprised by the cancelling \(\mathbf{i}\) components in part (f) and assuming an error. The numbers are chosen so that \(\overrightarrow{OX}\) is vertical.

How to check

With the given components, \(\overrightarrow{OM} = \tfrac{1}{2}(4\mathbf{i} + \mathbf{j}) + (-2\mathbf{i} + 5\mathbf{j}) = 0\mathbf{i} + \tfrac{11}{2}\mathbf{j}\), so \(OM\) itself lies along the \(\mathbf{j}\)-axis and \(\tfrac{2}{3}\) of it is \(\tfrac{11}{3}\mathbf{j}\), confirming part (f) independently. The ratio in part (e) is also verifiable numerically: \(A\) is at \((4, 1)\), \(B\) at \((-2, 5)\), and the point two thirds of the way along is \(\left(4 - \tfrac{2}{3}(6),\ 1 + \tfrac{2}{3}(4)\right) = \left(0,\ \tfrac{11}{3}\right)\), exactly the \(X\) found.

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.

Nine marks built on one idea: a quadratic inequality and the roots of the quadratic are two views of the same object. If \(x^2 + bx + c \lt 0\) exactly on \(-2 \lt x \lt 7\), then the endpoints of that interval must be the roots, because a positive quadratic is negative precisely between its roots. That single observation makes part (a) immediate.

(a) The values of \(b\) and \(c\) [3 marks]

The roots are \(-2\) and \(7\), and the coefficient of \(x^2\) is 1, so the quadratic factorises as \[x^2 + bx + c = (x + 2)(x - 7) \quad \textbf{M1}\] Expanding, \[= x^2 - 7x + 2x - 14 = x^2 - 5x - 14 \quad \textbf{A1}\] \[b = -5, \qquad c = -14 \quad \textbf{A1}\] The M1 is for using the interval endpoints as roots. An equally valid route uses sum and product of roots: \(-b = (-2) + 7 = 5\) so \(b = -5\), and \(c = (-2)(7) = -14\). Either way the final A1 needs both constants, with the signs right. The commonest slip is writing \((x - 2)(x + 7)\), which corresponds to the interval \(-7 \lt x \lt 2\) instead.

(b) Solve \(x^2 + bx + c \ge 4x - 14\) [3 marks]

Substitute the constants and collect everything on one side. The \(-14\) appears on both sides and cancels, which is the point of the numbers chosen: \[x^2 - 5x - 14 \ge 4x - 14 \ \Rightarrow\ x^2 - 9x \ge 0 \quad \textbf{M1}\] \[x(x - 9) \ge 0 \quad \textbf{A1}\] The critical values are \(0\) and \(9\), and since the parabola opens upwards it is at or above the axis outside its roots: \[x \le 0 \quad \text{or} \quad x \ge 9 \quad \textbf{A1}\] Note that \(x = 0\) is a genuine solution here, so the inequality is inclusive at both ends. Do not divide through by \(x\); that would lose the root \(x = 0\) and is invalid because \(x\) may be negative or zero.

(c) Solve \(\dfrac{x^2 + bx + c}{x + 1} \le 0\) [3 marks]

Use the factorised numerator from part (a). A rational expression changes sign at a zero of the numerator and at a zero of the denominator, so all three matter: \[\frac{(x + 2)(x - 7)}{x + 1} \le 0, \qquad \text{critical values } x = -2, \ -1, \ 7 \quad \textbf{M1}\] Build a sign table over the four regions. To the left of \(-2\) the two numerator factors are negative and the denominator is negative, giving an overall negative value; each subsequent critical value flips the sign: \[\lt 0 \ \text{for} \ x \lt -2; \quad \gt 0 \ \text{for} \ -2 \lt x \lt -1; \quad \lt 0 \ \text{for} \ -1 \lt x \lt 7; \quad \gt 0 \ \text{for} \ x \gt 7 \quad \textbf{A1}\] The two numerator roots may be included, because the fraction is then exactly zero; the denominator root must be excluded, because the expression is undefined there: \[x \le -2 \quad \text{or} \quad -1 \lt x \le 7 \quad \textbf{A1}\] Notice the asymmetry in the final answer: an inclusive bound at \(-2\) and at \(7\), but a strict one at \(-1\). Getting that right is exactly what the last A1 is testing.

Common wrong turns on this question

• Multiplying through by \((x + 1)\) in part (c). That expression is negative for \(x \lt -1\), so the inequality would have to reverse there; the safe move is a sign table, or multiplying by \((x + 1)^2\).
• Ignoring the denominator when finding critical values and reporting \(-2 \le x \le 7\).
• Including \(x = -1\) in the solution set.
• Dividing by \(x\) in part (b) and losing the region \(x \le 0\).
• Getting the sign of \(b\) wrong in part (a) by confusing the roots with the coefficients.

How to check

Part (a) is self-checking: \((-2)^2 - 5(-2) - 14 = 4 + 10 - 14 = 0\) and \(7^2 - 35 - 14 = 0\), so both endpoints are roots as required. For part (c), test one value per region: at \(x = -3\) the value is \(\dfrac{(-1)(-10)}{-2} = -5 \le 0\), so that region is in; at \(x = -1.5\) it is \(\dfrac{(0.5)(-8.5)}{-0.5} = 8.5\), so that region is out; at \(x = 0\) it is \(-14 \le 0\), in; at \(x = 8\) it is \(\dfrac{(10)(1)}{9} \gt 0\), out. That matches the answer exactly.

The purpose of part (a) is to produce an exact value for \(\cos 15^\circ\), which no calculator display can give in surd form. The route is to write \(15^\circ\) as a difference of two angles whose sine and cosine are known exactly, and \(45^\circ - 30^\circ\) is the natural choice. Part (b) then feeds that exact value into the cosine rule, which is the correct rule here because two sides and the angle between them are known and the side opposite that angle is wanted; the sine rule cannot be started, since it would need a side and its opposite angle as a matched pair.

(a) Show that \(\cos 15^\circ = \dfrac{\sqrt{6} + \sqrt{2}}{4}\) [3]

The result is printed, so the marks are for the derivation. Quote the addition formula:

\[\cos(A - B) = \cos A\cos B + \sin A\sin B \qquad \textbf{[M1]}\]

The [M1] is for the correct formula, and the sign is the thing to get right: for \(\cos(A - B)\) the connective is \(+\), the opposite of the sign in the bracket. Now substitute \(A = 45^\circ\) and \(B = 30^\circ\), as the question directs:

\[\cos 15^\circ = \cos 45^\circ\cos 30^\circ + \sin 45^\circ\sin 30^\circ \qquad \textbf{[M1]}\]

The second [M1] is for this substitution. Insert the exact values \(\cos 45^\circ = \sin 45^\circ = \dfrac{\sqrt{2}}{2}\), \(\cos 30^\circ = \dfrac{\sqrt{3}}{2}\) and \(\sin 30^\circ = \dfrac{1}{2}\):

\[= \frac{\sqrt{2}}{2} \times \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \times \frac{1}{2} = \frac{\sqrt{6}}{4} + \frac{\sqrt{2}}{4} = \frac{\sqrt{6} + \sqrt{2}}{4} \qquad \textbf{[A1]}\]

The [A1] is for reaching the printed form. Note \(\sqrt{2} \times \sqrt{3} = \sqrt{6}\), using \(\sqrt{m}\sqrt{n} = \sqrt{mn}\). The wrong turn this part is built to catch is writing \(\cos 15^\circ = \cos 45^\circ - \cos 30^\circ\); the cosine of a difference is not the difference of the cosines, and a quick decimal test exposes it, since \(0.7071 - 0.8660 = -0.1589\) while \(\cos 15^\circ \approx 0.9659\).

(b) Hence find the exact value of \(BC^2\) [3]

In the triangle \(ABC\) shown, \(AB = 4\) cm and \(AC = 6\) cm are the two sides enclosing the known angle \(BAC = 15^\circ\), and \(BC\) is opposite it. That is precisely the configuration the cosine rule handles:

\[BC^2 = AB^2 + AC^2 - 2 \times AB \times AC \times \cos(BAC) \qquad \textbf{[M1]}\]

The [M1] is for a correct statement of the cosine rule with the right angle paired to the right side. Substituting, and using the exact value from part (a) rather than a decimal:

\[BC^2 = 4^2 + 6^2 - 2 \times 4 \times 6 \times \frac{\sqrt{6} + \sqrt{2}}{4} = 16 + 36 - 48 \times \frac{\sqrt{6} + \sqrt{2}}{4} \qquad \textbf{[M1]}\] \[= 52 - 12\left(\sqrt{6} + \sqrt{2}\right) = 52 - 12\sqrt{6} - 12\sqrt{2} \qquad \textbf{[A1]}\]

so \(p = 52\), \(q = -12\) and \(r = -12\). The second [M1] is for the substitution, and it is a follow-through mark: a candidate whose part (a) went wrong can still earn it by substituting their own surd correctly. The [A1] needs the three integers, and \(48 \div 4 = 12\) is the arithmetic that must not slip.

Two errors are common. The first is switching to a decimal for \(\cos 15^\circ\), which gives \(BC^2 \approx 5.63\) and cannot be written in the required form \(p + q\sqrt{6} + r\sqrt{2}\), so the accuracy mark is lost however correct the number is; the word "exact" forbids it. The second is failing to distribute the minus sign across the bracket, producing \(52 - 12\sqrt{6} + 12\sqrt{2}\). Note also that the question asks for \(BC^2\), not \(BC\), so no square root should be taken at the end.

Check: evaluate both forms numerically. \(52 - 12(2.44949) - 12(1.41421) = 52 - 29.394 - 16.971 = 5.635\), and directly \(16 + 36 - 48\cos 15^\circ = 52 - 46.365 = 5.635\). They agree, so the surd expression is right. A magnitude check also reassures: \(BC = \sqrt{5.635} \approx 2.37\) cm, which is sensibly small because the \(15^\circ\) angle is narrow, and it satisfies the triangle inequality with sides \(4\) and \(6\), since \(6 - 4 = 2 \lt 2.37 \lt 10\).