Further Pure Mathematics 4PM1 | Paper 2 Mock 01 | Written Paper 2

The three parts exercise three different uses of logarithms: treating logarithms as unknowns in a pair of simultaneous equations, combining logarithms into one before undoing them, and using a substitution to turn an exponential equation into a quadratic. All of them rest on the definition \(\log_{2} N = m\) meaning \(N = 2^{m}\).

(a) Finding \(a\) and \(b\) [4 marks]

Do not expand the logarithms into a product. Treat \(\log_{2}a\) and \(\log_{2}b\) as two unknowns in a linear system, which is what makes elimination the right method. Adding the two equations removes \(\log_{2}b\):

\[\left(\log_{2}a + \log_{2}b\right) + \left(\log_{2}a - \log_{2}b\right) = 5 + 1 \quad \Longrightarrow \quad 2\log_{2}a = 6 \quad \Longrightarrow \quad \log_{2}a = 3 \qquad \textbf{M1 A1}\]

The M1 is for a correct elimination and survives an arithmetic slip; the A1 needs the value \(3\). Converting from logarithmic to index form:

\[a = 2^{3} = 8 \qquad \textbf{A1}\]

Subtracting the equations instead removes \(\log_{2}a\):

\[2\log_{2}b = 5 - 1 = 4 \quad \Longrightarrow \quad \log_{2}b = 2 \quad \Longrightarrow \quad b = 2^{2} = 4 \qquad \textbf{A1}\]

Check in the original statements: \(\log_{2}8 + \log_{2}4 = 3 + 2 = 5\) and \(3 - 2 = 1\), both correct. The characteristic error is stopping at \(\log_{2}a = 3\) and offering that as the answer for \(a\); the question asks for \(a\) itself, so the conversion to \(8\) is required. A second error is dividing \(6\) by \(2\) after taking a logarithm of both sides, which confuses \(2\log_{2}a\) with \(\log_{2}\left(2a\right)\).

(b) Solving \(\log_{2}\left(3x + 2\right) - \log_{2}\left(x - 1\right) = 3\) [4 marks]

A difference of two logarithms with the same base is the logarithm of a quotient, so combine them into a single logarithm. That is the step that allows the logarithm to be removed:

\[\log_{2}\left(\frac{3x + 2}{x - 1}\right) = 3 \qquad \textbf{M1}\] \[\frac{3x + 2}{x - 1} = 2^{3} = 8 \qquad \textbf{M1}\]

The second M1 is for correctly undoing the logarithm; writing the right-hand side as \(3\), or as \(3^{2}\), is a common way to lose it. Multiply up and solve the linear equation:

\[3x + 2 = 8\left(x - 1\right) = 8x - 8 \quad \Longrightarrow \quad 10 = 5x \qquad \textbf{M1}\] \[x = 2 \qquad \textbf{A1}\]

A logarithm is defined only for a positive argument, so both \(3x + 2 \gt 0\) and \(x - 1 \gt 0\) are needed, which together require \(x \gt 1\). Since \(x = 2\) satisfies that, it is valid. Substituting back confirms it: \(\log_{2}8 - \log_{2}1 = 3 - 0 = 3\). The wrong turn here is subtracting the arguments rather than dividing them, that is writing \(\log_{2}\left(3x + 2 - x + 1\right)\), which destroys the equation.

(c) Solving \(2^{2x+1} - 17\left(2^{x}\right) + 8 = 0\) [5 marks]

The equation contains \(2^{x}\) and \(2^{2x+1}\), which look different but are related, so a substitution reduces it to a quadratic. Put \(y = 2^{x}\). The key manipulation is splitting the index using the laws of indices:

\[2^{2x+1} = 2^{1} \times 2^{2x} = 2 \times \left(2^{x}\right)^{2} = 2y^{2} \qquad \textbf{M1}\]

This is the mark the question is built around. Writing \(2^{2x+1}\) as \(y^{2}\), or as \(2y^{2x}\), or as \(\left(2^{x}\right)^{2x+1}\), all fail here. With that substitution:

\[2y^{2} - 17y + 8 = 0 \qquad \textbf{A1}\] \[\left(2y - 1\right)\left(y - 8\right) = 0 \quad \Longrightarrow \quad y = \frac{1}{2} \text{ or } y = 8 \qquad \textbf{M1 A1}\]

Now reverse the substitution. Because \(2^{x} \gt 0\) for every real \(x\), both roots are admissible, unlike the logarithm questions where a root is often rejected:

\[2^{x} = \frac{1}{2} = 2^{-1} \quad \Longrightarrow \quad x = -1; \qquad 2^{x} = 8 = 2^{3} \quad \Longrightarrow \quad x = 3 \qquad \textbf{A1}\]

Both must be given: offering only \(x = 3\) loses the final accuracy mark. Verify the awkward one directly. With \(x = -1\), \(2^{2x+1} = 2^{-1} = \tfrac{1}{2}\) and \(2^{x} = \tfrac{1}{2}\), so the expression is \(\tfrac{1}{2} - \tfrac{17}{2} + 8 = 0\). With \(x = 3\), \(2^{7} = 128\) and \(128 - 17\left(8\right) + 8 = 128 - 136 + 8 = 0\). The final trap is forgetting to return from \(y\) to \(x\) and quoting \(\tfrac{1}{2}\) and \(8\) as the answers.

Examination takeaway. Combine logarithms into a single one before removing them, and always convert a result such as \(\log_{2}a = 3\) into the value of \(a\). In an exponential equation, look for a term whose index is double another and split it with the index laws, remembering that a substituted exponential is always positive so neither root of the quadratic can be dismissed on those grounds.

This is the standard method for finding where two lines meet in vector form: express the same point by two different routes and equate. The statement that \(\mathbf{a}\) and \(\mathbf{c}\) are not parallel is not incidental. It is what makes them a basis for the plane, so that a vector has only one expression in terms of them and the coefficients can be compared. Without that condition the whole method collapses.

(a) The two diagonals [2 marks]

Build each vector by walking along routes that are already known.

(i) From \(O\) to \(B\) via \(A\):

\[\overrightarrow{OB} = \overrightarrow{OA} + \overrightarrow{AB} = \mathbf{a} + 3\mathbf{c} \qquad \textbf{[1]}\]

(ii) From \(A\) to \(C\) via \(O\), remembering that reversing a vector reverses its sign:

\[\overrightarrow{AC} = \overrightarrow{AO} + \overrightarrow{OC} = -\mathbf{a} + \mathbf{c} = \mathbf{c} - \mathbf{a} \qquad \textbf{[1]}\]

Each is one independent mark. The order matters: \(\mathbf{a} - \mathbf{c}\) is \(\overrightarrow{CA}\) and does not score. Note that \(\overrightarrow{AB} = 3\mathbf{c}\) is given, so \(AB\) is parallel to \(OC\) and three times as long, which is what makes \(OABC\) a trapezium and explains the answer to part (c) in advance.

(b) Showing \(\lambda = \dfrac{1}{4}\) and finding \(\mu\) [4 marks]

The point \(X\) lies on both diagonals, so reach it twice. Along \(OB\):

\[\overrightarrow{OX} = \lambda\,\overrightarrow{OB} = \lambda\left(\mathbf{a} + 3\mathbf{c}\right) = \lambda\mathbf{a} + 3\lambda\mathbf{c} \qquad \textbf{M1}\]

Along \(AC\), starting from \(A\) and therefore adding the position vector of \(A\) first:

\[\overrightarrow{OX} = \overrightarrow{OA} + \mu\,\overrightarrow{AC} = \mathbf{a} + \mu\left(\mathbf{c} - \mathbf{a}\right) = \left(1 - \mu\right)\mathbf{a} + \mu\mathbf{c} \qquad \textbf{M1}\]

These two M marks are for the two routes and survive the algebra that follows. The commonest error is writing the second route as \(\mu\left(\mathbf{c} - \mathbf{a}\right)\) alone, forgetting that \(\overrightarrow{AX}\) is measured from \(A\) and so must be added to \(\overrightarrow{OA}\) to give a position vector from \(O\).

Because \(\mathbf{a}\) and \(\mathbf{c}\) are not parallel, the two expressions can only agree if the coefficients of \(\mathbf{a}\) match and the coefficients of \(\mathbf{c}\) match separately:

\[\lambda = 1 - \mu, \qquad 3\lambda = \mu \qquad \textbf{M1}\]

This is the step to state explicitly, together with its justification, since the third M mark is for equating components. Substituting the second equation into the first:

\[\lambda = 1 - 3\lambda \quad \Longrightarrow \quad 4\lambda = 1 \quad \Longrightarrow \quad \lambda = \frac{1}{4}, \qquad \mu = 3\lambda = \frac{3}{4} \qquad \textbf{A1}\]

The value of \(\lambda\) is printed in the question, so it functions as a check: a candidate who obtains anything else knows at once that a slip has occurred, and one who writes \(\lambda = \tfrac{1}{4}\) down without the two routes earns nothing. The single A1 covers both constants, so \(\mu\) must be right as well. Substituting into either expression gives the same point, which is the final check:

\[\tfrac{1}{4}\mathbf{a} + \tfrac{3}{4}\mathbf{c} \quad \text{and} \quad \left(1 - \tfrac{3}{4}\right)\mathbf{a} + \tfrac{3}{4}\mathbf{c} = \tfrac{1}{4}\mathbf{a} + \tfrac{3}{4}\mathbf{c}.\]

(c) The ratio \(AX : XC\) and the position vector of \(X\) [2 marks]

Since \(\overrightarrow{AX} = \dfrac{3}{4}\overrightarrow{AC}\), the point \(X\) lies three quarters of the way from \(A\) to \(C\). The remaining quarter is \(XC\), so

\[AX : XC = \frac{3}{4} : \frac{1}{4} = 3 : 1 \qquad \textbf{[1]}\]

The trap is quoting \(3 : 4\), which is the ratio \(AX : AC\) rather than \(AX : XC\). The part after the colon is the remaining piece, not the whole. From part (b), using \(\lambda = \dfrac{1}{4}\):

\[\overrightarrow{OX} = \tfrac{1}{4}\left(\mathbf{a} + 3\mathbf{c}\right) = \tfrac{1}{4}\mathbf{a} + \tfrac{3}{4}\mathbf{c} \qquad \textbf{[1]}\]

The two answers are consistent with the geometry. Because \(AB\) is three times \(OC\) and parallel to it, triangles \(OXC\) and \(BXA\) are similar with a scale factor of \(3\), so the diagonals cut each other in the ratio \(3 : 1\) with the longer part next to the longer parallel side. That similar-triangles view is a quick way to predict the answer before doing any algebra, and a useful way to confirm it afterwards. It also explains \(\lambda = \dfrac{1}{4}\): \(X\) is one quarter of the way along \(OB\) from \(O\), the same quarter that \(OC\) represents of \(OC\) plus \(AB\) taken together.

Examination takeaway. To find an intersection in vector form, write the same position vector by two routes and equate the coefficients of the two non-parallel base vectors, always adding the position vector of the starting point when a displacement begins away from the origin. When converting a fraction of a segment into a ratio, remember that the second number is what is left over.

This question tests the second Pythagorean identity, the one connecting the cotangent and the cosecant, and then asks you to read the number of solutions off the printed graph rather than count them by trial. The graph is not decoration: part (b) explicitly requires it as the justification.

(a) Showing that the equation may be written as \(2\operatorname{cosec}^{2}x + 5\operatorname{cosec} x - 12 = 0\) [2 marks]

The equation \(2\cot^{2}x + 5\operatorname{cosec} x = 10\) contains two different functions, so it cannot be solved as it stands. Dividing \(\sin^{2}x + \cos^{2}x \equiv 1\) through by \(\sin^{2}x\) gives \(1 + \cot^{2}x \equiv \operatorname{cosec}^{2}x\), which rearranges to

\[\cot^{2}x \equiv \operatorname{cosec}^{2}x - 1.\]

That is the identity to use, because it converts the squared term into the same function as the linear term. Substituting:

\[2\left(\operatorname{cosec}^{2}x - 1\right) + 5\operatorname{cosec} x = 10 \qquad \textbf{M1}\] \[2\operatorname{cosec}^{2}x - 2 + 5\operatorname{cosec} x = 10 \quad \Longrightarrow \quad 2\operatorname{cosec}^{2}x + 5\operatorname{cosec} x - 12 = 0 \qquad \textbf{A1}\]

Since the result is printed, the M1 is for quoting and using the correct identity and the A1 for reaching the given form exactly, constant term included. The characteristic error is using the wrong Pythagorean relation, for instance \(\cot^{2}x = 1 - \operatorname{cosec}^{2}x\), which flips a sign and gives \(-2\operatorname{cosec}^{2}x + 5\operatorname{cosec} x - 8 = 0\). The other is arithmetic: the constant is \(-2 - 10 = -12\), not \(-8\).

(b) Solving the equation and explaining why there are exactly four solutions [4 marks]

Treat \(\operatorname{cosec} x\) as a single unknown; the expression is a quadratic in it and factorises:

\[\left(2\operatorname{cosec} x - 3\right)\left(\operatorname{cosec} x + 4\right) = 0 \qquad \textbf{M1}\]

The M1 is for a correct method of solution, so the quadratic formula earns it equally, and it is a follow-through from part (a). The two roots are \(\operatorname{cosec} x = \dfrac{3}{2}\) and \(\operatorname{cosec} x = -4\). Now invert, since \(\operatorname{cosec} x = \dfrac{1}{\sin x}\) and so \(\sin x = \dfrac{1}{\operatorname{cosec} x}\):

\[\sin x = \frac{2}{3} \qquad \text{or} \qquad \sin x = -\frac{1}{4} \qquad \textbf{A1}\]

Both are admissible because each has modulus at most \(1\). This inversion is where marks are most often lost: writing \(\sin x = \tfrac{3}{2}\), which is impossible, or \(\sin x = -4\), reverses the reciprocal. A useful guard is that the cosecant never takes a value strictly between \(-1\) and \(1\), so a root of the quadratic in that band would have to be rejected, whereas here both roots are outside it and both survive.

For \(\sin x = \dfrac{2}{3}\) the sine is positive, so \(x\) is in the first or second quadrant. The principal value is \(41.81^{\circ}\) and its partner is \(180^{\circ} - 41.81^{\circ}\):

\[x = 41.8^{\circ},\ 138.2^{\circ} \qquad \textbf{A1}\]

For \(\sin x = -\dfrac{1}{4}\) the sine is negative, so \(x\) is in the third or fourth quadrant. With an acute reference angle of \(14.48^{\circ}\), the solutions are \(180^{\circ} + 14.48^{\circ}\) and \(360^{\circ} - 14.48^{\circ}\):

\[x = 194.5^{\circ},\ 345.5^{\circ} \qquad \textbf{A1}\]

All four values are given to one decimal place as instructed, and all four lie strictly inside \(0^{\circ} \lt x \lt 360^{\circ}\). Note that \(x = 180^{\circ}\) is excluded in the question because \(\operatorname{cosec} x\) is undefined there, which is why the interval is written the way it is.

Reading the count from Figure 3. The explanation is part of what these four marks pay for, so a bare list of values is not a complete answer to the part. Solving \(\operatorname{cosec} x = \tfrac{3}{2}\) and \(\operatorname{cosec} x = -4\) amounts to intersecting the curve \(C\) with the two horizontal lines \(y = \tfrac{3}{2}\) and \(y = -4\). Figure 3 shows that the branch on \(0^{\circ} \lt x \lt 180^{\circ}\) lies entirely at or above \(y = 1\), dipping to a minimum of \(1\), so a line at height \(\tfrac{3}{2}\) lies above that minimum and cuts the branch twice, while a line at \(-4\) misses it altogether. The branch on \(180^{\circ} \lt x \lt 360^{\circ}\) lies entirely at or below \(y = -1\), rising to a maximum of \(-1\), so the line \(y = -4\) cuts it twice and the line \(y = \tfrac{3}{2}\) misses it. Two intersections plus two intersections gives exactly four solutions, which is precisely the list found algebraically.

That is also the check on the algebra. If the working had produced a value of \(\operatorname{cosec} x\) between \(-1\) and \(1\), the graph would show no intersection at all at that height, revealing the slip at once.

Examination takeaway. Learn \(1 + \cot^{2}x \equiv \operatorname{cosec}^{2}x\) alongside the more familiar Pythagorean identity, and take the reciprocal carefully when returning from the cosecant to the sine. When a printed graph is offered, use it to justify the number of solutions: the horizontal line \(y = k\) meets the curve as often as the equation has roots.

(a) Equations of the two asymptotes of \(C\) [2]

An asymptote is a line the curve approaches without ever meeting. For \(y = \dfrac{12}{x}\) there are two:

• As \(x\) approaches \(0\) the value of \(\dfrac{12}{x}\) grows without bound, so the vertical asymptote is \(x = 0\), the \(y\) axis. [B1]
• As \(x\) grows large in either direction \(\dfrac{12}{x}\) approaches \(0\), so the horizontal asymptote is \(y = 0\), the \(x\) axis. [B1]

Two independent B marks. Both answers must be given as equations of lines: writing "the \(x\) axis and the \(y\) axis" in words, or writing \(0\) alone, does not satisfy "write down the equations". The exclusion \(x \neq 0\) in the question is the hint for the vertical one.

(b) Show that the \(x\) coordinates of \(A\) and \(B\) satisfy \(x^{2} - 8x + 12 = 0\) [2]

At a point of intersection the two \(y\) values agree, so equate the equations of \(C\) and \(l\):

\[\frac{12}{x} = 8 - x \quad \textbf{[M1]}\]

Clear the fraction by multiplying every term by \(x\). This is legitimate because \(x \neq 0\) on the curve, and saying so is good practice since multiplying by a possibly zero quantity can introduce spurious roots:

\[12 = 8x - x^{2} \quad \Rightarrow \quad x^{2} - 8x + 12 = 0 \quad \textbf{[A1]}\]

The target is printed, so both marks are for the derivation and quoting the equation earns nothing. Multiplying only the left side by \(x\), or dropping the \(x\) from \(8x\), are the slips to guard against.

(c) Coordinates of \(A\) and of \(B\) [2]

The quadratic factorises, since \(-2\) and \(-6\) multiply to \(12\) and add to \(-8\):

\[(x - 2)(x - 6) = 0 \quad \Rightarrow \quad x = 2 \ \text{ or } \ x = 6 \quad \textbf{[M1]}\]

Find each \(y\) from either equation, the line being the simpler: at \(x = 2\), \(y = 8 - 2 = 6\); at \(x = 6\), \(y = 8 - 6 = 2\). The question states that \(A\) has the smaller \(x\) coordinate, so

\[A(2,\, 6) \qquad \text{and} \qquad B(6,\, 2) \quad \textbf{[A1]}\]

Assigning the labels the other way round is the error the question's own wording is there to prevent, and the accuracy mark depends on getting them the right way about.

(d) Exact length of \(AB\) in the form \(k\sqrt{2}\) [3]

Use the distance formula, which is Pythagoras applied to the horizontal and vertical separations:

\[AB^{2} = (6 - 2)^{2} + (2 - 6)^{2} \quad \textbf{[M1]}\] \[AB^{2} = 4^{2} + (-4)^{2} = 16 + 16 = 32 \quad \textbf{[M1]}\]

Now simplify the surd by extracting the largest square factor, \(32 = 16 \times 2\):

\[AB = \sqrt{32} = \sqrt{16}\sqrt{2} = 4\sqrt{2} \quad \textbf{[A1]}\]

The two method marks survive arithmetic slips; the accuracy mark requires the exact surd form, so \(5.66\) does not answer the question as set. Note that squaring removes the sign, so \((2 - 6)^{2} = 16\), not \(-16\); a negative contribution under the root is a sure sign of an error.

Checks. Both points must lie on the curve as well as the line: \(\dfrac{12}{2} = 6\) and \(\dfrac{12}{6} = 2\), so \(A(2,\,6)\) and \(B(6,\,2)\) are correct. The symmetry is a further check: \(A\) and \(B\) are reflections of each other in the line \(y = x\), which is expected because both \(y = \dfrac{12}{x}\) and \(y = 8 - x\) are unchanged when \(x\) and \(y\) are swapped. Finally, since the line \(l\) has gradient \(-1\), the horizontal and vertical separations between \(A\) and \(B\) must be equal, and both are \(4\), which is exactly why the answer comes out as a multiple of \(\sqrt{2}\).

This question is about symmetric functions of the roots of a quadratic. The instruction "Without solving the equation" in part (b) is the key: the roots of \(2x^{2} - 5x + 4 = 0\) are not real, since the discriminant is \(25 - 32 = -7\), so any attempt to find them individually would fail. Yet the sum and the product of the roots are perfectly ordinary numbers, and every symmetric expression in the roots can be built from those two.

(a) The sum and product of the roots [2 marks]

For \(ax^{2} + bx + c = 0\) with roots \(\alpha\) and \(\beta\):

\[\alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a}.\]

Here \(a = 2\), \(b = -5\) and \(c = 4\), so

\[\alpha + \beta = -\frac{-5}{2} = \frac{5}{2} \qquad \textbf{B1}\] \[\alpha\beta = \frac{4}{2} = 2 \qquad \textbf{B1}\]

These are independent B marks, one each, so a slip in one does not cost the other. The sign is the trap in the sum: the formula carries a minus, and \(b\) is itself negative, so the two negatives give a positive \(\dfrac{5}{2}\). Writing \(-\dfrac{5}{2}\) is the commonest error, and it propagates through both remaining parts. The other frequent slip is forgetting to divide by \(a\), giving \(5\) and \(4\) instead.

(b) The value of \(\alpha^{2} + \beta^{2}\) [2 marks]

Squaring the sum produces the wanted squares plus an unwanted cross term:

\[\left(\alpha + \beta\right)^{2} = \alpha^{2} + 2\alpha\beta + \beta^{2},\]

so subtracting twice the product isolates what is needed. This identity is the whole method:

\[\alpha^{2} + \beta^{2} = \left(\alpha + \beta\right)^{2} - 2\alpha\beta \qquad \textbf{M1}\]

The M1 is for quoting a correct identity and stands even if the substituted values are wrong, which is why the identity should be written before the numbers go in. Substituting:

\[\alpha^{2} + \beta^{2} = \left(\frac{5}{2}\right)^{2} - 2\left(2\right) = \frac{25}{4} - 4 = \frac{25 - 16}{4} = \frac{9}{4} \qquad \textbf{A1}\]

The wrong turn here is writing \(\alpha^{2} + \beta^{2} = \left(\alpha + \beta\right)^{2}\), that is \(\dfrac{25}{4}\), which ignores the cross term entirely. A second error is subtracting \(\alpha\beta\) rather than \(2\alpha\beta\). Notice that the answer is positive even though the roots themselves are not real; a sum of squares of complex conjugates can be positive, negative or zero, so a positive value here is no cause for alarm.

(c) The value of \(\alpha^{3} + \beta^{3}\) [2 marks]

The same strategy applies one degree higher. Cubing the sum gives

\[\left(\alpha + \beta\right)^{3} = \alpha^{3} + 3\alpha^{2}\beta + 3\alpha\beta^{2} + \beta^{3} = \alpha^{3} + \beta^{3} + 3\alpha\beta\left(\alpha + \beta\right),\]

since the two middle terms share the factor \(3\alpha\beta\). Rearranging:

\[\alpha^{3} + \beta^{3} = \left(\alpha + \beta\right)^{3} - 3\alpha\beta\left(\alpha + \beta\right) \qquad \textbf{M1}\]

Substituting the two known values:

\[\alpha^{3} + \beta^{3} = \left(\frac{5}{2}\right)^{3} - 3\left(2\right)\left(\frac{5}{2}\right) = \frac{125}{8} - 15 = \frac{125 - 120}{8} = \frac{5}{8} \qquad \textbf{A1}\]

An equally valid route uses the factorisation \(\alpha^{3} + \beta^{3} = \left(\alpha + \beta\right)\left(\alpha^{2} - \alpha\beta + \beta^{2}\right)\) together with part (b), giving \(\dfrac{5}{2}\left(\dfrac{9}{4} - 2\right) = \dfrac{5}{2} \times \dfrac{1}{4} = \dfrac{5}{8}\), and the agreement of the two methods is a good check. The errors to watch are cubing only the numerator, so that \(\left(\dfrac{5}{2}\right)^{3}\) becomes \(\dfrac{125}{2}\), and dropping the factor \(\left(\alpha + \beta\right)\) from the correction term, which would give \(\dfrac{125}{8} - 6\).

It is worth noting how small the final answer is compared with \(\dfrac{125}{8} \approx 15.6\). The two terms nearly cancel, which means an arithmetic slip anywhere produces a wildly wrong result rather than a slightly wrong one. Keeping everything over the denominator \(8\) rather than converting to decimals is the safest way through.

Examination takeaway. Read the sum and product of the roots straight off the coefficients, remembering the minus sign on the sum and the division by \(a\). Then express any symmetric function of the roots in terms of those two quantities using an identity written down before substitution. This works whether or not the roots are real, which is exactly why the question forbids solving the equation.

(a) \(\dfrac{dy}{dx}\) and the gradient at \(P\) [2]

The gradient of a curve at a point is the value of its derivative there, so differentiate \(y = x^{2} - 6x + 13\) and substitute \(x = 4\).

\[\frac{dy}{dx} = 2x - 6 \quad \textbf{[M1]}\] \[\text{At } x = 4: \quad \frac{dy}{dx} = 8 - 6 = 2,\]

so the gradient of \(C\) at \(P\) is \(2\). [A1] The M mark for a correct derivative survives an evaluation slip; the A mark is for the value \(2\).

(b) Show that the tangent at \(P\) has equation \(y = 2x - 3\) [3]

A straight line needs a gradient and a point, so first find the \(y\) coordinate of \(P\) from the equation of the curve:

\[\text{At } x = 4: \quad y = 16 - 24 + 13 = 5, \quad \text{so } P(4,\, 5). \quad \textbf{[B1]}\]

This is a B mark, given on sight for the correct point. Then use \(y - y_{1} = m(x - x_{1})\) with \(m = 2\):

\[y - 5 = 2(x - 4) \quad \textbf{[M1]}\] \[y = 2x - 8 + 5 = 2x - 3 \quad \textbf{[A1]}\]

The equation is printed in the question, so the marks are for the derivation. Do not substitute \(x = 4\) into the derivative to get the \(y\) coordinate: the derivative gives the gradient, the curve gives the height.

(c) Equation of the normal at \(P\) in the form \(ax + by + c = 0\) [3]

The normal is perpendicular to the tangent, and perpendicular gradients multiply to \(-1\), so the normal has gradient

\[m_{n} = -\frac{1}{2} \quad \textbf{[M1]}\] \[y - 5 = -\frac{1}{2}(x - 4) \quad \textbf{[M1]}\]

Multiply through by \(2\) to clear the fraction and collect everything on one side, because the question demands integer coefficients:

\[2y - 10 = -(x - 4) = -x + 4 \quad \Rightarrow \quad x + 2y - 14 = 0 \quad \textbf{[A1]}\]

Both method marks are independent of the arithmetic; the accuracy mark needs the required form with integers, so \(y = -0.5x + 7\) does not satisfy the instruction.

(d) Area of triangle \(ABP\) [2]

The two lines cross the \(y\) axis at \(A\) and \(B\), so put \(x = 0\) in each equation:

• Tangent: \(y = 2(0) - 3 = -3\), so \(A(0,\, -3)\).
• Normal: \(0 + 2y - 14 = 0 \Rightarrow y = 7\), so \(B(0,\, 7)\).

Because \(A\) and \(B\) both lie on the \(y\) axis, \(AB\) is a vertical segment and is the natural base:

\[AB = 7 - (-3) = 10.\]

The corresponding perpendicular height is the horizontal distance from \(P\) to the \(y\) axis, which is simply the \(x\) coordinate of \(P\), namely \(4\). [M1]

\[\text{Area} = \frac{1}{2} \times 10 \times 4 = 20 \ \text{square units} \quad \textbf{[A1]}\]

Common wrong turns on this question. Taking the gradient of the normal as \(-2\) or as \(\dfrac{1}{2}\) rather than the negative reciprocal \(-\dfrac{1}{2}\). Computing \(AB\) as \(7 + (-3) = 4\) instead of subtracting the coordinates. In part (d), using the length \(AP\) or \(BP\) as the height: the height must be measured perpendicular to the chosen base, and here the base is vertical, so the height is horizontal and equals \(4\). Using \(\dfrac{1}{2}ab\sin C\) with the angle at \(P\) also works but is far more effort, and the right angle at \(P\) gives a third valid route: \(AP = \sqrt{16 + 64} = 4\sqrt{5}\), \(BP = \sqrt{16 + 4} = 2\sqrt{5}\), and \(\dfrac{1}{2} \times 4\sqrt{5} \times 2\sqrt{5} = 20\).

Check. That third route agreeing with the base-and-height calculation is the strongest available check, and it also confirms the perpendicularity: the tangent and normal meet at right angles at \(P\), so \(AP\) and \(BP\) are the two legs of a right-angled triangle. A quick reading off the figure also confirms \(A\) below the origin and \(B\) above it.

This is a standard optimisation chain: use the fixed volume to eliminate one variable from the surface area, differentiate, solve for the stationary point, and test its nature with the second derivative. The reason a constraint is needed is that \(S\) depends on both \(x\) and \(h\), and calculus in one variable requires a function of \(x\) alone.

(a) Showing that \(S = 2x^{2} + \dfrac{108}{x}\) [4 marks]

The tank is a cuboid of base \(2x\) by \(x\) and height \(h\), so its volume is the product of the three dimensions. The given volume of \(36\,\mathrm{m}^{3}\) is the constraint:

\[2x \times x \times h = 2x^{2}h = 36 \quad \Longrightarrow \quad h = \frac{36}{2x^{2}} = \frac{18}{x^{2}} \qquad \textbf{M1 A1}\]

The M1 is for forming a correct volume equation and the A1 for rearranging it to give \(h\) in terms of \(x\). Now the surface area. The top is open, so there is one base of area \(2x \times x = 2x^{2}\), plus four vertical faces: two of them measure \(2x\) by \(h\) and two measure \(x\) by \(h\):

\[S = 2x^{2} + 2\left(2xh\right) + 2\left(xh\right) = 2x^{2} + 6xh \qquad \textbf{M1}\]

Substituting the expression for \(h\) eliminates it:

\[S = 2x^{2} + 6x \times \frac{18}{x^{2}} = 2x^{2} + \frac{108}{x} \qquad \textbf{A1}\]

as required. Since the result is printed, the marks belong to the derivation. Two errors are frequent. The first is including a top, which adds a second \(2x^{2}\) and gives \(4x^{2} + \dfrac{108}{x}\); the phrase "open top" and the request for the area "of the base and the four vertical faces" both rule it out. The second is counting only two vertical faces, or taking all four as \(2x\) by \(h\), which gives \(8xh\) instead of \(6xh\). Check the units: \(2x^{2}\) and \(\dfrac{108}{x}\) are both areas in \(\mathrm{m}^{2}\) because the \(108\) carries units of \(\mathrm{m}^{3}\).

(b) The first and second derivatives [3 marks]

Rewrite the fraction as a negative power so the power rule applies:

\[S = 2x^{2} + 108x^{-1}\] \[\frac{\mathrm{d}S}{\mathrm{d}x} = 4x - 108x^{-2} = 4x - \frac{108}{x^{2}} \qquad \textbf{M1 A1}\] \[\frac{\mathrm{d}^{2}S}{\mathrm{d}x^{2}} = 4 + 216x^{-3} = 4 + \frac{216}{x^{3}} \qquad \textbf{A1}\]

The M1 is for differentiating a negative power at all and survives a coefficient slip; the two A1 marks require each derivative to be exactly right. Watch the signs: differentiating \(108x^{-1}\) gives \(-108x^{-2}\), and differentiating that in turn gives \(+216x^{-3}\), because multiplying by the index \(-2\) turns the negative coefficient positive. Losing that second sign change is the usual route to a wrong conclusion in part (d).

(c) The stationary value [4 marks]

A stationary value occurs where the first derivative is zero, which is the definition to quote:

\[4x - \frac{108}{x^{2}} = 0 \qquad \textbf{M1}\]

Multiply through by \(x^{2}\), legitimate because \(x \gt 0\) for a physical tank:

\[4x^{3} = 108 \quad \Longrightarrow \quad x^{3} = 27 \qquad \textbf{M1}\] \[x = 3 \qquad \textbf{A1}\]

Only the real cube root is relevant, and \(x = 3\) is the value the question states, so this part is a "show that" as far as \(x\) is concerned. Substituting back into the expression for \(S\), not into a derivative:

\[S = 2\left(3\right)^{2} + \frac{108}{3} = 18 + 36 = 54 \qquad \textbf{A1}\]

so the stationary value is \(S = 54\,\mathrm{m}^{2}\). The commonest loss is substituting \(x = 3\) into \(\dfrac{\mathrm{d}S}{\mathrm{d}x}\) and reporting zero, which answers a different question.

(d) Maximum or minimum [2 marks]

Evaluate the second derivative at the stationary point:

\[\left.\frac{\mathrm{d}^{2}S}{\mathrm{d}x^{2}}\right|_{x=3} = 4 + \frac{216}{27} = 4 + 8 = 12 \qquad \textbf{M1}\] \[12 \gt 0, \quad \text{so } S = 54 \text{ is a minimum} \qquad \textbf{A1}\]

The M1 is for substituting into the second derivative, the A1 for the correct conclusion drawn from its sign. A positive second derivative means the gradient is increasing, so the curve is concave upwards and the stationary point is a minimum. The A1 requires the sign to be stated as the reason; simply asserting "minimum" does not show the test was used. The answer also makes physical sense: with a fixed volume, a very thin tall tank and a very wide flat tank both have huge surface area, so the stationary value between them must be the least material, never the most.

(e) The height [1 mark]

Use the constraint from part (a) with \(x = 3\):

\[h = \frac{18}{3^{2}} = \frac{18}{9} = 2\,\mathrm{m} \qquad \textbf{B1}\]

This is a B mark for the correct value, and the unit should be given. Check the whole answer against both original conditions: the volume is \(2\left(3\right)\left(3\right)\left(2\right) = 36\,\mathrm{m}^{3}\), and the surface area is \(2\left(9\right) + 6\left(3\right)\left(2\right) = 18 + 36 = 54\,\mathrm{m}^{2}\). Both match, so the optimisation is confirmed.

Examination takeaway. In an optimisation question, use the fixed quantity to express the redundant variable in terms of the one you will differentiate, and read the surface description carefully, since an open top removes a face. Substitute the optimal value into the original function for the stationary value, and always quote the sign of the second derivative as the reason for calling it a maximum or a minimum.

(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.

(a) Value of \(\dfrac{dx}{dt}\) [1]

The edge length increases at the constant rate \(0.02\) cm s\(^{-1}\), and a rate of change of \(x\) with respect to time is precisely \(\dfrac{dx}{dt}\), so

\[\frac{dx}{dt} = 0.02 \quad \textbf{[B1]}\]

A B mark: it is awarded for the correct value alone and is independent of every other part.

(b) \(V\) in terms of \(x\), and \(\dfrac{dV}{dx}\) [2]

A cube of edge \(x\) has volume

\[V = x^{3}, \qquad \text{so} \qquad \frac{dV}{dx} = 3x^{2} \quad \textbf{[B1][B1]}\]

Two independent B marks, one for each result.

(c) Rate of increase of the volume when \(x = 5\) [3]

The question links a rate with respect to time to a rate with respect to length, which is exactly what the chain rule is for. Nothing gives \(V\) as a function of \(t\) directly, so connect the two derivatives already found:

\[\frac{dV}{dt} = \frac{dV}{dx} \times \frac{dx}{dt} = 3x^{2} \times 0.02 = 0.06x^{2} \quad \textbf{[M1][A1]}\] \[\text{At } x = 5: \quad \frac{dV}{dt} = 0.06 \times 25 = 1.5 \ \mathrm{cm^{3}\,s^{-1}} \quad \textbf{[A1]}\]

The M mark is for a correct chain-rule statement and survives an arithmetic slip; the two accuracy marks require the expression and then the value with its unit. Writing the chain the wrong way round, as \(\dfrac{dV}{dx} \div \dfrac{dx}{dt}\), is the standard error, and a unit check catches it: cm\(^{2}\) multiplied by cm s\(^{-1}\) gives cm\(^{3}\) s\(^{-1}\), which is a volume rate, whereas dividing does not.

(d) Show that \(\dfrac{dS}{dt} = 0.24x\) [4]

A cube has six congruent square faces, each of area \(x^{2}\), so

\[S = 6x^{2} \quad \textbf{[B1]}\] \[\frac{dS}{dx} = 12x \quad \textbf{[B1]}\]

Then apply the chain rule again:

\[\frac{dS}{dt} = \frac{dS}{dx} \times \frac{dx}{dt} \quad \textbf{[M1]}\] \[= 12x \times 0.02 = 0.24x \quad \textbf{[A1]}\]

The result is given, so all four marks are for the derivation; a candidate who writes \(\dfrac{dS}{dt} = 0.24x\) and stops earns nothing. The two B marks are independent, the M mark is for the chain-rule structure, and the final A mark requires the exact constant. Using \(S = x^{2}\) or \(S = 4x^{2}\) rather than \(6x^{2}\) is the wrong turn here: a solid cube has a top and a bottom as well as four sides.

(e) Rate of increase of the surface area when \(V = 216\) cm\(^{3}\) [3]

The condition is given as a volume, but the formula from part (d) needs \(x\), so convert first:

\[V = 216 \ \Rightarrow \ x^{3} = 216 \ \Rightarrow \ x = 6 \ \mathrm{cm} \quad \textbf{[M1][A1]}\] \[\frac{dS}{dt} = 0.24 \times 6 = 1.44 \ \mathrm{cm^{2}\,s^{-1}} \quad \textbf{[A1]}\]

The M mark is for recognising that \(x^{3} = 216\) must be solved, and it survives a slip in the cube root; the final accuracy mark requires the value with its unit.

The wrong turn this part catches. Substituting \(216\) straight into \(\dfrac{dS}{dt} = 0.24x\) to get \(51.84\). The variable in that formula is the edge length, not the volume, and mixing the two is the single most common loss on this question. A magnitude check settles it: the edge of the cube is only \(6\) cm and it grows at \(0.02\) cm s\(^{-1}\), so a surface-area rate of \(51.84\) cm\(^{2}\) s\(^{-1}\) is far too large for so slow a growth.

Further check. At \(x = 6\) the volume rate is \(0.06 \times 36 = 2.16\) cm\(^{3}\) s\(^{-1}\), which exceeds the value \(1.5\) cm\(^{3}\) s\(^{-1}\) found at \(x = 5\); both rates should increase with \(x\), and they do.

Part (a) establishes an identity, part (b) uses it to factorise an equation, and part (c) asks for an argument rather than a calculation. The instruction "Without carrying out any further solution" in part (c) means the marks are for reasoning from part (a), and any attempt to solve the equation from scratch misses the point.

(a) Showing that \(\tan^{2}\theta - \sin^{2}\theta \equiv \tan^{2}\theta\sin^{2}\theta\) [4 marks]

The right-hand side is a product, so work on the left and aim to factorise. The first move is to express the tangent in terms of sine and cosine, since that is the only way to combine two functions that are currently unrelated:

\[\tan^{2}\theta - \sin^{2}\theta = \frac{\sin^{2}\theta}{\cos^{2}\theta} - \sin^{2}\theta \qquad \textbf{M1}\]

Put both terms over the common denominator \(\cos^{2}\theta\) and take out the common factor \(\sin^{2}\theta\):

\[= \frac{\sin^{2}\theta - \sin^{2}\theta\cos^{2}\theta}{\cos^{2}\theta} = \frac{\sin^{2}\theta\left(1 - \cos^{2}\theta\right)}{\cos^{2}\theta} \qquad \textbf{M1}\]

Now the Pythagorean identity \(\sin^{2}\theta + \cos^{2}\theta \equiv 1\), rearranged as \(1 - \cos^{2}\theta \equiv \sin^{2}\theta\), replaces the bracket:

\[= \frac{\sin^{2}\theta}{\cos^{2}\theta} \times \sin^{2}\theta \qquad \textbf{M1}\] \[= \tan^{2}\theta\sin^{2}\theta \qquad \textbf{A1}\]

as required. Since the identity is printed, all four marks are for the derivation. The three M marks are for the three distinct ideas, converting the tangent, forming a single fraction and factorising, and applying the Pythagorean identity, so partial credit is available even if the work stalls; the A1 is for a complete and correct chain ending in the printed form. Two errors are common: writing \(\tan^{2}\theta = \dfrac{\sin^{2}\theta}{\cos^{2}\theta}\) but then failing to give the second term the same denominator, and cancelling \(\sin^{2}\theta\) from numerator and denominator, which is impossible since the denominator contains only \(\cos^{2}\theta\).

(b) Solving \(\tan^{2}\theta - \sin^{2}\theta = 3\sin^{2}\theta\) for \(0^{\circ} \le \theta \le 360^{\circ}\) [4 marks]

Replace the left-hand side by the product from part (a), then bring everything to one side and factorise rather than dividing:

\[\tan^{2}\theta\sin^{2}\theta = 3\sin^{2}\theta \quad \Longrightarrow \quad \sin^{2}\theta\left(\tan^{2}\theta - 3\right) = 0 \qquad \textbf{M1}\]

This is the mark the question is built around. Dividing both sides by \(\sin^{2}\theta\) is the wrong turn: \(\sin^{2}\theta\) may be zero, and dividing by it silently destroys a whole family of solutions. Factorising keeps both cases. The first factor gives

\[\sin\theta = 0: \quad \theta = 0^{\circ},\ 180^{\circ},\ 360^{\circ} \qquad \textbf{A1}\]

and all three lie in the closed interval, so all three count. The second factor gives

\[\tan^{2}\theta = 3 \quad \Longrightarrow \quad \tan\theta = \pm\sqrt{3} \qquad \textbf{M1}\]

Taking both square roots is essential. Since the tangent has period \(180^{\circ}\), the positive case gives \(60^{\circ}\) and \(240^{\circ}\), and the negative case gives \(120^{\circ}\) and \(300^{\circ}\):

\[\theta = 60^{\circ},\ 120^{\circ},\ 240^{\circ},\ 300^{\circ} \qquad \textbf{A1}\]

There are therefore seven solutions in all. Check the least obvious one: at \(\theta = 180^{\circ}\), \(\tan\theta = 0\) and \(\sin\theta = 0\), so the equation reads \(0 - 0 = 3 \times 0\), which is true. Check one of the others: at \(\theta = 120^{\circ}\), \(\tan^{2}\theta = 3\) and \(\sin^{2}\theta = \tfrac{3}{4}\), so the left side is \(3 \times \tfrac{3}{4} = \tfrac{9}{4}\) by part (a) and the right side is \(3 \times \tfrac{3}{4} = \tfrac{9}{4}\). Note that \(\theta = 90^{\circ}\) and \(\theta = 270^{\circ}\) are not solutions, because the tangent is undefined there.

(c) Explaining why only \(\sin\theta = 0\) satisfies \(\tan^{2}\theta - \sin^{2}\theta = -\sin^{2}\theta\tan^{2}\theta\) [2 marks]

By part (a) the left-hand side is identically \(\tan^{2}\theta\sin^{2}\theta\), so the equation becomes

\[\tan^{2}\theta\sin^{2}\theta = -\sin^{2}\theta\tan^{2}\theta \quad \Longrightarrow \quad 2\sin^{2}\theta\tan^{2}\theta = 0 \qquad \textbf{M1}\]

The M1 is for reaching this equation by substituting the identity, which is the "without further solution" route the question demands. A product is zero only when one of its factors is zero, so either \(\sin\theta = 0\) or \(\tan\theta = 0\). But \(\tan\theta = \dfrac{\sin\theta}{\cos\theta}\) vanishes exactly when its numerator vanishes, so \(\tan\theta = 0\) if and only if \(\sin\theta = 0\). The two conditions collapse into one, and the solutions are precisely those with \(\sin\theta = 0\) A1.

The instructive point is why the sign change on the right-hand side is so destructive. In part (b) the equation was a product equal to a non-zero multiple of \(\sin^{2}\theta\), which left room for \(\tan^{2}\theta = 3\). Here moving the right-hand side across doubles the same product instead of cancelling it, so nothing survives except the values that make the product itself zero. Those are \(\theta = 0^{\circ},\ 180^{\circ},\ 360^{\circ}\), the same three found in part (b).

Examination takeaway. Never divide a trigonometric equation by a factor that can be zero; move everything to one side and factorise, so the zero case is preserved. When a part says "without further solution", the expected answer is an argument built on the identity just proved, and a zero product is the tool that turns it into a complete one.