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

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.

This is a "show that" chain in which a definite integral produces a cubic, and the last part asks not merely for a root but for a proof that it is the only real one. That final demand is what distinguishes the question: finding \(a = 3\) is easy, and the marks are for closing off the other possibilities.

(a) Showing that \(\displaystyle\int_{1}^{a}\left(3x^{2} - 4x + 1\right)\mathrm{d}x = a\left(a - 1\right)^{2}\) [3 marks]

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

\[\int\left(3x^{2} - 4x + 1\right)\mathrm{d}x = \frac{3x^{3}}{3} - \frac{4x^{2}}{2} + x = x^{3} - 2x^{2} + x \qquad \textbf{M1 A1}\]

The M1 is for the integration process and survives a coefficient slip; the A1 requires the antiderivative to be exactly right. In a definite integral the constant of integration cancels between the limits, so it is omitted. Now substitute the limits, upper minus lower:

\[\left[x^{3} - 2x^{2} + x\right]_{1}^{a} = \left(a^{3} - 2a^{2} + a\right) - \left(1 - 2 + 1\right) = a^{3} - 2a^{2} + a\]

The lower limit contributes zero, which is a small gift but must still be shown. Since the target is a product, factorise rather than expand:

\[a^{3} - 2a^{2} + a = a\left(a^{2} - 2a + 1\right) = a\left(a - 1\right)^{2} \qquad \textbf{A1}\]

as required. Because the answer is printed, the marks are for the derivation: taking out the common factor \(a\) and recognising \(a^{2} - 2a + 1\) as the perfect square \(\left(a - 1\right)^{2}\). A candidate who stops at \(a^{3} - 2a^{2} + a\) has done the calculus but not the algebra the part asks for, and loses the final accuracy mark. Check the identity at a convenient value: with \(a = 2\), the left side is \(8 - 8 + 2 = 2\) and the right side is \(2\left(1\right)^{2} = 2\).

(b) Showing that \(a^{3} - 2a^{2} + a - 12 = 0\) [1 mark]

Set the result of part (a) equal to the stated value of the integral and expand:

\[a\left(a - 1\right)^{2} = 12 \quad \Longrightarrow \quad a^{3} - 2a^{2} + a = 12 \quad \Longrightarrow \quad a^{3} - 2a^{2} + a - 12 = 0 \qquad \textbf{[1]}\]

The single mark is for reversing the factorisation and moving the \(12\) across, so it is genuinely one line of work. Leaving the answer as \(a\left(a - 1\right)^{2} = 12\) does not score, because the requested form is the expanded cubic set equal to zero.

(c) Finding \(a\) and proving it is the only real value [2 marks]

A cubic with integer coefficients and leading coefficient \(1\) can only have an integer root that divides the constant term, so the candidates are the factors of \(12\). Testing \(a = 3\):

\[27 - 18 + 3 - 12 = 0,\]

so \(a = 3\) is a root and, by the factor theorem, \(\left(a - 3\right)\) is a factor. Dividing out:

\[a^{3} - 2a^{2} + a - 12 = \left(a - 3\right)\left(a^{2} + a + 4\right) = 0 \qquad \textbf{[1]}\]

The factorisation can be checked by expanding: \(a^{3} + a^{2} + 4a - 3a^{2} - 3a - 12 = a^{3} - 2a^{2} + a - 12\). It is worth noting the middle coefficient is \(+1\), not \(-1\), which is a common slip when dividing a cubic whose \(a^{2}\) coefficient is negative.

Now the uniqueness argument. The product is zero only if one factor is zero, so any further real root must satisfy \(a^{2} + a + 4 = 0\). Its discriminant is

\[b^{2} - 4ac = 1^{2} - 4\left(1\right)\left(4\right) = 1 - 16 = -15,\]

which is negative, so that quadratic has no real roots. Hence \(a = 3\) is the only real value satisfying the equation [1].

The discriminant is the point of this part. A candidate who simply asserts that \(a^{2} + a + 4\) "cannot be factorised" has not proved anything, since an unfactorisable quadratic may still have irrational real roots; the negative discriminant is what settles it. Completing the square gives the same conclusion in a different language, because \(a^{2} + a + 4 = \left(a + \tfrac{1}{2}\right)^{2} + \tfrac{15}{4}\) is a sum of a square and a positive number and so is always positive.

Finally, the value is consistent with the condition \(a \gt 1\) stated at the top of the question, and it checks against part (a): \(3\left(3 - 1\right)^{2} = 3 \times 4 = 12\), which is the given value of the integral.

Examination takeaway. When a "show that" asks for a factorised form, factorise rather than expand, and always take out a common factor before hunting for a perfect square. To prove that a cubic has exactly one real root, divide out the known linear factor and use the discriminant of the remaining quadratic; a claim that it "does not factorise" is not a proof.

(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) asks for a division, so it must be carried out; part (b) then rewards a candidate who recognises that the remainder theorem makes a second calculation unnecessary. The word "Hence" in part (b) is the instruction to reuse part (a) rather than to substitute afresh.

(a) The quotient and remainder on division by \(\left(x + 2\right)\) [3 marks]

Long division of polynomials works exactly like long division of numbers: at each stage divide the leading term of what is left by the leading term of the divisor, multiply back, and subtract. The divisor is \(x + 2\) and the dividend is \(\mathrm{f}(x) = 3x^{3} + 2x^{2} - 5x + 7\).

1. \(3x^{3} \div x = 3x^{2}\). Multiplying back, \(3x^{2}\left(x + 2\right) = 3x^{3} + 6x^{2}\). Subtracting from \(3x^{3} + 2x^{2} - 5x + 7\) leaves \(-4x^{2} - 5x + 7\), since \(2x^{2} - 6x^{2} = -4x^{2}\). M1
2. \(-4x^{2} \div x = -4x\). Multiplying back, \(-4x\left(x + 2\right) = -4x^{2} - 8x\). Subtracting leaves \(3x + 7\), since \(-5x - \left(-8x\right) = 3x\). M1
3. \(3x \div x = 3\). Multiplying back, \(3\left(x + 2\right) = 3x + 6\). Subtracting leaves \(1\), and since \(1\) has lower degree than \(x + 2\) the division stops.

\[\text{Quotient } = 3x^{2} - 4x + 3, \qquad \text{remainder } = 1 \qquad \textbf{A1}\]

The two M marks are for the division process, so they survive an arithmetic slip inside a subtraction; the single A1 requires both the quotient and the remainder to be completely correct, and it does not. The subtraction at each stage is where marks are lost, because subtracting a negative term changes its sign: at the second stage \(-5x - \left(-8x\right)\) is \(+3x\), not \(-13x\).

Verify by multiplying out, which takes only a moment and catches every slip:

\[\left(x + 2\right)\left(3x^{2} - 4x + 3\right) + 1 = 3x^{3} - 4x^{2} + 3x + 6x^{2} - 8x + 6 + 1 = 3x^{3} + 2x^{2} - 5x + 7 = \mathrm{f}(x).\]

Note also that the divisor is \(x + 2\), so the number used in a synthetic-division layout is \(-2\), not \(+2\). Using \(+2\) is the single most common route to a wrong remainder here.

(b) The value of \(\mathrm{f}(-2)\) [1 mark]

The remainder theorem says that when a polynomial \(\mathrm{f}(x)\) is divided by \(\left(x - a\right)\), the remainder equals \(\mathrm{f}(a)\). Writing the divisor as \(x - \left(-2\right)\) identifies \(a = -2\), so the remainder found in part (a) is precisely \(\mathrm{f}(-2)\):

\[\mathrm{f}(-2) = 1 \qquad \textbf{B1}\]

This is a B mark: a single independent mark for the correct value, with no method credit, so the answer must simply be right. The reason the theorem holds is visible in the identity above. Since

\[\mathrm{f}(x) = \left(x + 2\right)\left(3x^{2} - 4x + 3\right) + 1,\]

putting \(x = -2\) makes the bracket \(\left(x + 2\right)\) zero, which annihilates the whole quotient term and leaves only the remainder.

Direct substitution confirms it: \(\mathrm{f}(-2) = 3\left(-8\right) + 2\left(4\right) - 5\left(-2\right) + 7 = -24 + 8 + 10 + 7 = 1\). That calculation is a legitimate check but not what the part is testing, and a candidate who does it without ever linking the answer to the remainder has missed the point of the word "Hence". The sign errors to guard against in the check are \(3\left(-2\right)^{3} = -24\) rather than \(+24\), and \(-5 \times \left(-2\right) = +10\).

Examination takeaway. Convert the divisor into the form \(x - a\) before using the remainder theorem, so that \(x + 2\) gives \(a = -2\). When a question asks first for a division and then for a value of the function, the remainder is the answer, and the marks reward spotting that rather than repeating the work.

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.

Every part of this question rests on one habit: locate a right-angled triangle inside the solid, then use Pythagoras for a length or a trigonometric ratio for an angle. In a cuboid every vertical edge is perpendicular to the horizontal base, and every face is a rectangle, which is what supplies the right angles.

(a) Show that \(AC = 15\,\mathrm{cm}\) [2]

\(AC\) is a diagonal of the rectangle \(ABCD\), and in a rectangle angle \(ABC = 90^{\circ}\), so Pythagoras applies in triangle \(ABC\):

\[AC^{2} = AB^{2} + BC^{2} = 12^{2} + 9^{2} = 144 + 81 = 225 \quad \textbf{[M1]}\] \[AC = \sqrt{225} = 15\,\mathrm{cm} \quad \textbf{[A1]}\]

(b) Show that \(AG = 17\,\mathrm{cm}\) exactly [2]

\(G\) is vertically above \(C\), so \(CG\) is perpendicular to the whole base plane and in particular to \(AC\). That makes angle \(ACG = 90^{\circ}\), and \(CG = AE = 8\,\mathrm{cm}\) since opposite vertical edges of a cuboid are equal.

\[AG^{2} = AC^{2} + CG^{2} = 15^{2} + 8^{2} = 225 + 64 = 289 \quad \textbf{[M1]}\] \[AG = \sqrt{289} = 17\,\mathrm{cm} \quad \textbf{[A1]}\]

Both targets are printed, so these four marks are for the working. Note how (a) feeds (b): the space diagonal is built from the base diagonal and the height, never from two edges directly.

(c) Angle between \(AG\) and the plane \(ABCD\), to \(0.1^{\circ}\) [3]

The angle between a line and a plane is the angle between the line and its projection onto that plane. Dropping a perpendicular from \(G\) to the base lands at \(C\), so the projection of \(AG\) is \(AC\) and the required angle is \(\angle GAC\). [M1]

Triangle \(GAC\) is right-angled at \(C\), with \(CG = 8\) opposite the angle and \(AC = 15\) adjacent to it, so use the tangent ratio:

\[\tan(\angle GAC) = \frac{CG}{AC} = \frac{8}{15} \quad \textbf{[M1]}\] \[\angle GAC = 28.072\ldots^{\circ} = 28.1^{\circ} \quad \textbf{[A1]}\]

Identifying the correct angle carries its own method mark, which is the real content of this part; using \(\sin^{-1}\dfrac{8}{17}\) is an equally valid route and gives the same value.

(d) Show \(BG = \sqrt{145}\,\mathrm{cm}\) and find angle \(AGB\), to \(0.1^{\circ}\) [3]

In the rectangular face \(BCGF\), angle \(BCG = 90^{\circ}\) because \(CG\) is vertical, so

\[BG^{2} = BC^{2} + CG^{2} = 9^{2} + 8^{2} = 81 + 64 = 145, \qquad BG = \sqrt{145}\,\mathrm{cm} \quad \textbf{[M1][A1]}\]

For the angle, note that \(AB\) is perpendicular to the plane \(BCGF\), so \(AB\) is perpendicular to \(BG\) and triangle \(ABG\) is right-angled at \(B\). With the angle at \(G\), \(AB = 12\) is opposite and \(BG = \sqrt{145}\) is adjacent:

\[\tan(\angle AGB) = \frac{12}{\sqrt{145}} \quad \Rightarrow \quad \angle AGB = 44.900\ldots^{\circ} = 44.9^{\circ} \quad \textbf{[A1]}\]

(e) Area of triangle \(ACG\) [3]

Part (b) established angle \(ACG = 90^{\circ}\). [B1] The two perpendicular sides therefore serve as base and height, so no trigonometry is needed:

\[\text{Area} = \frac{1}{2} \times AC \times CG = \frac{1}{2} \times 15 \times 8 \quad \textbf{[M1]}\] \[= 60\,\mathrm{cm^{2}} \quad \textbf{[A1]}\]

Using \(\tfrac{1}{2} \times AC \times AG\) is the error to avoid: \(AG\) is the hypotenuse of this triangle, not a height.

(f) Length of \(AM\), where \(M\) is the midpoint of \(GH\), to 3 s.f. [2]

Coordinates make this straightforward. Take \(A\) as the origin with \(AB\), \(AD\) and \(AE\) along the axes, so \(A(0,\,0,\,0)\), \(G(12,\,9,\,8)\) and \(H(0,\,9,\,8)\). The midpoint of \(GH\) is then

\[M\left(\frac{12 + 0}{2},\ 9,\ 8\right) = M(6,\, 9,\, 8).\] \[AM^{2} = 6^{2} + 9^{2} + 8^{2} = 36 + 81 + 64 = 181 \quad \textbf{[M1]}\] \[AM = \sqrt{181} = 13.4536\ldots = 13.5\,\mathrm{cm} \ (3\ \mathrm{s.f.}) \quad \textbf{[A1]}\]

(g) Angle between \(AM\) and the plane \(ABCD\), to \(0.1^{\circ}\) [2]

Apply the same projection principle as in part (c). The foot of the perpendicular from \(M\) to the base is \(N(6,\, 9,\, 0)\), so the required angle is \(\angle MAN\), and

\[AN = \sqrt{6^{2} + 9^{2}} = \sqrt{117}, \qquad MN = 8,\] \[\tan(\angle MAN) = \frac{8}{\sqrt{117}} \quad \textbf{[M1]}\] \[\angle MAN = 36.486\ldots^{\circ} = 36.5^{\circ} \quad \textbf{[A1]}\]

Common wrong turns on this question. Measuring the angle in part (c) or (g) between the line and a vertical edge instead of the base, which gives the complement, \(61.9^{\circ}\) and \(53.5^{\circ}\) respectively; a quick sanity check is that these space diagonals lie closer to the base than to the vertical, so their angles with the base must be under \(45^{\circ}\). Working in radian mode, which produces nonsense values. Rounding \(AC\) or \(AN\) before the final trigonometric step. Assuming \(M\) lies above the centre of the base, which it does not: \(N(6,\,9,\,0)\) is the midpoint of \(DC\), not the centre.

Check. \(AM = \sqrt{181} \approx 13.45\) must be shorter than the space diagonal \(AG = 17\) and longer than \(AH = \sqrt{81 + 64} = \sqrt{145} \approx 12.04\), since \(M\) sits between \(H\) and \(G\); it is. In part (d), \(AG^{2} = AB^{2} + BG^{2} = 144 + 145 = 289\), agreeing with part (b), and \(\angle GAB + \angle AGB\) should be \(90^{\circ}\): \(45.1^{\circ} + 44.9^{\circ} = 90^{\circ}\).

(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 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 is about the addition formulae, but the marks turn on something more elementary: choosing the correct sign when a Pythagorean identity is used to recover a second ratio. The words "obtuse" and "acute" are the whole reason the question is set this way, and ignoring them is what the question is built to catch.

(a) Showing \(\cos A = -\dfrac{4}{5}\) and finding \(\sin B\) [2 marks]

Rearranging \(\sin^{2} A + \cos^{2} A \equiv 1\) gives

\[\cos^{2} A = 1 - \sin^{2} A = 1 - \frac{9}{25} = \frac{16}{25}, \qquad \text{so } \cos A = \pm\frac{4}{5}.\]

The square root produces two candidates, and the context decides between them. An obtuse angle lies between \(90^{\circ}\) and \(180^{\circ}\), which is the second quadrant, where the cosine is negative. Hence

\[\cos A = -\frac{4}{5} \qquad \textbf{B1}\]

Because the target is printed in the question, the mark is for the reasoning, not the statement: a candidate who writes \(\cos A = -\tfrac{4}{5}\) with no identity and no appeal to the quadrant earns nothing. The same identity applied to \(B\), which is acute and therefore in the first quadrant where every ratio is positive, gives

\[\sin^{2} B = 1 - \frac{25}{169} = \frac{144}{169}, \qquad \sin B = +\frac{12}{13} \qquad \textbf{B1}\]

The two B marks are independent, so a sign error on \(\cos A\) does not cost the \(\sin B\) mark. Both results come from the triples \(3, 4, 5\) and \(5, 12, 13\), so untidy surds at this stage signal a slip.

(b) The exact value of \(\sin\left(A + B\right)\) [3 marks]

Use the addition formula for sine:

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

The M1 is for quoting the correct expansion, and it stands even if the substituted values are wrong, which is why writing the formula down before substituting is worth doing. Substituting all four ratios, including the negative sign on \(\cos A\):

\[\sin\left(A + B\right) = \frac{3}{5} \times \frac{5}{13} + \left(-\frac{4}{5}\right) \times \frac{12}{13} = \frac{15}{65} - \frac{48}{65} \qquad \textbf{M1}\] \[= -\frac{33}{65} \qquad \textbf{A1}\]

Since \(33\) and \(65\) share no factor, the fraction is already in lowest terms. A negative answer is exactly what should be expected: \(A\) is obtuse and \(B\) is acute, so \(A + B\) can exceed \(180^{\circ}\) and place the sum in the third quadrant, where the sine is negative. A candidate who takes \(\cos A = +\tfrac{4}{5}\) obtains \(+\dfrac{63}{65}\), which loses both the second method mark and the accuracy mark. The other classic error is confusing the formulae and writing \(\sin A\cos B - \cos A\sin B\), which is the expansion of \(\sin\left(A - B\right)\).

(c) The exact value of \(\tan\left(A - B\right)\) [3 marks]

The tangent addition formula needs the two tangents, obtained as the ratio of sine to cosine:

\[\tan A = \frac{3/5}{-4/5} = -\frac{3}{4}, \qquad \tan B = \frac{12/13}{5/13} = \frac{12}{5} \qquad \textbf{B1}\]

Both come out with the fifths and thirteenths cancelling, which is why building the tangent from the two ratios is safer than reaching for a triangle. Note the sign: \(\tan A\) is negative because \(A\) is obtuse. Now apply the formula:

\[\tan\left(A - B\right) = \frac{\tan A - \tan B}{1 + \tan A\tan B} = \frac{-\dfrac{3}{4} - \dfrac{12}{5}}{1 + \left(-\dfrac{3}{4}\right)\left(\dfrac{12}{5}\right)} \qquad \textbf{M1}\]

Work out numerator and denominator over a common denominator of \(20\):

\[\text{numerator} = -\frac{15}{20} - \frac{48}{20} = -\frac{63}{20}, \qquad \text{denominator} = 1 - \frac{36}{20} = -\frac{16}{20}\] \[\tan\left(A - B\right) = \frac{-63/20}{-16/20} = \frac{63}{16} \qquad \textbf{A1}\]

Because \(63\) and \(16\) are coprime, this is in lowest terms. Two sign traps sit in this part. The first is the sign pattern of the formula itself: for \(A - B\) the numerator subtracts and the denominator adds, the opposite arrangement from \(A + B\). The second is the double negative in the denominator, where \(1 + \left(-\tfrac{3}{4}\right)\left(\tfrac{12}{5}\right)\) is \(1 - \tfrac{9}{5}\), a negative quantity; two negatives divided give the positive answer \(\dfrac{63}{16}\). Check the size against the geometry: \(A \approx 143.1^{\circ}\) and \(B \approx 67.4^{\circ}\), so \(A - B \approx 75.8^{\circ}\), whose tangent is about \(3.94\), and \(\dfrac{63}{16} = 3.9375\).

Examination takeaway. Whenever an identity forces a square root, write \(\pm\) explicitly and then quote the quadrant that fixes the sign. Learn the addition formulae as a family and check the sign pattern before substituting, because in questions of this type almost every lost mark is a lost sign rather than lost arithmetic.