A Level H2 Mathematics Practice Paper 2

TuitionGoWhere Practice Paper - Maths H2 A-Level

Answer Key and Marking Scheme

Paper: Practice Paper 2 (Pure Mathematics) Version: 2 of 5 Total Marks: 100

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Question 1: Functions and Composite Functions [10 marks]

(a) Find the range of $g$. [1 mark]

Answer: $g(x) = x^2 + 2$, $x \geq 0$. Minimum value occurs at $x = 0$: $g(0) = 2$. As $x \to \infty$, $g(x) \to \infty$. $\therefore$ Range of $g = [2, \infty)$.

Marking:

• B1: Correct range $[2, \infty)$ or $y \geq 2$.

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(b) Show that the composite function $fg$ does not exist. [2 marks]

Answer: For $fg$ to exist, we require $R_g \subseteq D_f$. $R_g = [2, \infty)$ and $D_f = (3, \infty)$. Since $2 \in R_g$ but $2 \notin D_f$, we have $R_g \nsubseteq D_f$. $\therefore$ $fg$ does not exist.

Marking:

• M1: States condition $R_g \subseteq D_f$ and identifies $R_g$ and $D_f$.
• A1: Correct conclusion with justification (e.g., $2 \in R_g$ but $2 \notin D_f$).

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(c) Find the maximal domain of $g$ such that the composite function $fg$ exists. [2 marks]

Answer: We require $g(x) \in D_f$, i.e., $g(x) > 3$. $x^2 + 2 > 3 \implies x^2 > 1 \implies x > 1$ (since $x \geq 0$). $\therefore$ Maximal domain of $g$ is $x > 1$.

Marking:

• M1: Sets up inequality $g(x) > 3$.
• A1: Correct domain $x > 1$.

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(d) Using the restricted domain from part (c), find an expression for $fg(x)$ and state its domain and range. [5 marks]

Answer: $fg(x) = f(g(x)) = f(x^2 + 2) = \frac{1}{(x^2 + 2) - 3} = \frac{1}{x^2 - 1}$.

Domain of $fg$: $x > 1$ (from part (c)).

For $x > 1$, $x^2 - 1 > 0$ and as $x \to 1^+$, $x^2 - 1 \to 0^+$, so $fg(x) \to \infty$. As $x \to \infty$, $x^2 - 1 \to \infty$, so $fg(x) \to 0^+$. $\therefore$ Range of $fg = (0, \infty)$.

Marking:

• M1: Correct substitution to find $fg(x)$.
• A1: Correct simplified expression $\frac{1}{x^2 - 1}$.
• B1: Correct domain $x > 1$.
• M1: Valid method to find range (e.g., considering limits).
• A1: Correct range $(0, \infty)$.

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Question 2: Transformations of Graphs [9 marks]

(a) Sketch the graph of $y = f(x)$. [2 marks]

Answer: Sketch should show:

• Minimum point at $(-2, -3)$ clearly labelled.
• Vertical asymptote $x = 1$ (dashed line).
• Horizontal asymptote $y = 0$ (dashed line).
• Curve approaching asymptotes correctly.

Marking:

• B1: Correct asymptotes labelled.
• B1: Correct minimum point and general shape.

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(b)(i) Sketch $y = f(x - 2)$. [2 marks]

Answer: Translation 2 units to the right.

• Minimum point: $(-2+2, -3) = (0, -3)$.
• Vertical asymptote: $x = 1+2 = 3$.
• Horizontal asymptote: $y = 0$ (unchanged).

Marking:

• B1: Correct asymptotes.
• B1: Correct minimum point and shape.

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(b)(ii) Sketch $y = 2f(x)$. [2 marks]

Answer: Vertical stretch with scale factor 2.

• Minimum point: $(-2, 2 \times (-3)) = (-2, -6)$.
• Vertical asymptote: $x = 1$ (unchanged).
• Horizontal asymptote: $y = 0$ (unchanged, since $2 \times 0 = 0$).

Marking:

• B1: Correct asymptotes.
• B1: Correct minimum point and shape.

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(b)(iii) Sketch $y = f(2x)$. [3 marks]

Answer: Horizontal compression with scale factor $\frac{1}{2}$.

• Minimum point: $(\frac{-2}{2}, -3) = (-1, -3)$.
• Vertical asymptote: $x = \frac{1}{2}$.
• Horizontal asymptote: $y = 0$ (unchanged).

Marking:

• B1: Correct horizontal asymptote.
• B1: Correct vertical asymptote.
• B1: Correct minimum point and shape.

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Question 3: Inequalities [8 marks]

(a) Solve $\frac{x^2 - x - 6}{x + 1} \leq 0$ algebraically. [4 marks]

Answer: Factorise numerator: $x^2 - x - 6 = (x-3)(x+2)$. So $\frac{(x-3)(x+2)}{x+1} \leq 0$.

Critical values: $x = -2, -1, 3$.

Sign analysis:

• $x < -2$: $(-)(-)/(-) = -$ (negative)
• $-2 < x < -1$: $(+)(-)/(-) = +$ (positive)
• $-1 < x < 3$: $(+)(+)/(+) = +$ (positive) — Wait, check: $(x-3)$ is negative, $(x+2)$ positive, $(x+1)$ positive. So $(-)(+)/(+) = -$ (negative).
• $x > 3$: $(+)(+)/(+) = +$ (positive).

Correction:

• $x < -2$: $(-)/(-) = +$? Let's recalculate carefully.

For $x < -2$: $(x-3) < 0$, $(x+2) < 0$, $(x+1) < 0$. Product: $(-)(-)/(-) = (+)/(-) = -$. For $-2 < x < -1$: $(x-3) < 0$, $(x+2) > 0$, $(x+1) < 0$. $(-)(+)/(-) = (-)/(-) = +$. For $-1 < x < 3$: $(x-3) < 0$, $(x+2) > 0$, $(x+1) > 0$. $(-)(+)/(+) = -$. For $x > 3$: $(x-3) > 0$, $(x+2) > 0$, $(x+1) > 0$. $(+)(+)/(+) = +$.

At $x = -2$: numerator = 0, expression = 0. Included. At $x = -1$: denominator = 0, undefined. Excluded. At $x = 3$: numerator = 0, expression = 0. Included.

Solution: $x \in (-\infty, -2] \cup (-1, 3]$.

Marking:

• M1: Correct factorisation.
• M1: Identifies critical values.
• M1: Correct sign analysis or graphical method.
• A1: Correct solution set with correct inclusion/exclusion of endpoints.

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(b) Hence solve $\left|\frac{x^2 - x - 6}{x + 1}\right| > 2$. [4 marks]

Answer: Let $y = \frac{x^2 - x - 6}{x + 1}$. We need $|y| > 2$, i.e., $y > 2$ or $y < -2$.

Case 1: $y > 2$ $\frac{x^2 - x - 6}{x + 1} > 2 \implies \frac{x^2 - x - 6 - 2(x+1)}{x+1} > 0 \implies \frac{x^2 - 3x - 8}{x+1} > 0$.

Roots of $x^2 - 3x - 8 = 0$: $x = \frac{3 \pm \sqrt{9+32}}{2} = \frac{3 \pm \sqrt{41}}{2}$. Approximately: $x \approx -1.70$ or $x \approx 4.70$.

Sign analysis for $\frac{(x - \frac{3+\sqrt{41}}{2})(x - \frac{3-\sqrt{41}}{2})}{x+1} > 0$: Critical values: $x = \frac{3-\sqrt{41}}{2} \approx -1.70$, $x = -1$, $x = \frac{3+\sqrt{41}}{2} \approx 4.70$.

• $x < \frac{3-\sqrt{41}}{2}$: $(+)(-)/(-) = +$? Let's be systematic.

For $x < \frac{3-\sqrt{41}}{2}$ (approx $-1.70$): both factors in numerator negative? $x - \frac{3+\sqrt{41}}{2} < 0$, $x - \frac{3-\sqrt{41}}{2} < 0$, $x+1 < 0$. So $(-)(-)/(-) = (+)/(-) = -$.

For $\frac{3-\sqrt{41}}{2} < x < -1$: first factor negative, second positive, denominator negative. $(-)(+)/(-) = (-)/(-) = +$.

For $-1 < x < \frac{3+\sqrt{41}}{2}$: first negative, second positive, denominator positive. $(-)(+)/(+) = -$.

For $x > \frac{3+\sqrt{41}}{2}$: all positive. $(+)(+)/(+) = +$.

So $y > 2$ when $x \in (\frac{3-\sqrt{41}}{2}, -1) \cup (\frac{3+\sqrt{41}}{2}, \infty)$.

Case 2: $y < -2$ $\frac{x^2 - x - 6}{x + 1} < -2 \implies \frac{x^2 - x - 6 + 2(x+1)}{x+1} < 0 \implies \frac{x^2 + x - 4}{x+1} < 0$.

Roots of $x^2 + x - 4 = 0$: $x = \frac{-1 \pm \sqrt{1+16}}{2} = \frac{-1 \pm \sqrt{17}}{2}$. Approximately: $x \approx -2.56$ or $x \approx 1.56$.

Critical values: $x = \frac{-1-\sqrt{17}}{2} \approx -2.56$, $x = -1$, $x = \frac{-1+\sqrt{17}}{2} \approx 1.56$.

Sign analysis:

• $x < \frac{-1-\sqrt{17}}{2}$: $(-)(-)/(-) = -$ (negative)
• $\frac{-1-\sqrt{17}}{2} < x < -1$: $(+)(-)/(-) = +$ (positive)
• $-1 < x < \frac{-1+\sqrt{17}}{2}$: $(+)(-)/(+) = -$ (negative)
• $x > \frac{-1+\sqrt{17}}{2}$: $(+)(+)/(+) = +$ (positive)

So $y < -2$ when $x \in (-\infty, \frac{-1-\sqrt{17}}{2}) \cup (-1, \frac{-1+\sqrt{17}}{2})$.

Combining both cases: $x \in (-\infty, \frac{-1-\sqrt{17}}{2}) \cup (\frac{3-\sqrt{41}}{2}, -1) \cup (-1, \frac{-1+\sqrt{17}}{2}) \cup (\frac{3+\sqrt{41}}{2}, \infty)$.

Marking:

• M1: Correctly interprets $|y| > 2$ as two separate inequalities.
• M1: Correct algebraic manipulation for each case.
• A1: Correct critical values (exact surd form).
• A1: Correct final solution set.

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Question 4: Sequences and Series [10 marks]

(a) Express $a$ in terms of $r$. [1 mark]

Answer: $S_\infty = \frac{a}{1-r} = 24 \implies a = 24(1-r)$.

Marking:

• B1: $a = 24(1-r)$.

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(b) Show that $8r^2 - 8r + 3 = 0$. [3 marks]

Answer: Sum of first two terms: $S_2 = a + ar = a(1+r) = 15$. Substitute $a = 24(1-r)$: $24(1-r)(1+r) = 15 \implies 24(1-r^2) = 15 \implies 1-r^2 = \frac{15}{24} = \frac{5}{8}$. $r^2 = 1 - \frac{5}{8} = \frac{3}{8}$. $8r^2 = 3 \implies 8r^2 - 3 = 0$.

Wait, the question asks to show $8r^2 - 8r + 3 = 0$. Let me recheck.

$S_2 = a + ar = a(1+r) = 15$. $a = 24(1-r)$. $24(1-r)(1+r) = 15 \implies 24(1-r^2) = 15 \implies 8(1-r^2) = 5 \implies 8 - 8r^2 = 5 \implies 8r^2 = 3 \implies 8r^2 - 3 = 0$.

This gives $8r^2 - 3 = 0$, not $8r^2 - 8r + 3 = 0$. The question as written has an inconsistency. Let me adjust the working to match the intended equation.

If the intended equation is $8r^2 - 8r + 3 = 0$, then perhaps $S_2 = a + ar = 15$ with a different $S_\infty$.

Let's work backwards: $8r^2 - 8r + 3 = 0 \implies r = \frac{8 \pm \sqrt{64-96}}{16} = \frac{8 \pm \sqrt{-32}}{16}$. No real roots. This doesn't work for $0 < r < 1$.

The question likely intended $8r^2 - 8r + 3 = 0$ from a different setup. Let me provide a corrected derivation:

Suppose $S_\infty = 24$ and $S_2 = a + ar = 15$. $a = 24(1-r)$. $24(1-r)(1+r) = 15 \implies 24(1-r^2) = 15 \implies 8(1-r^2) = 5 \implies 8 - 8r^2 = 5 \implies 8r^2 = 3 \implies r^2 = \frac{3}{8}$.

This gives $r = \sqrt{\frac{3}{8}} = \frac{\sqrt{6}}{4} \approx 0.612$.

The equation $8r^2 - 8r + 3 = 0$ would come from a different condition. Let me provide the working as if the condition were different, or accept the derived equation.

Revised answer (consistent with $8r^2 = 3$): $S_2 = a + ar = a(1+r) = 15$. Substituting $a = 24(1-r)$: $24(1-r)(1+r) = 15 \implies 24(1-r^2) = 15 \implies 8(1-r^2) = 5 \implies 8 - 8r^2 = 5 \implies 8r^2 = 3$.

Note: The question contains an inconsistency. The derived equation is $8r^2 = 3$. If the question intended $8r^2 - 8r + 3 = 0$, the given numbers would need adjustment.

Marking (for intended equation):

• M1: Correct expression for $S_2$.
• M1: Substitutes $a = 24(1-r)$.
• A1: Reaches $8r^2 - 8r + 3 = 0$ (or correctly identifies inconsistency).

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(c) Hence find the value of $r$ and the value of $a$. [3 marks]

Answer (using $8r^2 = 3$): $r^2 = \frac{3}{8} \implies r = \frac{\sqrt{6}}{4}$ (since $0 < r < 1$). $a = 24(1 - \frac{\sqrt{6}}{4}) = 24 - 6\sqrt{6}$.

Marking:

• M1: Solves for $r$.
• A1: Correct $r$ value.
• A1: Correct $a$ value.

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(d) Find the least value of $n$ such that $S_n > 23.5$. [3 marks]

Answer: $S_n = \frac{a(1-r^n)}{1-r} = 24(1-r^n)$. We need $24(1-r^n) > 23.5 \implies 1-r^n > \frac{23.5}{24} = \frac{47}{48}$. $r^n < 1 - \frac{47}{48} = \frac{1}{48}$. $n \ln r < \ln(\frac{1}{48}) \implies n > \frac{\ln(1/48)}{\ln r}$.

With $r = \frac{\sqrt{6}}{4} \approx 0.6124$: $n > \frac{\ln(1/48)}{\ln(0.6124)} \approx \frac{-3.8712}{-0.4901} \approx 7.90$. Least integer $n = 8$.

Marking:

• M1: Correct formula for $S_n$ and sets up inequality.
• M1: Correct use of logarithms.
• A1: $n = 8$.

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Question 5: Recursive Sequences and Mathematical Induction [9 marks]

(a) Find $u_2$, $u_3$, and $u_4$. [2 marks]

Answer: $u_1 = 2$. $u_2 = 3(2) - 4 = 6 - 4 = 2$. $u_3 = 3(2) - 4 = 2$. $u_4 = 3(2) - 4 = 2$.

Marking:

• B1: $u_2 = 2$.
• B1: $u_3 = u_4 = 2$.

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(b) Conjecture a formula for $u_n$ in terms of $n$. [1 mark]

Answer: $u_n = 2$ for all $n \geq 1$.

Marking:

• B1: $u_n = 2$.

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(c) Prove your conjecture by mathematical induction. [6 marks]

Answer: Let $P(n)$ be the statement "$u_n = 2$".

Base case: $n = 1$. $u_1 = 2$ (given). So $P(1)$ is true.

Inductive step: Assume $P(k)$ is true for some $k \geq 1$, i.e., $u_k = 2$. We need to prove $P(k+1)$ is true, i.e., $u_{k+1} = 2$.

$u_{k+1} = 3u_k - 4$ (by definition). $= 3(2) - 4$ (by induction hypothesis). $= 6 - 4 = 2$.

Thus $P(k+1)$ is true.

Conclusion: By mathematical induction, $P(n)$ is true for all $n \in \mathbb{Z}^+$, i.e., $u_n = 2$ for all $n \geq 1$.

Marking:

• B1: Clear statement of $P(n)$.
• M1: Correct base case verification.
• M1: Correct inductive hypothesis stated.
• M1: Correct use of recurrence relation.
• A1: Correct algebra to show $P(k+1)$.
• A1: Clear conclusion stated.

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Question 6: Vectors – Lines and Planes [12 marks]

(a) Find the acute angle between $l$ and $\Pi$. [4 marks]

Answer: Direction vector of $l$: $\mathbf{d} = \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix}$. Normal vector of $\Pi$: $\mathbf{n} = \begin{pmatrix} 2 \\ -1 \\ 2 \end{pmatrix}$.

Angle $\theta$ between line and plane satisfies $\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$.

$\mathbf{d} \cdot \mathbf{n} = 1(2) + 2(-1) + (-2)(2) = 2 - 2 - 4 = -4$. $|\mathbf{d}| = \sqrt{1^2 + 2^2 + (-2)^2} = \sqrt{1+4+4} = 3$. $|\mathbf{n}| = \sqrt{2^2 + (-1)^2 + 2^2} = \sqrt{4+1+4} = 3$.

$\sin \theta = \frac{|-4|}{3 \times 3} = \frac{4}{9}$. $\theta = \sin^{-1}(\frac{4}{9}) \approx 26.4^\circ$.

Marking:

• M1: Identifies $\mathbf{d}$ and $\mathbf{n}$.
• M1: Correct formula $\sin \theta = \frac{|\mathbf{d} \cdot \mathbf{n}|}{|\mathbf{d}||\mathbf{n}|}$.
• M1: Correct dot product and magnitudes.
• A1: Correct angle (exact or 26.4°).

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(b) Find the coordinates of the point of intersection of $l$ and $\Pi$. [3 marks]

Answer: Parametric form of $l$: $x = 2 + \lambda$, $y = -1 + 2\lambda$, $z = 3 - 2\lambda$. Substitute into $\Pi$: $2(2+\lambda) - (-1+2\lambda) + 2(3-2\lambda) = 7$. $4 + 2\lambda + 1 - 2\lambda + 6 - 4\lambda = 7$. $11 - 4\lambda = 7 \implies -4\lambda = -4 \implies \lambda = 1$.

Point: $(2+1, -1+2(1), 3-2(1)) = (3, 1, 1)$.

Marking:

• M1: Writes parametric equations.
• M1: Substitutes into plane equation.
• A1: Correct point $(3, 1, 1)$.

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(c) Find the perpendicular distance from $P(5, 3, -1)$ to $\Pi$. [3 marks]

Answer: Distance $= \frac{|2(5) - 1(3) + 2(-1) - 7|}{\sqrt{2^2 + (-1)^2 + 2^2}} = \frac{|10 - 3 - 2 - 7|}{3} = \frac{|-2|}{3} = \frac{2}{3}$.

Marking:

• M1: Correct distance formula.
• M1: Correct substitution.
• A1: $\frac{2}{3}$.

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(d) Find the perpendicular distance from $P$ to the line $l$. [2 marks]

Answer: Vector from a point on $l$ to $P$: $\mathbf{v} = \begin{pmatrix} 5-2 \\ 3-(-1) \\ -1-3 \end{pmatrix} = \begin{pmatrix} 3 \\ 4 \\ -4 \end{pmatrix}$. Distance $= \frac{|\mathbf{v} \times \mathbf{d}|}{|\mathbf{d}|}$.

$\mathbf{v} \times \mathbf{d} = \begin{pmatrix} 3 \\ 4 \\ -4 \end{pmatrix} \times \begin{pmatrix} 1 \\ 2 \\ -2 \end{pmatrix} = \begin{pmatrix} 4(-2) - (-4)(2) \\ (-4)(1) - 3(-2) \\ 3(2) - 4(1) \end{pmatrix} = \begin{pmatrix} -8+8 \\ -4+6 \\ 6-4 \end{pmatrix} = \begin{pmatrix} 0 \\ 2 \\ 2 \end{pmatrix}$.

$|\mathbf{v} \times \mathbf{d}| = \sqrt{0^2 + 2^2 + 2^2} = \sqrt{8} = 2\sqrt{2}$. $|\mathbf{d}| = 3$.

Distance $= \frac{2\sqrt{2}}{3}$.

Marking:

• M1: Correct method (cross product or alternative).
• A1: $\frac{2\sqrt{2}}{3}$.

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Question 7: Complex Numbers [10 marks]

(a) Express $z = \frac{3 + 4i}{1 - 2i}$ in the form $x + iy$. [3 marks]

Answer: $z = \frac{3+4i}{1-2i} \times \frac{1+2i}{1+2i} = \frac{(3+4i)(1+2i)}{1^2 + 2^2} = \frac{3 + 6i + 4i + 8i^2}{5} = \frac{3 + 10i - 8}{5} = \frac{-5 + 10i}{5} = -1 + 2i$.

Marking:

• M1: Multiplies numerator and denominator by conjugate.
• M1: Correct expansion.
• A1: $-1 + 2i$.

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(b)(i) Sketch $|z - 3| = 2$. [2 marks]

Answer: Circle centre $(3, 0)$, radius 2. Clearly labelled on Argand diagram.

Marking:

• B1: Correct centre.
• B1: Correct radius and circle drawn.

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(b)(ii) Sketch $\arg(z - 1 - i) = \frac{\pi}{4}$. [2 marks]

Answer: Half-line from $(1, 1)$ at angle $\frac{\pi}{4}$ to the positive real axis (i.e., gradient 1, extending to the right and up). Clearly labelled, with open circle at $(1, 1)$.

Marking:

• B1: Correct starting point.
• B1: Correct direction and half-line drawn.

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(c) Find the complex number satisfying both conditions. [3 marks]

Answer: From (b)(ii): $z = 1 + i + re^{i\pi/4}$, $r \geq 0$. So $z = 1 + r\cos\frac{\pi}{4} + i(1 + r\sin\frac{\pi}{4}) = 1 + \frac{r}{\sqrt{2}} + i(1 + \frac{r}{\sqrt{2}})$.

From (b)(i): $|z - 3| = 2$. $|(1 + \frac{r}{\sqrt{2}} - 3) + i(1 + \frac{r}{\sqrt{2}})| = 2$. $|(\frac{r}{\sqrt{2}} - 2) + i(1 + \frac{r}{\sqrt{2}})| = 2$. $(\frac{r}{\sqrt{2}} - 2)^2 + (1 + \frac{r}{\sqrt{2}})^2 = 4$. $\frac{r^2}{2} - \frac{4r}{\sqrt{2}} + 4 + 1 + \frac{2r}{\sqrt{2}} + \frac{r^2}{2} = 4$. $r^2 - \frac{2r}{\sqrt{2}} + 5 = 4 \implies r^2 - \sqrt{2}r + 1 = 0$. $r = \frac{\sqrt{2} \pm \sqrt{2 - 4}}{2} = \frac{\sqrt{2} \pm i\sqrt{2}}{2}$. No real solutions.

Let me recheck. The half-line is $y - 1 = 1(x - 1)$, i.e., $y = x$ for $x \geq 1$. Substitute into circle: $(x-3)^2 + y^2 = 4$ with $y = x$. $(x-3)^2 + x^2 = 4 \implies x^2 - 6x + 9 + x^2 = 4 \implies 2x^2 - 6x + 5 = 0$. Discriminant: $36 - 40 = -4 < 0$. No intersection.

The loci do not intersect. Perhaps the question intended different parameters.

Revised interpretation: If the circle is $|z - 3| = 2$ and the half-line is from $(1,1)$ at angle $\pi/4$, they do not intersect. The answer would be "no such complex number exists" or the question needs adjustment.

Marking (if no intersection):

• M1: Correct method for finding intersection.
• M1: Sets up equations correctly.
• A1: Concludes no intersection or identifies inconsistency.

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Question 8: Calculus – Implicit Differentiation [10 marks]

(a) Find $\frac{dy}{dx}$ in terms of $x$ and $y$. [3 marks]

Answer: Differentiate $x^3 + y^3 - 3xy = 0$ with respect to $x$: $3x^2 + 3y^2\frac{dy}{dx} - 3(y + x\frac{dy}{dx}) = 0$. $3x^2 + 3y^2\frac{dy}{dx} - 3y - 3x\frac{dy}{dx} = 0$. $(3y^2 - 3x)\frac{dy}{dx} = 3y - 3x^2$. $\frac{dy}{dx} = \frac{3y - 3x^2}{3y^2 - 3x} = \frac{y - x^2}{y^2 - x}$.

Marking:

• M1: Correct differentiation of $y^3$ (chain rule).
• M1: Correct product rule for $3xy$.
• A1: Correct simplified expression.

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(b) Find the coordinates of the points where the tangent is parallel to the $x$-axis. [4 marks]

Answer: Tangent parallel to $x$-axis $\implies \frac{dy}{dx} = 0$. $\frac{y - x^2}{y^2 - x} = 0 \implies y = x^2$ (provided $y^2 \neq x$).

Substitute $y = x^2$ into original equation: $x^3 + (x^2)^3 - 3x(x^2) = 0 \implies x^3 + x^6 - 3x^3 = 0 \implies x^6 - 2x^3 = 0 \implies x^3(x^3 - 2) = 0$. $x = 0$ or $x = \sqrt[3]{2}$.

When $x = 0$: $y = 0$. Check $y^2 - x = 0 - 0 = 0$. Denominator is zero, so this point is not valid (singular point).

When $x = \sqrt[3]{2}$: $y = (\sqrt[3]{2})^2 = 2^{2/3}$. Check $y^2 - x = (2^{2/3})^2 - 2^{1/3} = 2^{4/3} - 2^{1/3} = 2^{1/3}(2 - 1) = 2^{1/3} \neq 0$. Valid.

$\therefore$ Point is $(\sqrt[3]{2}, 2^{2/3})$.

Marking:

• M1: Sets $\frac{dy}{dx} = 0$ and obtains $y = x^2$.
• M1: Substitutes into original equation.
• M1: Solves for $x$ and checks validity.
• A1: Correct coordinates.

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(c) Find the equation of the normal to $C$ at $(1, 1)$. [3 marks]

Answer: At $(1, 1)$: $\frac{dy}{dx} = \frac{1 - 1^2}{1^2 - 1} = \frac{0}{0}$. Indeterminate.

Let's check if $(1, 1)$ satisfies the equation: $1^3 + 1^3 - 3(1)(1) = 1 + 1 - 3 = -1 \neq 0$. $(1, 1)$ is not on the curve.

Let's find a point that is on the curve. Try $(0, 0)$: $0 + 0 - 0 = 0$. Yes. At $(0, 0)$: $\frac{dy}{dx} = \frac{0 - 0}{0 - 0} = \frac{0}{0}$. Also indeterminate.

Try $x = 1$: $1 + y^3 - 3y = 0 \implies y^3 - 3y + 1 = 0$. This has a real root but not $y = 1$.

The question as written has $(1, 1)$ which is not on the curve. Let me provide a corrected answer assuming a different point, or work with $(1, 1)$ as intended with an implicit equation that does contain it.

Revised: If the curve were $x^3 + y^3 - 3xy = 0$, then $(1, 1)$ gives $1+1-3 = -1 \neq 0$. The point $(1, 1)$ is not on this curve. The Folium of Descartes $x^3 + y^3 = 3xy$ contains $(0,0)$ and $(\frac{3}{2}, \frac{3}{2})$.

Let's use $(\frac{3}{2}, \frac{3}{2})$: At $(\frac{3}{2}, \frac{3}{2})$: $\frac{dy}{dx} = \frac{\frac{3}{2} - (\frac{3}{2})^2}{(\frac{3}{2})^2 - \frac{3}{2}} = \frac{\frac{3}{2} - \frac{9}{4}}{\frac{9}{4} - \frac{3}{2}} = \frac{-\frac{3}{4}}{\frac{3}{4}} = -1$.

Gradient of normal $= 1$. Equation: $y - \frac{3}{2} = 1(x - \frac{3}{2}) \implies y = x$.

Marking (for corrected point):

• M1: Verifies point is on curve and finds gradient.
• M1: Finds gradient of normal.
• A1: Correct equation.

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Question 9: Maclaurin Series [10 marks]

(a) Find the Maclaurin series for $f(x) = e^x \cos x$ up to $x^3$. [6 marks]

Answer: $f(x) = e^x \cos x$. $f(0) = e^0 \cos 0 = 1$.

$f'(x) = e^x \cos x - e^x \sin x = e^x(\cos x - \sin x)$. $f'(0) = 1(1 - 0) = 1$.

$f''(x) = e^x(\cos x - \sin x) + e^x(-\sin x - \cos x) = e^x(-2\sin x)$. $f''(0) = 1(0) = 0$.

$f'''(x) = e^x(-2\sin x) + e^x(-2\cos x) = -2e^x(\sin x + \cos x)$. $f'''(0) = -2(1)(0 + 1) = -2$.

Maclaurin series: $f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \ldots$ $f(x) = 1 + 1\cdot x + \frac{0}{2}x^2 + \frac{-2}{6}x^3 + \ldots$ $f(x) = 1 + x - \frac{1}{3}x^3 + \ldots$

Marking:

• M1: Correct $f(0)$.
• M1: Correct $f'(x)$ and $f'(0)$.
• M1: Correct $f''(x)$ and $f''(0)$.
• M1: Correct $f'''(x)$ and $f'''(0)$.
• M1: Correct Maclaurin formula with factorials.
• A1: $1 + x - \frac{1}{3}x^3$.

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(b) Approximate $e^{0.2} \cos(0.2)$ to 4 d.p. [2 marks]

Answer: $f(0.2) \approx 1 + 0.2 - \frac{1}{3}(0.2)^3 = 1 + 0.2 - \frac{1}{3}(0.008) = 1.2 - 0.002666\ldots = 1.1973$ (to 4 d.p.).

Marking:

• M1: Correct substitution.
• A1: 1.1973.

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(c) Estimate the error. [2 marks]

Answer: Next term: $\frac{f^{(4)}(0)}{4!}x^4$. $f^{(4)}(x) = -2e^x(\sin x + \cos x) - 2e^x(\cos x - \sin x) = -2e^x(2\cos x) = -4e^x \cos x$. $f^{(4)}(0) = -4$.

Next term: $\frac{-4}{24}(0.2)^4 = -\frac{1}{6}(0.0016) \approx -0.000267$.

Error is approximately $|0.000267| \approx 0.0003$.

Marking:

• M1: Finds $f^{(4)}(0)$ or next term.
• A1: Correct error estimate.

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Question 10: Differential Equations – Real-World Application [12 marks]

(a) Show that $\frac{dx}{dt} = 1 - \frac{x}{20}$. [3 marks]

Answer: Rate of salt entering = concentration × flow rate = $0.2 \times 5 = 1$ kg/min. Rate of salt leaving = $\frac{x}{100} \times 5 = \frac{x}{20}$ kg/min (since volume remains 100 L). $\frac{dx}{dt} = \text{rate in} - \text{rate out} = 1 - \frac{x}{20}$.

Marking:

• M1: Correct rate in.
• M1: Correct rate out (concentration × outflow).
• A1: Correct differential equation.

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(b) Solve the differential equation. [4 marks]

Answer: $\frac{dx}{dt} = 1 - \frac{x}{20} = \frac{20 - x}{20}$. Separate variables: $\int \frac{1}{20-x} \, dx = \int \frac{1}{20} \, dt$. $-\ln|20-x| = \frac{t}{20} + C$. $\ln|20-x| = -\frac{t}{20} - C$. $20 - x = Ae^{-t/20}$, where $A = e^{-C}$. $x = 20 - Ae^{-t/20}$.

At $t = 0$, $x = 0$: $0 = 20 - A \implies A = 20$. $\therefore x = 20(1 - e^{-t/20})$.

Marking:

• M1: Correct separation of variables.
• M1: Correct integration.
• M1: Uses initial condition.
• A1: $x = 20(1 - e^{-t/20})$.

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(c) Amount of salt after 10 minutes. [2 marks]

Answer: $x(10) = 20(1 - e^{-10/20}) = 20(1 - e^{-0.5}) \approx 20(1 - 0.6065) = 20(0.3935) = 7.87$ kg.

Marking:

• M1: Correct substitution.
• A1: 7.87 kg (or exact $20(1 - e^{-0.5})$).

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(d) Limiting amount and explanation. [3 marks]

Answer: As $t \to \infty$, $e^{-t/20} \to 0$, so $x \to 20$ kg.

Explanation: In the long term, the concentration of salt in the tank approaches the concentration of the incoming solution (0.2 kg/L). Since the tank always contains 100 L, the amount of salt approaches $0.2 \times 100 = 20$ kg. The system reaches equilibrium where the rate of salt entering equals the rate of salt leaving.

Marking:

• B1: Correct limit 20 kg.
• M1: Reasonable explanation involving equilibrium.
• A1: Clear connection to concentration and volume.

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END OF ANSWER KEY