FAD1014: MATHEMATICS II — Hard Simulation Solution Guide

Full step-by-step worked solutions with reasoning, traps, and alternative methods.

[!tip] How to Use This Guide

  1. Attempt the question first before reading the solution.
  2. If stuck, read only the Methodology hint — then try again.
  3. Use the Traps & Tips at the end of each solution to avoid repeating mistakes.
  4. Cross-references in brackets link to concept pages for deeper study.

PART A — Answer ALL Questions (24 Marks)

Question 1 [12 marks]

1(a) — Maclaurin series for $\cos x$ from definition [3]

Methodology: The Maclaurin series definition is $f(x) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$. Compute successive derivatives, evaluate at $x=0$, then plug into the formula.

Step 1: Compute derivatives until we reach $x^4$ term (need $f^{(4)}$):

$n$ $f^{(n)}(x)$ $f^{(n)}(0)$
0 $\cos x$ $1$
1 $-\sin x$ $0$
2 $-\cos x$ $-1$
3 $\sin x$ $0$
4 $\cos x$ $1$

Step 2: Assemble the series:

$$\cos x = \frac{1}{0!}x^0 + \frac{0}{1!}x^1 + \frac{(-1)}{2!}x^2 + \frac{0}{3!}x^3 + \frac{1}{4!}x^4 + \cdots$$

$$\boxed{\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} + \cdots = 1 - \frac{x^2}{2} + \frac{x^4}{24} + \cdots}$$

Traps & Tips:

  • ⚠️ Trap: Forgetting that $n!$ goes in the denominator. Each term is $\frac{f^{(n)}(0)}{n!}x^n$, not $f^{(n)}(0)x^n$.
  • ⚠️ Trap: The pattern alternates: $+, 0, -, 0, +, 0, \dots$ for $\cos x$ at $x=0$.
  • 💡 Pattern: $\cos x$ has only even powers. $\sin x$ has only odd powers. This is a consistency check.

1(b) — Ellipse from general form [5]

Methodology: Group $x$ and $y$ terms, complete the square for each, factor into standard form $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$.

Step 1: Group $x$ and $y$ terms:

$$9x^2 + 4y^2 - 18x + 16y - 11 = 0$$ $$9(x^2 - 2x) + 4(y^2 + 4y) = 11$$

Step 2: Complete the square for each group:

$$x^2 - 2x = (x-1)^2 - 1$$ $$y^2 + 4y = (y+2)^2 - 4$$

Step 3: Substitute back:

$$9[(x-1)^2 - 1] + 4[(y+2)^2 - 4] = 11$$ $$9(x-1)^2 - 9 + 4(y+2)^2 - 16 = 11$$ $$9(x-1)^2 + 4(y+2)^2 = 36$$

Step 4: Divide by 36 to get standard form:

$$\boxed{\frac{(x-1)^2}{4} + \frac{(y+2)^2}{9} = 1}$$

Step 5: Read off features:

  • Centre: $(1, -2)$
  • $a^2 = 4$ (under $x$), $b^2 = 9$ (under $y$)
  • Since $b > a$, the ellipse is vertical (major axis along $y$)
  • Semi-major axis $b = 3$, semi-minor axis $a = 2$
  • Vertices (centre $\pm$ 3 in $y$): $(1, 1)$ and $(1, -5)$

Traps & Tips:

  • ⚠️ Trap: Sign of centre. $(x-1)^2$ means $h = 1$, $(y+2)^2 = (y-(-2))^2$ means $k = -2$.
  • ⚠️ Trap: Don't confuse $a$ and $b$. For a vertical ellipse, $b > a$, so vertices are at $(h, k \pm b)$.
  • ⚠️ Trap: $9$ under $x$ and $4$ under $y$ in the original doesn't directly tell you orientation — complete the square first.
  • 💡 Check: The RHS must equal 1. If it's not 1 after completing the square, divide through.

1(c) — Separable DE with initial condition [4]

Methodology: Separate variables so all $y$ terms are with $dy$ and all $x$ terms with $dx$. Integrate both sides. Apply initial condition.

Step 1: Separate variables:

$$\frac{dy}{dx} = \frac{2x}{y+1}$$ $$(y+1),dy = 2x,dx$$

Step 2: Integrate both sides:

$$\int (y+1),dy = \int 2x,dx$$ $$\frac{y^2}{2} + y = x^2 + C$$

Step 3: Apply initial condition $y(0) = 1$:

$$\frac{1}{2} + 1 = 0 + C \Rightarrow C = \frac{3}{2}$$

Step 4: Solve for $y$ explicitly:

$$\frac{y^2}{2} + y = x^2 + \frac{3}{2}$$ Multiply by 2: $y^2 + 2y = 2x^2 + 3$ Complete the square: $(y+1)^2 - 1 = 2x^2 + 3$ $$(y+1)^2 = 2x^2 + 4$$ $$y + 1 = \sqrt{2x^2 + 4}$$

We take the positive root because $y(0) = 1 > -1$ means $y+1 > 0$.

$$\boxed{y = -1 + \sqrt{2x^2 + 4}}$$

Traps & Tips:

  • ⚠️ Trap: Forgetting the $+C$ after integration.
  • ⚠️ Trap: Choosing the wrong sign when taking the square root. Always check with the initial condition.
  • 💡 Alternative: Complete the square on the left: $y^2 + 2y = (y+1)^2 - 1$.
  • 💡 Domain check: $2x^2 + 4 > 0$ for all real $x$, so the solution is valid for all $x$.

Question 2 [12 marks]

2(a) — Maclaurin series by substitution [3]

Step 1: Recall the standard series for $\ln(1+u)$:

$$\ln(1+u) = u - \frac{u^2}{2} + \frac{u^3}{3} - \frac{u^4}{4} + \cdots$$ Valid for $-1 < u \leq 1$.

Step 2: Substitute $u = 3x^2$:

$$\ln(1 + 3x^2) = 3x^2 - \frac{(3x^2)^2}{2} + \frac{(3x^2)^3}{3} - \frac{(3x^2)^4}{4} + \cdots$$ $$= 3x^2 - \frac{9x^4}{2} + \frac{27x^6}{3} - \frac{81x^8}{4} + \cdots$$ $$= 3x^2 - \frac{9}{2}x^4 + 9x^6 - \frac{81}{4}x^8 + \cdots$$

Up to $x^6$: $\displaystyle \boxed{3x^2 - \frac{9}{2}x^4 + 9x^6}$

Step 3: Find range of validity: The original series converges for $-1 < u \leq 1$. With $u = 3x^2$: $$-1 < 3x^2 \leq 1$$ Since $3x^2 \geq 0$, the left inequality is automatic. So: $$3x^2 \leq 1 \Rightarrow |x| \leq \frac{1}{\sqrt{3}}$$ $$\boxed{|x| \leq \frac{1}{\sqrt{3}}}$$

Traps & Tips:

  • ⚠️ Trap: Forgetting the validity range. The exam often asks for it explicitly.
  • ⚠️ Trap: Forgetting to square/cube the coefficient when substituting. $(3x^2)^2 = 9x^4$, not $3x^4$.
  • 💡 Memory aid: $\ln(1+x)$ has alternating signs and no factorials in denominators.

2(b) — Sequence convergence [2]

Methodology: For rational sequences, divide numerator and denominator by the highest power of $n$.

Step 1: Divide by $n^2$:

$$a_n = \frac{5n^2 - 2n + 1}{2n^2 + 3n - 4} = \frac{5 - \frac{2}{n} + \frac{1}{n^2}}{2 + \frac{3}{n} - \frac{4}{n^2}}$$

Step 2: Take the limit as $n \to \infty$:

$$\lim_{n\to\infty} a_n = \frac{5 - 0 + 0}{2 + 0 - 0} = \frac{5}{2}$$

The sequence converges to $\boxed{\dfrac{5}{2}}$.

Traps & Tips:

  • ⚠️ Trap: Forgetting to check the leading coefficients. The limit of a rational function $\frac{an^p + \dots}{bn^p + \dots}$ is $\frac{a}{b}$ when degrees are equal.
  • 💡 If numerator degree < denominator degree, limit = 0. If numerator degree > denominator degree, diverges.

2(c) — Method of differences [4]

Methodology: Decompose the fraction into partial fractions, write out terms, notice telescoping cancellation.

Step 1: Partial fraction decomposition:

$$\frac{1}{(2k-1)(2k+1)} = \frac{A}{2k-1} + \frac{B}{2k+1}$$

Multiply both sides by $(2k-1)(2k+1)$: $$1 = A(2k+1) + B(2k-1)$$

Step 2: Solve for $A$ and $B$:

  • Let $k = \frac{1}{2}$: $1 = A(2) \Rightarrow A = \frac{1}{2}$
  • Let $k = -\frac{1}{2}$: $1 = B(-2) \Rightarrow B = -\frac{1}{2}$

$$\frac{1}{(2k-1)(2k+1)} = \frac{1}{2}\left(\frac{1}{2k-1} - \frac{1}{2k+1}\right)$$

Step 3: Write out the sum and telescope:

$$\sum_{k=1}^n \frac{1}{(2k-1)(2k+1)} = \frac{1}{2}\sum_{k=1}^n \left(\frac{1}{2k-1} - \frac{1}{2k+1}\right)$$

Write terms explicitly:

  • $k=1$: $\frac{1}{1} - \frac{1}{3}$
  • $k=2$: $\frac{1}{3} - \frac{1}{5}$
  • $k=3$: $\frac{1}{5} - \frac{1}{7}$
  • $\vdots$
  • $k=n$: $\frac{1}{2n-1} - \frac{1}{2n+1}$

All interior terms cancel. What remains:

$$\frac{1}{2}\left(1 - \frac{1}{2n+1}\right) = \frac{1}{2}\left(\frac{2n+1-1}{2n+1}\right) = \frac{n}{2n+1}$$

Step 4: Sum to infinity:

$$\lim_{n\to\infty} \frac{n}{2n+1} = \frac{1}{2}$$

Final: $\displaystyle \boxed{\sum_{k=1}^n \frac{1}{(2k-1)(2k+1)} = \frac{n}{2n+1}, \quad \sum_{k=1}^\infty = \frac{1}{2}}$

Traps & Tips:

  • ⚠️ Trap: Forgetting the factor $\frac{1}{2}$ from the decomposition. Without it, you'll get double the correct answer.
  • ⚠️ Trap: Writing $\frac{1}{2k-1} - \frac{1}{2k+1}$ directly without the $\frac{1}{2}$ factor.
  • 💡 Pattern: $\frac{1}{(k+a)(k+b)}$ where $b = a+2$ always gives $\frac{1}{2}\left(\frac{1}{k+a} - \frac{1}{k+b}\right)$.

2(d) — Ellipse equation from geometric data [3]

Methodology: Determine orientation from the shared coordinate between centre-vertex and centre-focus. Find $b$ (distance to vertex), $c$ (distance to focus), then compute $a^2 = b^2 - c^2$.

Step 1: Determine orientation:

  • Centre: $(2, -1)$
  • Vertex: $(2, 4)$ — same $x$-coordinate as centre, so major axis is vertical
  • Distance from centre to vertex: $b = 4 - (-1) = 5$
  • Focus: $(2, 2)$ — also same $x$-coordinate, vertical orientation confirmed
  • Distance from centre to focus: $c = 2 - (-1) = 3$

Step 2: Compute $a$ (semi-minor axis for vertical ellipse): For vertical ellipse: $c^2 = b^2 - a^2$ $$a^2 = b^2 - c^2 = 25 - 9 = 16 \Rightarrow a = 4$$

Step 3: Write standard form: $$\boxed{\frac{(x-2)^2}{16} + \frac{(y+1)^2}{25} = 1}$$

Traps & Tips:

  • ⚠️ Trap: Using $c^2 = a^2 + b^2$ (hyperbola formula) instead of $c^2 = b^2 - a^2$ (ellipse formula).
  • ⚠️ Trap: Putting the larger denominator under the wrong variable. For vertical ellipse, $b$ (the larger) goes under $(y-k)^2$.
  • 💡 Check: $a = 4, b = 5$ means $b > a$, confirming vertical orientation.

PART B — Answer Any FOUR Questions (56 Marks)

Question 3 [14 marks] — Bernoulli DE → Maclaurin Series

3(a) — Identify Bernoulli [2]

Step 1: Compare with standard form $\frac{dy}{dx} + P(x)y = Q(x)y^n$:

$$\frac{dy}{dx} + y = xy^2$$

  • $P(x) = 1$ (coefficient of $y$)
  • $Q(x) = x$ (coefficient of $y^2$)
  • $n = 2$ (power of $y$ on RHS, $n \neq 0, 1$)

Therefore it's a Bernoulli equation.

3(b) — Reduce to linear via substitution [4]

Step 1: Apply the Bernoulli substitution $v = y^{1-n} = y^{-1}$.

Step 2: Differentiate: $$\frac{dv}{dx} = -y^{-2}\frac{dy}{dx} \quad\Rightarrow\quad \frac{dy}{dx} = -y^2\frac{dv}{dx}$$

Step 3: Substitute into the original DE:

$$-y^2\frac{dv}{dx} + y = xy^2$$

Step 4: Divide by $y^2$: $$-\frac{dv}{dx} + y^{-1} = x$$

Step 5: Since $y^{-1} = v$: $$-\frac{dv}{dx} + v = x$$ $$\boxed{\frac{dv}{dx} - v = -x}$$

This is a first-order linear DE in $v$.

Key reasoning: The substitution $v = y^{1-n}$ always turns Bernoulli into linear. The power $1-n = -1$ means $v = 1/y$.

3(c) — Solve linear DE [5]

Step 1: Compute integrating factor: $$\frac{dv}{dx} - v = -x$$ $$P(x) = -1, \quad I = e^{\int P,dx} = e^{\int -1,dx} = e^{-x}$$

Step 2: Apply integrating factor: $$\frac{d}{dx}(v e^{-x}) = (-x)e^{-x}$$

Step 3: Integrate both sides: $$v e^{-x} = \int -x e^{-x},dx$$

Integrate by parts: let $u = x$, $dw = e^{-x},dx$ → $du = dx$, $w = -e^{-x}$: $$\int x e^{-x},dx = x(-e^{-x}) - \int (-e^{-x}),dx = -xe^{-x} - e^{-x} + C$$ Wait — let me redo this carefully.

$$\int -x e^{-x},dx = -\int x e^{-x},dx$$

For $\int x e^{-x},dx$: let $u = x$, $dv = e^{-x}dx$ → $du = dx$, $v = -e^{-x}$.

$$\int x e^{-x},dx = x(-e^{-x}) - \int (-e^{-x}),dx = -xe^{-x} + \int e^{-x},dx = -xe^{-x} - e^{-x}$$

Therefore: $$\int -x e^{-x},dx = -(-xe^{-x} - e^{-x}) = xe^{-x} + e^{-x}$$

So: $$v e^{-x} = xe^{-x} + e^{-x} + C$$ $$v = x + 1 + Ce^{x}$$

Step 4: Back-substitute $v = y^{-1}$: $$y^{-1} = x + 1 + Ce^{x}$$ $$\boxed{y = \frac{1}{x + 1 + Ce^{x}}}$$

3(d) — Initial condition and Maclaurin by implicit differentiation [3]

Step 1: Apply $y(0) = 1$: $$1 = \frac{1}{0 + 1 + C} \Rightarrow C = 0$$ $$\boxed{y = \frac{1}{x+1}}$$

This is the geometric series function — but we're asked to not expand it directly.

Step 2: Use the original DE to find derivatives: The original DE is $\frac{dy}{dx} + y = xy^2$. At $x = 0$: $y'(0) + y(0) = 0 \cdot y(0)^2 \Rightarrow y'(0) + 1 = 0 \Rightarrow y'(0) = -1$.

Step 3: Differentiate the DE implicitly to get $y''$: $$\frac{d}{dx}\left(\frac{dy}{dx} + y\right) = \frac{d}{dx}(xy^2)$$ $$y'' + y' = y^2 + 2xyy'$$

At $x = 0$: $y''(0) + y'(0) = y(0)^2 + 0$ $$y''(0) + (-1) = 1 \Rightarrow y''(0) = 2$$

Step 4: Differentiate again to get $y'''$: $$\frac{d}{dx}(y'' + y') = \frac{d}{dx}(y^2 + 2xyy')$$ $$y''' + y'' = 2yy' + 2yy' + 2x(y')^2 + 2xyy''$$ $$y''' + y'' = 4yy' + 2x[(y')^2 + yy'']$$

At $x = 0$: $y'''(0) + y''(0) = 4y(0)y'(0) + 0$ $$y'''(0) + 2 = 4(1)(-1) = -4 \Rightarrow y'''(0) = -6$$

Step 5: Assemble the Maclaurin series: $$y(x) = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \cdots$$ $$= 1 + (-1)x + \frac{2}{2}x^2 + \frac{(-6)}{6}x^3 + \cdots$$ $$\boxed{y(x) = 1 - x + x^2 - x^3 + \cdots}$$

Why this is level 7-8: The student must differentiate the DE three times using the product rule, tracking which terms survive at $x=0$. Most would just expand $y = (1-x)^{-1}$ directly — the point is to test understanding of implicit differentiation for Maclaurin series when the explicit form isn't available.

Verification: $y = \frac{1}{1+x} = (1+x)^{-1} = 1 - x + x^2 - x^3 + \cdots$ ✓

Traps & Tips:

  • ⚠️ Trap: Forgetting the product rule when differentiating $xy^2$: $\frac{d}{dx}(xy^2) = y^2 + 2xyy'$ (not just $2xyy'$).
  • ⚠️ Trap: The $n!$ in the denominator: $y''(0)/2!$, not $y''(0)/2$.
  • 💡 Key insight: The implicit differentiation method works even when you cannot solve the DE explicitly. This is its power.

Question 4 [14 marks] — Separable DE → Ellipse

4(a) — Solve separable DE [3]

Step 1: Rearrange to separate variables:

$$\frac{dy}{dx} = -\frac{9x}{4y}$$ $$4y,dy = -9x,dx$$

Step 2: Integrate: $$\int 4y,dy = -\int 9x,dx$$ $$2y^2 = -\frac{9}{2}x^2 + C$$

Step 3: Multiply by 2: $$4y^2 = -9x^2 + 2C$$ $$9x^2 + 4y^2 = 2C$$

Let $K = 2C$: $\boxed{9x^2 + 4y^2 = K}$ (general solution).

Key reasoning: The integration constant is named differently at each step — what matters is that $9x^2 + 4y^2$ equals a constant. This already looks like an ellipse.

4(b) — Find particular solution [3]

Step 1: Apply $y(2) = 0$: $$9(2)^2 + 4(0)^2 = K \Rightarrow K = 36$$

$$9x^2 + 4y^2 = 36$$

Step 2: Divide by 36: $$\boxed{\frac{x^2}{4} + \frac{y^2}{9} = 1}$$

4(c) — Identify ellipse features [4]

Step 1: Compare with standard form:

  • $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with $a^2 = 4$, $b^2 = 9$
  • $a = 2$, $b = 3$
  • Since $b > a$, this is a vertical ellipse (major axis along $y$)

Step 2: Compute features:

  • Centre: $(0, 0)$
  • Vertices: $(0, \pm 3)$
  • For vertical ellipse: $c^2 = b^2 - a^2 = 9 - 4 = 5$
  • $c = \sqrt{5}$
  • Foci: $(0, \pm\sqrt{5})$

Final: $\boxed{\text{Centre }(0,0),\ \text{Vertices }(0,\pm3),\ \text{Foci }(0,\pm\sqrt{5})}$

Traps & Tips:

  • ⚠️ Trap: For a vertical ellipse, $c^2 = b^2 - a^2$, not $a^2 - b^2$. The larger denominator gives $b$.
  • ⚠️ Trap: Foci are on the major axis. Since $y$ has the larger denominator, foci are at $(0, \pm c)$ not $(\pm c, 0)$.
  • 💡 The equation came from a DE — always check if your solution actually satisfies the DE and the IC.

4(d) — Tangent at a point [4]

Step 1: Verify the point lies on the curve ($9x^2 + 4y^2 = 36$): $$9\left(\frac{16}{9}\right) + 4(5) = 16 + 20 = 36 \quad\checkmark$$

Step 2: Find the gradient at this point using the DE: $$\left.\frac{dy}{dx}\right|_{(4/3, \sqrt{5})} = -\frac{9(4/3)}{4\sqrt{5}} = -\frac{12}{4\sqrt{5}} = -\frac{3}{\sqrt{5}}$$

Step 3: Equation of tangent line using point-slope form $y - y_0 = m(x - x_0)$: $$y - \sqrt{5} = -\frac{3}{\sqrt{5}}\left(x - \frac{4}{3}\right)$$

Step 4: Simplify: $$y = -\frac{3}{\sqrt{5}}x + \frac{4}{\sqrt{5}} + \sqrt{5}$$ $$y = -\frac{3}{\sqrt{5}}x + \frac{4}{\sqrt{5}} + \frac{5}{\sqrt{5}}$$ $$\boxed{y = -\frac{3}{\sqrt{5}}x + \frac{9}{\sqrt{5}}}$$

Alternative method: Implicit differentiation of $9x^2 + 4y^2 = 36$: $$18x + 8y\frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{9x}{4y}$$ Same result. ✓

Traps & Tips:

  • ⚠️ Trap: Not verifying the point lies on the curve first. If the point isn't on the curve, the tangent doesn't exist.
  • ⚠️ Trap: Rationalizing: $\frac{9}{\sqrt{5}} = \frac{9\sqrt{5}}{5}$ is often preferred in final answers.
  • 💡 Check: The gradient is negative (point is in Q1 with positive $x$, positive $y$), which makes sense for an ellipse.

Question 5 [14 marks] — Binomial Series → Ellipse Arc Length

5(a) — Binomial expansion [2]

Step 1: Apply $(1+u)^n = 1 + nu + \frac{n(n-1)}{2!}u^2 + \frac{n(n-1)(n-2)}{3!}u^3 + \cdots$ with $n = -\frac{1}{2}$:

$$(1+u)^{-1/2} = 1 + \left(-\frac{1}{2}\right)u + \frac{(-\frac{1}{2})(-\frac{3}{2})}{2!}u^2 + \frac{(-\frac{1}{2})(-\frac{3}{2})(-\frac{5}{2})}{3!}u^3 + \cdots$$

Step 2: Simplify each coefficient: $$= 1 - \frac{1}{2}u + \frac{\frac{3}{4}}{2}u^2 - \frac{-\frac{15}{8}}{6}u^3 + \cdots$$ $$= \boxed{1 - \frac{1}{2}u + \frac{3}{8}u^2 - \frac{5}{16}u^3 + \cdots}$$

Valid for $|u| < 1$.

Traps & Tips:

  • ⚠️ Trap: $\binom{-1/2}{2} = \frac{(-1/2)(-3/2)}{2} = \frac{3/4}{2} = 3/8$, not just $3/4$.
  • 💡 Memory: The coefficients of $(1+u)^{-1/2}$ appear in many physics contexts (relativity, elliptic integrals).

5(b) — Ellipse arc length integral [4]

Step 1: Compute derivatives: $$\frac{dx}{d\theta} = -4\sin\theta, \qquad \frac{dy}{d\theta} = 3\cos\theta$$

Step 2: Square and sum: $$\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = 16\sin^2\theta + 9\cos^2\theta$$ $$= 16(1 - \cos^2\theta) + 9\cos^2\theta$$ $$= 16 - 7\cos^2\theta$$

Step 3: Take square root and factor: $$\sqrt{16 - 7\cos^2\theta} = \sqrt{16\left(1 - \frac{7}{16}\cos^2\theta\right)} = 4\sqrt{1 - \frac{7}{16}\cos^2\theta}$$

Step 4: Write the integral: $$\boxed{s = \int_0^{\pi/2} 4\sqrt{1 - \frac{7}{16}\cos^2\theta},d\theta}$$

Key insight: This integral has no closed-form elementary solution — it's an elliptic integral of the second kind. That's why series expansion is the right approach.

5(c) — Series expansion and approximation [8]

Step 1: Adapt the binomial expansion. We need $\sqrt{1-u} = (1-u)^{1/2}$, so set $n = 1/2$:

$$(1+u)^{1/2} = 1 + \frac{1}{2}u - \frac{1}{8}u^2 + \frac{1}{16}u^3 - \frac{5}{128}u^4 + \cdots$$

(Or use $(1-u)^{1/2} = 1 - \frac{u}{2} - \frac{u^2}{8} - \frac{u^3}{16} - \cdots$ directly — same result.)

Step 2: Substitute $u = -\frac{7}{16}\cos^2\theta$:

$$\sqrt{1 - \frac{7}{16}\cos^2\theta} = 1 + \frac{1}{2}\left(-\frac{7}{16}\cos^2\theta\right) - \frac{1}{8}\left(\frac{49}{256}\cos^4\theta\right) + \cdots$$

Wait — let me be more careful. Using $(1+u)^{1/2} = 1 + \frac{u}{2} - \frac{u^2}{8} + \cdots$ and $u = -\frac{7}{16}\cos^2\theta$:

$$\sqrt{1 - \frac{7}{16}c^2} = 1 + \frac{1}{2}\left(-\frac{7}{16}c^2\right) - \frac{1}{8}\left(\frac{49}{256}c^4\right) + \cdots$$ $$= 1 - \frac{7}{32}c^2 - \frac{49}{2048}c^4 + \cdots$$

where $c = \cos\theta$.

Step 3: Integrate term by term: $$s \approx 4\int_0^{\pi/2} \left(1 - \frac{7}{32}\cos^2\theta - \frac{49}{2048}\cos^4\theta\right) d\theta$$

Using the given integrals: $$\int_0^{\pi/2} 1,d\theta = \frac{\pi}{2}, \quad \int_0^{\pi/2} \cos^2\theta,d\theta = \frac{\pi}{4}, \quad \int_0^{\pi/2} \cos^4\theta,d\theta = \frac{3\pi}{16}$$

Step 4: Substitute: $$s \approx 4\left[\frac{\pi}{2} - \frac{7}{32}\cdot\frac{\pi}{4} - \frac{49}{2048}\cdot\frac{3\pi}{16}\right]$$ $$= 4\pi\left[\frac{1}{2} - \frac{7}{128} - \frac{147}{32768}\right]$$

Step 5: Common denominator 32768: $$\frac{1}{2} = \frac{16384}{32768}, \quad \frac{7}{128} = \frac{1792}{32768}, \quad \frac{147}{32768} = \frac{147}{32768}$$

$$s = 4\pi\left[\frac{16384 - 1792 - 147}{32768}\right] = 4\pi\left[\frac{14445}{32768}\right]$$

$$\boxed{s = \frac{14445\pi}{8192} \approx 5.54 \text{ (3 s.f.)}}$$

Why this is level 7-8: The student must:

  1. Recognize arc-length parametric formula
  2. Manipulate trigonometric identities
  3. Know when to use $n = 1/2$ vs $n = -1/2$ (by carefully reading the question)
  4. Integrate $\cos^4\theta$ (using given result)
  5. Handle rational arithmetic with large denominators

Traps & Tips:

  • ⚠️ Trap: The expansion asks for $\sqrt{1 - \frac{7}{16}\cos^2\theta} = (1 + u)^{1/2}$ with $u = -\frac{7}{16}\cos^2\theta$, not $(1+u)^{-1/2}$.
  • ⚠️ Trap: $\cos^4\theta$ integration requires $\int_0^{\pi/2} \cos^4\theta,d\theta = \frac{3\pi}{16}$, not $\frac{\pi}{8}$.
  • 💡 Verification: The exact arc length of a quarter ellipse with $a=4, b=3$ is $4E(\sqrt{7}/4)$ where $E$ is the complete elliptic integral. Our approximation $5.54$ is close to the true value $\approx 5.53$.

Question 6 [14 marks] — Pure Differential Equations

6(a) — Test for exactness [3]

Step 1: Identify $M$ and $N$: $$M = 3x^2y^2 + 2xy, \qquad N = 2x^3y + x^2$$

Step 2: Compute partial derivatives: $$\frac{\partial M}{\partial y} = 6x^2y + 2x$$ $$\frac{\partial N}{\partial x} = 6x^2y + 2x$$

Step 3: Compare: $$\frac{\partial M}{\partial y} = \frac{\partial N}{\partial x} \quad\checkmark$$ The equation is exact.

Key reasoning: Exactness means the DE came from differentiating a function $F(x,y)$. If we can find $F$, the solution is $F(x,y) = C$.

6(b) — Solve the exact DE [5]

Step 1: Integrate $M$ with respect to $x$: $$F(x,y) = \int M,dx = \int (3x^2y^2 + 2xy),dx = x^3y^2 + x^2y + g(y)$$

where $g(y)$ is an unknown function of $y$ only.

Step 2: Differentiate $F$ with respect to $y$ and equate to $N$: $$\frac{\partial F}{\partial y} = 2x^3y + x^2 + g'(y) = N = 2x^3y + x^2$$ $$g'(y) = 0$$

Step 3: Integrate to find $g(y)$: $$g(y) = \text{constant}$$

This constant gets absorbed into $C$. So: $$\boxed{F(x,y) = x^3y^2 + x^2y = C}$$

6(c) — Non-homogeneous DE (linearly dependent) [6]

Step 1: Check linear dependence: $$(x + 2y + 1),dx + (2x + 4y - 3),dy = 0$$ $a_1 = 1$, $b_1 = 2$, $a_2 = 2$, $b_2 = 4$

$$a_1b_2 - a_2b_1 = 1(4) - 2(2) = 4 - 4 = 0$$

Since $a_1b_2 - a_2b_1 = 0$, the equation is linearly dependent.

Key reasoning: When $a_1b_2 - a_2b_1 = 0$, the coefficients of $x$ and $y$ in the $dx$ and $dy$ terms are proportional. This means we can express everything in terms of a single new variable $u = a_1x + b_1y$.

Step 2: Apply the substitution $u = x + 2y$: $$du = dx + 2,dy \Rightarrow dx = du - 2,dy$$

Step 3: Substitute into the DE: $$(u + 1)(du - 2,dy) + (2u - 3)dy = 0$$

Step 4: Expand and collect: $$(u+1)du - 2(u+1)dy + (2u - 3)dy = 0$$ $$(u+1)du + [-2u - 2 + 2u - 3]dy = 0$$ $$(u+1)du - 5,dy = 0$$

Step 5: Rearrange and integrate: $$\frac{du}{dy} = \frac{5}{u+1}$$ $$(u+1),du = 5,dy$$ $$\int (u+1),du = \int 5,dy$$ $$\frac{u^2}{2} + u = 5y + C$$

Step 6: Back-substitute $u = x + 2y$: $$\frac{(x+2y)^2}{2} + (x+2y) = 5y + C$$

Multiply by 2: $$\boxed{(x+2y)^2 + 2(x+2y) = 10y + K}$$ where $K = 2C$.

Alternative simplification: The implicit solution can be left as: $$(x+2y)^2 + 2x - 6y = K$$ (Expanding: $x^2 + 4xy + 4y^2 + 2x + 4y = 10y + K \Rightarrow x^2 + 4xy + 4y^2 + 2x - 6y = K$)

Traps & Tips:

  • ⚠️ Trap: Skipping the linear dependence test. If you try coordinate translation on a dependent system, $h$ and $k$ will be inconsistent.
  • ⚠️ Trap: Getting the sign wrong on $dx = du - 2dy$. Double-check: $u = x+2y \Rightarrow du = dx + 2dy \Rightarrow dx = du - 2dy$.
  • ⚠️ Trap: Forgetting the factor of 2 when multiplying through by 2.
  • 💡 Identifying dependence: Compute $a_1b_2 - a_2b_1$. Zero means dependent; non-zero means independent (use coordinate translation instead).

Question 7 [14 marks] — Pure Maclaurin Series

7(a) — $\sin x$ from definition [4]

Step 1: Compute derivatives:

$n$ $f^{(n)}(x)$ $f^{(n)}(0)$
0 $\sin x$ $0$
1 $\cos x$ $1$
2 $-\sin x$ $0$
3 $-\cos x$ $-1$
4 $\sin x$ $0$
5 $\cos x$ $1$

Step 2: Assemble: $$\sin x = 0 + \frac{1}{1!}x + \frac{0}{2!}x^2 + \frac{(-1)}{3!}x^3 + \frac{0}{4!}x^4 + \frac{1}{5!}x^5 + \cdots$$ $$\boxed{\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} + \cdots = x - \frac{x^3}{6} + \frac{x^5}{120} + \cdots}$$

Traps & Tips:

  • ⚠️ Trap: $\sin x$ has alternating signs and only odd powers. If your series has an even power, you've made an error.
  • ⚠️ Trap: $3! = 6$, $5! = 120$. These are common denominators in the $\sin x$ series.

7(b) — $e^{x^2}$ by substitution [2]

Step 1: Recall $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \cdots$

Step 2: Substitute $u = x^2$: $$e^{x^2} = 1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \cdots = \boxed{1 + x^2 + \frac{x^4}{2} + \frac{x^6}{6} + \cdots}$$

7(c) — Multiply series [4]

Step 1: Write both series up to needed order: $$e^{x^2} = 1 + x^2 + \frac{x^4}{2} + \cdots \qquad \sin x = x - \frac{x^3}{6} + \frac{x^5}{120} + \cdots$$

Step 2: Multiply term by term, collecting up to $x^5$:

Product $x$ term $x^3$ term $x^5$ term
$1 \cdot \sin x$ $1 \cdot x$ $1(-\frac{x^3}{6})$ $1(\frac{x^5}{120})$
$x^2 \cdot \sin x$ — $x^2 \cdot x$ $x^2(-\frac{x^3}{6})$
$\frac{x^4}{2} \cdot \sin x$ — — $\frac{x^4}{2} \cdot x$

Step 3: Sum contributions:

$x^1$ term: $x$ $x^3$ term: $-\frac{x^3}{6} + x^3 = \frac{5}{6}x^3$ $x^5$ term: $\frac{x^5}{120} - \frac{x^5}{6} + \frac{x^5}{2}$

Step 4: Simplify $x^5$ coefficient to common denominator 120: $$\frac{1}{120} - \frac{20}{120} + \frac{60}{120} = \frac{41}{120}$$

$$\boxed{h(x) = x + \frac{5}{6}x^3 + \frac{41}{120}x^5 + \cdots}$$

Traps & Tips:

  • ⚠️ Trap: Missing cross-terms. Draw a multiplication table to systematically track all contributions.
  • ⚠️ Trap: The $x^5$ term has three sources: $1 \cdot \frac{x^5}{120}$, $x^2 \cdot (-\frac{x^3}{6})$, and $\frac{x^4}{2} \cdot x$.
  • 💡 Verification: At small $x$, $e^{x^2}\sin x \approx x$ (since both $e^{x^2} \approx 1$ and $\sin x \approx x$). Our series starts with $x$ ✓.

7(d) — Term-by-term integration [4]

Step 1: Integrate the series term by term: $$\int_0^{0.5} h(x),dx \approx \int_0^{0.5} \left(x + \frac{5}{6}x^3 + \frac{41}{120}x^5\right) dx$$

Step 2: Apply $\int x^n,dx = \frac{x^{n+1}}{n+1}$: $$= \left[\frac{x^2}{2} + \frac{5}{24}x^4 + \frac{41}{720}x^6\right]_0^{0.5}$$

Step 3: Evaluate at $x = 0.5$: $$\frac{(0.5)^2}{2} = \frac{0.25}{2} = 0.125$$ $$\frac{5}{24}(0.5)^4 = \frac{5}{24}(0.0625) = \frac{0.3125}{24} = 0.0130208$$ $$\frac{41}{720}(0.5)^6 = \frac{41}{720}(0.015625) = \frac{0.640625}{720} = 0.0008898$$

Step 4: Sum: $$0.125 + 0.0130208 + 0.0008898 = 0.1389106$$ $$\boxed{\approx 0.1389 \text{ (4 d.p.)}}$$

Why series integration is necessary: The function $e^{x^2}\sin x$ has no elementary antiderivative. Its integral cannot be expressed as a finite combination of polynomials, trig functions, exponentials, logarithms, or inverse trig functions. Maclaurin series allow us to approximate such integrals to any desired accuracy by taking enough terms. This is the standard method in physics and engineering for integrals with no closed form.

Traps & Tips:

  • ⚠️ Trap: Forgetting to change the integration limits. The series is valid near $x=0$, and $x=0.5$ is within its radius of convergence.
  • ⚠️ Trap: Arithmetic errors in decimal multiplication. Write fractions first, then evaluate numerically.
  • 💡 Accuracy: The next term ($x^7$) would contribute $\approx 10^{-5}$ to the integral — our 4 d.p. accuracy is reliable with just 3 terms.
  • 💡 Error bound: For an alternating series (this one is not strictly alternating but converges rapidly), the error is less than the first omitted term.

Question 8 [14 marks] — Pure Ellipse

8(a) — Completing the square and features [6]

Step 1: Group $x$ and $y$ terms: $$4x^2 + 9y^2 - 16x + 18y - 11 = 0$$ $$4(x^2 - 4x) + 9(y^2 + 2y) = 11$$

Step 2: Complete the square: $$x^2 - 4x = (x-2)^2 - 4$$ $$y^2 + 2y = (y+1)^2 - 1$$

Step 3: Substitute: $$4[(x-2)^2 - 4] + 9[(y+1)^2 - 1] = 11$$ $$4(x-2)^2 - 16 + 9(y+1)^2 - 9 = 11$$ $$4(x-2)^2 + 9(y+1)^2 = 36$$

Step 4: Divide by 36: $$\boxed{\frac{(x-2)^2}{9} + \frac{(y+1)^2}{4} = 1}$$

Step 5: Read off features:

  • Centre: $(2, -1)$
  • $a^2 = 9$ (under $x$), $a = 3$, $b^2 = 4$ (under $y$), $b = 2$
  • Since $a > b$, this is a horizontal ellipse
  • $c^2 = a^2 - b^2 = 9 - 4 = 5$, $c = \sqrt{5}$
  • Vertices: $(2 \pm 3, -1) = (5, -1)$ and $(-1, -1)$
  • Foci: $(2 \pm \sqrt{5}, -1)$
  • Eccentricity: $e = \frac{c}{a} = \frac{\sqrt{5}}{3}$

Traps & Tips:

  • ⚠️ Trap: Centre sign. $(x-2)^2$ means $h = 2$, $(y+1)^2 = (y-(-1))^2$ means $k = -1$.
  • ⚠️ Trap: $a$ is always the semi-major axis. Here $9 > 4$, so $a^2 = 9$ goes under $(x-h)^2$.
  • ⚠️ Trap: Eccentricity $e = c/a$, not $c/b$.
  • 💡 Eccentricity range: For an ellipse, $0 \leq e < 1$. $e = \sqrt{5}/3 \approx 0.745$ — this is a noticeably elongated ellipse.

8(b) — Tangents parallel to a given line [5]

Step 1: Find the gradient of the given line: $$4x + 3y = 5 \Rightarrow y = -\frac{4}{3}x + \frac{5}{3}$$ Gradient $m = -\dfrac{4}{3}$.

Step 2: Translate the ellipse to centre at origin: $$X = x - 2, \quad Y = y + 1 \Rightarrow \frac{X^2}{9} + \frac{Y^2}{4} = 1$$

Step 3: Use the tangent formula for an ellipse centred at origin: Tangents with gradient $m$ are $Y = mX \pm \sqrt{a^2m^2 + b^2}$.

Compute $a^2m^2 + b^2$: $$a^2m^2 + b^2 = 9\left(\frac{16}{9}\right) + 4 = 16 + 4 = 20$$ $$\sqrt{a^2m^2 + b^2} = \sqrt{20} = 2\sqrt{5}$$

Step 4: Write the tangents in translated coordinates: $$Y = -\frac{4}{3}X \pm 2\sqrt{5}$$

Step 5: Translate back to original coordinates: $$y + 1 = -\frac{4}{3}(x - 2) \pm 2\sqrt{5}$$ $$y = -\frac{4}{3}x + \frac{8}{3} - 1 \pm 2\sqrt{5}$$ $$y = -\frac{4}{3}x + \frac{5}{3} \pm 2\sqrt{5}$$

The two tangents are: $$\boxed{y = -\frac{4}{3}x + \frac{5}{3} + 2\sqrt{5}}$$ $$\boxed{y = -\frac{4}{3}x + \frac{5}{3} - 2\sqrt{5}}$$

Alternative method — discriminant approach (if the formula isn't memorized): A line $y = mx + c$ is tangent to $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ if the quadratic in $x$ from substitution has discriminant $\Delta = 0$. This gives $c^2 = a^2m^2 + b^2$.

Traps & Tips:

  • ⚠️ Trap: Forgetting to translate back. The tangent lines must be in original $(x,y)$ coordinates.
  • ⚠️ Trap: Getting the sign wrong on $Y = y+1$ (be careful: $y+1$ not $y-1$).
  • 🧠 Derivation of tangent formula: Substitute $y = mx + c$ into $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Set discriminant to zero → $c^2 = a^2m^2 + b^2$.

8(c) — Distance sum verification [3]

Step 1: Find $y$ when $x = 3$: $$\frac{(3-2)^2}{9} + \frac{(y+1)^2}{4} = 1$$ $$\frac{1}{9} + \frac{(y+1)^2}{4} = 1$$ $$\frac{(y+1)^2}{4} = \frac{8}{9}$$ $$(y+1)^2 = \frac{32}{9}$$ $$y+1 = \pm\frac{4\sqrt{2}}{3}$$ $$y = -1 \pm \frac{4\sqrt{2}}{3}$$

Take $P = \left(3, -1 + \dfrac{4\sqrt{2}}{3}\right)$ (the upper point).

Step 2: Compute distances to the foci: Foci: $F_1 = (2 - \sqrt{5}, -1)$, $F_2 = (2 + \sqrt{5}, -1)$

$$PF_1 = \sqrt{(3 - (2 - \sqrt{5}))^2 + \left(\frac{4\sqrt{2}}{3}\right)^2} = \sqrt{(1 + \sqrt{5})^2 + \frac{32}{9}}$$ $$= \sqrt{1 + 2\sqrt{5} + 5 + \frac{32}{9}} = \sqrt{6 + 2\sqrt{5} + \frac{32}{9}}$$ $$= \sqrt{\frac{54 + 18\sqrt{5} + 32}{9}} = \frac{1}{3}\sqrt{86 + 18\sqrt{5}}$$

$$PF_2 = \sqrt{(3 - (2 + \sqrt{5}))^2 + \left(\frac{4\sqrt{2}}{3}\right)^2} = \sqrt{(1 - \sqrt{5})^2 + \frac{32}{9}}$$ $$= \sqrt{1 - 2\sqrt{5} + 5 + \frac{32}{9}} = \sqrt{6 - 2\sqrt{5} + \frac{32}{9}}$$ $$= \sqrt{\frac{54 - 18\sqrt{5} + 32}{9}} = \frac{1}{3}\sqrt{86 - 18\sqrt{5}}$$

Step 3: Verify the hidden perfect squares: $$86 + 18\sqrt{5} = 81 + 5 + 18\sqrt{5} = 9^2 + (\sqrt{5})^2 + 2(9)(\sqrt{5}) = (9 + \sqrt{5})^2$$ $$86 - 18\sqrt{5} = 81 + 5 - 18\sqrt{5} = 9^2 + (\sqrt{5})^2 - 2(9)(\sqrt{5}) = (9 - \sqrt{5})^2$$

Step 4: Sum the distances: $$PF_1 + PF_2 = \frac{1}{3}(9 + \sqrt{5} + 9 - \sqrt{5}) = \frac{1}{3}(18) = 6$$

Step 5: Verify against $2a$: $$2a = 2(3) = 6$$

$$\boxed{PF_1 + PF_2 = 6 = 2a \text{ — confirms the geometric definition of an ellipse}}$$

Why this is level 7-8: The student must:

  1. Spot that $86 \pm 18\sqrt{5}$ are perfect squares of binomials involving $\sqrt{5}$
  2. Work with nested radicals
  3. Connect back to the fundamental definition of an ellipse (sum of distances to foci is constant and equals $2a$)
  4. This final verification closes the loop — it's the mathematical signature of an ellipse

Traps & Tips:

  • ⚠️ Trap: Giving up when the radicals look messy. Always check if they simplify — $(p+q\sqrt{r})^2$ is a common pattern.
  • ⚠️ Trap: Forgetting that $F_1$ and $F_2$ are different points: $2-\sqrt{5}$ vs $2+\sqrt{5}$.
  • 💡 Pattern recognition: $86 = 81 + 5 = 9^2 + (\sqrt{5})^2$, and $18 = 2\cdot9\cdot\sqrt{5}$. This is $(9+\sqrt{5})^2$.
  • 🧠 Verification: If your sum doesn't equal $2a$, one of your earlier calculations is wrong. This is a built-in check.

Summary of Difficulty Mechanisms (Level 7-8)

Question Difficulty Mechanism
Q3(d) Implicit differentiation of DE to find Maclaurin coefficients — requires tracking product rule through multiple derivatives, isolating surviving terms at $x=0$
Q4(d) Point verification + tangent from DE gradient formula, rationalizing radicals
Q5(c) Elliptic integral (no closed form), binomial expansion with $n=1/2$, integrating $\cos^4\theta$, rational arithmetic with large denominators
Q6(c) Must test $a_1b_2 - a_2b_1 = 0$ to identify dependent type, then apply $u = a_1x + b_1y$ substitution — wrong substitution means dead end
Q7(d) Concept question: why series integration is necessary (no elementary antiderivative exists)
Q8(c) Nested radical simplification, recognizing $(9\pm\sqrt{5})^2$ pattern, connecting to ellipse definition

Cross-References