FAD1014: MATHEMATICS II — Simulation UAS Paper

Academic Session 2025/2026
Duration 2 Hours
Total Marks 80
Instructions Answer ALL questions in Part A. Answer any FOUR of the six questions in Part B.

Formula Sheet (Provided)

Standard Integrals

$$\int x^n , dx = \frac{x^{n+1}}{n+1} + C, \quad n \neq -1$$ $$\int e^{ax} , dx = \frac{e^{ax}}{a} + C$$ $$\int \frac{1}{x} , dx = \ln \vert x \vert + C$$ $$\int \sin x , dx = -\cos x + C, \quad \int \cos x , dx = \sin x + C$$ $$\int \sec^2 x , dx = \tan x + C, \quad \int \sec x \tan x , dx = \sec x + C$$ $$\int \frac{1}{\sqrt{a^2 - x^2}} , dx = \sin^{-1}\left(\frac{x}{a}\right) + C, \quad \int \frac{1}{a^2 + x^2} , dx = \frac{1}{a}\tan^{-1}\left(\frac{x}{a}\right) + C$$

Summation Formulas

$$\sum_{r=1}^{n} r = \frac{n(n+1)}{2}, \quad \sum_{r=1}^{n} r^2 = \frac{n(n+1)(2n+1)}{6}$$

Binomial Theorem (Positive Integer $n$)

$$(a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r$$

Maclaurin Series

$$f(x) = f(0) + f'(0)x + \frac{f''(0)}{2!}x^2 + \frac{f'''(0)}{3!}x^3 + \cdots$$

Standard Expansions

$$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots$$ $$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$ $$\cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots$$ $$\ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots, \quad -1 < x \leq 1$$

Conic Sections

Shape Standard Equation Key Properties
Parabola $(x-h)^2 = 4a(y-k)$ Vertex: $(h,k)$, Focus: $(h,k+a)$, Directrix: $y = k-a$
Parabola $(y-k)^2 = 4a(x-h)$ Vertex: $(h,k)$, Focus: $(h+a,k)$, Directrix: $x = h-a$
Ellipse $\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$ $c^2 = a^2 - b^2$, Foci: $(h\pm c,k)$ or $(h,k\pm c)$
Hyperbola $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$ $c^2 = a^2 + b^2$, Foci: $(h\pm c,k)$, Asymptotes: $y-k = \pm\frac{b}{a}(x-h)$
Hyperbola $\frac{(y-k)^2}{a^2} - \frac{(x-h)^2}{b^2} = 1$ $c^2 = a^2 + b^2$, Foci: $(h,k\pm c)$, Asymptotes: $y-k = \pm\frac{a}{b}(x-h)$

PART A — Answer ALL Questions (24 Marks)

Question 1 (12 marks)

(a) Find the following indefinite integrals:

 (i) $\displaystyle \int \left(5e^{3x} - \frac{2}{x} + 4\cos x\right) dx$  [3]

 (ii) $\displaystyle \int x^2(x^3 - 1)^5 , dx$  [3]

(b) Evaluate the definite integral:

$$\int_{0}^{\pi/4} \sec^2 x , e^{\tan x} , dx$$  [3]

(c) Find the area bounded by the curve $y = x^2 + 1$, the $x$-axis, and the lines $x = 0$ and $x = 2$.  [3]

Question 2 (12 marks)

(a) Determine whether the sequence $a_k = \dfrac{3k^2 - 2k + 1}{5k^2 + 4k - 3}$ converges or diverges. If it converges, find its limit.  [3]

(b) Using the method of differences, evaluate:

$$\sum_{k=1}^{n} \frac{2}{(2k-1)(2k+1)}$$  [4]

(c) Solve the separable differential equation:

$$\frac{dy}{dx} = \frac{3x^2}{2y}, \quad y(0) = 2$$

Find $y$ explicitly in terms of $x$.  [3]

(d) A parabola has equation $(x-2)^2 = 12(y+1)$. State the vertex, focus, and directrix of this parabola.  [2]

PART B — Answer Any FOUR Questions (56 Marks)

Question 3 (14 marks) — Integration by Parts & Standard Integration

(a) Evaluate the following integrals using integration by parts:

 (i) $\displaystyle \int x e^{2x} , dx$  [3]

 (ii) $\displaystyle \int \ln x , dx$  [3]

(b) Find $\displaystyle \int x^2 \cos x , dx$.  [4]

(c) Hence, or otherwise, evaluate $\displaystyle \int_{0}^{\pi} x^2 \cos x , dx$.  [4]

Question 4 (14 marks) — Trigonometric Substitution

(a) State the trigonometric substitution appropriate for an integrand containing $\sqrt{a^2 - x^2}$. Hence evaluate:

$$\int \frac{1}{x^2 \sqrt{9 - x^2}} , dx$$  [7]

(b) Using the substitution $x = 2\tan\theta$, evaluate:

$$\int_{0}^{2} \frac{1}{\sqrt{x^2 + 4}} , dx$$  [7]

Question 5 (14 marks) — Maclaurin Series

(a) Find the Maclaurin series for $f(x) = \sin(2x)$ up to the term in $x^5$ by computing derivatives at $x = 0$.  [4]

(b) Write down the first four non-zero terms of the Maclaurin series for $e^{x^2}$.  [3]

(c) Use your answers to parts (a) and (b) to find the series for $e^{x^2} \sin(2x)$ up to the term in $x^5$.  [4]

(d) Hence, approximate $\displaystyle \int_{0}^{0.5} e^{x^2} \sin(2x) , dx$ by integrating term-by-term up to $x^5$.  [3]

Question 6 (14 marks) — Homogeneous Differential Equations

(a) Show that the differential equation

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

is homogeneous.  [2]

(b) Use the substitution $y = vx$ to reduce the equation to a separable DE in $v$ and $x$.  [4]

(c) Solve the differential equation to find the general solution. Give your answer in the form $y^2 = f(x)$.  [5]

(d) Find the particular solution given that $y(1) = 2$.  [3]

Question 7 (14 marks) — Parametric Equations

A curve is defined by the parametric equations:

$$x = t^2 - 1, \quad y = t^3 - t$$

(a) Find $\dfrac{dy}{dx}$ in terms of $t$.  [3]

(b) Find the equation of the tangent to the curve at the point where $t = 2$.  [4]

(c) Find the values of $t$ for which the tangent to the curve is horizontal.  [3]

(d) Find the Cartesian equation of the curve by eliminating the parameter $t$.  [4]

Question 8 (14 marks) — Hyperbola & Series

(a) A hyperbola has equation $\dfrac{x^2}{9} - \dfrac{y^2}{16} = 1$.

 (i) State the centre, vertices, and foci.  [3]

 (ii) Find the equations of the asymptotes.  [2]

 (iii) Sketch the hyperbola, labelling all key features.  [3]

(b) Expand $\sqrt{1 + x}$ as a binomial series up to the term in $x^3$, stating the range of validity.  [3]

(c) The right branch of the hyperbola can be written as $y = \frac{4}{3}\sqrt{x^2 - 9}$. Use your series from part (b) to expand $\sqrt{x^2 - 9}$ for $x > 3$ by writing $x^2 - 9 = x^2\left(1 - \frac{9}{x^2}\right)$, and hence find the first two non-zero terms of the expansion for $y$ in descending powers of $x$.  [3]

END OF QUESTION PAPER

Answer Key

Part A

Q1 (a)(i) $\displaystyle \frac{5}{3}e^{3x} - 2\ln|x| + 4\sin x + C$ (a)(ii) $\displaystyle \frac{1}{18}(x^3 - 1)^6 + C$ (b) $e - 1$ (c) $\displaystyle \frac{14}{3}$ square units

Q2 (a) Converges to $\dfrac{3}{5}$ (b) $\displaystyle \sum_{k=1}^{n} \frac{2}{(2k-1)(2k+1)} = \frac{2n}{2n+1}$ (c) $y = \sqrt{x^3 + 4}$ (d) Vertex: $(2, -1)$, Focus: $(2, 2)$, Directrix: $y = -4$

Part B

Q3 (a)(i) $\displaystyle \frac{x}{2}e^{2x} - \frac{1}{4}e^{2x} + C$ (a)(ii) $x\ln x - x + C$ (b) $x^2\sin x + 2x\cos x - 2\sin x + C$ (c) $-\pi^2 + 2$

Q4 (a) $x = 3\sin\theta$; $\displaystyle -\frac{\sqrt{9-x^2}}{9x} + C$ (b) $\ln(1 + \sqrt{2})$

Q5 (a) $2x - \frac{4}{3}x^3 + \frac{4}{15}x^5 + \cdots$ (b) $1 + x^2 + \frac{x^4}{2!} + \frac{x^6}{3!} + \cdots$ (c) $2x + \left(2 - \frac{4}{3}\right)x^3 + \left(\frac{4}{15} + 1 - \frac{2}{3}\right)x^5 + \cdots = 2x + \frac{2}{3}x^3 + \frac{3}{5}x^5 + \cdots$ (d) $\displaystyle \int_0^{0.5} \approx \left[x^2 + \frac{1}{6}x^4 + \frac{1}{10}x^6\right]_0^{0.5} = 0.25 + \frac{1}{96} + \frac{1}{640} \approx 0.2615$

Q6 (c) $y^2 = x^2\ln|x| + Cx^2$ (d) $y^2 = x^2(\ln|x| + 4)$

Q7 (a) $\displaystyle \frac{dy}{dx} = \frac{3t^2 - 1}{2t}$ (b) $11x - 4y = 7$ (c) $t = \pm\frac{1}{\sqrt{3}}$ (d) $y^2 = x(x+1)^2$ $($ or $y^2 = x^3 + 2x^2 + x$ $)$

Q8 (a)(i) Centre: $(0,0)$, Vertices: $(\pm 3, 0)$, Foci: $(\pm 5, 0)$ (a)(ii) $y = \pm\frac{4}{3}x$ (c) $y = \frac{4}{3}x\left(1 - \frac{9}{2x^2} - \cdots\right) = \frac{4}{3}x - \frac{6}{x} - \cdots$

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