FAD1014: MATHEMATICS II — Part A Practice Paper

Tailored for Qashrina Firqina


Focus	Part A — 2 questions per topic
Difficulty	PASUM UAS Exam Level

PART A

Answer ALL questions. Each question is worth 4–6 marks.

Integration — Antiderivative, Standard, Substitution, Definite

Q1. Evaluate:

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

(b) $\displaystyle \int \frac{6x^2 + 2}{x^3 + x} , dx$

Q2. Evaluate:

(a) $\displaystyle \int 3x^2 \sqrt{x^3 + 4} , dx$

(b) $\displaystyle \int_{-1}^{2} |x| , dx$

(c) The region bounded by $y = 4 - x^2$ and the $x$-axis is rotated about the $x$-axis. Find the volume of the solid generated.

Area Under the Curve

Q3.

(a) Find the area under the curve $y = \sin x$ from $x = 0$ to $x = \pi$.

(b) Find the area of the region bounded by $y = x^2$ and $y = 2x + 3$.

Q4.

(a) Find the area of the region bounded by $y = \sqrt{x}$, the $x$-axis, and $x = 4$.

(b) Find the area between $y = \cos x$ and $y = \sin x$ from $x = 0$ to $x = \frac{\pi}{4}$.

Differential Equations — Introduction & Separable

Q5.

(a) State the order and degree of the DE: $\displaystyle \left(\frac{d^2y}{dx^2}\right)^3 + 5\frac{dy}{dx} - 4y = e^x$

(b) Verify that $y = Ae^{2x} + Be^{-3x}$ is a solution to $\displaystyle \frac{d^2y}{dx^2} + \frac{dy}{dx} - 6y = 0$.

Q6.

(a) Solve $\displaystyle \frac{dy}{dx} = \frac{x}{y}$, $y(0) = 2$.

(b) Solve $\displaystyle \frac{dy}{dx} = \frac{x^2 + 1}{y^2}$, $y(1) = 2$.

Series — Introduction & Method of Differences

Q7.

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

(b) Evaluate $\displaystyle \sum_{r=1}^{20} (3r^2 - 2r + 1)$ using standard formulas.

Q8.

(a) Express $\displaystyle \frac{1}{r(r+1)}$ in partial fractions.

(b) Hence find $\displaystyle \sum_{r=1}^{n} \frac{1}{r(r+1)}$ using the method of differences.

Binomial Expansion I — $(a+b)^n$, $n \in \mathbb{Z}^+$

Q9.

(a) Expand $(2x - 3y)^4$ using the binomial theorem.

(b) Find the term independent of $x$ in the expansion of $\displaystyle \left( x^2 + \frac{2}{x} \right)^9$.

Q10.

(a) Find the coefficient of $x^5$ in the expansion of $(1 + 2x)^8$.

(b) Use the binomial theorem to approximate $(1.98)^6$ correct to 3 decimal places.

Hint: Write $1.98 = 2 - 0.02$.

Parabola

Q11.

(a) A parabola has vertex $(2, 3)$ and focus $(2, 5)$. Find its equation, directrix, and the length of the latus rectum.

(b) Sketch the parabola.

Q12.

(a) Find the vertex, focus, and directrix of the parabola $y^2 + 8x - 6y + 25 = 0$.

(b) Find the equation of the parabola with focus $(-1, 2)$ and directrix $x = 3$.

Ellipse

Q13.

(a) An ellipse has equation $\displaystyle \frac{x^2}{25} + \frac{y^2}{9} = 1$. Find the centre, vertices, foci, eccentricity, and length of latus rectum.

(b) Determine whether the ellipse is horizontal or vertical, and sketch it.

Q14.

(a) Express $4x^2 + 9y^2 - 24x + 36y + 36 = 0$ in standard form. Hence find the centre, vertices, and foci.

(b) Find the equation of the ellipse with centre $(1, -2)$, focus $(1, 0)$, and vertex $(1, 3)$.

Hyperbola

Q15.

(a) A hyperbola has equation $\displaystyle \frac{x^2}{16} - \frac{y^2}{9} = 1$. Find the centre, vertices, foci, asymptotes, and length of latus rectum.

(b) Sketch the hyperbola showing all key features.

Q16.

(a) Express $9x^2 - 4y^2 - 36x - 32y - 64 = 0$ in standard form. Hence determine the centre, vertices, foci, and asymptotes.

(b) Find the equation of the hyperbola with foci $(\pm 5, 0)$ and vertices $(\pm 3, 0)$.

SOLUTIONS

Q1. (a) $\displaystyle \int (4x^3 - 3e^{2x} + \frac{5}{x}),dx = \frac{4x^4}{4} - \frac{3e^{2x}}{2} + 5\ln|x| + C = x^4 - \frac{3}{2}e^{2x} + 5\ln|x| + C$

(b) $\displaystyle \int \frac{6x^2 + 2}{x^3 + x},dx$ Let $u = x^3 + x$, $du = (3x^2 + 1),dx$ Note $6x^2 + 2 = 2(3x^2 + 1) = 2,du/dx$ $$\displaystyle \int \frac{2,du}{u} = 2\ln|x^3 + x| + C$$

(c) $\displaystyle \int_1^2 \frac{x}{\sqrt{2x^2 + 1}},dx$ Let $u = 2x^2 + 1$, $du = 4x,dx$, $x,dx = du/4$ When $x = 1$, $u = 3$; when $x = 2$, $u = 9$ $$\displaystyle \int_3^9 \frac{1}{\sqrt{u}} \cdot \frac{du}{4} = \frac14 \int_3^9 u^{-1/2},du = \frac14 [2u^{1/2}]_3^9 = \frac12(3 - \sqrt{3})$$

Q2. (a) $\displaystyle \int 3x^2\sqrt{x^3 + 4},dx$ Let $u = x^3 + 4$, $du = 3x^2,dx$ $$\displaystyle \int \sqrt{u},du = \frac{2}{3}u^{3/2} + C = \frac{2}{3}(x^3 + 4)^{3/2} + C$$

(b) $\displaystyle \int_{-1}^2 |x|,dx = \int_{-1}^0 (-x),dx + \int_0^2 x,dx = \left[-\frac{x^2}{2}\right]_{-1}^0 + \left[\frac{x^2}{2}\right]_0^2$ $$= (0 + \frac12) + (2 - 0) = \frac12 + 2 = \frac52$$

(c) $y = 4 - x^2$ cuts $x$-axis at $x = \pm 2$. $$V = \pi \int_{-2}^2 (4 - x^2)^2,dx = \pi \int_{-2}^2 (16 - 8x^2 + x^4),dx$$ $$= \pi \left[16x - \frac{8x^3}{3} + \frac{x^5}{5}\right]_{-2}^2$$ $$= \pi\left[(32 - \frac{64}{3} + \frac{32}{5}) - (-32 + \frac{64}{3} - \frac{32}{5})\right]$$ $$= \pi\left[64 - \frac{128}{3} + \frac{64}{5}\right] = \pi\left[\frac{960 - 640 + 192}{15}\right] = \frac{512\pi}{15}$$

Q3. (a) Area $= \int_0^{\pi} \sin x,dx = [-\cos x]_0^{\pi} = (-\cos\pi) - (-\cos 0) = 1 + 1 = 2$

(b) Intersection: $x^2 = 2x + 3$ → $x^2 - 2x - 3 = 0$ → $(x-3)(x+1) = 0$ → $x = -1, 3$ Area $= \int_{-1}^3 [(2x+3) - x^2],dx = \left[x^2 + 3x - \frac{x^3}{3}\right]_{-1}^3$ $$= (9 + 9 - 9) - (1 - 3 + \frac13) = 9 - (-\frac53) = 9 + \frac53 = \frac{32}{3}$$

Q4. (a) Area $= \int_0^4 \sqrt{x},dx = \left[\frac{2}{3}x^{3/2}\right]_0^4 = \frac{2}{3}(8 - 0) = \frac{16}{3}$

(b) Area $= \int_0^{\pi/4} (\cos x - \sin x),dx = [\sin x + \cos x]_0^{\pi/4}$ $$= (\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}) - (0 + 1) = \frac{2}{\sqrt{2}} - 1 = \sqrt{2} - 1$$

Q5. (a) Order $= 2$ (highest derivative is $\frac{d^2y}{dx^2}$), Degree $= 3$ (power of highest derivative after removing radicals)

(b) $y = Ae^{2x} + Be^{-3x}$ $$y' = 2Ae^{2x} - 3Be^{-3x}$$ $$y'' = 4Ae^{2x} + 9Be^{-3x}$$ LHS: $y'' + y' - 6y = (4Ae^{2x} + 9Be^{-3x}) + (2Ae^{2x} - 3Be^{-3x}) - 6(Ae^{2x} + Be^{-3x})$ $= (4A + 2A - 6A)e^{2x} + (9B - 3B - 6B)e^{-3x} = 0$ ✓

Q6. (a) $y,dy = x,dx$ $\int y,dy = \int x,dx$ → $\frac{y^2}{2} = \frac{x^2}{2} + C$ → $y^2 = x^2 + 2C$ $y(0) = 2$ → $4 = 0 + 2C$ → $C = 2$ $y^2 = x^2 + 4$ → $y = \sqrt{x^2 + 4}$

(b) $y^2,dy = (x^2 + 1),dx$ $\int y^2,dy = \int (x^2 + 1),dx$ → $\frac{y^3}{3} = \frac{x^3}{3} + x + C$ $y(1) = 2$ → $\frac{8}{3} = \frac13 + 1 + C$ → $\frac{8}{3} = \frac43 + C$ → $C = \frac43$ $y^3 = x^3 + 3x + 4$ → $y = \sqrt[3]{x^3 + 3x + 4}$

Q7. (a) $\displaystyle \lim_{k\to\infty} \frac{3k^2 + 1}{2k^2 + 5} = \lim_{k\to\infty} \frac{3 + 1/k^2}{2 + 5/k^2} = \frac{3}{2}$ Since the limit exists and is finite, the sequence converges to $\frac32$.

(b) $\displaystyle \sum_{r=1}^{20} (3r^2 - 2r + 1) = 3\sum r^2 - 2\sum r + \sum 1$ $$= 3\cdot\frac{20\cdot21\cdot41}{6} - 2\cdot\frac{20\cdot21}{2} + 20$$ $$= 3\cdot\frac{17220}{6} - 420 + 20 = 8610 - 420 + 20 = 8210$$

Q8. (a) $\displaystyle \frac{1}{r(r+1)} = \frac{1}{r} - \frac{1}{r+1}$

(b) $\displaystyle \sum_{r=1}^{n} \left(\frac{1}{r} - \frac{1}{r+1}\right) = \left(1 - \frac12\right) + \left(\frac12 - \frac13\right) + \cdots + \left(\frac1n - \frac{1}{n+1}\right)$ $$= 1 - \frac{1}{n+1} = \frac{n}{n+1}$$

Q9. (a) $(2x - 3y)^4 = \binom{4}{0}(2x)^4 + \binom{4}{1}(2x)^3(-3y) + \binom{4}{2}(2x)^2(-3y)^2 + \binom{4}{3}(2x)(-3y)^3 + \binom{4}{4}(-3y)^4$ $$= 16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4$$

(b) General term: $\binom{9}{r}(x^2)^{9-r}\left(\frac{2}{x}\right)^r = \binom{9}{r} x^{18-2r} \cdot 2^r x^{-r} = \binom{9}{r} 2^r x^{18-3r}$ For term independent of $x$: $18 - 3r = 0$ → $r = 6$ Term: $\binom{9}{6} 2^6 = 84 \cdot 64 = 5376$

Q10. (a) General term: $\binom{8}{r}(1)^{8-r}(2x)^r = \binom{8}{r} 2^r x^r$ Coefficient of $x^5$: $\binom{8}{5} 2^5 = 56 \cdot 32 = 1792$

(b) $(1.98)^6 = (2 - 0.02)^6 = 2^6\left(1 - \frac{0.02}{2}\right)^6 = 64(1 - 0.01)^6$ $$(1 - 0.01)^6 = 1 + 6(-0.01) + 15(-0.01)^2 + 20(-0.01)^3 + \cdots$$ $$= 1 - 0.06 + 0.0015 - 0.00002 + \cdots$$ $$= 0.94148$$ $$(1.98)^6 \approx 64 \times 0.94148 = 60.255$$

Q11. (a) Vertex $(2,3)$, focus $(2,5)$ — vertical parabola opening upward $a = 2$ (distance from vertex to focus) Standard form: $(x - 2)^2 = 8(y - 3)$ Directrix: $y = 3 - 2 = 1$ Length of latus rectum $= |4a| = 8$

(b) Vertex $(2,3)$, opens upward. Axis $x = 2$. Latus rectum endpoints at $x = 2 \pm 4 = -2, 6$ at $y = 5$.

Q12. (a) $y^2 + 8x - 6y + 25 = 0$ $$y^2 - 6y + 9 = -8x - 25 + 9$$ $$(y - 3)^2 = -8x - 16 = -8(x + 2)$$ $$(y - 3)^2 = -8(x + 2)$$ Vertex: $(-2, 3)$ $4a = 8$ → $a = 2$ Opens left Focus: $(-2 - 2, 3) = (-4, 3)$ Directrix: $x = -2 + 2 = 0$

(b) Focus $(-1, 2)$, directrix $x = 3$ Vertex is midpoint between focus and directrix: $x = \frac{-1 + 3}{2} = 1$ Vertex: $(1, 2)$ $a =$ distance from vertex to focus $= |1 - (-1)| = 2$ Opens left $$(y - 2)^2 = -8(x - 1)$$

Q13. (a) $\frac{x^2}{25} + \frac{y^2}{9} = 1$ Centre: $(0, 0)$, $a = 5$, $b = 3$ $c^2 = a^2 - b^2 = 25 - 9 = 16$ → $c = 4$ Horizontal (since $a > b$) Vertices: $(\pm 5, 0)$ Foci: $(\pm 4, 0)$ Eccentricity: $e = \frac{c}{a} = \frac{4}{5}$ Latus rectum: $\frac{2b^2}{a} = \frac{2(9)}{5} = \frac{18}{5}$

(b) Horizontal ellipse.

Q14. (a) $4x^2 + 9y^2 - 24x + 36y + 36 = 0$ $$4(x^2 - 6x) + 9(y^2 + 4y) = -36$$ $$4[(x - 3)^2 - 9] + 9[(y + 2)^2 - 4] = -36$$ $$4(x - 3)^2 - 36 + 9(y + 2)^2 - 36 = -36$$ $$4(x - 3)^2 + 9(y + 2)^2 = 36$$ $$\frac{(x - 3)^2}{9} + \frac{(y + 2)^2}{4} = 1$$ Centre: $(3, -2)$, $a = 3$, $b = 2$ $c^2 = a^2 - b^2 = 9 - 4 = 5$ → $c = \sqrt{5}$ Vertices: $(3 \pm 3, -2)$ → $(0, -2)$ and $(6, -2)$ Foci: $(3 \pm \sqrt{5}, -2)$

(b) Centre $(1, -2)$, focus $(1, 0)$, vertex $(1, 3)$ — vertical $a = 3$, $c = 2$ $$b^2 = a^2 - c^2 = 9 - 4 = 5$$ $$\frac{(x - 1)^2}{5} + \frac{(y + 2)^2}{9} = 1$$

Q15. (a) $\frac{x^2}{16} - \frac{y^2}{9} = 1$ Centre: $(0, 0)$, $a = 4$, $b = 3$ $c^2 = a^2 + b^2 = 16 + 9 = 25$ → $c = 5$ Horizontal Vertices: $(\pm 4, 0)$ Foci: $(\pm 5, 0)$ Asymptotes: $y = \pm \frac{b}{a}x = \pm \frac{3}{4}x$ Latus rectum: $\frac{2b^2}{a} = \frac{2(9)}{4} = \frac{9}{2}$

(b) Sketch showing centre at origin, vertices at $(\pm 4, 0)$, asymptotes $y = \pm \frac34 x$, two branches.

Q16. (a) $9x^2 - 4y^2 - 36x - 32y - 64 = 0$ $$9(x^2 - 4x) - 4(y^2 + 8y) = 64$$ $$9[(x - 2)^2 - 4] - 4[(y + 4)^2 - 16] = 64$$ $$9(x - 2)^2 - 36 - 4(y + 4)^2 + 64 = 64$$ $$9(x - 2)^2 - 4(y + 4)^2 = 36$$ $$\frac{(x - 2)^2}{4} - \frac{(y + 4)^2}{9} = 1$$ Centre: $(2, -4)$, $a = 2$, $b = 3$ $c^2 = a^2 + b^2 = 4 + 9 = 13$ → $c = \sqrt{13}$ Horizontal Vertices: $(2 \pm 2, -4)$ → $(0, -4)$ and $(4, -4)$ Foci: $(2 \pm \sqrt{13}, -4)$ Asymptotes: $y + 4 = \pm \frac{3}{2}(x - 2)$ → $y = \frac{3}{2}x - 7$ and $y = -\frac{3}{2}x - 1$

(b) Foci $(\pm 5, 0)$, vertices $(\pm 3, 0)$ — centre at $(0, 0)$, horizontal $a = 3$, $c = 5$ $b^2 = c^2 - a^2 = 25 - 9 = 16$ → $b = 4$ $$\frac{x^2}{9} - \frac{y^2}{16} = 1$$