My suggestion is that you don't check your answers. Checking can't add a single mark, but a mistake you find can stay on your mind for the rest of your O Levels.

— Kenneth

(a)

\begin{align*} u & = x^2 &&& v & = \sin x \\ \frac{du}{dx} & = 2x &&& \frac{dv}{dx} & = \cos x \end{align*} \begin{align*} \frac{d}{dx} (x^2 \sin x) & = (x^2)(\cos x) + (\sin x)(2x) \qquad [\text{Product rule}] \\ & = x^2 \cos x + 2x \sin x \end{align*}

(b)

\begin{align*} \int \frac{3}{2x + 1} \, dx & = 3 \int \frac{1}{2x + 1} \, dx \\ & = 3 \left[ \frac{ \ln (2x + 1) }{2} \right] \qquad \left[ \int \frac{1}{f(x)} \, dx = \frac{ \ln [f(x)] }{ f'(x) } \right] \\ & = \frac{3}{2} \ln (2x + 1) + c \end{align*}

(a)(i)

\begin{align*} x^2 + 5x & = k \\ \\ \text{Let } x = \frac{-5 + \sqrt{57}}{2},& \\ \left( \frac{-5 + \sqrt{57}}{2} \right)^2 + 5 \left(\frac{-5 + \sqrt{57}}{2}\right) & = k \\ \frac{ (-5 + \sqrt{57} )^2 } { (2)^2 } + \frac{-25 + 5 \sqrt{57}}{2} & = k \\ \frac{ (-5)^2 + 2(-5)(\sqrt{57}) + (\sqrt{57})^2 } {4} + \frac{ 2(-25 + 5 \sqrt{57})}{4} & = k \\ \frac{ 25 - 10 \sqrt{57} + 57 - 50 + 10 \sqrt{57} }{4} & = k \\ \frac{ 32 } {4} & = k \\ 8 & = k \end{align*}

(a)(ii)

\begin{align*} x & = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a } \\ \\ \text{First solution: } x & = \frac{-5 + \sqrt{57}}{2} \\ \\ \text{Other solution: } x & = \frac{-5 - \sqrt{57}}{2} \end{align*}

(b)

\begin{align*} \text{Gradient} & = \frac{y_2 - y_1}{x_2 - x_1} \\ & = \frac{ 4 \sqrt{2} - \sqrt{2} }{ \sqrt{2} - 1 } \\ & = \frac{ 3 \sqrt{2} }{ \sqrt{2} - 1} \times \frac{ \sqrt{2} + 1 }{ \sqrt{2} + 1 } \qquad [\text{Rationalise denominator}] \\ & = \frac{ 3 \sqrt{2} (\sqrt{2} + 1) }{ (\sqrt{2} - 1)(\sqrt{2} + 1) } \qquad [ (a - b)(a + b) = a^2 - b^2 ] \\ & = \frac{ 3(2) + 3 \sqrt{2} } { (\sqrt{2})^2 - (1)^2 } \\ & = \frac{ 6 + 3 \sqrt{2} }{ 2 - 1} \\ & = \frac{ 6 + 3 \sqrt{2} } {1} \\ & = 6 + 3 \sqrt{2} \end{align*}

(a)

(b)

\begin{align*} \underbrace{2^{n - 1}}_\text{Current level} + \underbrace{2^{n - 2}}_\text{Previous level} & > 8000 \\ \frac{2^n}{2^1} + \frac{2^n}{2^2} & > 8000 \qquad \left[ a^{m - n} = \frac{a^m}{a^n} \right] \\ \frac{2^n}{2} + \frac{2^n}{4} & > 8000 \\ 4 \left( \frac{2^n}{2} + \frac{2^n}{4} \right) & > 4(8000) \\ 2(2^n) + 2^n & > 32 \, 000 \\ 2^n (2 + 1) & > 32 \, 000 \\ 2^n (3) & > 32 \, 000 \\ 2^n & > \frac{32 \, 000}{3} \\ \lg 2^n & > \lg \frac{32 \, 000}{3} \\ n \lg 2 & > \lg \frac{32 \, 000}{3} \qquad [\text{Power law (logarithms)}] \\ n & > \frac{ \lg \frac{32 \, 000}{3} }{ \lg 2} \qquad [\text{Note } \lg 2 \text{ is positive}] \\ n & > 13.381 \\ \\ \therefore & \, \text{ Level } 14 \qquad [\text{Smallest integer satisfying } n > 13.381] \end{align*}

\begin{align*} \text{Substitute } y = 2x - 3 & \text{ into } \frac{x}{y} + \frac{2}{x} = 1, \\ \frac{x}{2x - 3} + \frac{2}{x} & = 1 \\ \frac{x(x)}{x(2x - 3)} + \frac{2(2x - 3)}{x(2x - 3)} & = 1 \\ \frac{x(x) + 2(2x - 3)}{x(2x - 3)} & = 1 \\ \frac{x^2 + 4x - 6}{2x^2 - 3x} & = \frac{1}{1} \\ x^2 +4 x - 6 & = 2x^2 - 3x \\ 0 & = x^2 - 7x + 6 \\ 0 & = (x - 1)(x - 6) \end{align*} \begin{align*} x - 1 & = 0 & \text{ or } && x - 6 & = 0 \\ x & = 1 &&&x & = 6 \\ \\ \text{Substitute } & x = 1 \text{ into } y = 2x - 3, &&& \text{Substitute } & x = 6 \text{ into } y = 2x - 3, \\ y & = 2(1) - 3 &&& y & = 2(6) - 3 \\ y & = -1 &&& y & = 9 \end{align*} \begin{align*} \therefore & \phantom{.} (1, -1) \text{ and } (6, 9) \end{align*}

(a)

\begin{align*} 5x^2 - 10x + k & = 5(x^2 - 2x) + k \\ & = 5\left[ x^2 - 2x + \left( \frac{2}{2} \right)^2 - \left( \frac{2}{2} \right)^2 \right] + k \qquad [\text{Complete the square}] \\ & = 5 [ (x - 1)^2 - 1] + k \\ & = 5(x - 1)^2 - 5 + k \\ & = 5(x - 1)^2 + k - 5 \end{align*}

(b)(i)

\begin{align*} y & = 5x^2 - 10x + k \\ y & = 5(x - 1)^2 + k - 5 \qquad [\text{General form: } y = a(x - h)^2 + k] \\ \\ \text{Turning} & \text{ point: } (1, k - 5) \end{align*}

(b)(ii)

\begin{align*} y = 5(x - 1)^2 & - 5 + k \\ \\ \text{For all} & \text{ real values of } x, \\ (x - 1)^2 & \ge 0 \\ 5(x - 1)^2 & \ge 0 \\ 5(x - 1)^2 + k - 5 & \ge k - 5 \\ \\ \text{Since the minimum value of } & y \text{ is } k - 5, \text{ the curve has a minimum point} \end{align*}

(b)(iii)

\begin{align*} \text{Minimum} \text{ point } & (1, k - 5) \text{ must lie on or above } y = 3 \\ \\ k - 5 & \ge 3 \\ k & \ge 8 \end{align*}

(a)

\begin{align*} [ \text{Distance between two points} & = \sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 }] \\ \\ \text{Radius} & = \text{Distance between centre (2, 3) } \\ & \qquad \text{and (5, -4) on circumference of outer circle} \\ & = \sqrt{(5 - 2)^2 + (-4 - 3)^2} \\ & = \sqrt{58} \\ \\ \therefore (x - 2)^2 & + (y - 3)^2 = (\sqrt{58})^2 \\ (x - 2)^2 & + (y - 3)^2 = 58 \end{align*}

(b)

\begin{align*} \text{Let radius of inner circle} & = r \\ \\ (x - 2)^2 + (y - 3)^2 & = r^2 \\ x^2 - 2(x)(2) + 2^2 + y^2 - 2(y)(3) + 3^2 & = r^2 \\ x^2 - 4x + 4 + y^2 - 6y + 9 & = r^2 \\ x^2 + y^2 - 4x - 6y + 13 - r^2 & = 0 \\ \\ \text{Comparing coefficients with } x^2+ y^2 - px - qy &- 23 = 0, \end{align*} \begin{align*} -px & = -4x &&& -qy & = -6y &&& 13 - r^2 & = -23 \\ p & = 4 &&& q & = 6 &&& -r^2 & = -36 \\ & &&& & &&& r^2 & = 36 \\ & &&& & &&& r & = \pm \sqrt{36} \\ & &&& & &&& r & = 6 \text{ or } - 6 \text{ (NA)} \end{align*} $$ \therefore p = 4, q = 6 \text{ and radius} = 6 \text{ units} $$

(a)

\begin{align*} V & = \underbrace{\frac{1}{2} \left( \frac{4}{3} \pi r^3 \right)}_\text{Hemisphere} + \underbrace{\pi r^2 h}_\text{Cylinder} \\ & = \frac{2}{3} \pi r^3 + \pi r^2 \underline{h} \\ \\ [\text{Need to } & \text{form an equation linking } h \text{ and } r] \\\\ \text{Curved surface area of hemisphere} & = \frac{1}{2} \left( 4 \pi r^2 \right) \\ & = 2 \pi r^2 \\ \\ \text{Curved surface area of cylinder} & = 2 \pi r h \\ \\ \text{Surface area} & = 2 \pi r^2 + 2\pi rh + \underbrace{\pi r^2}_\text{Base of cylinder} \\ 80\pi & = 3\pi r^2 + 2\pi rh \\ - 2\pi rh & = 3 \pi r^2 - 80 \pi \\ 2 \pi r h & = 80\pi - 3 \pi r^2 \\ h & = \frac{80 \pi - 3 \pi r^2}{2\pi r} \\ h & = \frac{80\pi}{2 \pi r} - \frac{3 \pi r^2}{2 \pi r} \\ h & = \frac{40}{r} - \frac{3}{2}r \\ \\ \therefore V & = \frac{2}{3}\pi r^3 + \pi r^2 \left(\frac{40}{r} - \frac{3}{2}r \right) \\ & = \frac{2}{3}\pi r^3 + \frac{40\pi r^2}{r} - \frac{3}{2} \pi r^3 \\ & = 40 \pi r - \frac{5}{6} \pi r^3 \text{ (Shown)} \end{align*}

(b)

\begin{align*} V & = 40\pi r - \frac{5}{6}\pi r^3 \\ \frac{dV}{dr} & = 40\pi - \frac{5}{6}\pi(3r^2) \\ & = 40\pi - \frac{5}{2}\pi r^2 \\ \\ \text{Let } & \frac{dV}{dr} = 0, \qquad [\text{Maximum volume} \implies \text{Stationary value}] \\ 40\pi - \frac{5}{2}\pi r^2 & = 0 \\ - \frac{5}{2}\pi r^2 & = -40 \pi \\ \frac{5}{2}\pi r^2 & = 40\pi \\ r^2 & = \frac{40\pi}{ \frac{5}{2} \pi} \\ r^2 & = 16 \\ r & = \pm\sqrt{16} \\ r & = 4 \text{ or } -4 \text{ (NA)} \\ \\ \text{Substitute } r = 4 & \text{ into } h = \frac{40}{r} - \frac{3}{2}r \\ h & = \frac{40}{4} - \frac{3}{2}(4) \\ & = 4 \end{align*} $$ \therefore \text{Radius} = 4 \text{ m and height of cylinder} = 4 \text{ m} $$

(a)

\begin{align*} \require{cancel} \text{LHS} & = \tan x(\sin x + \cos x)(\operatorname{cosec} x + \sec x) \\ & = \frac{\sin x}{\cos x}(\sin x + \cos x)\left(\frac{1}{\sin x} + \frac{1}{\cos x}\right) \\ & = \frac{\sin x}{\cos x} \left( \frac{\sin x + \cos x}{1} \right)\left(\frac{\cos x + \sin x}{\sin x\cos x}\right) \\ & = \frac{ \sin x (\sin x + \cos x)^2 }{ \cos x( \sin x \cos x)} \\ & = \frac{ \cancel{\sin x} (\sin x + \cos x)^2 }{ \cancel{\sin x} \cos^2 x} \\ & = \frac{(\sin x + \cos x)^2}{\cos^2 x} \\ & = \frac{ \sin^2 x + 2 \sin x \cos x + \cos^2 x }{\cos^2 x} \\ & = \frac{\sin^2 x}{\cos^2 x} + \frac{2 \sin x \cancel{\cos x}}{\cos^\cancel{2} x} + \frac{\cos^2 x}{\cos^2 x} \\ & = \tan^2 x + \frac{2 \sin x}{\cos x} + 1 \\ & = \tan^2 x + 2 \tan x + 1 \\ & = (\tan x + 1)(\tan x + 1) \\ & = (\tan x + 1)^2 \\ & = \text{RHS} \end{align*}

(b)

\begin{align*} (\sin x + \cos x)(\operatorname{cosec} x + \sec x) & = 3(\tan x + 1) \\ \frac{(\tan x + 1)^2}{\tan x} & = 3(\tan x + 1) \qquad [\text{Using (a)}] \\ (\tan x + 1)^2 & = 3 \tan x(\tan x + 1) \\ (\tan x + 1)^2 - 3 \tan x(\tan x + 1) & = 0 \\ (\tan x + 1)[ (\tan x + 1) - 3 \tan x ] & = 0 \\ (\tan x + 1)(1 - 2 \tan x) & = 0 \\ \\ \tan x + 1 & = 0 \qquad \text{or} \qquad 1 - 2 \tan x = 0 \\ \\ \\ \tan x + 1 & = 0 \\ \tan x & = -1 \qquad [\text{2nd or 4th quadrants}] \\ \\ \alpha & = \tan^{-1}(1) \\ & = 45^\circ \end{align*}

\begin{align*} x & = 180^\circ - \alpha \text{ or } 360^\circ - \alpha \\ x & = 135^\circ \text{ or } 315^\circ \\ \\ \\ 1 - 2 \tan x & = 0 \\ -2 \tan x & =- 1 \\ \tan x & = \frac{-1}{-2} \\ \tan x & = \frac{1}{2} \qquad [\text{1st or 3rd quadrants}] \\ \\ \alpha & = \tan^{-1}\left(\frac{1}{2}\right) \\ & \approx 26.565^\circ \\ \end{align*}

\begin{align*} x & = \alpha \text{ or } 180^\circ + \alpha \\ x & \approx 26.565^\circ \text{ or } 206.565^\circ \\ x & \approx 26.6^\circ \text{ or } 206.6^\circ \text{ (1 d.p.)} \end{align*}
$$ \therefore x = 26.6^\circ,\ 135^\circ,\ 206.6^\circ \text{ or } 315^\circ $$

(a)

\begin{align*} \text{Let } & y = 0, \\ 2\sin 3x + 3\cos 3x & = 0 \\ 2\sin 3x & = -3\cos 3x \\ \frac{\sin 3x}{\cos 3x} & = -\frac{3}{2} \\ \tan 3x & = -\frac{3}{2} \qquad [\text{2nd or 4th quadrants}] \\ \\ \alpha & = \tan^{-1}\left(\frac{3}{2}\right) \\ & \approx 0.98279 \text{ radians} \\ \end{align*}

\begin{align*} 3x & \approx \pi - 0.98279 \text{ or } 2\pi - 0.98279 \\ 3x & \approx 2.1588 \text{ or } 5.3004 \\ x & \approx 0.7196 \text{ or } 1.7668 \\ \\ x \text{-coor} & \text{dinate of } B \approx 1.767 \text{ (4 sf)} \end{align*}

(b)

\begin{align*} \text{Area of } P & = \int_0^{x_A} 2 \sin 3x + 3 \cos 3x \, dx \\ \\ \text{Area of } Q & = - \int_{x_A}^{x_B} 2 \sin 3x + 3 \cos 3x \, dx \qquad [Q \text{ is below } x \text{-axis}] \\ \\ P - Q & = \int_0^{x_A} 2 \sin 3x + 3 \cos 3x \, dx - \left( - \int_{x_A}^{x_B} 2 \sin 3x + 3 \cos 3x \, dx \right) \\ & = \int_0^{x_A} 2 \sin 3x + 3 \cos 3x \, dx + \int_{x_A}^{x_B} 2 \sin 3x + 3 \cos 3x \, dx \\ & = \int_0^{x_B} 2 \sin 3x + 3 \cos 3x \, dx \\ \\ \therefore x &\text{-coordinate} \text{ of } A \text{ is not necessary} \end{align*}

(c)

\begin{align*} \text{From (b), } P - Q & = \int_0^{x_B} 2 \sin 3x + 3 \cos 3x \, dx \\ & \approx \int_0^{1.767} 2\sin 3x + 3 \cos 3x \, dx \\ & = \left[ \frac{-2\cos 3x}{3} + \frac{3\sin 3x}{3} \right]_0^{1.767} \\ & = \left[ -\frac{2}{3}\cos 3x + \sin 3x \right]_0^{1.767} \\ & = -\frac{2}{3} \cos [3(1.767)] + \sin [3(1.767)] - \left( -\frac{2}{3} \cos 0 + \sin 0 \right) \\ & \approx -0.53518 \\ & \approx -0.535 \end{align*}

(d)

Since \(P - Q < 0\), \(P < Q\). The area of \(P\) is smaller than the area of \(Q\).

(a)

(b)(i)

\begin{align*} \text{Since } L \text{ is the } & \text{line of symmetry of } ABCD, \\ L \text{ is perpen} & \text{dicular to both } AD \text{ and } BC. \\ \\ L : & \, y = x + 8 \text{ --- (1)} \\ \\ \text{Gradient of } L & = 1 \qquad [y = mx + c] \\ \\ \text{Gradient of } AD & = \frac{-1}{1} \qquad [m_1 \times m_2 = -1] \\ & = -1 \\ \\ y & = mx + c \\ y & = -x + c \\ \\ \text{Using } & A(-3,13), \\ 13 & = -(-3) + c \\ 13 & = 3 + c \\ 10 & = c \\ \\ AD: & \, y = -x + 10 \text{ --- (2)} \\ \\ \text{Substitute } & (1) \text{ into (2),} \\ x + 8 & = -x + 10 \\ 2x & = 2 \\ x & = 1 \\ \\ \text{Substitute } & x = 1 \text{ into (1),} \\ y & = 1 + 8 \\ y & = 9 \\ \\ \implies \text{Midpoint of } AD & = (1, 9) \\ \\ \text{Let coordinates of } & D \text{ be } (x, y) \\ \\ \left( \frac{-3 + x}{2}, \frac{13 + y}{2} \right) & = (1, 9) \qquad [\text{Midpoint formula}] \end{align*} \begin{align*} \frac{-3 + x}{2} & = 1 &&& \frac{13 + y}{2} & = 9 \\ -3 + x & = 2 &&& 13 + y & = 18 \\ x & = 5 &&& y & = 5 \end{align*}
$$ \therefore D(5,5) $$

(b)(ii)

\begin{align*} A(-3,13) & \text{ and } AB = 2 \text{ units} \\ \\ \implies & B(-1, 13) \\ \\ D(5,5) & \text{ and } CD = AB = 2 \text{ units} \\ \\ \implies & C(5, 7) \\ \\ \text{Area of trapezium} & = \frac{1}{2}\left|\begin{array}{rrrrr} -1 & -3 & 5 & 5 & -1 \\ 13 & 13 & 5 & 7 & 13 \end{array}\right| \qquad [\text{Anti-clockwise order: } B, A, D, C, B]\\ & = \frac{1}{2}[(-1)(13)+(-3)(5)+(5)(7)+(5)(13)] \\ & \qquad - \frac{1}{2} [(13)(-3)+(13)(5)+(5)(5)+(7)(-1)] \\ & = 36 - 22 \\ & = 14 \text{ units}^2 \end{align*}