(a)

(b)

(c)

(a)

(b)

(c)

(a)

\begin{align*} \frac{d}{dx} (y) & = \frac{d}{dx}(vx) \qquad \qquad [\text{Since } y = vx] \\ \frac{dy}{dx} & = v(1) + (x)\frac{dv}{dx} \qquad [\text{Product rule}] \\ \frac{dy}{dx} & = v + x \frac{dv}{dx} \\ \\ x^2 \frac{dy}{dx} & = y^2 + xy \\ \implies x^2 \left( v + x \frac{dv}{dx} \right) & = (vx)^2 + x(vx) \\ vx^2 + x^3 \frac{dv}{dx} & = v^2 x^2 + vx^2 \\ x^3 \frac{dv}{dx} & = v^2 x^2 \\ \frac{ x^3 \frac{dv}{dx} }{x^2 } & = v^2 \\ x \frac{dv}{dx} & = v^2 \qquad (x \ne 0) \text{ (Shown)} \end{align*}

(b)

\begin{align*} x \frac{dv}{dx} & = v^2 \\ \frac{1}{v^2} \frac{dv}{dx} & = \frac{1}{x} \qquad (v \ne 0) \\ \int \frac{1}{v^2} \frac{dv}{dx} \, dx & = \int \frac{1}{x} \, dx \\ \int v^{-2} \, dv & = \int \frac{1}{x} \, dx \\ \frac{v^{-1}}{-1} & = \ln |x| + c \\ - \frac{1}{v} & = \ln |x| + c \\ - \frac{1}{\frac{y}{x}} & = \ln |x| + c \qquad [\text{Since } y = vx, \text{ then } v = \frac{y}{x}] \\ - \frac{x}{y} & = \ln |x| + c \\ - x & = y(\ln |x| + c) \\ \\ \therefore y & = - \frac{x}{\ln|x| + c}, \text{ where } c \text{ is an arbitrary constant} \end{align*}

\begin{align*} 2z + 7i & = -iw \text{ --- (1)} \\ \\ z - 4 & = w - 4i \\ z & = w - 4i + 4 \text{ --- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 2(w - 4i + 4) + 7i & = -iw \\ 2w - 8i + 8 + 7i & = -iw \\ 2w + iw & = i - 8 \\ w(2 + i) & = i - 8 \\ w & = \frac{i - 8}{2 + i} \times \frac{2 - i}{2 - i} \\ w & = \frac{ 2i - i^2 - 16 + 8i }{ 2^2 - i^2 } \\ w & = \frac{ 10i + 1 - 16 } {4 + 1} \\ w & = \frac{ 10i - 15 } {5} \\ w & = \frac{10i}{5} - \frac{15}{5} \\ w & = -3 + 2i \\ \\ \text{Substitute } & w = -3 + 2i \text{ into (2),} \\ z & = -3 + 2i - 4i + 4 \\ z & = 1 - 2i \\ \\ \therefore z = 1 - 2i, & \, w = -3 + 2i \end{align*}

(a)

(b)

(a)

\begin{align*} [ \frac{d}{dx} [ e^{f(x)}] = f'(x) . e^{f(x)}, &\, \frac{d}{dx}[ \cos f(x)] = - f'(x) . \sin f(x) ] \\ \\ \frac{d}{dx} (e^{\cos 2x}) & = -2 \sin 2x . e^{\cos 2x} \\ & = -2 e^{\cos 2x} \sin 2x \end{align*}

(b)

\begin{align*} \int e^{\cos 2x} \sin 4x \, dx & = \int e^{\cos 2x} (2 \sin 2x \cos 2x) \, dx \qquad [\sin 2A = 2\sin A \cos A] \\ & = \int \underbrace{(-2e^{\cos 2x} \sin 2x)}_\text{Part a}(-\cos 2x) \, dx \\ \\ u & = - \cos 2x \\ \frac{du}{dx} & = - (-2 \sin 2x) = 2 \sin 2x \\ \\ \frac{dv}{dx} & = -2e^{\cos 2x} \sin 2x \\ v & = e^{\cos 2x} \qquad \qquad \qquad [\text{Reverse of differentiation from part a}] \\ \\ \int u \frac{dv}{dx} \, dx & = uv - \int v \frac{du}{dx} \, dx \qquad [\text{Integration by parts}] \\ \int (-\cos 2x)(-2e^{\cos 2x} \sin 2x) \, dx & = (- \cos 2x)(e^{\cos 2x}) - \int (e^{\cos 2x})(2 \sin 2x) \, dx \\ & = - e^{\cos 2x} \cos 2x + \int \underbrace{- 2 e^{\cos 2x} \sin 2x}_\text{Part a} \, dx \\ & = - e^{\cos 2x} \cos 2x + e^{\cos 2x} + c \\ \\ \therefore \int_\frac{\pi}{4}^\frac{\pi}{2} e^{\cos 2x} \sin 4x \, dx & = \left[ - e^{\cos 2x} \cos 2x + e^{\cos 2x} \right]_\frac{\pi}{4}^\frac{\pi}{2} \\ & = - e^{\cos \pi} \cos \pi + e^{\cos \pi} + e^{\cos \frac{\pi}{2}} \cos \frac{\pi}{2} - e^{\cos \frac{\pi}{2}} \\ & = -e^{-1} (-1) + e^{-1} + e^0 (0) - e^0 \\ & = e^{-1} + e^{-1} - 1 \\ & = 2e^{-1} - 1 \end{align*}

(a)

\begin{align*} \frac{y}{1} & = \frac{2x^2}{2x - 1} \\ y(2x - 1) & = 2x^2 \\ 2yx - y & = 2x^2 \\ 0 & = 2x^2 - 2yx + y \\ \\ b^2 - 4ac & = (-2y)^2 - 4(2)(y) \\ & = 4y^2 - 8y \\ \\ b^2 - 4ac & < 0 \qquad [\text{Values that } y \text{ cannot take} \implies \text{No real roots}] \\ 4y^2 - 8y & < 0 \\ y^2 - 2y & < 0 \\ y(y - 2) & < 0 \\ \\ 0 & < y < 2 \\ \\ \therefore \{ y \in \mathbb{R} : & \, 0 < y < 2 \} \end{align*}

(b)(i)

$$ \begin{array}{rr} & 2x + 1 \phantom{00000.} \\ \phantom{} & 2x - 1 ) \overline{4x^2 + 0x + 1} \phantom{0} \\ & -\underline{(4x^2 - 2x)\phantom{{}+1}} \\ & 2x + 1 \\ & -\underline{(2x - 1)} \\ & 2 \end{array} $$ \begin{align*} y & = \frac{4x^2 + 1}{2x - 1} \\ & = 2x + 1 + \frac{2}{2x - 1} \\ \\ \therefore a & = 2,\quad b = 1,\quad c = 2 \\ \\ \text{Oblique } & \text{asymptote: } y = 2x + 1 \\ \\ \text{Vertical } & \text{asymptote: } x = 0.5 \end{align*}

(b)(ii)

\begin{align*} y & = x + \frac{1}{2x - 1} \\ \\ (1) & \text{ Replace } y \text{ by } \frac{y}{2} \\ \frac{y}{2} & = x + \frac{1}{2x - 1} \\ y & = 2 \left( x + \frac{1}{2x - 1} \right) \\ y & = 2x + \frac{2}{2x - 1} \\ \\ (2) & \text{ Replace } y \text{ by } y - 1 \\ y - 1 & = 2x + \frac{2}{2x - 1} \\ \text{Curve } D: \, y & = 2x + 1 + \frac{2}{2x - 1} \\ \\ \\ \therefore (1) & \text{ Scale parallel to } y \text{-axis by factor of 2} \\ (2) & \text{ Translate by 1 unit in positive } y \text{-direction} \end{align*}

(a)

\begin{align*} f(x) & = \ln (a + a \sin x) \\ \\ f'(x) & = \frac{a \cos x}{a + a \sin x} \qquad [ \frac{d}{dx}[ \ln f(x)] = \frac{f'(x)}{f(x)} ] \\ & = \frac{a \cos x}{a(1 + \sin x)} \\ & = \frac{\cos x}{1 + \sin x} \\ \\ f''(x) & = \frac{ (1 + \sin x)(-\sin x) - (\cos x)(\cos x) }{ (1 + \sin x)^2 } \qquad [\text{Quotient rule}] \\ & = \frac{ -\sin x - \sin^2 x - \cos^2 x}{ (1 + \sin x)^2 } \\ & = \frac{ - \sin x - 1}{ (1 + \sin x)^2 } \qquad [\sin^2 A + \cos^2 A = 1] \\ & = \frac{ -(1 + \sin x)}{(1 + \sin x)^2} \\ & = \frac{ - 1 }{ 1 + \sin x } \text{ (Shown)} \\ \\ \therefore k & = -1 \end{align*}

(b)

\begin{align*} y|_{x = 0} = f(0) & = \ln (a + a \sin 0) \\ & = \ln a \\ \\ \frac{dy}{dx}|_{x = 0} = f'(0) & = \frac{\cos 0}{1 + \sin 0} \\ & = 1 \\ \\ \frac{d^2y}{dx^2}|_{x = 0} = f''(0) & = \frac{-1}{1 + \sin 0} \\ & = -1 \\ \\ f''(x) & = -(1 + \sin x)^{-1} \\ \\ f'''(x) & = -(-1)(1 + \sin x)^{-2} . (\cos x) \qquad [\text{Chain rule}] \\ & = \frac{ \cos x}{ (1 + \sin x)^2 } \\ \\ \frac{d^3y}{dx^3}|_{x = 0} = f'''(0) & = \frac{\cos 0}{ (1 + \sin 0)^2 } \\ & = 1 \\ \\ f(x) & = f(0) + f'(0) x + \frac{ f''(0)}{2!} x^2 + \frac{ f'''(0)}{3!}x^3 + ... \\ & = \ln a + (1)(x) + \frac{(-1)}{2} x^2 + \frac{(1)}{6}x^3 + ... \\ & = \ln a + x - \frac{1}{2}x^2 + \frac{1}{6}x^3 + ... \end{align*}

(c)(i)

\begin{align*} \int_0^{1.2} f(x) \, dx & \approx \int_0^{1.2} \left( \ln 2 + x - \frac{1}{2}x^2 + \frac{1}{6}x^3 \right) \, dx \\ & = [ (\ln 2)x + \frac{1}{2}x^2 - \frac{1}{6}x^3 + \frac{1}{24}x^4]_0^{1.2} \\ & = [ (\ln 2)(1.2) + \frac{1}{2}(1.2)^2 - \frac{1}{6}(1.2)^3 + \frac{1}{24}(1.2)^4 ] - 0 \\ & = 1.350176\ldots \\ & \approx 1.35 \text{ (2 d.p.)} \end{align*}

(c)(ii)

\begin{align*} \int_0^{1.2} f(x) \, dx & = \int_0^{1.2} \ln(2 + 2\sin x) \, dx \\ & = 1.322161\ldots \qquad [\text{From GC in radian mode}] \\ & \approx 1.32 \text{ (2 d.p.)} \end{align*}

(c)(iii)

(a)

(b)

\begin{align*} D_f & = R_{f^{-1}} = (-\infty, 3] \\ \\ R_f & = D_{f^{-1}} = [1, \infty) \\ \\ \text{Let } y & = e^{|3 - x|} \\ \ln y & = \ln e^{|3 - x|} \\ \ln y & = |3 - x| \ln e \\ \ln y & = |3 - x| (1) \\ \ln y & = |3 - x| \\ \pm \ln y & = 3 - x \\ x & = 3 \pm \ln y \\ \\ \text{Since } & x \le 3, \\ x & = 3 - \ln y \\ \\ f^{-1}(x) & = 3 - \ln x, x \in \mathbb{R}, x \ge 1 \end{align*}

(c)

\begin{align*} \text{From (b), } D_{f^{-1}} & = [1, \infty) \\ \\ \text{For } f[ f^{-1} (|x|) ] & \text{ to be defined,} \\ |x| & \in D_{f^{-1}} \\ |x| & \in [1, \infty) \\ |x| & \ge 1 \\ \\ \text{Since } f[f^{-1} (x)] & = x, \\ f[ f^{-1} (|x|) ] & = |x| \\ \\ \therefore f[ f^{-1} (|x|) ] \ge 1 & \iff |x| \ge 1 \\ \\ x \le -1 & \text{ or } x \ge 1 \end{align*}

(d)

\begin{align*} \text{For composite function } & f^{-1} g \text{ to exist, } R_g \subseteq D_{f^{-1}} \\ \\ \text{For } & x \ge 4, \\ \sqrt{x - 4} & \ge 0 \\ n + \sqrt{x - 4} & \ge n \\ \\ \implies R_g & = [n, \infty) \\ \\ \text{From (b), } D_{f^{-1}} & = [1, \infty) \\ \\ \text{To satisfy } R_g \subseteq D_{f^{-1}}, & \, n \ge 1 \\ \\ \text{Smallest value of } n & = 1 \end{align*}

(a)

(b)(i)

\begin{align*} \frac{dx}{d\theta} & = 2 + 2 \cos 2 \theta \\ \\ \frac{dy}{d\theta} & = 2 \cos \theta \\ \\ \frac{dy}{dx} & = \frac{dy}{d\theta} \div \frac{dx}{d\theta} \\ & = 2 \cos \theta \div (2 + 2 \cos 2 \theta) \\ & = \frac{2 \cos \theta}{1} \times \frac{1}{2 + 2 \cos 2 \theta} \\ & = \frac{2 \cos \theta}{2 + 2 \cos 2 \theta} \\ & = \frac{2 \cos \theta}{2 + 2(2 \cos^2 \theta - 1)} \qquad [ \cos 2A = 2 \cos^2 A - 1] \\ & = \frac{2 \cos \theta}{4 \cos^2 \theta} \\ & = \frac{1}{2\cos \theta} \\ & = \frac{1}{2} \left( \frac{1}{\cos \theta} \right) \\ & = \frac{1}{2} \sec \theta \qquad (\cos \theta \ne 0) \text{ (Shown)} \end{align*}

(b)(ii)

(b)(iii)

\begin{align*} x|_{\theta = \frac{\pi}{3}} & = 2 \left( \frac{\pi}{3}\right) + \sin \frac{2\pi}{3} \\ & = \frac{2\pi}{3} + \frac{\sqrt{3}}{2} \\ \\ y|_{\theta = \frac{\pi}{3}} & = 2 \sin \frac{\pi}{3} \\ & = 2 \left( \frac{\sqrt{3}}{2} \right) \\ & = \sqrt{3} \\ \\ \frac{dy}{dx}|_{\theta = \frac{\pi}{3}} & = \frac{1}{2\cos \frac{\pi}{3}} \\ & = 1 \\ \\ \text{The tangent at } & \left( \frac{2\pi}{3} + \frac{\sqrt{3}}{2}, \sqrt{3} \right) \text{ has gradient } 1 \\ \\ y - y_1 & = m(x - x_1) \\ y - \sqrt{3} & = (1)\left(x - \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \right) \\ y & = x - \frac{2\pi}{3} - \frac{\sqrt{3}}{2} + \sqrt{3} \\ \text{Equation of tangent: } y & = x - \frac{2\pi}{3} + \frac{\sqrt{3}}{2} \\ \\ \text{Let } & x = 0, \\ y & = - \frac{2\pi}{3} + \frac{\sqrt{3}}{2} \\ \\ \implies & \, S \left(0, - \frac{2\pi}{3} + \frac{\sqrt{3}}{2} \right) \\ \\ \text{Let } & y = 0, \\ 0 & = x - \frac{2\pi}{3} + \frac{\sqrt{3}}{2} \\ -x & = - \frac{2\pi}{3} + \frac{\sqrt{3}}{2} \\ x & = \frac{2\pi}{3} - \frac{\sqrt{3}}{2} \\ \\ \implies & \, R \left( \frac{2\pi}{3} - \frac{\sqrt{3}}{2}, 0 \right) \\ \\ RS & = \sqrt{ \left[0 - \left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right)\right]^2 + \left( - \frac{2\pi}{3} + \frac{\sqrt{3}}{2} - 0 \right)^2 } \\ & = \sqrt{2}\left(\frac{2\pi}{3} - \frac{\sqrt{3}}{2}\right) \\ & \approx 1.74 \text{ units (3 s.f.)} \end{align*}

(a)

\begin{align*} \overrightarrow{AB} & = \overrightarrow{OB} - \overrightarrow{OA} \\ & = \begin{pmatrix} 5 \\ 8 \\ -40 \end{pmatrix} - \begin{pmatrix} 10 \\ 15 \\ -60 \end{pmatrix} \\ & = \begin{pmatrix} -5 \\ -7 \\ 20 \end{pmatrix} \\ \\ \overrightarrow{AC} & = \overrightarrow{OC} - \overrightarrow{OA} \\ & = \begin{pmatrix} -5 \\ 10 \\ -50 \end{pmatrix} - \begin{pmatrix} 10 \\ 15 \\ -60 \end{pmatrix} \\ & = \begin{pmatrix} -15 \\ -5 \\ 10 \end{pmatrix} \\ \\ \overrightarrow{AB} \times \overrightarrow{AC} & = \begin{pmatrix} -5 \\ -7 \\ 20 \end{pmatrix} \times \begin{pmatrix} -15 \\ -5 \\ 10 \end{pmatrix} \\ & = \begin{pmatrix} (-7)(10) - (20)(-5) \\ (20)(-15) - (-5)(10) \\ (-5)(-5) - (-7)(-15) \end{pmatrix} \\ & = \begin{pmatrix} 30 \\ -250 \\ -80 \end{pmatrix} \\ & = 10 \begin{pmatrix} 3 \\ -25 \\ -8 \end{pmatrix} \\ \\ \text{Normal vector of plane, } \textbf{n} & = \begin{pmatrix} 3 \\ -25 \\ -8 \end{pmatrix} \\ \\ d & = \overrightarrow{OA} \cdot \textbf{n} \\ & = \begin{pmatrix} 10 \\ 15 \\ -60 \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -25 \\ -8 \end{pmatrix} \\ & = (10)(3) + (15)(-25) + (-60)(-8) \\ & = 30 - 375 + 480 \\ & = 135 \\ \\ \text{Scalar-product form: } & \textbf{r} \cdot \begin{pmatrix} 3 \\ -25 \\ -8 \end{pmatrix} = 135 \\ \\ \begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 3 \\ -25 \\ -8 \end{pmatrix} & = 135 \\ 3x - 25y - 8z & = 135 \end{align*}

(b)

\begin{align*} [\text{Angle between two planes} & = \text{Angle between the normal of each plane}] \\ \\ \text{Ground: } z & = 0 \\ \begin{pmatrix} x \\ y \\ z \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} & = 0 \\ \textbf{r} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} & = 0 \qquad [xy \text{-plane that contains origin]} \\ \\ \cos \theta & = \frac{ | \textbf{n}_\text{Mineral} \cdot \textbf{n}_\text{Ground} | }{ | \textbf{n}_\text{Mineral} | | \textbf{n}_\text{Ground} | } \\ & = \frac{ \left| \begin{pmatrix} 3 \\ -25 \\ -8 \end{pmatrix} \cdot \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \right| }{ \sqrt{3^2 + (-25)^2 + (-8)^2} \sqrt{0^2 + 0^2 + 1^2} } \\ & = \frac{ |(3)(0) + (-25)(0) + (-8)(1)| }{ \sqrt{698}(1) } \\ & = \frac{|-8|}{\sqrt{698}} \\ & = \frac{8}{\sqrt{698}} \\ \\ \theta & = \cos^{-1} \left( \frac{8}{\sqrt{698}} \right) \\ & \approx 72.4^\circ \text{ (1 d.p.)} \\ \\ \therefore \text{The acute angle between } & \text{the mineral layer and the ground is } 72.4^\circ. \end{align*}

(c)

(d)

\begin{align*} \text{Required distance} = | \overrightarrow{PF} | & = \text{Length of projection of } \overrightarrow{PV} \text{ onto line } PF \\ & = | \overrightarrow{PV} \cdot \hat{ \overrightarrow{PF} } | \\ \\ \overrightarrow{PV} & = \overrightarrow{OV} - \overrightarrow{OP} \\ & = \begin{pmatrix} -17 \\ 188 \\ -610.75 \end{pmatrix} - \begin{pmatrix} -17 \\ 188 \\0 \end{pmatrix} \\ & = \begin{pmatrix} 0 \\ 0 \\ -610.75 \end{pmatrix} \\ \\ \overrightarrow{PF} & \, // \, \text{Normal vector of mineral layer} \\ \overrightarrow{PF} & = k \begin{pmatrix} 3 \\ -25 \\ -8 \end{pmatrix}, \text{ for a certain non-zero value of } k \\ & = \begin{pmatrix} 3k \\ -25k \\ -8k \end{pmatrix} \\ \\ | \overrightarrow{PF} | & = \sqrt{ (3k)^2 + (-25k)^2 + (-8k)^2 } \\ & = \sqrt{698k^2} \\ & = |k| \sqrt{ 698} \\ \\ | \overrightarrow{PV} \cdot \hat{ \overrightarrow{PF} } | & = \frac{ | \overrightarrow{PV} \cdot \overrightarrow{PF} | }{ | \overrightarrow{PF} | } \\ & = \frac{ \left| \begin{pmatrix} 0 \\ 0 \\ -610.75 \end{pmatrix} \cdot \begin{pmatrix} 3k \\ -25k \\ -8k \end{pmatrix} \right| }{ |k| \sqrt{ 698} } \\ & = \frac{ | 0 + 0 + 4886k | }{|k| \sqrt{698}} \\ & = \frac{ | 4886k | }{|k| \sqrt{698}} \\ & = \frac{|k||4886|}{|k| \sqrt{698} } \\ & = \frac{ 4886}{\sqrt{698}} \\ & = 184.937827\ldots \\ & \approx 185 \text{ m (3 s.f.)} \end{align*}