(a)

$$\begin{align} {d \over dx}(3x - 2 \cos x) & = 3(1) - 2 (-\sin x) \\ & = 3 + 2 \sin x \end{align}$$

(b)

$$\begin{align} u & = \sin 2x &&& v & = x \\ {du \over dx} & = 2 \cos 2x &&& {dv \over dx} & = 1 \end{align}$$ $$\begin{align} {d \over dx} \left( \sin 2x \over x \right) & = { (x)(2 \cos 2x) - (\sin 2x)(1) \over (x)^2 } \phantom{000000} [\text{Quotient rule}] \\ & = {2x \cos 2x - \sin 2x \over x^2} \end{align}$$

(c)

$$\begin{align} u & = 2x &&& v & = \tan 3x \\ {du \over dx} & = 2 &&& {dv \over dx} & = 3 \sec^2 3x \end{align}$$ $$\begin{align} {d \over dx} (2x \tan 3x) & = (2x)(3 \sec^2 3x) + (\tan 3x)(2) \phantom{000000} [\text{Product rule}] \\ & = 6 x \sec^2 3x + 2 \tan 3x \end{align}$$

(d)

$$\begin{align} {d \over dx} (3 \sin^4 8x) & = {d \over dx} [3 (\sin 8x)^4 ] \\ & = 3 (4) (\sin 8x)^3 . (8)(\cos 8x) \phantom{000000} [\text{Chain rule}] \\ & = 96 \sin^3 8x \cos 8x \end{align}$$

(e)

$$\begin{align} {d \over dx} [ \sin (x^2 + 2)] & = (2x) \cos (x^2 + 2) \\ & = 2x \cos (x^2 + 2) \end{align}$$

(a)

$$\begin{align} u & = x + 1 &&& v & = e^{-x} \\ {du \over dx} & = 1 &&& {dv \over dx} & = (-1)e^{-x} \\ & &&& & = - e^{-x} \end{align}$$ $$\begin{align} {d \over dx} [ (x + 1)e^{-x} ] & = (x + 1)(- e^{-x}) + (e^{-x})(1) \phantom{000000} [\text{Product rule}] \\ & = - e^{-x} (x + 1) + e^{-x} \\ & = e^{-x} [-(x + 1) + 1] \\ & = e^{-x} (- x - 1 + 1 ) \\ & = e^{-x} (-x) \\ & = - x e^{-x} \end{align}$$

(b)

$$\begin{align} u & = e^{2x} &&& v & = \sin x \\ {du \over dx} & = 2e^{2x} &&& {dv \over dx} & = \cos x \end{align}$$ $$\begin{align} {d \over dx} (e^{2x} \sin x) & = (e^{2x})(\cos x) + (\sin x)(2e^{2x}) \phantom{000000} [\text{Product rule}] \\ & = e^{2x} \cos x + 2 e^{2x} \sin x \\ & = e^{2x} (\cos x + 2 \sin x) \end{align}$$

(c)

$$\begin{align} u & = 3x^2 &&& v & = \ln x \\ {du \over dx} & = 3(2)x &&& {dv \over dx} & = {1 \over x} \\ & = 6x \end{align}$$ $$\begin{align} {d \over dx} (3x^2 \ln x) & = (3x^2) \left(1 \over x\right) + (\ln x)(6x) \phantom{000000} [\text{Product rule}] \\ & = {3x^2 \over x} + 6x \ln x \\ & = 3x + 6x \ln x \\ & = 3x(1 + 2 \ln x) \end{align}$$

(d)

$$\begin{align} u & = e^x &&& v & = 2x + 1 \\ {du \over dx} & = (1)e^x &&& {dv \over dx} & = 2 \\ & = e^x \end{align}$$ $$\begin{align} {d \over dx} \left(e^x \over 2x + 1\right) & = {(2x + 1)(e^x) - (e^x)(2) \over (2x + 1)^2 } \phantom{000000} [\text{Quotient rule}] \\ & = { e^x(2x + 1) - 2e^x \over (2x + 1)^2 } \\ & = { e^x [(2x + 1) - 2 ] \over (2x + 1)^2 } \\ & = { e^x (2x + 1 - 2) \over (2x + 1)^2 } \\ & = { e^x (2x - 1) \over (2x + 1)^2 } \end{align}$$

(e)

$$\begin{align} \text{Let } y & = \ln \left(1 - 2x \over 1 + x\right) \\ & = \ln (1 - 2x) - \ln (1 + x) \phantom{000000} [\text{Quotient law (logarithms)}] \\ \\ {dy \over dx} & = { - 2 \over 1 - 2x} - {1 \over 1 + x} \\ & = {2 \over 2x - 1} - {1 \over 1 + x} \\ & = {2(1 + x) \over (2x - 1)(1 + x) } - {2x - 1 \over (2x - 1)(1 + x)} \\ & = {2(1 + x) - (2x - 1) \over (2x - 1)(1 + x)} \\ & = {2 + 2x - 2x + 1 \over (2x - 1)(1 + x)} \\ & = {3 \over (2x - 1)(1 + x)} \end{align}$$

(i)

$$\begin{align} \cot x & = {\cos x \over \sin x} \\ \\ u & = \cos x &&& v & = \sin x \\ {du \over dx} & = - \sin x &&& {dv \over dx} & = \cos x \end{align}$$ $$\begin{align} {d \over dx}(\cot x) & = {(\sin x)(- \sin x) - (\cos x)(\cos x) \over (\sin x)^2 } \phantom{000000} [\text{Quotient rule}] \\ & = { - \sin^2 x - \cos^2 x \over \sin^2 x } \\ & = { - (\sin^2 x + \cos^2 x) \over \sin^2 x} \\ & = { - 1 \over \sin^2 x} \phantom{00000000000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1] \\ & = - \text{cosec}^2 x \phantom{00} \text{ (Shown)} \end{align}$$

(ii)

$$\begin{align} t & = 20 \cot \theta \\ \\ {dt \over d\theta} & = 20 (-\text{cosec}^2 \theta) \\ {dt \over d\theta} & = -20 \text{cosec}^2 \theta \\ {dt \over d\theta} & = {-20 \over \sin^2 \theta} \\ \\ {d\theta \over dt} & = {\sin^2 \theta \over -20} \\ \\ \text{When } & \theta = {\pi \over 4}, \\ {d \theta \over dt} & = { \sin^2 {\pi \over 4} \over -20} \\ & = { \left(1 \over \sqrt{2}\right)^2 \over -20} \\ & = { {1 \over 2} \over -20 } \\ & = -{1 \over 40} \text{ radians per second} \end{align}$$

(i)

$$\begin{align} y & = 1 - 2 \cos^2 x + \cos x \\ \\ \text{Let } & y = 0, \\ 0 & = 1 - 2 \cos^2 x + \cos x \\ 0 & = -2 \cos^2 x + \cos x + 1 \\ 0 & = 2 \cos^2 x - \cos x - 1 \\ 0 & = (\cos x - 1)(2 \cos x + 1) \\ \\ \cos x - 1 & = 0 \phantom{0} \text{ or } \phantom{0} 2 \cos x + 1 = 0 \\ \cos x & = 1 \phantom{00000000} 2 \cos x = - 1 \\ & \phantom{000000000000(} \cos x = -{1 \over 2} \end{align}$$

$$\cos x = 1$$

$$\begin{align} x & = 0 \text{ (Reject)}, 2 \pi \text{ (Reject)} \end{align}$$

$$\begin{align} \cos x & = -{1 \over 2} \phantom{000000} [\text{2nd & 3rd quadrants since } \cos x < 0] \end{align}$$

$$\begin{align} \text{Basic angle, } \alpha & = \cos^{-1} \left(1 \over 2\right) \\ & = 60^\circ \ & = {\pi \over 3} \\ \\ x & = \pi - {\pi \over 3}, \pi + {\pi \over 3} \\ & = {2 \pi \over 3}, {4 \pi \over 3} \end{align}$$

(ii)

$$\begin{align} y & = 1 - 2 \cos^2 x + \cos x \\ & = 1 - 2 (\cos x)^2 + \cos x \\ \\ {dy \over dx} & = - 2(2)(\cos x). (- \sin x) + (-\sin x) \\ & = 4 \sin x \cos x - \sin x \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = 4 \sin x \cos x - \sin x \\ 0 & = \sin x(4 \cos x - 1) \\ \\ \sin x & = 0 \phantom{0} \text{ or } \phantom{0} 4\cos x - 1 = 0 \\ & \phantom{00000000000(} 4 \cos x = 1 \\ & \phantom{00000000000000} \cos = {1 \over 4} \end{align}$$

$$\sin x = 0$$

$$\begin{align} x & = \pi \end{align}$$

$$\begin{align} \cos x & = {1 \over 4} \phantom{000000} [\text{1st & 4th quadrants since } \cos x > 0] \end{align}$$

$$\begin{align} \text{Basic angle, } \alpha & = \cos^{-1} \left(1 \over 4\right) \phantom{000000} [\text{Radian mode!}] \\ & = 1.3181 \\ \\ x & = 1.3181, 2\pi - 1.3181 \\ & = 1.3181, 4.9650 \\ & \approx 1.32, 4.97 \\ \\ \therefore x & = 1.32, \pi, 4.97 \end{align}$$

(i)

$$\begin{align} u & = x &&& v & = e^{x - 3} \\ {du \over dx} & = 1 &&& {dv \over dx} & = (1)e^{x - 3} \\ & &&& & = e^{x - 3} \end{align}$$ $$\begin{align} {dy \over dx} & = (x)(e^{x - 3}) + (e^{x - 3})(1) \phantom{000000} [\text{Product rule}] \\ & = x e^{x - 3} + e^{x - 3} \\ & = e^{x - 3} (x + 1) \end{align}$$

(ii)

$$\begin{align} {dy \over dx} & = e^{x - 3} (x + 1) \\ \\ \text{When } & x = 3, \\ {dy \over dx} & = e^{3 - 3} (3 + 1) \\ & = (1) (4) \\ & = 4 \end{align}$$

(i)

(ii)

$$\begin{align} x^2 e^x & = 12 \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln (x^2 e^x) & = \ln 12 \\ [\text{Product law}] \phantom{000000} \ln x^2 + \ln e^x & = \ln 12 \\ [\text{Power law }] \phantom{000000} 2 \ln x + x \ln e & = \ln 12 \\ 2 \ln x + x (1) & = \ln 12 \\ 2 \ln x + x & = \ln 12 \\ 2 \ln x & = -x + \ln 12 \\ \ln x & = -{1 \over 2}x + {1 \over 2} \ln 12 \\ \\ a & = -{1 \over 2} \\ \\ b & = {1 \over 2} \ln 12 \\ & \approx 1.24 \end{align}$$

(iii)

$$\begin{align} x^2 e^x & = 12 \\ \\ \text{From (ii), } \ln x & = -{1 \over 2}x + {1 \over 2} \ln 12 \\ \\ \text{L.H.S: } \phantom{.} y & = \ln x \phantom{000} \text{--- (1)} \\ \text{R.H.S: } \phantom{.} y & = -{1 \over 2} x + {1 \over 2} \ln 12 \phantom{000} \text{--- (2)} \\ \\ \therefore \text{Draw } & y = -{1 \over 2}x + {1 \over 2} \ln 12 \end{align}$$

(i)

$$\begin{align} \cos \theta \left({1 \over 1 + \sin \theta} + {1 \over 1 - \sin \theta}\right) & = \cos \theta \left[ {(1 - \sin \theta) \over (1 + \sin \theta)(1 - \sin \theta)} + {(1 + \sin \theta) \over (1 + \sin \theta)(1 - \sin \theta)} \right] \\ & = \cos \theta \left[ {1 - \sin \theta + 1 + \sin \theta \over (1 + \sin \theta)(1 - \sin \theta)} \right] \\ & = \cos \theta \left( 2 \over 1 - \sin^2 \theta \right) \\ & = {2 \cos \theta \over (1 - \sin^2 \theta)} \\ & = {2 \cos \theta \over \cos^2 \theta} \phantom{000000} [\text{Identity: } \sin^2 A + \cos^2 A = 1 ] \\ & = {2 \over \cos \theta} \\ & = 2 \left(1 \over \cos \theta\right) \\ & = 2 \sec \theta \\ \\ \therefore k & = 2 \end{align}$$

(ii)

$$\begin{align} {d \over dx} \left[ \cos \theta \left({1 \over 1 + \sin \theta} + {1 \over 1 - \sin \theta}\right) \right] & = {d \over dx} (2 \sec \theta) \\ & = {d \over dx} \left(2 \over \cos \theta \right) \\ & = {d \over dx} [ 2 (\cos \theta)^{-1} ] \\ & = 2(-1) (\cos \theta)^{-2} . (- \sin \theta) \\ & = 2 \left(1 \over \cos^2 \theta\right) (\sin \theta) \\ & = 2 \left(1 \over \cos \theta\right) \left(\sin \theta \over \cos \theta\right) \\ & = 2 \sec \theta \tan \theta \end{align}$$

(i)

$$\begin{align} n & = 18 e^{0.2 t} \\ \\ \text{When } & t = 10, \\ n & = 18 e^{0.2 (10)} \\ & = 133.00 \\ & \approx 133 \end{align}$$

(ii)

$$\begin{align} n & = 18 e^{0.2 t} \\ \\ \text{When } & n = 40, \\ 40 & = 18 e^{0.2 t } \\ {40 \over 18} & = e^{0.2 t} \\ {20 \over 9} & = e^{0.2 t} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln {20 \over 9} & = \ln e^{0.2 t} \\ \ln {20 \over 9} & = 0.2t \ln e \phantom{000000} [\text{Power law (logarithms)}] \\ \ln {20 \over 9} & = 0.2t (1) \\ \ln {20 \over 9} & = 0.2t \\ \\ t & = { \ln {20 \over 9} \over 0.2} \\ & = 3.9925 \\ & \approx 4 \end{align}$$

(a)

$$\begin{align} {d \over dx} (\sin 4x - 3 \cos 2x) & = (4)\cos 4x - 3 (2) (- \sin 2x) \\ & = 4 \cos 4x + 6 \sin 2x \end{align}$$

(b)

$$\begin{align} {d \over dx} \left(1 \over 1 + \cos 4x \right) & = {d \over dx} (1 + \cos 4x)^{-1} \\ & = (-1)(1 + \cos 4x)^{-2} . [0 + (4)(-\sin 4x)] \phantom{000000} [\text{Chain rule}] \\ & = (-1)\left[ 1 \over (1 + \cos 4x)^2 \right] (- 4 \sin 4x) \\ & = { 4 \sin 4x \over (1 + \cos 4x)^2 } \end{align}$$

(c)

$$\begin{align} u & = x &&& v & = \cos^2 3x \\ & &&& & = (\cos 3x)^2 \\ {du \over dx} & = 1 &&& {dv \over dx} & = 2(\cos 3x). (3)(-\sin 3x) \phantom{000000} [\text{Chain rule}] \\ & &&& & = - 6 \sin 3x \cos 3x \end{align}$$ $$\begin{align} {d \over dx} (x \cos^2 3x) & = (x)(- 6 \sin 3x \cos 3x) + (\cos^2 3x)(1) \\ & = - 6x \sin 3x \cos 3x + \cos^2 3x \\ & = - 3x (2 \sin 3x \cos 3x) + \cos^2 3x \\ & = - 3x \sin 6x + \cos^2 3x \phantom{00000} [\text{Double angle formula: } \sin 2A = 2 \sin A \cos A] \end{align}$$

(d)

$$\begin{align} {d \over dx} (x^2 - 3 \tan^2 4x) & = {d \over dx} [ x^2 - 3(\tan 4x)^2 ] \\ & = 2x - 3(2)(\tan 4x). (4)(\sec^2 4x) \\ & = 2x - 24 \tan 4x \sec^2 4x \end{align}$$

(e)

$$\begin{align} {d \over dx} [2 \cos (1 - x^2)] & = 2(0 - 2x) [- \sin (1 - x^2) ] \\ & = 2 (-2x) [- \sin (1 - x^2)] \\ & = 4x \sin (1 - x^2) \end{align}$$

(a)

$$\begin{align} u & = x^2 &&& v & = \ln 2x \\ {du \over dx} & = 2x &&& {dv \over dx} & = {2(1) \over 2x} \\ & &&& & = {1 \over x} \end{align}$$ $$\begin{align} {d \over dx} \left(x^2 \over \ln 2x\right) & = { (\ln 2x)(2x) - (x^2) \left(1 \over x\right) \over (\ln 2x)^2 } \phantom{000000} [\text{Quotient rule}] \\ & = { 2 x \ln 2x - x \over (\ln 2x)^2 } \end{align}$$

(b)

$$\begin{align} u & = x^2 - 1 &&& v & = \ln (2x + 1) \\ {du \over dx} & = 2x &&& {dv \over dx} & = {2(1) \over 2x + 1} \\ & &&& & = {2 \over 2x + 1} \end{align}$$ $$\begin{align} {d \over dx} [(x^2 - 1)\ln (2x + 1)] & = (x^2 - 1)\left(2 \over 2x + 1\right) + [\ln (2x + 1)](2x) \phantom{000000} [\text{Product rule}] \\ & = \left(x^2 - 1 \over 1\right) \left(2 \over 2x + 1\right) + 2x \ln (2x + 1) \\ & = {2(x^2 - 1) \over 2x + 1} + 2x \ln (2x + 1) \end{align}$$

(c)

$$\begin{align} u & = e^{{1 \over 2}x} &&& v & = \ln (5 - 4x) \\ {du \over dx} & = {1 \over 2}e^{{1 \over 2}x} &&& {dv \over dx} & = {-4 \over 5 - 4x} \end{align}$$ $$\begin{align} {d \over dx} [e^{{1 \over 2}x} \ln (5 - 4x)] & = (e^{{1 \over 2}x})\left(-4 \over 5 - 4x\right) + [\ln (5 - 4x)] \left({1 \over 2}e^{{1 \over 2}x} \right) \phantom{000000} [\text{Product rule}] \\ & = (e^{{1 \over 2}x})\left(-4 \over 5 - 4x\right) + {1 \over 2} e^{{1 \over 2}x} \ln (5 - 4x) \\ & = e^{{1 \over 2}x} \left[ {-4 \over 5 - 4x} + {1 \over 2} \ln (5 - 4x ) \right] \end{align}$$

(d)

$$\begin{align} {d \over dx} \left[ \ln (x + \sqrt{x^2 + 1}) \right] & = {d \over dx} \left\{ \ln \left[ x + (x^2 + 1)^{1 \over 2} \right] \right\} \\ & = {1 + {1 \over 2}(x^2 + 1)^{-{1 \over 2}} . (2x) \over x + (x^2 + 1)^{1 \over 2} } \\ & = {1 + x (x^2 + 1)^{-{1 \over 2}} \over x + \sqrt{x^2 + 1} } \\ & = {1 + {x \over \sqrt{x^2 + 1}} \over x + \sqrt{x^2 + 1} } \\ & = {1 + {x \over \sqrt{x^2 + 1}} \over x + \sqrt{x^2 + 1} } \times {\sqrt{x^2 + 1} \over \sqrt{x^2 + 1}} \\ & = {\sqrt{x^2 + 1} + x \over \sqrt{x^2 + 1} \left(x + \sqrt{x^2 + 1}\right) } \\ & = {1 \over \sqrt{x^2 + 1}} \end{align}$$

(e)

$$\begin{align} u & = x^3 &&& v & = \ln (\cos^2 x) \\ & &&& & = \ln (\cos x)^2 \\ & &&& & = 2 \ln (\cos x) \phantom{000000} [\text{Power law (logarithms)}] \\ {du \over dx} & = 3x^2 &&& {dv \over dx} & = 2 \left(-\sin x \over \cos x\right) \\ & &&& & = -2 \tan x \end{align}$$ $$\begin{align} {d \over dx} [x^3 \ln (\cos^2 x)] & = (x^3)(-2 \tan x) + [\ln (\cos^2 x)] (3x^2) \phantom{000000} [\text{Product rule}] \\ & = - 2 x^3 \tan x + 3x^2 \ln (\cos^2 x) \\ & = x^2 [ - 2x \tan x + 3 \ln (\cos^2 x) ] \end{align}$$

(i)

$$\begin{align} y & = \sin 3x + \cos^3 x \\ & = \sin 3x + (\cos x)^3 \\ \\ {dy \over dx} & = (3)\cos 3x + 3(\cos x)^2 . (- \sin x) \\ & = 3 \cos 3x - 3 \cos^2 x \sin x \\ \\ \text{When } & x = {\pi \over 6}, \\ {dy \over dx} & = 3 \cos \left[3 \left({\pi \over 6}\right) \right] - 3 \cos^2 \left(\pi \over 6\right) \sin \left(\pi \over 6\right) \\ & = 3(0) - 3\left(\sqrt{3} \over 2\right)^2 \left(1 \over 2\right) \\ & = -{9 \over 8} \end{align}$$

(ii)

$$\begin{align} {dx \over dt} & = 0.3 \text{ radians per second} \\ \\ {dy \over dt} & = {dy \over dx} \times {dx \over dt} \\ & = -{9 \over 8} \times 0.3 \\ & = -0.3375 \text{ units per second} \end{align}$$

Since the question is asking for turning points, we have to exclude the coordinates of any stationary point(s) of inflexion.

$$\begin{align} u & = \sin x &&& v & = \cos^3 x \\ & &&& & = (\cos x)^3 \\ {du \over dx} & = \cos x &&& {dv \over dx} & = 3(\cos x)^2 . (- \sin x) \\ & &&& & = - 3 \cos^2 x \sin x \end{align}$$ $$\begin{align} {dy \over dx} & = (\sin x)(-3 \cos^2 \sin x) + (\cos^3 x)(\cos x) \\ & = - 3 \sin^2 x \cos^2 x + \cos^4 x \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = - 3 \sin^2 x \cos^2 x + \cos^4 x \\ 0 & = \cos^2 x (- 3 \sin^2 x + \cos^2 x) \end{align}$$ $$\begin{align} \cos^2 x & = 0 && \text{ or } & - 3 \sin^2 x + \cos^2 x & = 0 \\ \cos x & = 0 &&& - 3 \sin^2 x & = - \cos^2 x \\ & &&& 3 \sin^2 x & = \cos^2 x \\ & &&& {\sin^2 x \over \cos^2 x} & = {1 \over 3} \\ & &&& \tan^2 x & = {1 \over 3} \\ & &&& \tan x & = \pm \sqrt{1 \over 3} \end{align}$$

$$\cos x = 0$$

$$\begin{align} x & = {\pi \over 2}, {3\pi \over 2} \text{ (Reject, since } 0 < x < \pi) \end{align}$$

$x$	$1.5$	${\pi \over 2}$	$1.6$
${dy \over dx}$	$-$	$0$	$-$
Slope	\	-	\

$$\text{Point where } x = {\pi \over 2} \text{ is a stationary point of inflexion}$$

$$\begin{align} \tan x & = \pm \sqrt{1 \over 3} \phantom{000000} [\text{All 4 quadrants}] \end{align}$$

$$\begin{align} \text{Basic angle, } \alpha & = \tan^{-1} \sqrt{1 \over 3} \\ & = 30^\circ \\ & = {\pi \over 6} \\ \\ x & = {\pi \over 6}, \pi - {\pi \over 6}, \pi + {\pi \over 6}, 2\pi - {\pi \over 6} \\ & = {\pi \over 6}, {5 \pi \over 6}, {7 \pi \over 6} \text{ (Reject)}, {11\pi \over 6} \text{ (Reject)} \end{align}$$

$x$	$0.5$	${\pi \over 6}$	$0.6$
${dy \over dx}$	$+$	$0$	$-$
Slope	/	-	\

$$\begin{align} \text{Substitute } & x = {\pi \over 6} \text{ into eqn of curve,} \\ y & = \sin {\pi \over 6} \cos^3 {\pi \over 6} \\ & = {1 \over 2} \left(\sqrt{3} \over 2\right)^3 \\ & = {1 \over 2} \left(3\sqrt{3} \over 8\right) \\ & = {3\sqrt{3} \over 16} \\ \\ \therefore \left({\pi \over 6}, {3\sqrt{3} \over 16}\right) & \text{ is a maximum point} \end{align}$$

$x$	$2.6$	${5\pi \over 6}$	$2.7$
${dy \over dx}$	$-$	$0$	$+$
Slope	\	-	/

$$\begin{align} \text{Substitute } & x = {5\pi \over 6} \text{ into eqn of curve,} \\ y & = \sin {5\pi \over 6} \cos^3 {5\pi \over 6} \\ & = {1 \over 2} \left(-{\sqrt{3} \over 2}\right)^3 \\ & = {1 \over 2} \left(-{3\sqrt{3} \over 8}\right) \\ & = -{3\sqrt{3} \over 16} \\ \\ \therefore \left({5\pi \over 6}, -{3\sqrt{3} \over 16}\right) & \text{ is a minimum point} \end{align}$$

(a)

(b)

$$\begin{align} u & = x &&& v & = \ln x \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over x} \end{align}$$ $$\begin{align} {dy \over dx} & = (x)\left(1 \over x\right) + (\ln x)(1) - 2(1) \\ & = 1 + \ln x - 2 \\ & = \ln x - 1 \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = \ln x - 1 \\ 1 & = \ln x \\ 1 & = \log_e x \\ e^1 & = x \\ e & = x \\ \\ \text{Substitute } & x = e \text{ into eqn of curve,} \\ y & = (e) \ln e - 2(e) \\ & = (e)(1) - 2e \\ & = e - 2e \\ & = - e \\ \\ \therefore \text{Stationary} & \text{ point is } (e, -e) \end{align}$$

$$\begin{align} 2y + 3x & = 5 \\ 2y & = -3x + 5 \\ y & = -{3 \over 2}x + {5 \over 2} \\ \\ \text{Gradient of line} & = -{3 \over 2} \\ \\ \text{Gradient of normal} & = -{3 \over 2} \phantom{000000} [\text{Parallel lines}] \\ \\ \text{Gradient of tangent} & = {-1 \over -{3 \over 2}} \\ & = {2 \over 3} \\ \\ \text{Eqn of curve: } & y = (2x + c) \ln x \end{align}$$ $$\begin{align} u & = 2x + c &&& v & = \ln x \\ {du \over dx} & = 2 &&& {dv \over dx} & = {1 \over x} \end{align}$$ $$\begin{align} {dy \over dx} & = (2x + c)\left(1 \over x\right) + (\ln x)(2) \phantom{000000} [\text{Product rule}] \\ & = {1 \over x}(2x + c) + 2 \ln x \\ & = 2 + {c \over x} + 2 \ln x \\ \\ \text{When } & x = 1 \text{ and } {dy \over dx} = {2 \over 3}, \phantom{000000} \left[\text{Gradient of tangent at } A(1, 0) \text{ is } {2 \over 3} \right] \\ {2 \over 3} & = 2 + {c \over 1} + 2 \ln 1 \\ {2 \over 3} & = 2 + c + 2(0) \\ {2 \over 3} & = 2 + c + 0 \\ {2 \over 3} - 2 & = c \\ -{4 \over 3} & = c \end{align}$$

(i)

$$\begin{align} m & = 24e^{-{t \over 8}} \\ \\ \text{When } & t = 7, \\ m & = 24e^{-{(7) \over 8}} \\ & = 10.004 \\ & \approx 10.0 \text{ g} \end{align}$$

(ii)

$$\begin{align} m & = 24e^{-{t \over 8}} \\ \\ m & = 24e^{-{(0) \over 8}} \\ & = 24(1) \\ & = 24 \\ \\ \therefore \text{Initial mass} & = 24 \text{ g} \\ \\ \text{Half of initial mass} & = {1 \over 2} \times 24 \\ & = 12 \text{ g} \\ \\ \text{When } & m = 12, \\ 12 & = 24e^{-{t \over 8}} \\ {12 \over 24} & = e^{-{t \over 8}} \\ 0.5 & = e^{-{t \over 8}} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln 0.5 & = \ln e^{-{t \over 8}} \\ \ln 0.5 & = -{t \over 8} (\ln e) \phantom{000000} [\text{Power law (logarithms)}] \\ \ln 0.5 & = -{t \over 8} (1) \\ \ln 0.5 & = -{t \over 8} \\ -\ln 0.5 & = {t \over 8} \\ \\ t & = -8\ln 0.5 \\ & = 5.5451 \\ & \approx 5.55 \text{ h} \end{align}$$

(iii)

$$\begin{align} m & = 24e^{-{t \over 8}} \\ \\ {dm \over dt} & = 24 \left(-{1 \over 8}\right)e^{-{t \over 8}} \\ & = - 3 e^{-{t \over 8}} \\ \\ \text{When } & t = 8, \\ {dm \over dt} & = -3e^{-{(8) \over 8}} \\ & = -3e^{-1} \\ & = -3 \left(1 \over e\right) \\ & = -{3 \over e} \\ \\ \therefore \text{Rate of decrease } & = {3 \over e} \text{ grams per hour} \end{align}$$

(i)

$$\begin{align} x & = 26 - 30e^{-0.2t} \\ \\ \text{When } & t = 9, \\ x & = 26 - 30e^{-0.2(9)} \\ & = 21.041 \\ & \approx 21.0 \end{align}$$

(ii)

$$\begin{align} x & = 26 - 30e^{-0.2t} \\ \\ \text{When } & t = 0, \\ x & = 26 - 30e^{-0.2(0)} \\ & = -4 \end{align}$$

(iii)

$$\begin{align} x & = 26 - 30e^{-0.2t} \\ \\ \text{When } & x = 25, \\ 25 & = 26 - 30e^{-0.2t} \\ -1 & = -30e^{-0.2t} \\ 1 & = 30e^{-0.2t} \\ {1 \over 30} & = e^{-0.2t} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln {1 \over 30} & = \ln e^{-0.2t} \\ \ln {1 \over 30} & = -0.2t(\ln e) \phantom{000000} [\text{Power law (logarithms)}] \\ \ln {1 \over 30} & = -0.2t(1) \\ \ln {1 \over 30} & = -0.2t \\ \\ t & = {\ln {1 \over 30} \over -0.2} \\ & = 17.005 \\ & \approx 17.0 \end{align}$$

(iv)

$$\begin{align} x & = 26 - 30e^{-0.2t} \\ \\ {dx \over dt} & = - 30(-0.2)e^{-0.2t} \\ & = 6 e^{-0.2t} \\ \\ \text{When } & t = 5, \\ {dx \over dt} & = 6e^{-0.2(5)} \\ & = 2.2072 \\ & \approx 2.21 ^\circ \text{C/min} \end{align}$$

Value of x as t becomes large

$$\begin{align} x & = 26 - 30e^{-0.2t} \\ & = 26 - 30 \left(1 \over e^{0.2t}\right) \\ & = 26 - {30 \over e^{0.2t}} \\ \\ \text{As } t \rightarrow & \infty, \phantom{.} {30 \over e^{0.2t}} \rightarrow 0 \\ \\ \therefore x & = 26 - (0) \\ & = 26^\circ \text{C} \end{align}$$