(a)

\begin{align} u & = 3 & & & v & = 2 - x \\ {du \over dx} & = 0 & & & {dv \over dx} & = -1 \end{align} \begin{align} {d \over dx} \left(3 \over 2 - x\right) & = {(2 - x)(0) - (3)(-1) \over (2 - x)^2} \\ & = { 0 + 3 \over (2 - x)^2 } \\ & = {3 \over (2 - x)^2} \end{align}

(b)

\begin{align} u & = 3x + 2 & & & v & = 1 + 4x \\ {du \over dx} & = 3 & & & {dv \over dx} & = 4 \end{align} \begin{align} {d \over dx} \left(3x + 2 \over 1 + 4x\right) & = {(1 + 4x)(3) - (3x + 2)(4) \over (1 + 4x)^2 } \\ & = {3 + 12x - 12x - 8 \over (1 + 4x)^2 } \\ & = {- 5 \over (1 + 4x)^2} \\ & = -{5 \over (1 + 4x)^2} \end{align}

(c)

\begin{align} u & = 2x + 8 & & & v & = 3x^2 - 2 \\ {du \over dx} & = 2 & & & {dv \over dx} & = 6x \end{align} \begin{align} {d \over dx} \left(2x + 8 \over 3x^2 - 2\right) & = {(3x^2 - 2)(2) - (2x + 8)(6x) \over (3x^2 - 2)^2} \\ & = {6x^2 - 4 - 12x^2 - 48x \over (3x^2 - 2)^2} \\ & = {-6x^2 - 48x - 4 \over (3x^2 - 2)^2} \\ & = {-(6x^2 + 48x + 4 \over (3x^2 - 2)^2} \\ & = -{6x^2 + 48x + 4 \over (3x^2 - 2)^2} \end{align}

(d)

\begin{align} u & = 4x + 7 & & & v & = 1 - 2x^2 \\ {du \over dx} & = 4 & & & {dv \over dx} & = -4x \end{align} \begin{align} {d \over dx} \left(4x + 7 \over 1 - 2x^2 \right) & = {(1 - 2x^2)(4) - (4x + 7)(-4x) \over (1 - 2x^2)^2} \\ & = {4 - 8x^2 + 16x^2 + 28x \over (1 - 2x^2)^2} \\ & = {8x^2 + 28x + 4 \over (1 - 2x^2)^2} \end{align}

(e)

\begin{align} u & = x^2 & & & v & = 2x - 1 \\ {du \over dx} & = 2x & & & {dv \over dx} & = 2 \end{align} \begin{align} {d \over dx} \left(x^2 \over 2x - 1\right) & = {(2x - 1)(2x) - (x^2)(2) \over (2x - 1)^2} \\ & = {4x^2 - 2x - 2x^2 \over (2x - 1)^2} \\ & = {2x^2 - 2x \over (2x - 1)^2} \end{align}

(f)

\begin{align} u & = x & & & v & = 4(x^2 + 3) \\ & & & & & = 4x^2 + 12 \\ {du \over dx} & = 1 & & & {dv \over dx} & = 8x \end{align} \begin{align} {d \over dx} \left( x \over 4x^2 + 12 \right) & = {(4x^2 + 12)(1) - (x)(8x) \over (4x^2 + 12)^2} \\ & = {4x^2 + 12 - 8x^2 \over (4x^2 + 12)^2} \\ & = {12 - 4x^2 \over (4x^2 + 12)^2} \\ & = {4(3 - x^2) \over [4(x^2 + 3)]^2 } \\ & = {4(3 - x^2) \over (4)^2 (x^2 + 3)^2 } \\ & = {3 - x^2 \over 4(x^2 + 3)^2} \end{align}

(a)

\begin{align} u & = \sqrt{x} & & & v & = 3 + x \\ & = x^{1 \over 2} \\ {du \over dx} & = {1 \over 2}x^{-{1 \over 2}} & & & {dv \over dx} & = 1 \end{align} \begin{align} {d \over dx} \left(x^{1 \over 2} \over 3 + x\right) & = {(3 + x)\left({1 \over 2}x^{-{1 \over 2}}\right) - \left(x^{1 \over 2}\right)(1) \over (3 + x)^2} \\ & = { {3 \over 2}x^{-{1 \over 2}} + {1 \over 2}x^{1 \over 2} - x^{1 \over 2} \over (3 + x)^2} \\ & = { {3 \over 2}x^{-{1 \over 2}} - {1 \over 2}x^{1 \over 2} \over (3 + x)^2 } \\ & = { {1 \over 2}x^{-{1 \over 2}} (3 - x) \over (3 + x)^2 } \\ & = {3 - x \over 2\sqrt{x} (3 + x)^2 } \end{align}

(b)

\begin{align} u & = 4\sqrt{x} & & & v & = x^2 - 5 \\ & = 4x^{1 \over 2} \\ {du \over dx} & = 4\left({1 \over 2}x^{-{1 \over 2}}\right) & & & {dv \over dx} & = 2x \\ & = 2 \left(1 \over \sqrt{x}\right) \\ & = {2 \over \sqrt{x}} \end{align} \begin{align} {d \over dx} \left(4\sqrt{x} \over x^2 - 5\right) & = {(x^2 - 5)\left(2 \over \sqrt{x}\right) - (4\sqrt{x})(2x) \over (x^2 - 5)^2} \\ & = { {2(x^2 - 5) \over \sqrt{x} } - {8x \sqrt{x} \over 1} \over (x^2 -5)^2} \\ & = { {2(x^2 - 5) \over \sqrt{x} } - {8x \sqrt{x} (\sqrt{x}) \over \sqrt{x}} \over (x^2 - 5)^2 } \\ & = { { 2(x^2 - 5) - 8x (x) \over \sqrt{x}} \over (x^2 - 5)^2} \\ & = { 2(x^2 - 5) - 8x^2 \over \sqrt{x} (x^2 - 5)^2} \\ & = { 2x^2 - 10 - 8x^2 \over \sqrt{x} (x^2 - 5)^2 } \\ & = { -6x^2 - 10 \over \sqrt{x} (x^2 - 5)^2 } \\ & = { -(6x^2 + 10) \over \sqrt{x} (x^2 - 5)^2} \\ & = - { 6x^2 + 10 \over \sqrt{x} (x^2 - 5)^2 } \end{align}

(c)

\begin{align} u & = 2x & & & v & = \sqrt{x - 3} \\ & & & & & = (x - 3)^{1 \over 2} \\ {du \over dx} & = 2 & & & {dv \over dx} & = {1 \over 2}(x - 3)^{-{1 \over 2}} . (1) \phantom{00000} [\text{Chain rule}] \\ & & & & & = {1 \over 2} \left(1 \over \sqrt{x - 3}\right) \\ & & & & & = {1 \over 2\sqrt{x - 3}} \end{align} \begin{align} {d \over dx} \left(2x \over \sqrt{x - 3}\right) & = {(\sqrt{x - 3})(2) - (2x) \left({1 \over 2\sqrt{x - 3}}\right) \over (\sqrt{x - 3})^2 } \\ & = { 2\sqrt{x - 3} - {2x \over 2\sqrt{x - 3}} \over x - 3 } \\ & = { {2\sqrt{x - 3} \over 1} - {x \over \sqrt{x - 3} } \over x - 3} \\ & = { {2\sqrt{x - 3}(\sqrt{x - 3}) \over \sqrt{x - 3}} - {x \over \sqrt{x - 3}} \over x- 3} \\ & = { {2(x - 3) - x \over \sqrt{x - 3}} \over x - 3} \\ & = { 2(x - 3) - x \over \sqrt{x - 3} (x - 3)} \\ & = { 2x - 6 - x \over (x - 3)^{1 \over 2} (x - 3)} \\ & = {x - 6 \over (x - 3)^{3 \over 2} } \\ & = {x - 6 \over \sqrt{(x - 3)^3} } \end{align}

(d)

\begin{align} u & = x^2 & & & v & = \sqrt{x + 1} \\ & & & & & = (x + 1)^{1 \over 2} \\ {du \over dx} & = 2x & & & {dv \over dx} & = {1 \over 2}(x + 1)^{-{1 \over 2}} . (1) \phantom{00000} [\text{Chain rule}] \\ & & & & & = {1 \over 2} \left(1 \over \sqrt{x + 1}\right) \\ & & & & & = {1 \over 2\sqrt{x + 1}} \end{align} \begin{align} {d \over dx} \left(x^2 \over \sqrt{x + 1}\right) & = {(\sqrt{x + 1}) (2x) - (x^2) \left(1 \over 2\sqrt{x + 1}\right) \over (\sqrt{x + 1})^2 } \\ & = { {2x \sqrt{x + 1} \over 1} - {x^2 \over 2\sqrt{x + 1}} \over x + 1} \\ & = { {2x \sqrt{x + 1} (2\sqrt{x + 1}) \over 2\sqrt{x + 1}} - {x^2 \over 2\sqrt{x + 1}} \over x + 1} \\ & = { {4x (x + 1) - x^2 \over 2\sqrt{x + 1}} \over x + 1} \\ & = { 4x(x + 1) - x^2 \over 2\sqrt{x + 1} (x + 1)} \\ & = { 4x^2 + 4x - x^2 \over 2(x + 1)^{1 \over 2}(x + 1)} \\ & = {3x^2 + 4x \over 2 (x + 1)^{3 \over 2} } \\ & = {3x^2 + 4x \over 2 \sqrt{(x + 1)^3} } \end{align}

(i)

\begin{align} u & = x + 4 & & & v & = 5 - x \\ {du \over dx} & = 1 & & & {dv \over dx} & = - 1 \end{align} \begin{align} {dy \over dx} & = {(5 - x)(1) - (x + 4)(-1) \over (5 - x)^2} \\ & = {5 - x + x + 4 \over (5 - x)^2 } \\ & = {9 \over (5 - x)^2} \\ \\ \text{Substitute } & y = 0 \text{ into equation of curve,} \\ 0 & = {x + 4 \over 5 - x} \\ 0 & = x + 4 \\ -4 & = x \\ \\ \implies x & \text{-intercept is } - 4 \\ \\ \text{Substitute } & x = 4 \text{ into } {dy \over dx}, \\ {dy \over dx} & = {9 \over [5 - (-4)]^2 } \\ & = {1 \over 9} \end{align}

(ii) The curve crosses the y-axis when x = 0

\begin{align} \text{Substitute } & x = 0 \text{ into } {dy \over dx}, \\ {dy \over dx} & = {9 \over (5 - 0)^2} \\ & = {9 \over 25} \end{align}

Since the y-axis is a vertical line and the tangent to the curve is perpendicular to the y-axis, the tangent is a horizontal line and it's gradient is equals to 0.

\begin{align} u & = hx - 2 & & & v & = kx + 1 \\ {du \over dx} & = h & & & {dv \over dx} & = k \end{align} \begin{align} {dy \over dx} & = {(kx + 1)(h) - (hx - 2)(k) \over (kx + 1)^2} \\ & = {hkx + h - hkx + 2k \over (kx + 1)^2 } \\ & = {h + 2k \over (kx + 1)^2} \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = {h + 2k \over (kx + 1)^2} \\ 0 & = h + 2k \\ \\ \text{Let } & h = 2, \\ 0 & = 2 + 2k \\ -2 & = 2k \\ {-2 \over 2} & = k \\ -1 & = k \\ \\ \text{Possible values: } & h = 2, k = -1 \end{align}

(i)

\begin{align} y & = {x^2 +2 \over 1 - ax} \\ \\ \text{Using } & \left(1, -{1 \over 2}\right), \\ -{1 \over 2} & = {(1)^2 + 2 \over 1 - a(1)} \\ -{1 \over 2} & = {3 \over 1 - a} \\ (-1)(1 - a) & = 2(3) \\ - 1 + a & = 6 \\ a & = 7 \end{align}

(ii)

$$ y = {x^2 + 2 \over 1 - 7x} $$ \begin{align} u & = x^2 + 2 & & & v & = 1 - 7x \\ {du \over dx} & = 2x & & & {dv \over dx} & = -7 \end{align} \begin{align} 9x & = 36y + 2 \\ -36y & = -9x + 2 \\ 36y & = 9x - 2 \\ y & = {1 \over 4}x - {1 \over 18} \\ \\ \text{Gradient of line} & = {1 \over 4} \\ \\ \\ {dy \over dx} & = {(1 - 7x)(2x) - (x^2 + 2)(-7) \over (1 - 7x)^2} \\ & = { 2x - 14x^2 + 7x^2 + 14 \over (1 - 7x)^2} \\ & = {- 7x^2 + 2x + 14 \over (1 - 7x)^2 } \\ \\ \text{Let } & x = 1, \\ {dy \over dx} & = { -7(1)^2 + 2(1) + 14 \over [1 - 7(1)]^2} \\ & = {1 \over 4} \\ \\ \therefore \text{Tangent at } \left(1, -{1 \over 2}\right) & \text{ is parallel to the line } 9x = 36y + 2 \end{align}

The distance between the two points can be found by $\sqrt{ (x_2 - x_1)^2 + (y_2 - y_1)^2 }$

\begin{align} u & = 6x &&& v & = x^2 + 1 \\ {du \over dx} & = 6 &&& {dv \over dx} & = 2x \end{align} \begin{align} {dy \over dx} & = {(x^2 + 1)(6) - (6x)(2x) \over (x^2 + 1)^2} \\ & = {6x^2 + 6 - 12x^2 \over (x^2 + 1)^2} \\ & = { 6 - 6x^2 \over (x^2 + 1)^2 } \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = {6 - 6x^2 \over (x^2 + 1)^2 } \\ 0 & = 6 - 6x^2 \\ 6x^2 & = 6 \\ x^2 & = 1 \\ x & = \pm \sqrt{1} \\ & = \pm 1 \end{align} \begin{align} \text{Substitute } & x = 1 \text{ into eqn of curve,} &&& \text{Substitute } & x = -1 \text{ into eqn of curve,} \\ y & = {6(1) \over (1)^2 + 1} &&& y & = {6(-1) \over (-1)^2 + 1} \\ & = 3 &&& & = -3 \\ \\ \therefore & \phantom{.} (1, 3) &&& \therefore & \phantom{.} (-1, -3) \end{align}
\begin{align} \text{Distance between two points} & = \sqrt{(-1 - 1)^2 + (-3 -3)^2} \\ & = \sqrt{40} \text{ units} \\ & \approx 6.32 \text{ units} \end{align}

(i)

\begin{align} \text{Let } f(x) & = x^3 - x^2 + 5x - 5 \\ \\ f(1) & = (1)^3 - (1)^2 + 5(1) - 5 \\ & = 0 \\ \\ \text{By Factor} & \text{ theorem, } x - 1 \text{ is a factor of } f(x) \end{align}
$$ \require{enclose} \begin{array}{rll} x^2 + 5 \phantom{00000}\\ x - 1 \enclose{longdiv}{x^3 - x^2 + 5x - 5\phantom{0}}\kern-.2ex \\ -\underline{(x^3 - x^2){\phantom{0000000(}}} \\ 5x - 5\phantom{0} \\ -\underline{(5x - 5){\phantom{}}} \\ 0\phantom{0} \end{array} $$
\begin{align} \text{Since Polynomial} & = \text{Divisor} \times \text{Quotient} + \text{Remainder,} \\ x^3 - x^2 + 5x - 5 & = (x - 1)(x^2 + 5) + 0 \\ & = (x - 1)(x^2 + 5) \end{align}

(ii) The denominator of the fraction is the same cubic polynomial in (i)

\begin{align} {6x^2 - 5x + 23 \over x^3 - x^2 + 5x - 5} & = {6x^2 - 5x + 23 \over (x - 1)(x^2 + 5)} \\ & = {A \over x - 1} + {Bx + C \over x^2 + 5} \\ & = {A(x^2 + 5) + (Bx + C)(x - 1) \over (x - 1)(x^2 + 5)} \\ \\ 6x^2 - 5x + 23 & = A(x^2 + 5) + (Bx + C)(x - 1) \\ \\ \text{Let } & x = 1, \\ 6(1)^2 - 5(1) + 23 & = A[(1)^2 + 5] + [B(1) + C](0) \\ 24 & = A(6) + 0 \\ 24 & = 6A \\ {24 \over 6} & = A \\ 4 & = A \\ \\ 6x^2 - 5x + 23 & = 4(x^2 + 5) + (Bx + C)(x - 1) \\ \\ \text{Let } & x = 0, \\ 6(0)^2 - 5(0) + 23 & = 4[(0)^2 + 5] + [B(0) + C](0 - 1) \\ 23 & = 20 + (C)(-1) \\ 23 & = 20 - C \\ C & = 20 - 23 \\ & = -3 \\ \\ 6x^2 - 5x + 23 & = 4(x^2 + 5) + (Bx - 3)(x - 1) \\ \\ \text{Let } & x = 2, \\ 6(2)^2 - 5(2) + 23 & = 4[(2)^2 + 5] + [B(2) - 3](2 - 1) \\ 37 & = 36 + (2B - 3)(1) \\ 37 & = 36 + 2B - 3 \\ 37 - 36 + 3 & = 2B \\ 4 & = 2B \\ {4 \over 2} & = B \\ 2 & = B \\ \\ \therefore {6x^2 - 5x + 23 \over x^3 - x^2 + 5x - 5} & = {4 \over x - 1} + {2x - 3 \over x^2 + 5} \end{align}

(iii)

\begin{align} u & = x + 1 &&& v & = \sqrt{x - 1} \\ &&&& & = (x - 1)^{1 \over 2} \\ {du \over dx} & = 1 &&& {dv \over dx} & = {1 \over 2}(x - 1)^{-{1 \over 2}} . (1) \\ &&&& & = {1 \over 2} \left(1 \over \sqrt{x - 1}\right) \\ &&&& & = {1 \over 2\sqrt{x - 1}} \end{align} \begin{align} {dy \over dx} & = {(\sqrt{x - 1})(1) - (x + 1) \left(1 \over 2\sqrt{x - 1}\right) \over (\sqrt{x - 1})^2 } \\ & = { \sqrt{x - 1} - {x + 1 \over 2\sqrt{x - 1}} \over x - 1} \\ & = { {\sqrt{x - 1}(2\sqrt{x - 1}) \over 2\sqrt{x - 1}} - {x + 1 \over 2\sqrt{x - 1}} \over x - 1} \\ & = { {2(x - 1) - (x + 1) \over 2\sqrt{x - 1}} \over x - 1} \\ & = { 2(x - 1) - (x + 1) \over 2\sqrt{x - 1} (x - 1) } \\ & = {2x - 2 - x - 1 \over 2\sqrt{x - 1}(x - 1) } \\ & = {x - 3 \over 2\sqrt{x - 1}(x - 1)} \\ \\ \\ \text{To show } & 2\sqrt{x - 1} \left(\sqrt{x - 1}{dy \over dx} - 1 \right) = - y, \\ \\ \text{L.H.S} & = 2\sqrt{x - 1} \left(\sqrt{x - 1}{dy \over dx} - 1 \right) \\ & = 2\sqrt{x - 1} \left\{ (\sqrt{x - 1}) \left[ {x - 3 \over 2\sqrt{x - 1}(x - 1)}\right] - 1 \right\} \\ & = 2\sqrt{x - 1} \left[ {x - 3 \over 2(x - 1)} - 1 \right] \\ & = 2\sqrt{x - 1} \left[ {x - 3 \over 2(x - 1)} - {2(x - 1) \over 2(x - 1)} \right] \\ & = 2\sqrt{x - 1} \left[ {x - 3 - 2(x - 1) \over 2(x - 1)} \right] \\ & = 2\sqrt{x - 1} \left[ x - 3 - 2x + 2 \over 2(x - 1) \right] \\ & = 2\sqrt{x - 1} \left[ -x - 1 \over 2(x - 1) \right] \\ & = {2\sqrt{x - 1}(-x-1) \over 2(x - 1)} \\ & = {(x - 1)^{1 \over 2} (-x - 1) \over x - 1} \\ & = {- x - 1 \over (x - 1)^{-{1 \over 2}} (x - 1)} \\ & = {-(x + 1) \over (x - 1)^{1 \over 2} } \\ & = - {x + 1 \over \sqrt{x - 1}} \\ & = -y \\ & = \text{R.H.S} \end{align}

(i)

$$ \require{enclose} \begin{array}{rll} -1 \phantom{000}\\ -x + k \enclose{longdiv}{x + h\phantom{0}}\kern-.2ex \\ -\underline{(x - k){\phantom{(}}} \\ h + k\phantom{0} \end{array} $$ $$ {x + h \over -x + k} = -1 + {h + k \over -x + k} $$

(ii) The line y = 3.5 is a horizontal line with gradient 0. Thus the tangent to the curve should have gradient 0 as well.

\begin{align} u & = h + k &&& v & = -x + k \\ {du \over dx} & = 0 &&& {dv \over dx} & = -1 \end{align} \begin{align} {d \over dx} \left( h + k \over -x + k\right) & = {(-x + k)(0) - (h + k)(-1) \over (-x + k)^2} \phantom{000000} [\text{Quotient rule}] \\ & = { h + k \over (-x + k)^2} \\ \\ \\ y & = \sqrt{x + h \over k - x} \\ y & = \left(x + h \over k - x\right)^{1 \over 2} \\ \\ {dy \over dx} & = {1 \over 2} \left(x + h \over k - x\right)^{-{1 \over 2}} \left[ { h + k \over (-x + k)^2} \right] \phantom{000000} [\text{Chain rule}] \\ & = {1 \over 2} \left(-x + k \over x + h\right)^{1 \over 2} \left[ {h + k \over (- x + k)^2} \right] \\ & = {1 \over 2} \left[ (-x + k)^{1 \over 2} \over (x + h)^{1 \over 2} \right] \left[ {h + k \over (-x + k)^2} \right] \\ & = { h + k \over 2(- x + k)^{3 \over 2} (x + h)^{1 \over 2} } \\ \\ \text{Let } & {dy \over dx} = 0, \\ 0 & = { h + k \over 2(-x + k)^{3 \over 2} (x + h)^{1 \over 2} } \\ \\ \implies 0 & = h + k \\ -k & = h \\ \\ \text{Substitute } & h = -k \text{ into } y = \sqrt{-1 + {h + k \over - x+ k}}, \\ y & = \sqrt{-1 + {-k + k \over -x + k}} \\ y & = \sqrt{-1 + 0} \\ y & = \sqrt{-1} \\ \\ \text{Since } \sqrt{-1} \text{ is not a } & \text{real number, no tangents to the curve is parallel to } y = 3.5 \end{align}