(i)

$x^2$	$1$	$4$	$9$	$16$	$25$
$y$	$6.2$	$5.6$	$4.6$	$3.2$	$1.4$

(ii)

\begin{align} y & = ax^2 + b \\ \\ \therefore Y & = aX + b \\ \\ \text{where } Y & = y \\ X & = x^2 \\ m & = a \\ c & = b \\ \\ \text{Gradient, } a & = {6 - 3 \over 2 - 17} \\ & = -0.2 \\ \\ \text{From } & \text{the graph,} \\ \text{Vertical intercept, } b & = 6.4 \end{align}

(i)

$x$	$1$	$2$	$3$	$4$	$5$
${1 \over y}$	$0.40$	$0.90$	$1.41$	$1.89$	$2.38$

(ii)

\begin{align} {1 \over y} & = ax + b \\ \\ \therefore Y & = aX + b \\ \\ \text{where } Y & = {1 \over y} \\ X & = x \\ m & = a \\ c & = b \\ \\ \text{Gradient, } a & = {2 - 1 \over 4.2 - 2.2} \\ & = 0.5 \\ \\ \text{From } & \text{the graph,} \\ \text{Vertical intercept, } b & = -0.1 \end{align}

(i)

$x^2$	$1$	$4$	$9$	$16$	$25$
$(y - x)$	$40.5$	$36.0$	$28.5$	$18.0$	$4.5$

(ii)

\begin{align} y & = ax^2 + x - b \\ y - x & = ax^2 - b \\ \\ \therefore Y & = aX - b \\ \\ \text{where } Y & = y - x \\ X & = x^2 \\ m & = a \\ c & = - b \\ \\ \text{Gradient, } a & = {30 - 12 \over 8 - 20} \\ & = -1.5 \\ \\ \text{From } & \text{the graph,} \\ \text{Vertical intercept, } -b & = 42 \\ b & = -42 \end{align}

(i)

${1 \over t}$	$1$	$0.5$	$0.33$	$0.25$	$0.2$
${1 \over N}$	$0.752$	$0.990$	$1.087$	$1.124$	$1.149$

(ii)

\begin{align} {a \over N} + {b \over t} & = 4 \\ {a \over N} & = - {b \over t} + 4 \\ {a \over N} & = -b \left(1 \over t\right) + 4 \\ {1 \over N} & = -{b \over a} \left(1 \over t\right) + {4 \over a} \\ \\ \therefore Y & = -{b \over a}X + {4 \over a} \\ \\ \text{where } Y & = {1 \over N} \\ X & = {1 \over t} \\ m & = -{b \over a} \\ c & = {4 \over a} \\ \\ \text{From } & \text{the graph,} \\ \text{Vertical intercept, } {4 \over a} & = 1.25 \\ 4 & = 1.25a \\ {4 \over 1.25} & = a \\ 3.2 & = a \\ \\ \text{Gradient, } -{b \over a} & = {1.05 - 0.85 \over 0.4 - 0.8} \\ -{b \over 3.2} & = -0.5 \\ -b & = 3.2(-0.5) \\ -b & = -1.6 \\ b & = 1.6 \end{align}

(i)

${1 \over x}$	$5$	$2.5$	$1.67$	$1.25$	$1$
$y$	$5.25$	$3.38$	$2.75$	$2.44$	$2.23$

(ii)

\begin{align} xy & = h(x + k) \\ xy & = hx + kh \\ y & = {hx \over x} + {kh \over x} \\ y & = h + kh\left(1 \over x\right) \\ y & = kh\left(1 \over x\right) + h \\ \\ \therefore Y & = khX + h \\ \\ \text{where } Y & = y \\ X & = {1 \over x} \\ m & = kh \\ c & = h \\ \\ \text{From } & \text{the graph,} \\ \text{Vertical intercept, } h & = 1.5 \\ \\ \text{Gradient, } kh & = {4.5 - 3 \over 4 - 2} \\ & = 0.75 \\ \\ \text{Since } & h = 1.5, \\ k(1.5) & = 0.75 \\ k & = {0.75 \over 1.5} \\ & = 0.5 \end{align}

(i)

\begin{align} y & = ax + {b \over x} \\ xy & = x\left(ax + {b \over x}\right) \\ xy & = ax^2 + b \\ \\ \therefore Y & = aX + b \\ \\ \text{where } Y & = xy \\ X & = x^2 \\ m & = a \\ c & = b \\ \\ \therefore \text{Plot } xy & \text{ against } x^2 \end{align}

$x^2$	$0.25$	$1$	$2.25$	$4$
$xy$	$7.3$	$6.8$	$6$	$4.8$

(ii)

\begin{align} \text{Gradient, } a & = {6.5 - 5.5 \over 1.5 - 3} \\ & = -0.66666 \\ & \approx -0.67 \\ \\ \\ \text{From } & \text{the graph,} \\ \text{Vertical intercept, } b & = 7.5 \end{align}

(iii) Method 1: Solve using the graph

\begin{align} \text{When } & x = 1.7, \\ x^2 & = (1.7)^2 \\ & = 2.89 \\ & \approx 2.9 \\ \\ \text{From the graph, } & \text{when } x^2 \approx 2.9, \\ xy & = 5.55 \\ \\ \text{Since } & x = 1.7, \\ (1.7)y & = 5.55 \\ y & = {5.55 \over 1.7} \\ y & = 3.2647 \\ y & \approx 3.3 \end{align}

(iii) Method 2: Use equation and values of a and b found in (ii)

\begin{align} y & = -0.66666x + {7.5 \over x} \\ \\ \text{When } & x = 1.7, \\ y & = -0.66666(1.7) + {7.5 \over 1.7} \\ & = 3.2784 \\ & \approx 3.3 \end{align}

(i)

$x$	$ 1 $	$ 2 $	$ 3 $	$ 4 $
${y \over x}$	$ 1.6 $	$ 3.6 $	$ 5.6 $	$ 7.6 $

(ii)

\begin{align} y & = ax^2 + bx \\ {1 \over x}(y) & = {1 \over x}(ax^2 + bx) \\ {y \over x} & = {ax^2 \over x} + {bx \over x} \\ {y \over x} & = ax + b \\ \\ \text{Linear-form: } \phantom{0} Y & = aX + b \\ \\ \text{where } Y & = {y \over x} \\ X & = x \\ m & = a \\ c & = b \\ \\ \\ \text{From } & \text{the graph,} \\ \text{Vertical intercept, } & = -0.4 \\ \\ \text{Gradient, } a & = {8 - 2.6 \over 4.2 - 1.5} \\ & = 2 \end{align}

(iii)

(i)

$\ln E$	$ 1.61 $	$ 2.30 $	$ 2.89 $	$ 3.00 $	$ 3.22 $	$ 3.40 $
$\ln P$	$ 0.92 $	$ 2.30 $	$ 3.48 $	$ 3.69 $	$ 4.14 $	$ 4.50 $

(ii)

Since a straight line is obtained, the relationship is true.

(iii)

\begin{align} P & = {E^n \over R} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln P & = \ln \left({E^n \over R}\right) \\ \ln P & = \ln E^n - \ln R \phantom{00000} [\text{Quotient law (Logarithms)}] \\ \ln P & = n(\ln E) - \ln R \phantom{0000} [\text{Power law (Logarithms)}] \\ \text{Linear-form: } Y & = nX - \ln R \\ \\ \text{where } Y & = \ln P \\ X & = \ln E \\ m & = n \\ c & = -\ln R \\ \\ \text{From } & \text{the graph,} \\ \\ c & = -2.305 \\ -\ln R & = -2.305 \\ \ln R & = 2.305 \\ \log_e R & = 2.305 \\ \\ R & = e^{2.305} \phantom{00000000} [\text{Change to index form}] \\ & = 10.0241 \\ & \approx 10.0 \\ \\ \text{Gradient, } n & = {4 - 0 \over 3.15 - 1.15} \\ & = 2 \\ \\ \text{Linear-form: } \ln P & = 2(\ln E) - \ln R \\ \\ \text{Using } (1.15, 0), & \text{ when } \ln E = 1.15 \text{ and } \ln P = 0, \\ 0 & = 2(1.15) - \ln R \\ 0 & = 2.3 - \ln R \\ \ln R & = 2.3 \\ \log_e R & = 2.3 \\ R & = e^{2.3} \\ & = 9.9741 \\ & \approx 10.0 \end{align}

(i)

$\lg \lambda$	$ 2.40 $	$ 2.67 $	$ 2.81 $	$ 3.08 $	$ 3.18 $
$\lg f$	$ 3.08 $	$ 2.81 $	$ 2.67 $	$ 2.40 $	$ 2.30 $

(ii)

\begin{align} f & = k\lambda ^n \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg f & = \lg (k \lambda^n) \\ \lg f & = \lg k + \lg \lambda^n \phantom{00000} [\text{Product law (logarithms)}] \\ \lg f & = \lg k + n\lg \lambda \phantom{0000/} [\text{Power law (logarithms)}] \\ \lg f & = n(\lg \lambda) + \lg k \\ \\ \text{Linear-form: } Y & = nX + \lg k \\ \\ \text{where } Y & = \lg f \\ X & = \lg \lambda \\ m & = n \\ c & = \lg k \\ \\ \text{From } & \text{the graph,} \\ \text{Vertical intercept, } \lg k & = 5.5 \\ \log_{10} k & = 5.5 \\ k & = 10^{5.5} \phantom{00000000} [\text{Change to index form}] \\ & \approx 316, \phantom{.} 227 \\ \\ \text{Gradient, } n & = {5 - 3.5 \over 0.5 - 2} \\ & = -1 \end{align}

(iii)

\begin{align} k\lambda^n & = 10^{-2.5n} \\ f & = 10^{-2.5n} \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg f & = \lg 10^{-2.5n} \\ \lg f & = (-2.5n) \lg 10 \\ \lg f & = (-2.5n)(1) \\ \lg f & = -2.5n \\ \lg f & = -2.5(-1) \phantom{00000} [\text{From part (ii), } n = -1] \\ \lg f & = 2.5 \\ \\ \text{Since } Y = \lg f, & \text{ draw } Y = \lg f = 2.5 \end{align}
\begin{align} \text{From } & \text{the graph,} \\ \lg \lambda & = 3 \\ \log_{10} \lambda & = 3 \\ \lambda & = 10^{3} \\ & = 1000 \end{align}

(i)

$ x^2 $	$ 0.04 $	$ 0.16 $	$ 0.36 $	$ 0.64 $
$ xy $	$ 1.548 $	$ 1.692 $	$ 1.932 $	$ 2.272 $

(ii)

\begin{align} y & = Cx + {D \over x} \\ x(y) & = x \left( Cx + {D \over x} \right) \\ xy & = Cx^2 + D \\ \\ \text{Linear-form: } Y & = CX + D \\ \\ \text{where } Y & = xy \\ X & = x^2 \\ m & = C \\ c & = D \\ \\ \text{From } & \text{the graph,} \\ \\ \text{Vertical intercept, } D & = 1.5 \\ \\ \text{Gradient, } C & = {2.22 - 1.62 \over 0.6 - 0.1} \\ & = 1.2 \end{align}

(iii) Note the linear form of the equation is $xy = Cx^2 + D$

\begin{align} Cx^2 + D & = 2 \\ xy & = 2 \\ \\ \text{Since } Y = xy, & \text{ draw } Y = xy = 2 \end{align}
\begin{align} \text{From } & \text{the graph,} \\ X & = 0.41 \\ x^2 & = 0.41 \\ x & = \sqrt{0.41} \\ & = 0.64031 \\ & \approx 0.6 \end{align}

(i)

\begin{align} y - x & = kx^n \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg (y - x) & = \lg (kx^n) \\ \lg (y - x) & = \lg k + \lg x^n \phantom{00000} [\text{Product law (logarithms)}] \\ \lg (y - x) & = \lg k + n\lg x \phantom{0000/} [\text{Power law (logarithms)}] \\ \lg (y - x) & = n(\lg x) + \lg k \\ \\ \text{Linear-form: } Y & = nX + \lg k \\ \\ \text{where } Y & = \lg (y - x) \\ X & = \lg x \\ m & = n \\ c & = \lg k \\ \\ \therefore \text{Plot } \lg (y - x) & \text{ against } \lg x \end{align}

$ \lg x $	$ 0.301 $	$ 0.477 $	$ 0.602 $	$ 0.699 $	$ 0.778 $
$ \lg (y - x) $	$ 0.996 $	$ 1.260 $	$ 1.447 $	$ 1.592 $	$ 1.711 $

(ii)

\begin{align} \text{From } & \text{the graph,} \\ \text{Vertical intercept, } \lg k & = 0.54 \\ \log_{10} k & = 0.54 \\ k & = 10^{0.54} \phantom{000000} [\text{Change to index form}] \\ & = 3.46736 \\ & \approx 3.5 \\ \\ \text{Gradient, } n & = {1.68 - 1.14 \over 0.76 - 0.4} \\ & = 1.5 \end{align}

(iii)

\begin{align} y & = x + 4.5 \\ y - x & = 4.5 \\ \lg (y - x) & = \lg 4.5 \\ \\ \text{Since } Y = \lg (y - x), & \text{ draw } Y = \lg (y - x) = \lg 4.5 \end{align}
\begin{align} \text{From } & \text{the graph,} \\ \lg x & = 0.07 \\ \log_{10} x & = 0.07 \\ x & = 10^{0.07} \\ & = 1.1748 \\ & \approx 1.17 \end{align}

(i)

\begin{align} y & = Ae^{-bx} \\ \\ \text{Take } \ln & \text{ of both sides,} \\ \ln y & = \ln (Ae^{-bx}) \\ \ln y & = \ln A + \ln e^{-bx} \phantom{00000} [\text{Product law (logarithms)}] \\ \ln y & = \ln A + (-bx)(\ln e) \phantom{0/} [\text{Power law (logarithms)}] \\ \ln y & = \ln A + (-bx)(1) \\ \ln y & = \ln A - bx \\ \ln y & = -bx + \ln A \\ \\ \text{Linear-form: } Y & = -bX + \ln A \\ \\ \text{where } Y & = \ln y \\ X & = x \\ m & = -b \\ c & = \ln A \\ \\ \therefore \text{Plot } \ln y & \text{ against } x \end{align}

$ x $	$ 1 $	$ 2 $	$ 3 $	$ 4 $	$ 5 $	$ 6 $
$ \ln y $	$ 2.588 $	$ 2.293 $	$ 1.988 $	$ 1.686 $	$ 1.386 $	$ 1.099 $

(ii)

\begin{align} \text{From } & \text{the graph,} \\ \text{Vertical intercept, } \ln A & = 2.9 \\ \log_e A & = 2.9 \\ A & = e^{2.9} \phantom{00000000} [\text{Change to index form}] \\ & = 18.174 \\ & \approx 18 \\ \\ \text{Gradient, } -b & = {2.75 - 1.55 \over 0.5 - 4.5} \\ - b & = -0.3 \\ b & = 0.3 \end{align}

(iii)

\begin{align} \text{When } & y = 6, \\ \ln y & = \ln 6 \\ & = 1.791759 \\ & \approx 1.8 \\ \\ \text{From the graph, } & \text{when } \ln y = 1.8, \\ x & \approx 3.65 \end{align}

(i)

\begin{align} y & = a + {b \over x} \\ x(y) & = x\left(a + {b \over x}\right) \\ xy & = ax + b \\ \\ \text{Linear-form: } Y & = aX + b \\ \\ Y & = xy \\ X & = x \\ m & = a \\ c & = b \\ \\ \therefore \text{Plot } xy & \text{ against } x \end{align}

$ x $	$ 0.4 $	$ 1.7 $	$ 2.0 $	$ 2.5 $	$ 3.2 $	$ 4.0 $	$ 5.0 $
$ xy $	$ -0.2 $	$ 1.36 $	$ 3 $	$ 4 $	$ 5.44 $	$ 7 $	$ 9 $

(ii)(a)

\begin{align} \text{Abnormal reading, } y & = 0.8 \\ \\ \text{From the graph } & \text{and when } x = 1.7, \\ xy & = 2.4 \\ (1.7)(y) & = 2.4 \\ 1.7y & = 2.4 \\ y & = {2.4 \over 1.7} \\ & = 1.4117 \\ & \approx 1.4 \end{align}

(ii)(b)

\begin{align} \text{From } & \text{the graph,} \\ \text{Vertical intercept, } b & = -1 \\ \\ \text{Gradient, } a & = {8 - 1 \over 4.5 - 1} \\ a & = 2 \end{align}

(i)

\begin{align} T & = kx^n \\ \\ \text{Take } \lg & \text{ of both sides,} \\ \lg T & = \lg (kx^n) \\ \lg T & = \lg k + \lg x^n \phantom{00000} [\text{Product law (logarithms)}] \\ \lg T & = \lg k + n\lg x \phantom{00000} [\text{Power law (logarithms)}] \\ \lg T & = n(\lg x) + \lg k \\ \\ \text{Linear-form: } Y & = nX + \lg k \\ \\ \text{where } Y & = \lg T \\ X & = \lg x \\ m & = n \\ c & = \lg k \\ \\ \therefore \text{Plot } \lg T & \text{ against } \lg x \end{align}

$ \lg x $	$ 1.76 $	$ 2.03 $	$ 2.36 $	$ 2.89 $	$ 3.15 $
$ \lg T $	$ -0.62 $	$ -0.21 $	$ 0.27 $	$ 1.07 $	$ 1.47 $

(ii)

\begin{align} \text{From } & \text{the graph,} \\ \text{Gradient, } n & = {1.25 - 0.5 \over 3 - 2.5} \\ & = 1.5 \\ \\ \text{Linear form: } \lg T & = 1.5 (\lg x) + \lg k \\ \\ \text{Using } (2.5, 0.5), & \text{ when } \lg x = 2.5 \text{ and } \lg T = 0.5, \\ 0.5 & = 1.5(2.5) + \lg k \\ -\lg k & = 3.75 - 0.5 \\ -\lg k & = 3.25 \\ \lg k & = -3.25 \\ \log_{10} k & = -3.25 \\ k & = 10^{-3.25} \phantom{00000000} [\text{Change to index form}] \\ & = 5.6234 \times 10^{-4} \\ & \approx 5.6 \times 10^{-4} \end{align}

(iii) Method 1: Using equation of graph & values of k and n from (ii)

\begin{align} \text{Since } & k = 5.6234 \times 10^{-4} \text{ and } n = 1.5, \\ \\ T & = (5.6234 \times 10^{-4})x^{1.5} \\ \\ \text{When } & x = 149.6, \\ T & = (5.6234 \times 10^{-4})(149.6)^{1.5} \\ & = 1.0289 \\ & \approx 1 \text{ Earth years} \end{align}

(iii) Method 2: Using graph

\begin{align} \text{When } & x = 149.6, \\ \lg x & = \lg 149.6 \\ & \approx 2.17 \\ \\ \text{From the graph, } & \text{when } \lg x = 2.17, \\ \lg T & = 0 \\ \log_{10} T & = 0 \\ T & = 10^0 \phantom{00000} [\text{Change to index form}] \\ & = 1 \text{ Earth years} \end{align}

(i)

(ii)

\begin{align} & \text{From (i), shape resembles graph of } y = \ln x \\ \\ & \therefore \text{Plot } P \text{ against } \ln t \end{align}

$\ln t$	$0$	$0.693$	$1.609$	$2.302$	$2.708$	$2.995$	$3.401$
$ P $	$30$	$42$	$50$	$65$	$73$	$77$	$85$

\begin{align} m & = {61 - 30 \over 2 - 0} \\ & = 15.5 \\ \\ c & = 30 \\ \\ Y & = mX + c \\ Y & = 15.5X + 30 \\ \\ \text{Since } & Y = P \text{ and } X = \ln t, \\ P & = 15.5 \ln t + 30 \end{align}

(iii)

\begin{align} P & = 15.5 \ln t + 30 \\ \\ \text{When } & P = 95, \\ 95 & = 15.5 \ln t + 30 \\ 65 & = 15.5 \ln t \\ {65 \over 15.5} & = \ln t \\ {65 \over 15.5} & = \log_e t \\ \\ t & = e^{65 \over 15.5} \\ t & = 66.257 \\ t & \approx 66.3 \text{ s} \end{align}

(i)

\begin{align} \text{Quadratic equation, since the points resemble the shape of a quadratic graph } (\cup) \end{align}

(ii)

\begin{align} y & = ax^2 + bx \\ y & = x(ax + b) \\ {y \over x} & = ax + b \\ \\ \text{Plot } & {y \over x} \text{ against } x \end{align}

$x$	$2$	$3$	$4$	$5$	$6$	$7$
$ {y \over x} $	$9$	$11$	$13$	$15$	$17$	$19$

\begin{align} {y \over x} & = ax + b \\ \\ \text{Vertical intercept, } b & = 5 \\ \\ \text{Gradient, } a & = {18 - 5 \over 6.5 - 0} \\ & = 2 \\ \\ \\ y & = 2x^2 + 5x \end{align}

(iii)

\begin{align} y & = 2x^2 + 5x \\ \\ \text{When } & x = 4.5, \\ y & = 2(4.5)^2 + 5(4.5) \\ y & = 63 \end{align}

(iv)

\begin{align} & \text{No, since }x = 1 \text{ lies outside the experimental values of } x, 2 \le x \le 7 \\ \\ & \text{Thus, equation may not apply for } x = 1 \end{align}