To solve for \(C\) and \(a\), we start by considering the equation \(y = C(a^x)\). Taking the natural logarithm of both sides gives:

\(\ln y = \ln C + x \ln a\)

This equation is in the form of a straight line \(y = mx + c\), where \(\ln y\) is plotted against \(x\), \(\ln C\) is the y-intercept, and \(\ln a\) is the slope.

Plot the given points \((x, \ln y)\):

• (0.9, 1.7)
• (1.6, 1.9)
• (2.4, 2.3)
• (3.2, 2.6)

Draw a straight line through these points. From the line, determine the slope \(\ln a\) and the y-intercept \(\ln C\).

Using the mark scheme, we find:

\(\ln C = \ln 3.7\)

\(\ln a = \ln 1.5\)

Thus, \(C = 3.7\) and \(a = 1.5\).

(i) Start with the equation \(3^y = 4^{2-x}\).

Take the natural logarithm of both sides: \(\ln(3^y) = \ln(4^{2-x})\).

Using the logarithm power rule, this becomes \(y \ln 3 = (2-x) \ln 4\).

Rearrange to get \(y \ln 3 = 2 \ln 4 - x \ln 4\).

This can be written as \(y = -\frac{\ln 4}{\ln 3} x + \frac{2 \ln 4}{\ln 3}\), which is in the form \(y = mx + c\), showing it is a straight line.

The gradient \(m\) is \(-\frac{\ln 4}{\ln 3}\).

(ii) To find the intersection with \(y = 2x\), set \(-\frac{\ln 4}{\ln 3} x + \frac{2 \ln 4}{\ln 3} = 2x\).

Rearrange to get \(\frac{2 \ln 4}{\ln 3} = 2x + \frac{\ln 4}{\ln 3} x\).

Factor out \(x\): \(x \left(2 + \frac{\ln 4}{\ln 3}\right) = \frac{2 \ln 4}{\ln 3}\).

Solve for \(x\): \(x = \frac{\ln 4}{\ln 6}\).

\(Given the equation y = Ae^-kx2, taking the natural logarithm of both sides gives:\)

\(\ln y = \ln A - kx^2\)

This is in the form of a straight line \(y = mx + c\) with \(\ln y\) as the dependent variable and \(x^2\) as the independent variable. The slope \(m = -k\) and the intercept \(c = \ln A\).

Using the points (0.64, 0.76) and (1.69, 0.32), calculate the slope:

\(m = \frac{0.32 - 0.76}{1.69 - 0.64} = \frac{-0.44}{1.05} = -0.419\)

Thus, \(k = 0.42\).

Substitute one of the points into the equation \(\ln y = \ln A - kx^2\) to find \(\ln A\):

Using (0.64, 0.76):

\(0.76 = \ln A - 0.42 \times 0.64^2\)

\(0.76 = \ln A - 0.42 \times 0.4096\)

\(0.76 = \ln A - 0.172032\)

\(\ln A = 0.76 + 0.172032 = 0.932032\)

Therefore, \(A = e^{0.932032} \approx 2.80\).