Hey
12a)i)
- Find the gradient for the line AC. Recall that for lines \(l_1\) and \(l_2\), the respective gradients \(m_{l_1}\) and \(m_{l_2}\) must have product -1.
- Use the point C (which has coordinates given), and substitute the previous result into the point-gradient form of the line.
12a)ii).
- You can find the x-intercept of the line calculated above, then find the length of BC (reasonably easy since they both have y coordinate 0), then use formula \(A = \frac{bh}{2}\) (think about where you can find the height!)
12d)
Basically asking you to evaluate \(\int_0^3 \frac{3x}{x^2+1} \ dx\). Recall that \(\int \frac{f'(x)}{f(x)} \ dx = f(x) + C\) - try manipulating this integral into a better form so you can use this formula.
15b)i)
Note that \(M_0 = 0\), and that \(M_1 = X(1+\frac{\frac{4.2}{12}}{100})-2500 = (1.0035)X - 2500\). Always start with \(M_0\), and find \(M_1\) after, executing each successive action in order. Things to note here are that \(M_0 = 0\) because the withdrawal happens the day before the deposit (withdrawal is on the last day, deposit on the first day of the next month), and \(M_1\) is that expression because the month's deposit has been placed, interest has been tacked on, then the withdrawal occurs.
Executing instructions in this order on \(M_1\) means that \(M_2 = 1.0035(M_1) - 2500\). You can work it out from there - take away is the setup is so important and order matters.
ii)
After this, you can generalise the formula for \(M_n\) based on the pattern. Try working out \(M_3\) and even \(M_4\) if you have to, so you can see what it is. This can be applied for any question.
here, they want you to use n = 4, and \(M_4 = 80000\). After this, you have an equation in X that you can rearrange and solve.
Hope this helps