| Date | May 2015 | Marks available | 4 | Reference code | 15M.2.hl.TZ2.6 |
| Level | HL only | Paper | 2 | Time zone | TZ2 |
| Command term | Find | Question number | 6 | Adapted from | N/A |
Question
The graph of \(y = \ln (5x + 10)\) is obtained from the graph of \(y = \ln x\) by a translation of \(a\) units in the direction of the \(x\)-axis followed by a translation of \(b\) units in the direction of the \(y\)-axis.
Find the value of \(a\) and the value of \(b\).
The region bounded by the graph of \(y = \ln (5x + 10)\), the \(x\)-axis and the lines \(x = {\text{e}}\) and \(x = 2{\text{e}}\), is rotated through \(2\pi \) radians about the \(x\)-axis. Find the volume generated.
Markscheme
EITHER
\(y = \ln (x - a) + b = \ln (5x + 10)\) (M1)
\(y = \ln (x - a) + \ln c = \ln (5x + 10)\)
\(y = \ln \left( {c(x - a)} \right) = \ln (5x + 10)\) (M1)
OR
\(y = \ln (5x + 10) = \ln \left( {5(x + 2)} \right)\) (M1)
\(y = \ln (5) + \ln (x + 2)\) (M1)
THEN
\(a = - 2,{\text{ }}b = \ln 5\) A1A1
Note: Accept graphical approaches.
Note: Accept \(a = 2,{\text{ }}b = 1.61\)
[4 marks]
\(V = \pi {\int_e^{2e} {\left[ {\ln (5x + 10)} \right]} ^2}{\text{d}}x\) (M1)
\( = 99.2\) A1
[2 marks]
Total [6 marks]
