Date | May 2015 | Marks available | 5 | Reference code | 15M.2.sl.TZ1.4 |
Level | SL only | Paper | 2 | Time zone | TZ1 |
Command term | Find and Write down | Question number | 4 | Adapted from | N/A |
Question
Let \(f(x) = \frac{{2x - 6}}{{1 - x}}\), for \(x \ne 1\).
For the graph of \(f\)
(i) find the \(x\)-intercept;
(ii) write down the equation of the vertical asymptote;
(iii) find the equation of the horizontal asymptote.
Find \(\mathop {\lim }\limits_{x \to \infty } f(x)\).
Markscheme
(i) valid approach (M1)
eg\(\;\;\;\)sketch, \(f(x) = 0,{\text{ }}0 = 2x - 6\)
\(x = 3\) or \((3,{\text{ }}0)\) A1 N2
(ii) \(x = 1\;\;\;\)(must be equation) A1 N1
(iii) valid approach (M1)
eg\(\;\;\;\)sketch, \(\frac{{2x}}{{ - 1x}}\), inputting large values of \(x\), L’Hopital’s rule
\(y = - 2\;\;\;\)(must be equation) A1 N2
[5 marks]
valid approach (M1)
eg\(\;\;\;\)recognizing that \(\mathop {\lim }\limits_{x \to \infty } \) is related to the horizontal asymptote, table with large values of \(x\), their \(y\) value from (a)(iii), L’Hopital’s rule
\(\mathop {\lim }\limits_{x \to \infty } f(x) = - 2\) A1 N2
[2 marks]
Total [7 marks]
Examiners report
Part (a) was generally well done with candidates using both algebraic and graphical approaches to obtain solutions. There are still some who do not identify their asymptotes using equations.
Candidates rarely appreciated the relevance of the horizontal asymptote in (b), and often attempted a long, and often unsuccessful, algebraic approach to find the limit.