Date | May 2014 | Marks available | 6 | Reference code | 14M.2.hl.TZ1.4 |
Level | HL only | Paper | 2 | Time zone | TZ1 |
Command term | Find and Show that | Question number | 4 | Adapted from | N/A |
Question
A system of equations is given below.
\(x + 2y - z = 2\)
\(2x + y + z = 1\)
\( - x + 4y + az = 4\)
(a) Find the value of a so that the system does not have a unique solution.
(b) Show that the system has a solution for any value of a.
Markscheme
(a) \(\left[ \begin{array}{l}x + 2y - z = 2\\2x + y + z = 1\\ - x + 4y + az = 4\end{array} \right.\)
\( \to \left[ \begin{array}{l}x + 2y - z = 2\\ - 3y + 3z = - 3\\6y + (a - 1)z = 6\end{array} \right.\) M1A1
\( \to \left[ \begin{array}{l}x + 2y - z = 2\\ - 3y + 3z = - 3\\(a + 5)z = 0\end{array} \right.\) A1
(or equivalent)
if not a unique solution then \(a = - 5\) A1
Note: The first M1 is for attempting to eliminate a variable, the first A1 for obtaining two expression in just two variables (plus a), and the second A1 for obtaining an expression in just a and one other variable
[4 marks]
(b) if \(a = - 5\) there are an infinite number of solutions as last equation always true R1
and if \(a \ne - 5\) there is a unique solution R1
hence always a solution AG
[2 marks]
Total [6 marks]