Date | November 2018 | Marks available | 5 | Reference code | 18N.1.AHL.TZ0.H_4 |
Level | Additional Higher Level | Paper | Paper 1 (without calculator) | Time zone | Time zone 0 |
Command term | Find | Question number | H_4 | Adapted from | N/A |
Question
Consider the following system of equations where a∈R.
2x+4y−z=10
x+2y+az=5
5x+12y=2a.
Find the value of a for which the system of equations does not have a unique solution.
Find the solution of the system of equations when a=2.
Markscheme
* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.
an attempt at a valid method eg by inspection or row reduction (M1)
2×R2=R1⇒2a=−1
⇒a=−12 A1
[2 marks]
using elimination or row reduction to eliminate one variable (M1)
correct pair of equations in 2 variables, such as
5x+10y=255x+12y=4} A1
Note: Award A1 for z = 0 and one other equation in two variables.
attempting to solve for these two variables (M1)
x=26, y=−10.5, z=0 A1A1
Note: Award A1A0 for only two correct values, and A0A0 for only one.
Note: Award marks in part (b) for equivalent steps seen in part (a).
[5 marks]