| Date | November 2014 | Marks available | 4 | Reference code | 14N.3srg.hl.TZ0.2 |
| Level | HL only | Paper | Paper 3 Sets, relations and groups | Time zone | TZ0 |
| Command term | Prove that | Question number | 2 | Adapted from | N/A |
Question
Define \(f:\mathbb{R}\backslash \{ 0.5\} \to \mathbb{R}\) by \(f(x) = \frac{{4x + 1}}{{2x - 1}}\).
Prove that \(f\) is an injection.
Prove that \(f\) is not a surjection.
Markscheme
METHOD 1
\(f(x) = f(y) \Rightarrow \frac{{4x + 1}}{{2x - 1}} = \frac{{4y + 1}}{{2y - 1}}\) M1A1
for attempting to cross multiply and simplify M1
\((4x + 1)(2y - 1) = (2x - 1)(4y + 1)\)
\( \Rightarrow 8xy + 2y - 4x - 1 = 8xy + 2x - 4y - 1 \Rightarrow 6y = 6x\)
\( \Rightarrow x = y\) A1
hence an injection AG
METHOD 2
\(f'(x) = \frac{{4(2x - 1) - 2(4x + 1)}}{{{{(2x - 1)}^2}}} = \frac{{ - 6}}{{{{(2x - 1)}^2}}}\) M1A1
\( < 0\;\;\;{\text{(for all }}x \ne 0.5{\text{)}}\) R1
therefore the function is decreasing on either side of the discontinuity
and \(f(x) < 2\) and \(x < 0.5\) for \(f(x) > 0.5\) R1
hence an injection AG
Note: If a correct graph of the function is shown, and the candidate states this is decreasing in each part (or horizontal line test) and hence an injection, award M1A1R1.
[4 marks]
METHOD 1
attempt to solve \(y = \frac{{4x + 1}}{{2x - 1}}\) M1
\(y(2x - 1) = 4x + 1 \Rightarrow 2xy - y = 4x + 1\) A1
\(2xy - 4x = 1 + y \Rightarrow x = \frac{{1 + y}}{{2y - 4}}\) A1
no value for \(y = 2\) R1
hence not a surjection AG
METHOD 2
consider \(y = 2\) A1
attempt to solve \(2 = \frac{{4x + 1}}{{2x - 1}}\) M1
\(4x - 2 = 4x + 1\) A1
which has no solution R1
hence not a surjection AG
Note: If a correct graph of the function is shown, and the candidate states that because there is a horizontal asymptote at \(y = 2\) then the function is not a surjection, award M1R1.
[4 marks]
Total [8 marks]
Examiners report
Most students indicated an understanding of the concepts of Injection and Surjection, but many did not give rigorous proofs. Even where graphs were used, it was very common for a sketch to be so imprecise with no asymptotes marked that it was difficult to award even partial credit. Some candidates mistakenly stated that the function was not surjective because 0.5 was not in the domain.
Most students indicated an understanding of the concepts of Injection and Surjection, but many did not give rigorous proofs. Even where graphs were used, it was very common for a sketch to be so imprecise with no asymptotes marked that it was difficult to award even partial credit. Some candidates mistakenly stated that the function was not surjective because 0.5 was not in the domain.
