Prove that $f$ is an injection.

Prove that $f$ is not a surjection.

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]

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.