A-LevelMathematicsFunctions9709/12 · Oct/Nov 2024 · Q510 Marks

The function f is defined by f(x) = (2x+1)/(2x-1) for x < 1/2.

(a)(i)State the value of f(-1).[1]
(a)(ii)The diagram shows the graph of y = f(x). Sketch the graph of y = f⁻¹(x) on this diagram. Show any relevant mirror line. [Figure X.X][2]
(a)(iii)Find an expression for f⁻¹(x) and state the domain of the function f⁻¹.[4]
(b)The function g is defined by g(x) = 3x+2 for x ∈ R. Solve the equation f(x) = gf(1/4).[3]

This question uses a diagram from the original paper, which isn't shown here. Open the full paper from the past papers archive to see it.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2024). The mark or mark type is shown in brackets.

[B1] [f(-1)=] 1/3 Condone 0.333.
[B1] For showing the correct mirror line.
[B1] For correct shape: the curves should intersect in the first square in the third quadrant. To the left of the point of intersection, the reflection is below the original and crosses the x-axis. To the right of the point of intersection, the reflection is to the right the original.
[M1*] Equating y to the given function and clearing of fractions. x and y may be interchanged at this stage.
[DM1] Condone ± errors during simplification.
[A1] (x+1)/(2x-2) or (-x-1)/(2-2x) OE Allow 'f-1'or 'y =' but NOT 'x =', nor fractions within fractions.
[B1] [Domain of f-1 is] x < 1 Accept - ∞ < x <1 or (-∞, 1), condone [-∞, 1).
[B1] gf(1/4) = -7
[M1] Equating (2x+1)/(2x-1) to their gf(1/4)
[A1] [x=] 3/8 OE

On Oracle Prep you can write or photograph your own answer and have it marked by AI against this mark scheme, with feedback on what you missed.