It is given that f : x → (3x+1)²−4 for x ≥ a, and that f⁻¹ exists. (i) Find the least possible value of a. (ii) Using this value of a, write down the range of f. (iii) Using this value of a, sketch the graphs of y = f(x) and y = f⁻¹(x) on the axes, stating the intercepts with the coordinate axes.

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2023). The mark or mark type is shown in brackets.

1. [1] (a)(i) B1 Allow -0.333 or better Allow a correct recurring decimal
2. [1] (a)(ii) B1
3. [4] (a)(iii) B1 for y = f(x), must have a correct shape (right hand side from the vertex of a quadratic curve), must be a 1:1 function, intersecting each of the x and y axes once, in quadrants 1, 3 and 4.
4. [4] (a)(iii) B1, dependent on previous B for passing through (0, -3) and (1/3, 0).
5. [4] (a)(iii) B1 dependent on first B1 for y=f⁻¹(x), being a correct reflection of their y = f(x), intersecting each of the x and y axes and y = f (x) once.
6. [4] (a)(iii) B1 dependent on previous B for passing through (-3, 0) and (0, 1/3).
7. [3] (b) M1 For correct order
8. [3] (b) M1 dep for a correct attempt to deal with logarithms and obtain x = ... Allow one arithmetic or sign slip.