A-LevelMathematicsFunctions9709/12 · May/June 2016 · Q1111 Marks

The function f is defined by f : x → 6x − x² – 5 for x ∈ R.

(i)Find the set of values of x for which f(x) ≤ 3.[3]
(ii)Given that the line y = mx + c is a tangent to the curve y = f(x), show that 4c = m² – 12m + 16.[3]
(iii)Express 6x – x² – 5 in the form a – (x – b)², where a and b are constants.[2]
(iv)State the smallest value of k for which g has an inverse.[1]
(v)For this value of k, find an expression for g⁻¹(x).[2]

Mark scheme

Marking points from the official mark scheme (May/June 2016). The mark or mark type is shown in brackets.

[M1] 6x - x² - 5 ≤ 3 → x² - 6x + 8 ≥ 0 ±(6x-x²-8)=, <, > 0 and attempts to solve
[A1] → x = 2, x = 4 Needs both values whether =2, <2, >2
[A1] x < 2, x ≥ 4 Accept all recognisable notation. condone < and/or >
[M1] Equate mx + c and 6x - x² - 5 Equates, sets to 0.
[DM1] Use of "b² - 4ac" Use of discriminant with values of a,b,c independent of x.
[A1] 4c = m²-12m+16. AG = (0) must appear before last line.
OR
[M1] dy/dx = 6-2x = m → x = (6-m)/2 Equates dy/dx to m and rearrange
[M1] m(6-m)/2 + c = 6(6-m)/2 - ((6-m)/2)² - 5 Equates mx + c and 6x – x² -5 and substitutes for x
[A1] 4c = m²-12m+16. AG
[B1 B1] 6x – x² – 5 = 4 - (x – 3)² B1 - (x - 3)2 B1
[B1] k = 3. B1 for “b”.
[M1 A1] g⁻¹(x) = √4-x +3 Correct order of operations.
±√4-x+3 M1A0
√x-4+3 M1A0
√4-y+3 M1A0

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