A-LevelMathematicsFunctions9709/12 · Oct/Nov 2019 · Q912 Marks

Functions f and g are defined by f(x) = 2x² + 8x + 1 for x ∈ R, g(x) = 2x - k for x ∈ R, where k is a constant.

(i)Find the value of k for which the line y = g(x) is a tangent to the curve y = f(x).[3]
(ii)In the case where k = −9, find the set of values of x for which f(x) < g(x).[3]
(iii)In the case where k = −1, find g⁻¹f(x) and solve the equation g⁻¹f(x) = 0.[3]
(iv)Express f(x) in the form 2(x + a)² + b, where a and b are constants, and hence state the least value of f(x).[3]

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2019). The mark or mark type is shown in brackets.

[*M1] f(x) = g(x) → 2x²+6x+1+k (= 0) Forms a quadratic with all terms on same side.
[DM1] Use of b² = 4ac Uses the discriminant = 0.
[A1] (k =) 3½ OE, WWW
[*M1] 4x+8=2 (x = -1½) Differentiating, equating gradients and solving to give x =
[DM1] Substitutes their x value into either 2x² + 6x + 1 + k = 0 OR into the curve to find y = -13/2 then both values into the line. Substituting appropriately for their x and proceeding to find a value of k.
[A1] (k =) 3½ OE, WWW
[M1] 2x²+6x-8 (0) Forms a quadratic with all terms on same side
[A1] -4 and 1
[A1] -4 < x < 1 CAO
[B1] (g⁻¹(x)) = (x-1)/2 Needs to be in terms of x.
[M1] (g⁻¹f(x)) = (2x²+8x+1-1)/2 = 0 (2x²+8x = 0) → x = Substitutes f into g⁻¹ and attempts to solve it = 0 as far as x =
[A1] 0, -4 CAO
[B1 B1] 2(x+2)²-7 or a = +2, b = -7
[B1FT] (Least value of f(x) or y =) -7 or > -7 FT for their b from a correct form of the expression.

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