Functions f and g are defined by f : x → 2x – 3, x ∈ R, g : x → x² + 4x, x ∈ R. Function h is defined by h : x → x² + 4x for x ≥ k, and it is given that h has an inverse.
- (i)Solve the equation ff(x) = 11.[2]
- (ii)Find the range of g.[2]
- (iii)Find the set of values of x for which g(x) > 12.[3]
- (iv)Find the value of the constant p for which the equation gf(x) = p has two equal roots.[3]
- (v)State the smallest possible value of k.[1]
- (vi)Find an expression for h⁻¹(x).[4]
Mark scheme
Marking points from the official mark scheme (May/June 2014). The mark or mark type is shown in brackets.
- [M1] ff = 2(2x – 3) – 3 Either forms ff correctly, or solves 2 equations co
- [A1] Solves = 11 → x = 5 (or 2x – 3 = 11, x = 7. 2x – 3 = 7 → x = 5)
- [M1 A1] min at x = –2 → Range ≥ –4 Any valid method – could be guesswork.
- [M1] x² + 4x – 12 (> 0) Makes quadratic = 0 + 2 solutions
- [A1] → x = 2 or –6 Correct limits – even if >, <, ≥, ≤, =
- [A1] → x < –6, x > 2. co
- [B1] gf(x) = (2x – 3)² + 4(2x – 3) = p co unsimplified
- [M1] → 4x² – 4x – 3 – p = 0 Uses "b² – 4ac"
- [A1] 16 = 16(–3 – p) → p = –4 co
- [B1] –2 co
- [B2,1] y = (x + 2)² – 4 –1 for each error
- [M1] √y + 4 = x + 2 Correct order of operations
- [A1] h⁻¹(x) = √x + 4 – 2 co with x, not y. ± left A0.
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