A-LevelMathematicsFunctions9709/12 · May/June 2015 · Q1112 Marks
The function f is defined by f : x → 2x² – 6x + 5 for x ∈ R.
- (i)Find the set of values of p for which the equation f(x) = p has no real roots.[3]
- (ii)The function g is defined by g : x → 2x² – 6x + 5 for 0 ≤ x ≤ 4. Express g(x) in the form a(x + b)² + c, where a, b and c are constants.[3]
- (iii)Find the range of g.[2]
- (iv)The function h is defined by h : x → 2x² – 6x + 5 for k ≤ x ≤ 4, where k is a constant. State the smallest value of k for which h has an inverse.[1]
- (v)For this value of k, find an expression for h⁻¹(x).[3]
Mark scheme
Marking points from the official mark scheme (May/June 2015). The mark or mark type is shown in brackets.
- [M1] (i) 2x² - 6x + 5 - p = 0 has no real roots Sets to 0 with p on LHS.
- [DM1] (i) Uses b² – 4ac → 36 – 8(5 – p) Uses discriminant.
- [A1] (i) Sets to 0 → p < 1/2 co – must be "<", not "≤".
- [B1 B1 B1] (ii) 2x² - 6x + 5 = 2(x - 3/2)² + 1/2 co
- [B1 B1] (iii) Range of g: 1/2 ≤ g(x) ≤ 13 √ on (ii) co from sub of x = 4
- [B1] (iv) Smallest k = 3/2 √ on (ii)
- [M1] (v) Using comp square form to try and get x as subject or y if transposed.
- [DM1] (v) Order of operations ±1/2, ÷2, √, ±1/2 Order must be correct
- [A1] (v) → Inverse = 3/2 + √((x - 1/2)/2) co (without ±)
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