A-LevelMathematicsQuadratics9709/12 · Oct/Nov 2015 · Q89 Marks
The function f is defined, for x ∈ R, by f : x → x² + ax + b, where a and b are constants.
- (i)In the case where a = 6 and b = −8, find the range of f.[3]
- (ii)In the case where a = 5, the roots of the equation f(x) = 0 are k and −2k, where k is a constant. Find the values of b and k.[3]
- (iii)Show that if the equation f(x + a) = a has no real roots, then a² < 4(b − a).[3]
Mark scheme
Marking points from the official mark scheme (Oct/Nov 2015). The mark or mark type is shown in brackets.
- [B1] x² + 6x − 8 = (x + 3)² −17 (B1 for (x+3)², B1 for -17)
- [B1] → Range f(x) ≥ − 17 (Following through visible method)
- [M1] (x-k)(x + 2k) = 0 (Realises the link between roots and the equation)
- [A1] → k = 5 (comparing coefficients of x)
- [A1] → b = -2k² = -50
- [M1] (x + a)² + a(x + a) + b = a (Replaces "x" by “x + a” in 2 terms)
- [DM1] Uses b² - 4ac → 9a² - 4(2a² + b − a) (Any use of discriminant)
- [A1] → a² < 4(b − a)
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