The function f is such that f(x) = a + b cos x for 0 ≤ x ≤ 2π. It is given that f(π/3) = 5 and f(π) = 11.

Mark scheme

Marking points from the official mark scheme (May/June 2018). The mark or mark type is shown in brackets.

1. [B1] (i) f(π/3) = a + b cos(π/3) = a + ½b = 5.
2. [B1] (i) f(π) = a + b cos(π) = a - b = 11.
3. [B1] (i) Solving the simultaneous equations: a = 7 and b = -4. (Alternatively these marks can be awarded when ½ and –1 appear after a or b has been eliminated.)
4. [B1] (ii) The range of f(x) is determined by the maximum and minimum values of a + b cos x. With a=7, b=-4, f(x) = 7 - 4 cos x. The maximum value occurs when cos x = -1, so f(x) = 7 - 4(-1) = 11 (which is a + b or their a + their b (3)).
5. [B1] (ii) The minimum value occurs when cos x = 1, so f(x) = 7 - 4(1) = 3 (which is a - b or their a - their b (11). Note: Use of b² – 4ac scores 0/3).
6. [B1] (ii) The range of f(x) is [3, 11]. The equation f(x) = k has no solution when k is outside this range. So, k < 3 or k > 11. (Both inequalities correct. Allow combined statement as long as correct inequalities if taken separately. Both answers correct from T & I or guesswork 3/3 otherwise 0/3)