A-LevelMathematicsTrigonometry9709/12 · May/June 2022 · Q1110 Marks
The function f is given by f(x) = 4cos⁴x + cos²x − k for 0 ≤ x ≤ 2π, where k is a constant.
- (a)Given that k = 3, find the exact solutions of the equation f(x) = 0.[5]
- (b)Use the quadratic formula to show that, when k > 5, the equation f(x) = 0 has no solutions.[5]
Mark scheme
Marking points from the official mark scheme (May/June 2022). The mark or mark type is shown in brackets.
- [M1] 4cos⁴x + cos²x - 3 = 0 => (4cos²x - 3)(cos²x + 1) = 0
- [A1] [cos²x =] 3/4 [and cos²x = -1]
- [M1] cosx = [±]√3/2
- [A1 A1 FT] [x =] π/6, 5π/6, 7π/6, 11π/6
- [B1] cos²x = (-1 - √(1+16k))/8 < 0 [.. no solutions].
- [M1] [cos²x] = (-1 ± √(1+16k))/8
- [DM1] Substituting k = 5 and obtain 1 from the formula
- [A1] cos²x = 1 or cos²x > or ≥ 1
- [A1] Concluding statement having considered both ± cases. .. no solutions
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