f(x) = ax4 + bx3 + cx2 + dx + e, where a, b, c, d, and e are real numbers.
Suppose f(x) = 1 has p distinct real solutions, f(x) = 2 has q distinct real solutions, f(x) = 3 has r distinct real solutions, and f(x) = 4 has s distinct real solutions.
Which one of the following is not possible?
- Ap=1, q=2, r=4 and s=3
- Bp=1, q=3, r=2 and s=4
- Cp=1, q=4, r=3 and s=2
- Dp=2, q=4, r=3 and s=1
- Ep=4, q=3, r=2 and s=1
Show the answer and worked solution
answer · B
- Ap=1, q=2, r=4 and s=3
- Bp=1, q=3, r=2 and s=4
- Cp=1, q=4, r=3 and s=2
- Dp=2, q=4, r=3 and s=1
- Ep=4, q=3, r=2 and s=1
Think of counting where the horizontal lines y=1, 2, 3, 4 cut the curve. For a quartic with positive leading coefficient the count as k increases runs 0, 1, 2, 3, 4, 3, 2; with negative leading coefficient it runs 2, 3, 4, 3, 2, 1, 0. Either way the count rises to a peak and then falls — it never dips and then climbs again. Options A, C, D and E all follow that up-then-down shape, and each can be realised by choosing where the turning-point heights sit. Option B goes 1, 3, 2, 4, which falls from 3 to 2 and then rises to 4, so no quartic can produce it.