Since h(s) ≤ 0 for s ∈ [0, π], we c

Since h(s) ≤ 0 for s ∈ [0, π], we conclude from (6) that g(t) ≤ 0 for t ∈ [0, 1]. Thus, g
is concave. We have g(0) = g(1) = 0. Hence, g(t) ≥ 0 for t ∈ [0, 1].

ผลลัพธ์ (แอฟริกา) 1:

Sedert h (s) ≤ 0 vir s ∈ [0, π], ons aflei uit (6) wat g ?? (t) ≤ 0 vir t ∈ [0, 1]. So, G
konkaaf. Ons het g (0) = g (1) = 0. Dus, g (t) ≥ 0 vir t ∈ [0, 1].