“…x) = 15(β 2 + 6β + 7)x4 (1 − x)3 (1 + x) −[120x4 (1 − x)3 (1 + x) + 15(6α + 5)x 4 (1 − x) 2 (1 + x)](β + 3) +45x 5 (1 − x) 2 (1 + x)(β + 3) 2 + 130x 3 (1 − x) 4 (1 + x) +10(12α + 5)x 3 (1 − x) 3 (1 + x) + 5(9α 2 + 15α + 8)x 3 (1 − x) 2 (1 + x) +45x 4 (1 − x) 2 + 15(β + 5)x 3 (1 − x) 3 4 and A 2 (α, β, x) = 15(β 2 + 6β + 7)x 4 (1 − x) 3 (1 + x) −[120x 4 (1 − x) 3 (1 + x) + 15(6α + 5)x 4 (1 − x) 2 (1 + x)](β + 3) +45x 5 (1 − x) 2 (1 + x)(β + 3) 2 + 130x 3 (1 − x) 4 (1 + x) +10(12α + 5)x 3 (1 − x) 3 (1 + x) + 5(9α 2 + 15α + 8)x 3 (1 − x) 2 (1 + x) +45x 3 (1 − x) 2 − 15(β + 1)x 3 (1 − x) 3Because ρ is a continuous function, there exists an M > 0 such that |ρ(t; x)| < M , ∀ x, t ∈ [0, 1]. Using the above results for K(α,β) n,i ((t − x) 6 ; x), we obtain | K (α,β) n,i (ρ(t; x)(t − x) 6 ; x)| ≤ M 15x 4 (1 − x) 3 (1 + x)n 4 (n + β + 1) 6 + , i = 1, 2.So, (3.4) is proved.…”