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Question:

Let f(x) = (1-x)²sin(2x) + x² for all x ∈ ℝ, and let g(x) = ∫₁ˣ (2√t + 1 - ln t)f(t)dt for all x ∈ (1, ∞). Which of the following is true?

g is increasing on (1, 2) and decreasing on (2, ∞)

g is decreasing on (1, ∞)

g is increasing on (1, ∞)

g is decreasing on (1, 2) and increasing on (2, ∞)

Solution:

g'(x) = [2√x + 1 - ln x]f(x), For x ∈ (1, ∞), f(x) > 0 (Using Newton-Leibnitz rule) Let h(x) = (2√x + 1 - ln x) ⇒ h'(x) = [4(x+1)²√x] = -(√x)²/(x+1)² < 0 Also h(1) = 0. So, h(x) < 0 ∀x > 1 ⇒ g(x) is decreasing on (1, ∞)