The correct option (a) [– 1, 1]
Explanation:
f(x) = [(a2 – 1)/(a2 + 1)] x3 – 3x + 5 loge 2
∴ f'(x) = [(a2 – 1)/(a2 + 1)] 3x2 – 3x + 0
as f(x) is decreasing function
⇒ f'(x) < 0
∴ [(a2 – 1)/(a2 + 1)] 3x2 – 3x < 0
∴ [(a2 – 1)/(a2 + 1)]x < 0
∴ [(a2 – 1)/(a2 + 1)] < 0
a2 – 1 < 0
(a + 1)(a – 1) > 0
∴ a ∈ [– 1, 1]