Answer : (c) [2loge 3, ∞)
Given,
f (x) = loge (3x2 + 4)
Let y = loge (3x2 + 4)
⇒ 3x2 + 4 = ey
⇒ x = \(\sqrt{\frac{e^y-4}{3}}\)
For x to be defined \(\frac{e^y-4}{3}\) ≥ 0
ey – 4 ≥ 0
⇒ ey ≥ 4
⇒ y ≥ loge 4 ⇒ y ≥ 2 loge 2
∴ Range of the function is [2 loge 2, ∞)