The correct option is (d) (-∞, loge2)
Easy explanation: Let, g(t) = 2t^3 – 15t^2 + 36t -25
g’(t) = 6t^2 – 30t + 36 = 0
=> 6(t^2 – 5t + 6) = 0
=> t = 2, 3
For, 2 ≤ t ≤ 4, at t = 3, g”(t) > 0
So, g(t) is minimum.
g(t)min = g(3) = 2*27 – 15*9 + 36*3 – 25 = 2
Also, 2 + |sint| ≥ 2
Hence, minimum Φ(t) = 2
Therefore, 3^x + 3^f(x) = 2
=>3^f(x) = 2 – 3^x
Thus, 3^f(x) > 0
=> 3^x > 0 and 3^x < 2
Therefore, x € (-∞, loge2)