The function $f(x) = kx^3+6kx^2 + 18x + 17$ is increasing on (set of real numbers) if:

$k ∈ (0,3/2)$

The correct answer is Option (2) → $k ∈ (0,3/2)$

$f(x)=kx^3+6kx^2+18x+17$

$f'(x)=3kx^2+12kx+18$

$f'(x)>0\;$ for the function to be increasing.

$3kx^2+12kx+18=3\big(kx^2+4kx+6\big)$

$kx^2+4kx+6>0$ for all real $x$.

The discriminant must be negative:

$\Delta=(4k)^2-4(k)(6)=16k^2-24k$

$16k^2-24k<0$

$8k(2k-3)<0$

This inequality holds when $k$ lies between $0$ and $\frac{3}{2}$.