For the function f(x)=ax^3-9ax^2+9x+3,x∈ R , then which of the follow…

The correct answer is Option (2) → (A) and (B) only

Differentiating,

    \( f'(x)=3ax^2-18ax+9 \)

    \( =3(ax^2-6ax+3) \)

f is increasing for all x when \( ax^2-6ax+3\ge 0 \) for all x.

For a = 0, \( f'(x)=9\gt 0 \).

For a > 0, the quadratic stays non-negative when its discriminant is not positive:

    \( 36a^2-12a\le 0 \)

    \( 12a(3a-1)\le 0 \)

    \( \displaystyle 0\lt a\le\frac{1}{3} \)

For a < 0, the quadratic opens downwards and becomes negative for large |x|.

So \( f'(x)=3(ax^2-6ax+3) \), and f is increasing for every x exactly when \( 0\le a\le\frac{1}{3} \).

Checking the statements

(A) \( f'(x)=3[ax^2-6ax+3] \) — Correct. \( 3ax^2-18ax+9=3(ax^2-6ax+3) \).

(B) It is increasing for \( a\in\left[0,\displaystyle\frac{1}{3}\right]\ \forall x\in R \) — Correct. This is exactly the interval \( 0\le a\le\frac{1}{3} \).

(C) It is increasing for \( a\in(-\infty,0)\forall x\in R \) — Incorrect. For a < 0, f′(x) is negative for large |x|.

(D) It is increasing for \( \forall a\in R \) — Incorrect. For a = 1, \( f'(3)=3(9-18+3)=-18\lt 0 \).