Correct Answer - Option 3 : 122
From question, the function given is:
f(x) = 3x3 – 18x2 + 27x – 40
Now,
⇒ f'(x) = 9x2 – 36x + 27
⇒ f'(x) = 9(x2 – 4x + 3)
⇒ f'(x) = 9(x – 1)(x – 3)
From question,
S = {x ∈ R : x2 + 30 ≤ 11x}
From which,
⇒ x2 + 30 ≤ 11x
⇒ x2 – 11x + 30 ≤ 0
⇒ (x – 5)(x – 6) ≤ 0
∴ x ∈ [5, 6]
∴ S = [5, 6]
Since, f'(x) > 0 when x ∈ [5, 6].
So, f(x) is an increasing function is [5, 6] interval.
Now,
f(x) is maximum when x = 6.
⇒ f(6) = 3(6)3 – 18(6)2 + 27(6) – 40
∴ f(6) = 122