The function \( f(x)=17-9x+6x^2-x^3 \) is strictly increasing in the interval, is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( x\in(1,3) \)
A function is strictly increasing where its derivative is positive:
\( \displaystyle f'(x)=-9+12x-3x^2 \)
\( \displaystyle =-3(x^2-4x+3) \)
\( \displaystyle =-3(x-1)(x-3) \)
The product (x − 1)(x − 3) is negative between 1 and 3, and the factor −3 makes f'(x) positive there:
\( \displaystyle 1\lt x\lt 3:\ f'(x)\gt 0 \)
\( \displaystyle x\lt 1 \text{ or } x\gt 3:\ f'(x)\lt 0 \)
So f is strictly increasing in (1, 3).
Why the other options are incorrect
Option (1) \( \forall\ x\in\mathrm{R} \) — Incorrect. f'(0) = −9 < 0, so f is not increasing for every real x.
Option (2) \( x\in(-\infty,-1]\cup[3,\infty) \) — Incorrect. f'(x) < 0 for x < −1 and for x > 3, so f is decreasing on both parts.
Option (3) \( x\in(-\infty,1]\cup[3,\infty) \) — Incorrect. This is where f is decreasing, the opposite of what is asked.
Note: NTA dropped this question. For the benefit of students and to give them additional practice, the question has been suitably modified so that it has one clear answer.