The interval in which the function \( f(x)=x^3-12x^2+36x+20 \) is increasing, is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \( (-\infty,2)\cup(6,\infty) \)
f is increasing where \( f'(x)\gt 0 \):
\( \displaystyle f'(x)=3x^2-24x+36 \)
\( \displaystyle f'(x)=3(x-2)(x-6) \)
A quadratic with positive leading coefficient is positive outside its roots, so \( f'(x)\gt 0 \) for x < 2 and for x > 6.
So f is increasing on \( (-\infty,2)\cup(6,\infty) \).
Note: It seems that there is a mistake in the original NTA paper, which prints Option (2) as \( [-\infty,-3)\cup(2,\infty) \) when it should read \( (-\infty,-3)\cup(2,\infty) \), because an interval can never be closed at \( -\infty \). The question can still be solved, because Option (2) is incorrect in either form.