The function f(x) = 2log_e(x-2) - x^2 + 4x + 1, (x > 2) is increasing…

The correct answer is Option (3) → (2, 3)

Given $f(x)=2\log_e(x-2)-x^2+4x+1,\;x>2$

Derivative: $f'(x)=\frac{2}{x-2}-2x+4$

Simplify numerator over common denominator:

$f'(x)=\frac{2-2x(x-2)+4(x-2)}{x-2}=\frac{-2x^2+8x-6}{x-2}=\frac{-2(x-1)(x-3)}{x-2}$

Sign requirement $f'(x)>0$ gives

$\frac{-2(x-1)(x-3)}{x-2}>0 \Longleftrightarrow \frac{(x-1)(x-3)}{x-2}<0$

Critical points: $1,2,3$. With domain $x>2$, the sign is negative only on $(2,3)$.

The function is increasing on the interval $(2,3)$.