सही विकल्प है (B) R पर निरंतर वर्धमान है
\(f(x) = x^3 - 3x^2 + 12 x - 18\)
\(f'(x) = 3x^2 - 6x +12 = 3(x^2 - 2x +4)\)
Case I: f is strictly increasing (f'(x) > 0)
\(3(x^2 - 2x + 4) > 0 \rightarrow x^2 - 2x+ 4>0 \rightarrow x^2 - 2x + 1 + 3 > 0\)
\((x-1)^2 + 3> 0\)
\((x-1)^2 > -3 \rightarrow\) Always True because positive real number is always greater than negative real number.