Let \(f(x)=\left\{\begin{array}3-x, & 0≤x<1 \\ x^2+lnb, & x ≥ 0\end{array}\right.\). Then the set of values of b for which f(x) has the least value at x = 1, is given by

(0, 1] ( -e, 0] (-∞, e) none of these.

(-∞, e)

$\underset{x→1^-}{lim}f(x)≥f(1)⇒2≥1+lnb≤1⇒b≤e$ Hence (C) is the correct answer.