Question

Let$f(x) = \left\{ \begin{matrix} 3 - x0 \leq x < 1 \\ x^{2} + lnbx \geq 1 \end{matrix} \right.\$. Then the set of values of b for which f(x) has the least value at x = 1, is given by

• A. (0, 1]
• B. (–e, 0]
• C. (–∞, 0)
• D. None of these

Answer 1

Answer: (C)

(–∞, 0)

Answer 2

$\underset{x \rightarrow 1^{-}}{Lim}f(x) \geq f(1) \Rightarrow 2 \geq 1 + \ln b$ $\Rightarrow \ln b \leq 1 \Rightarrow b \leq e$