Question

Let $f\left(x\right)=\{\begin{array}{cc}-x^3+lo{g}_{2}b,& 0\le x<1\\ 3x,& 1\le x\le 3\end{array}$ The set of real values of b for which f(x) has smallest value at $x=1$, is?

• A. $(0,8]$
• B. $[8,\infty )$
• C. $[16,\infty ]$
• D. $(0,16]$

Answer 1

Answer: (C)

$[16,\infty ]$

If the function f(x) has smallest value at $x=1$, then
$\underset{x\to 1^{-}}{lim}f\left(x\right)\ge f\left(1\right)\Rightarrow -1+lo{g}_{2}b\ge 3$
$\Rightarrow lo{g}_{2}b\ge 4\Rightarrow b\ge 16$
$\therefore b\in [16,\infty )$.