Question

Let $f\left(x\right)=\{\begin{array}{cc}{x}^3-x^2+10x-5,& x\le 1\\ -2x+{\log }_2(b^2-2),& x>1\end{array}.$
If $f\left(x\right)$ has the greatest value at $x=1,$ then the set of values of $b$ is

• A. $(-\sqrt{130},-\sqrt{2})\cup (\sqrt{2},\sqrt{130})$
• B. $(0,\sqrt{130})$
• C. $(-\sqrt{130},\sqrt{130})$
• D. $(-\sqrt{130},0)$

Answer 1

The correct option is A $(-\sqrt{130},-\sqrt{2})\cup (\sqrt{2},\sqrt{130})$

Given:
$f\left(x\right)=\{\begin{array}{cc}{x}^3-x^2+10x-5,& x\le 1\\ -2x+{\log }_2(b^2-2),& x>1\end{array}.$
Since $f\left(x\right)$ has the greatest value at $x=1.$
Then, $f\left(1\right)>f(1+h)$ as $h\to 0^{+}$
$\Rightarrow 5>-2(1+h)+{\log }_2(b^2-2)$ as $h\to 0^{+}$
$\Rightarrow 5>-2+{\log }_2(b^2-2)$
$\Rightarrow {\log }_2(b^2-2)<7$
$\Rightarrow b^2<2^7+2$
$\Rightarrow b^2<130$
$\Rightarrow -\sqrt{130}<b<\sqrt{130}\cdots \left(i\right)$
For ${\log }_2(b^2-2),$ it is defined when $(b^2-2)>0$
$\Rightarrow (b-\sqrt{2})(b+\sqrt{2})>0$
$\Rightarrow b\in (-\infty ,-\sqrt{2})\cup (\sqrt{2},\infty )\cdots \left(ii\right)$
From $\left(i\right)$ and $\left(ii\right)$
$\therefore b\in (-\sqrt{130},-\sqrt{2})\cup (\sqrt{2},\sqrt{130})$