Question

Let $f\left(x\right)=\{\begin{array}{c}{x}^3-x^2+10x-5,x\le 1\\ -2x+{log}_2(b^2-2),x>1\end{array}$ the set of values of b for which $f\left(x\right)$ have greatest value at $x=1$ is given by

• A. $1\le b\le 2$
• B. $b=1,2$
• C. $b$ $ϵ$ $(-\infty ,-1)$
• D. $[-\sqrt{130},\sqrt{2}]\cup [\sqrt{2},\sqrt{130}]$

Answer 1

The correct option is D $[-\sqrt{130},\sqrt{2}]\cup [\sqrt{2},\sqrt{130}]$

For all $x\le 1$

$f\left(x\right)=x^3-x^2+10x-5$

$f^{\prime }\left(x\right)=3x^2-2x+10$

Discriminant$=\Delta =2^2-4\times 3\times 10=4-120=-116<0$

$\therefore f\left(x\right)$ is an increasing function for $x\le 1$

For $x>1,f\left(x\right)=-2x+{\log }_2(b^2-2)$

$f^{\prime }\left(x\right)=-2+0=-2$

$\therefore f\left(x\right)$ is an decreasing function for $x>1$

${lim}_{x\to 1}f\left(x\right)=f\left(1\right)$

$\Rightarrow {lim}_{x\to 1}(-2x+{\log }_2(b^2-2))=f\left(1\right)$

$\Rightarrow -2+{\log }_2(b^2-2)\le 1^3-1^2+10\times 1-5$

$\Rightarrow -2+{\log }_2(b^2-2)\le 1-1+10-5$

$\Rightarrow -2+{\log }_2(b^2-2)\le 5$

$\Rightarrow {\log }_2(b^2-2)\le 5+2$

$\Rightarrow {\log }_2(b^2-2)\le 7$

$\Rightarrow b^2-2\le 2^7=128$

$\Rightarrow b^2\le 128+2=130$

$\Rightarrow b\ge \pm \sqrt{130}$

$\Rightarrow b\in (-\infty ,-\sqrt{130}]\cup [\sqrt{130},\infty )$

We have $b^2-2\ge 0$

$\Rightarrow b\ge \pm \sqrt{2}$

$\Rightarrow b\in (-\infty ,-\sqrt{2}]\cup [\sqrt{2},\infty )$

Hence the set of values of $b$ for which $f\left(x\right)$ have greatest value at $x=1$ is $[-\sqrt{130},-\sqrt{2}]\cup [\sqrt{130},\sqrt{2}]$