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 $f\left(x\right)$ has greatest

value at $x=1$ is given by:

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

Answer 1

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

Given:$f\left(x\right)=\{\begin{array}{c}{x}^3-x^2+10x-5;x\le 1\\ -2x+{\log }_2(b^2-2);x>1\end{array}$

To find: Set of values of b for which $f\left(x\right)$ has greatest value at x=1

Sol:${lim}_{x\to 1^{-}}f\left(x\right)={lim}_{x\to 1}(x^3-x^2+10x-5)$

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

$=-2+{\log }_2(b^2-2)$

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

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

$⟹f\left(x\right)$ is a strictly increasing function.

$⟹f\left(1\right)>f\left(x\right)$ for all $x<1$

$⟹f\left(1\right)=5$

$⟹-2+{\log }_2(b^2-2)\le 5$

$⟹{\log }_2(b^2-2)\le 7$

$⟹b^2-2\le 2^7-128$

$⟹b^2\le 130$

$⟹b^2-130\le 0$

$⟹b\epsilon [-\sqrt{130,}\sqrt{130}]$

$But{b}^2\ne 2⟹b\ne \{-\sqrt{2,}\sqrt{2}\}$

Hence, correct answer is

$b\epsilon [-\sqrt{130,}-\sqrt{2}]U\left(\sqrt{2,}\sqrt{130}\right)$