Question

If the function $f\left(x\right)=(1-\frac{\sqrt{21-4b-b^2}}{b+1})x^3+5x+\sqrt{16}$ increases for all $x$, then

• A. $b\in (-1,2)$
• B. $b\in (-7,3)-\{-1\}$
• C. $b\in (-7,-1)\cup (2,3)$
• D. none of these

Answer 1

The correct option is C $b\in (-7,-1)\cup (2,3)$

since the function $f\left(x\right)$ increases for all $x,$ therefore,

$f^{\prime }\left(x\right)=(1-\frac{\sqrt{21-4b-b^2}}{b+1})3x^2+5>0,\forall x\in R$

$\Rightarrow (1-\frac{\sqrt{21-4b-b^2}}{b+1})>0$

and, ${\left(0\right)}^2-4\times 3\left(\frac{1-\sqrt{21-4b-b^2}}{b+1}\right)5<0$

$[\because {ax}^2+bx+c>0\forall x\in R\Rightarrow a>0$ and $D>0]$

$\Rightarrow 1-\frac{\sqrt{21-4b-b^2}}{b+1}>0$

The above inequality holds, when (1) $b+1<0$ and (2) $21-4b-b^2>0$

$\therefore b<-1$ and $b^2+4b-21<0$

$\Rightarrow b<-1$ and $(b+7)(b-3)<0$

$\Rightarrow b<-1$ and $-7<b<3$

$\therefore b\in (-7,-1)$

Again, when $b+1>0,f\left(x\right)$ will be increasing for all $x,$

if, $21-4b-b^2>0$ and $1>\frac{\sqrt{21-4b-b^2}}{b+1}$

or,$b^2+4b-21<0$

and, ${(b+1)}^2>(21-4b-b^2)$ $[$ as $b+1>0]$

or, $(b+7)(b-3)<0$ and $b^2+3b-10>0$

$\Rightarrow (-7<b<3)$ and $(b<-5$ or $b>2)$

$\Rightarrow 2<b<3$

From (1) and (2), we have

$b\in (-7,-1)\cup (2,3)$