Question

The interval to which b may belong so that the function $f\left(x\right)=(1-\frac{\sqrt{21-4b-b^2}}{b+1})x^3+5x+\sqrt{6}$ is increasing at every point of its domain is

• A. $[-7,-1]$
• B. $[-6,-2]$
• C. $[2,2.5]$
• D. $[2,3]$

Answer 1

The correct option is D $[2,3]$

$f\left(x\right)=(1-\frac{\sqrt{21-4b-b^2}}{b+1})x^3+5x+\sqrt{6}$

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

$\Rightarrow x^2>\frac{-5}{3(1-\frac{\sqrt{21-4b-b^2}}{b+1})}\Rightarrow 1-\frac{\sqrt{21-4b-b^2}}{b+1}>0$

Which is trivially true if $b+1<0\Rightarrow b<-1$

If $b+1>0$ then

${(b+1)}^2>21-4b-b^2\Rightarrow (b-2)(b+5)>0$

$\Rightarrow b>2$ or $b<-5$

So $b\in [2,3]$