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

all the above

For f(x) to be increasing, we must have

$f'(x)>0$ for all x

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

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

This inequality holds for

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

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

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

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

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

Therefore, answer (a), which includes answer (b) also, is correct.

Hence, all the options are correct.

For $b+1>0, f(x)$ will be increasing for all x, if

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

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

$\Rightarrow (b+7)(b-3)<0$ and $b^2+3 b-10>0$

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

$\Rightarrow b \in(2,3)$             [∵ b + 1 > 0 ∴ b > -1]

Therefore, answer (d), which includes answer (c) also, is correct.