If $f(x)=\left(a b+b^2+1\right) x+\int\limits_0^x\left(\cos ^4 \theta+\sin ^4 \theta\right) d \theta$ is an increasing function of x for all $x \in R$ and $b \in R$, b being independent of x, then

$a \in(-\sqrt{6}, \sqrt{6})$

We have,

$f(x) =\left(a b+b^2+1\right) x+\int\limits_0^x\left(\cos ^4 \theta+\sin ^4 \theta\right) d \theta$

$\Rightarrow f^{\prime}(x)=\left(a b+b^2+1\right)+\cos ^4 x+\sin ^4 x$ for all $x \in R$

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

$f'(x)>0$  for all $x \in R$

$\Rightarrow \left(a b+b^2+1\right)+\left(\sin ^4 x+\cos ^4 x\right)>0$ for all $x \in R$

$\Rightarrow a b+b^2+1+\frac{1}{2}>0$                  $\left[\begin{array}{l}∵ \text { Minimum value of } \\ \sin ^4 x+\cos ^4 x \text { is } 1 / 2\end{array}\right]$

$\Rightarrow 2 a b+2 b^2+3>0$ for all $b \in R$

$\Rightarrow 2 b^2+2 a b+3>0$ for all $b \in R$

$\Rightarrow 4 a^2-24<0 \Rightarrow a^2-6<0 \Rightarrow-\sqrt{6}<a<\sqrt{6}$