The value of a for which the function  $f\left(x\right)=(4a-3)(x+\log 5)+2(a-7)\cot \frac{x}{2}{\sin }^2\frac{x}{2}$ does not possess critical points, is

We have,

$\begin{array}{l}f\left(x\right)=(4a-3)(x+\log 5)+2(a-7)\cot \frac{x}{2}{\sin }^2\frac{x}{2}\\ \Rightarrow f\left(x\right)=(4a-3)(x+\log 5)+(a-7)\sin x\\ \therefore f^{\mathrm{\prime }}\left(x\right)=(4a-3)+(a-7)\cos x\end{array}$

If $f \left(x\right)$ does not have critical point, then $f\text{'} \left(x\right)=0$ does not have any solution in $R.$

Now,

$\begin{array}{l}{f}^{\mathrm{\prime }}\left(x\right)=0\\ \Rightarrow \cos x=\frac{4a-3}{7-a}\\ \Rightarrow \left|\frac{4a-3}{7-a}\right|\le 1\\ \\ \Rightarrow -1\le \frac{4a-3}{7-a}\le 1 [\because |\cos x|\le 1]\\ \Rightarrow a-7\le 4a-3\le 7-a\Rightarrow a\ge -4/3\text{ or }a\le 2.\end{array}$

Thus, $f\text{'} \left(x\right)=0$ has solution in R If a $a\ge -4/3\text{ or }a\le 2$.

So, $f\text{'} \left(x\right)=0$ is of solvable in R if $a<-4/3\text{ or }a>2$

i.e. $a\in (-\mathrm{\infty },-4/3)\cup (2,\mathrm{\infty })$