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

(a) $(-\infty,-4 / 3)$
(b) $(-\infty,-1)$
(c) $[1, \infty)$
(d) $(2, \infty)$

(a) and (d)

We have,

$f(x) =(4 a-3)(x+\log 5)+2(a-7) \cot \frac{x}{2} \sin ^2 \frac{x}{2}$

$\Rightarrow f(x)=(4 a-3)(x+\log 5)+(a-7) \sin x$

∴  $f'(x) =(4 a-3)+(a-7) \cos x$

If f(x) does not have critical points, then f'(x) = 0 does not have any solution in R.

Now,

f'(x) = 0

$\Rightarrow \cos x =\frac{4 a-3}{7-a}$

$\Rightarrow\left|\frac{4 a-3}{7-a}\right| \leq 1$                [∵ |cos x| ≤ 1]

$\Rightarrow -1 \leq \frac{4 a-3}{7-a} \leq 1$

$\Rightarrow a-7 \leq 4 a-3 \leq 7-a \Rightarrow a \geq-4 / 3$ or $a \leq 2$

Thus, f'(x) = 0 has solutions in R if $a \geq-4 / 3$ or $a \leq 2$.

So, f'(x) = 0 is not solvable in R if $a<-4 / 3$ or $a>2$ $a \in(-\infty-4 / 3) \cup(2, \infty)$