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) $(2, \infty)$
Answer & explanation
Correct answer: option 2
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 $a>2$ $a \in(-\infty-4 / 3) \cup(2, \infty)$