Question

The set of values of $a$ for which the function $f\left(x\right)=(4a-3)(x+\ln 5)+2(a-7)\cot \frac{x}{2}{\sin }^2\frac{x}{2}$ does not posses critical points in its domain is

• A. $(-\infty ,-\frac{4}{3})\cup (2,\infty )$
• B. $(-\infty ,2)$
• C. $[1,\infty )$
• D. $(1,\infty )$

Answer 1

Answer: (A)

$(-\infty ,-\frac{4}{3})\cup (2,\infty )$

$f\left(x\right)=(4a-3)(x+\log 5)+2(a-7)\cot \left(\frac{x}{2}\right){\sin }^2\left(\frac{x}{2}\right)$
$\Rightarrow f\left(x\right)=(4a-3)(x+\log 5)+(a-7)\sin x$
$f\left(x\right)$ posses critical points when $f^{\prime }\left(x\right)=(4a-3)+(a-7)\cos x=0$
$\Rightarrow \cos x=-\frac{4a-3}{a-7}$
$\Rightarrow -1\le -\frac{4a-3}{a-7}\le 1$
$\Rightarrow -\frac{4}{3}\le a\le 2$
Therefore, $f\left(x\right)$ does not posses critical points when $a\in (-\infty ,-\frac{4}{3})\cup (2,\infty )$
Ans: A