Question

The range of$a\in R$ for which the function $f\left(x\right)=(4a–3)(x+{\log }_{e}5)+2(a–7)cot\left(\frac{x}{2}\right){\sin }^2\left(\frac{x}{2}\right),x\ne 2n\pi ,n\in N$has critical points is

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

Answer 1

The correct option is A

$\left[-\frac{4}{3},2\right]$

The explanation for the correct answer.

Solve the critical points of $f\left(x\right)=(4a–3)(x+{\log }_{e}5)+2(a–7)\times cot\left(\frac{x}{2}\right)\times {\sin }^2\left(\frac{x}{2}\right)$

$f\left(x\right)=(4a–3)(x+{\log }_{e}5)+(a–7)\sin x$

$$\begin{gathered} f\left(x\right)=(4a–3)(x+{\log }_{e}5)+2(a–7)\times \left[\frac{\cos \left(\frac{x}{2}\right)}{\sin \left(\frac{x}{2}\right)}\right]\times {\sin }^2\left(\frac{x}{2}\right) \\ \Rightarrow f\text{'}\left(x\right)=(4a-3)+(a-7)\cos x=0 \\ \Rightarrow \cos x=-\left(\frac{4a-3}{a-7}\right) \end{gathered}$$

$\Rightarrow -1\le -\left(\frac{4a-3}{a-7}\right)\le 1$(because $–1\le \cos x\le 1$)

$\Rightarrow –1<\left(\frac{4a-3}{a-7}\right)\le 1$

$\frac{4a-3}{a-7}–1\le 0$ and$\frac{4a-3}{a-7}+1>0$

$a.\in [\frac{4}{3},7)$and $a\in (-\infty ,2)\cup (7,\infty )$

$\therefore \left(-\frac{4}{3}\right)\le a<2$

Hence, option(A) is the correct answer.