Question

The set of values of $\lambda$ for which the function $f\left(x\right)=(4\lambda -3)(x+\log 5)+2(\lambda -7).\cot \frac{x}{2}{\sin }^2\frac{x}{2}.$ does not posses critical point is:

• A. $(1,\infty )$
• B. $(2,\infty )$
• C. $(-\infty ,-4/3)$
• D. $(-\infty ,-1)$

Answer 1

Answer: (B), (C)

The correct options are
B $(2,\infty )$
C $(-\infty ,-4/3)$

$2\cot \frac{x}{2}{\sin }^2\frac{x}{2}=2\cos \frac{x}{2}\sin \frac{x}{2}=\sin x$
$\therefore f\left(x\right)=(4\lambda -3)(x+\log 5)+(\lambda -7)\sin x$
$f^{\prime }x=(4\lambda -3)+(\lambda -7)\cos x=0$ ...(1)$\therefore \cos x=\frac{4\lambda -3}{\lambda -7}$ ...(2)Now $-1\le \cos x\le 1\therefore -1\le \frac{4\lambda -3}{\lambda -7}\le 1$ Above gives us two
inequalities $-1-\frac{4\lambda -3}{\lambda -7}\le 0$
or $\frac{4\lambda -3}{\lambda -7}\le 0$
$\frac{-5\lambda +10}{\lambda -7}\le 0$ or $\frac{3\lambda +4}{\lambda -7}\le 0$ or $\frac{5(\lambda -2)}{\lambda -7}\ge 0$ or $\frac{3(\lambda +4/3)}{\lambda -7}\le 0$ Above are the
conditions for $f\left(x\right)$ to have critical points. But the function does not possess critical points. Therefore we must have $\frac{5(\lambda -2)}{\lambda -7}<0$ or
$\frac{3(\lambda +4/3)}{\lambda -7}>0$ or $\frac{5(\lambda -2)(\lambda -7)}{{(\lambda -7)}^2}<0$ or
$\frac{3(\lambda \lambda +4/3)(\lambda -7)}{{(\lambda -7)}^2}>0$
$\therefore \lambda ϵ(2,7)$ or $\lambda <-4/3$ or $>7$
$\therefore \lambda ϵ(2,7)$ and $\lambda ϵ(-\infty ,-4/3)or(7,\infty )$
$\therefore \lambda ϵ(2,\infty )$ or $\lambda ϵ(-\infty ,-4/3)$