Question

If the function $f\left(x\right)=(a^2-3a+2)\cos \frac{x}{2}+(a-1)x$ possesses critical points, then $a$ belongs to the interval

• A. $(-\infty ,0)\cup (4,\infty )$
• B. $(-\infty ,0]\cup [4,\infty )$
• C. $(-\infty ,0]\cup \left\{1\right\}\cup [4,\infty )$
• D. None of these

Answer 1

Answer: (C)

$(-\infty ,0]\cup \left\{1\right\}\cup [4,\infty )$

Given

$f\left(x\right)=(a^2-3a+3)\cos \frac{x}{2}+(a-1)x$

$\Rightarrow f^{\prime }\left(x\right)=\frac{-1}{2}(a-1)(a-2)\sin \left(\frac{x}{2}\right)+(a-1)$

$\Rightarrow f^{\prime }\left(x\right)=(a-1)[1-\frac{1}{2}(a-2)\sin \left(\frac{x}{2}\right)]$

If $f\left(x\right)$ possesses critical points, then

$f^{\prime }\left(x\right)=0$ $\Rightarrow (a-1)[1-\left(\frac{a-2}{2}\right)\sin \frac{x}{2}]=0$

$\Rightarrow a=1$ and $1-\left(\frac{a-2}{2}\right)\sin \frac{x}{2}=0$

$\Rightarrow a=1$ and $\sin \left(\frac{x}{2}\right)=\frac{2}{a-2}$

$\Rightarrow a=1$ and $\left|\frac{2}{a-2}\right|\le 1$

$\Rightarrow a=1$ and $|a-2|\ge 2$

$\Rightarrow a=1$ and $a-2\ge 2$ or $a-2\le -2$

$\Rightarrow a=1$ and $a\ge 4$ or $a\le 0$

$\Rightarrow a=1$ and $a\in (-\infty ,0]\cup [4,\infty )$

Therefore, $a\in (-\infty ,0]\cup \left\{1\right\}\cup [4,\infty )$