Question

The value of ${}^{\prime }{a}^{\prime }$ for which the function $f\left(x\right)=(4a-3)(x+\log 5)+2(a-7)\cot \frac{x}{2}\cdot {\sin }^2\frac{x}{2}$ does not possess critical points is

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

Answer 1

The correct option is D $(-\infty ,\frac{-4}{3})\cup (2,\infty )$

We have, $f\left(x\right)=(4a-3)(x+\log 5)+2(a-7)\cot \frac{x}{2}\cdot {\sin }^2\frac{x}{2}$

$=(4a-3)(x+\log 5)+(a-7)\sin x$

$\Rightarrow f^{\prime }\left(x\right)=(4a-3)+(a-7)\cos x$

If $f\left(x\right)$ does not have critical points, then $f^{\prime }\left(x\right)=0$ does not have any solution in $R$.

Now, $f^{\prime }\left(x\right)=0\Rightarrow \cos x=\frac{4a-3}{7-a}$

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

$\Rightarrow -1\le \frac{4a-3}{7-a}\le 1$

$\Rightarrow -1\le \frac{4a-3}{7-a}\le 1$

$\Rightarrow a-7\le 4a-3\le 7-a$

$\Rightarrow a\ge -4/3$and $a\le 2$

Thus, $f^{\prime }\left(x\right)=0$ has solution in $R$, if $-\frac{4}{3}\le a\le 2$

So, $f^{\prime }\left(x\right)=0$ is not solvable in $R$, if $a<-\frac{4}{3}$ or $a>2$

i.e., $a\in (-\infty ,-\frac{4}{3})\cup (2,\infty )$.