Question

The set of values of 'a' for which the function $f\left(x\right)=(4a-3)(x+ln5)+(a-7)sinx$ does not posses critical points is _____________________.

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

Answer 1

Answer: (A)

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

$f\left(x\right)=(4a-3)(x+\ln 5)+(a-7)\sin x$

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

$\Rightarrow (4a-3)+(a-7)\cos x=0$

$\Rightarrow (4a-3)=-(a-7)\cos x$

$\Rightarrow \cos x=\frac{3-4a}{a-7}$ where $a\ne 7$

We know that range of $\cos x$ is $[-1,1]$

$\Rightarrow \left|\frac{3-4a}{a-7}\right|\le 1$ is a solution

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

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

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

$\Rightarrow a-7+3\ge 4a\ge 7-a+3$

$\Rightarrow a-4\ge 4a\ge -a+10$

$\Rightarrow a-4\ge 4a\ge -a+10$

$\Rightarrow a\le -\frac{4}{3},a\ge 2$

$\therefore a\in (-\infty ,-\frac{4}{3}]\cup [2,\infty )$