Question

The values of $x$ for which the inequality ${\log }_{\cos \theta }(x^2+14)\ge {\log }_{\cos \theta }\left(9x\right),\theta \in (0,\frac{\pi }{2})$ holds true :

• A. $(-\infty ,2]\cup [7,\infty )$
• B. $[7,\infty ]$
• C. $(0,2]\cup [7,\infty )$
• D. $[2,7]$

Answer 1

The correct option is D $[2,7]$

${\log }_{\cos \theta }(x^2+14)\ge {\log }_{\cos \theta }\left(9x\right)$
${\log }_{a}x$ is decreasing for $a<1$
$\because \cos \theta <1$ for $\theta \in (0,\frac{\pi }{2})$
$\therefore x^2+14\le 9x\text{and}9x>0$
$\Rightarrow x^2-9x+14\le 0\text{and}x>0$
$\Rightarrow (x-2)(x-7)\le 0\text{and}x>0$
$\Rightarrow 2\le x\le 7\text{and}x>0$
$\Rightarrow x\in [2,7]$