Question

If $5^x+{\left(2\sqrt{3}\right)}^{2x}\ge {13}^x,$ then the solution set for $x$ is

• A. $[2,\infty ]$
• B. $\left\{2\right\}$
• C. $(-\infty ,2]$
• D. $[0,2]$

Answer 1

The correct option is C $(-\infty ,2]$

${\left(\frac{5}{13}\right)}^x+{\left(\frac{12}{13}\right)}^x\ge 1$

$\therefore {\cos }^x\alpha +{\sin }^x\alpha \ge 1,$ where $\cos \alpha =\frac{5}{13}$

Equality holds for $x=2.$

If $x<2,$ both $\cos \alpha$ and $\sin \alpha$ increase (being positive fractions).

So, ${\cos }^x\alpha +{\sin }^x\alpha \ge 1$ if $x\le 2.$

Thus, $x\le 2.$