Question

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

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

Answer 1

Answer: (C)

$(-\infty ,2]$

Given, $5^x+(2\sqrt{3}{)}^{2x}\ge {13}^x\Rightarrow 5^x+{12}^x\ge 13^x\Rightarrow {\left(\frac{5}{13}\right)}^x+{\left(\frac{12}{13}\right)}^x\ge 1$
Let $\frac{5}{13}=\sin \alpha \Rightarrow \cos \alpha =\frac{12}{13}$
${\sin }^x\alpha +{\cos }^x\alpha \ge 1$
And as $-1\le \sin \alpha \le 1\Rightarrow {\sin }^x\alpha ϵ(-1,1)x>1\Rightarrow {\sin }^x\alpha ϵ(-\infty ,1]\cup [1,\infty )x\le 1$
Similarly ${\cos }^x\alpha ϵ(-\infty ,1)\cup (1,\infty )x\le 1$
Therefore, ${\sin }^x\alpha +{\cos }^x\alpha \ge 1$ for $x\le 2$