Question

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

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

Answer 1

Answer: (C)

$(-\infty ,2]$

Simplifying, we get
$5^x+(12{)}^x\ge {13}^x$
Now,
$(\frac{5}{13}{)}^x+(\frac{12}{13}{)}^x\ge 1$
Let,
$\frac{5}{13}=\cos A$
Therefore,
$\frac{12}{13}=\sin A$
Hence, the inequality becomes
${\cos }^{x}A+{\sin }^{x}A\ge 1$
Now, we know that $\sin x,\cos x\in [-1,1]$ and ${\sin }^{2}x+{\cos }^{2}x=1$
Hence, the inequality satisfies for $x\le 2$
$\Rightarrow x\in (-\infty ,2]$