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

A) [$2,\infty)$
B) {2}
C) ($- \infty,2\rbrack$
D) [0, 2]
Explanation

(c)
Given, inequality can be rewritten as $\left( \frac{5}{13} \right)^{x} + \left( \frac{12}{13} \right)^{x} \geq 1$
$\therefore\cos^{x}{\alpha + {\sin^{x}\alpha}{\geq 1}}$
Where, $\cos{\alpha = \frac{5}{13}}$
If $x = 2,$ the above equality holds
If $x < 2$ both $\cos\alpha\$and $\sin\alpha\$increases in positive fraction
Hence, above inequality holds for $x \in ( - \infty,2\rbrack$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.