Question

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

A
[2,]
B
{2}
C
(,2]
D
[0,2]

Solution

## The correct option is C $$(-\infty,2]$$$$\displaystyle { \left( \frac { 5 }{ 13 } \right) }^{ x }+{ \left( \frac { 12 }{ 13 } \right) }^{ x }\ge 1$$$$\displaystyle \therefore \cos ^{ x }{ \alpha } +\sin ^{ x }{ \alpha } \ge 1,$$ where $$\displaystyle\cos { \alpha } =\frac { 5 }{ 13 }$$Equality holds for $$x=2.$$If $$x<2,$$ both $$\cos { \alpha }$$ and $$\sin { \alpha }$$ increase (being positive fractions).So, $$\displaystyle \cos ^{ x }{ \alpha } +\sin ^{ x }{ \alpha } \ge 1$$ if $$x\le 2.$$ Thus, $$x\le 2.$$ Mathematics

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