CameraIcon
CameraIcon
SearchIcon
MyQuestionIcon


Question

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


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

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

Suggest Corrections
thumbs-up
 
0


similar_icon
Similar questions
View More


similar_icon
People also searched for
View More



footer-image