Question

For all real number $x$, $2^x+3^x-4^x+6^x-9^x$ is always less than

• A. 0.0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (C), (D)

The correct options are
C 2
D 3

$2^x=a,3^x=b$
$\Rightarrow -(a^2-ab+b^2-a-b+1)+1$
$\Rightarrow -\frac{1}{2}\left(\right(a-b{)}^2+(b-1{)}^2+(a-1{)}^2)+1$
$\le 1$