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

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Solution

The correct option is D $3$
$2^{x} = a, 3^{x} = b$

$\Rightarrow - (a^{2} - a b + b^{2} - a - b + 1) + 1$

$\Rightarrow - \frac{1}{2} ((a - b)^{2} + (b - 1)^{2} + (a - 1)^{2}) + 1$

$\leq 1$

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