The solution set contained in $R$ of the inequation $3^x+3^{1-x}-4 < 0$, is :
The solution set contained in $R$ of the inequation $3^x+3^{1-x}-4 < 0$, is :
- $(1,3)$
- $(0,1)$
- $(1,2)$
- $(0,2)$
Solution
We have,
$3^x+3^{1-x}-4 < 0$
$3^x+\frac{3}{3^x}-4 < 0$
$3^{2 x}+3-4 \cdot 3^x < 0$
$3^{2 x}-4 \cdot 3^x+3 < 0$
$\left(3^x-1\right)\left(3^x-3\right) < 0$
$1 < 3^x < 3 \Rightarrow 0 < x < 1$
$\therefore \quad$ The solution set is $(0,1)$.
Asked in: AP EAMCET 2003
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