Question

The values of $ x $ for which $ 4^x + 4^{1-x} - 5 < 0 $ , is given by

• A. $ x = 1 $
• B. $ x = 0, 1 $
• C. $ x = 0 $
• D. $ 0 < x < 1 $

Answer 1

Answer: (D)

We have , $4^{x}+4^{1-x}-5<\,0 \dots (i)$
Let $4^{x}=y$
$\therefore (i)$ becomes, $y+\frac{4}{y}-5<\,0$
$\Rightarrow y^{2}-5y+4<\,0$
$\Rightarrow y^{2}-4y-y+4<\,0$
$\Rightarrow \left(y-4\right)\left(y-1\right)<\,0$
$\Rightarrow 1<\,y <\,4$
$\Rightarrow 1 <\,4^{*} <\,4$
$\Rightarrow 4^{\circ}<\,4^{*} <\,4^{1}$
$\therefore 0 <\,x<\,1$