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, $y+\frac{4}{y}-5<0$
$\Rightarrow y^2-5y+4<0$
$\Rightarrow y^2-4y-y+4<0$
$\Rightarrow \left(y-4\right)\left(y-1\right)<0$
$\Rightarrow 1<y<4$
$\Rightarrow 1<4^{\ast }<4$
$\Rightarrow 4^{\circ }<4^{\ast }<4^1$
$\therefore 0<x<1$