Question

The solution of the inequation $4^{-x+0.5}-7\ast 2^{-x}<4,xϵR$ is :

• A. $(-2,\infty )$
• B. $(2,\infty )$
• C. $(2,\frac{7}{2})$
• D. $(-\infty ,-1)$

Answer 1

The correct option is A $(-2,\infty )$

Best way to solve this problem is by checking different options by taking different values of x.
For x = 0,
$4^{-x+0.5}-7\ast 2^{-x}=2-7=-5<4,$ which is true.
Hence, option (a) is the correct answer.

Alternatively:
The given inequation is
$4^{-x+0.5}-7\ast 2^{-x}<4,xϵR$
Let $2^{-x}=t$
$\therefore 2t^2-7t<4$
$\Rightarrow 2t^2-7t-4<0$
$\Rightarrow (2t+1)(t-4)<0$
$\Rightarrow -\frac{1}{2}<t<4$
$\Rightarrow 0<t<4$ $[\because t=2^{-x}>0]$
$\Rightarrow 0<2^{-x}<4$
$\Rightarrow -2<x<\infty$
$\therefore xϵ(-2,\infty )$