Question

The solution set of the inequality $4^{-x+\frac{1}{2}}-7.(2^{-x})-4<0$ for $x\in R$ is

• A. $(-\infty ,2)$
• B. $(-2,\infty )$
• C. $(-\infty ,\infty )$
• D. $(2,\infty )$

Answer 1

Answer: (B)

Let $2^{-x}=t$
$\therefore 2t^2-7t-4<0$
$\Rightarrow (2t+1)(t-4)<0$
$\Rightarrow (2t+1)(t-4)<0$
$\Rightarrow -\frac{1}{2}<t<4$
Since, $2^{-x}$ is always positive,
$\therefore 0<2^{-x}<4$
$\Rightarrow -2<x<\infty$