Question

If $x\in R$, the solution set of the in-equation $4^{-x+0.5}-7\times 2^{-x}-4<0$ is equal to

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

Answer 1

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

$4^{-x+0.5}-7\times 2^{-x}-4<0⟹2\times 2^{-2x}-7\times 2^{-x}-4<0$
$⟹2\times 2^{-2x}-8\times 2^{-x}+2^{-x}-4<0$
$⟹(2^{-x}-4)(2\times 2^{-x}+1)<0$
$-\frac{1}{2}<2^{-x}<4$
since $2^{-x}>0$ therefore $0<2^{-x}<4$
$⟹-2<x<\infty$