Question

The solution set of the inequality $4^{-x+\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}-7·\left(2^{-x}\right)-4<0,$for all $x$belongs to R, is

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

Answer 1

The correct option is B

$(-2,\infty )$

Explanation for the correct option:

Solution of in-equality:

$$\begin{gathered} 4^{-x+\left(\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.\right)}-7·\left(2^{-x}\right)-4<0 \\ \Rightarrow 2^{-2x+1}-7\times \left(2^{-x}\right)<4 \\ \Rightarrow 22^{-2x}-7\times \left(2^{-x}\right)-4<0 \end{gathered}$$

Let $2^{-x}=t>0$

$$\begin{gathered} 2t^2-7t-4<0 \\ \Rightarrow 2t^2-8t+t-4<0 \\ \Rightarrow 2t(t-4)+1(t-4)<0 \\ \Rightarrow (2t+1)(t-4)<0 \\ \Rightarrow t\in \left(\frac{-1}{2},4\right)andt>0 \\ \Rightarrow 0<2^{-x}<4 \\ \Rightarrow 2^{-x}<2^2 \\ \Rightarrow -x<2 \\ \Rightarrow x>-2 \\ \Rightarrow x\in (-2,\infty ) \end{gathered}$$

Hence the correct option is option(B)