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,
$\therefore 0 < 2^{-x} < 4$
$\Rightarrow -2 < x < \infty$