Question

If $xϵR$, the solution set of the equation
$4^{-x+0.5}-{7.2}^{-x}-4<0$ is equal to

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

Answer 1

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

Put $2^{-x}=t$ and $4^{0.5}=4^{\frac{1}{2}}=2$
$\therefore 2t^2-7t-4<0\Rightarrow (t-4)(2t+1)<0$
$-\frac{1}{2}<t<4\Rightarrow 0<2^{-x}<2^2$
or ${\left(\frac{1}{2}\right)}^{\infty }<{\left(\frac{1}{2}\right)}^x<{\left(\frac{1}{2}\right)}^{-2}$
Since $\frac{1}{2}<1\therefore -2<x<\infty$
$\therefore xϵ(-2,\infty )\Rightarrow \left(b\right)$