Question

If the function $f\left(x\right)=\sqrt{{0.25}^{4-3x}-4^{2x}}$ is defined, then the possible of $x$ is/are

• A. $2.5$
• B. $3.8$
• C. $4.7$
• D. $6$

Answer 1

Answer: (C), (D)

$6$

$f\left(x\right)=\sqrt{{0.25}^{4-3x}-4^{2x}}$
For $f\left(x\right)$ to be defined
${0.25}^{4-3x}-4^{2x}\ge 0\Rightarrow {0.25}^{4-3x}\ge 4^{2x}\Rightarrow {\left(\frac{1}{4}\right)}^{4-3x}\ge 4^{2x}\Rightarrow 4^{3x-4}\ge 4^{2x}\Rightarrow 3x-4\ge 2x\Rightarrow x\ge 4$