The domain of definition of the function $f(x)=\sqrt[6]{4^x+8^{\frac{2}{3}(x-2)}-52-2^{2(x-1)}}$ is: |
$[1, \infty)$ $[2, \infty)$ $[3, \infty)$ None of these |
$[3, \infty)$ |
$4^x+8^{\frac{2}{3}(x-2)}-52-2^{2(x-1)}>0$ $⇒2^{2x}+2^{2(x-2)}-52-2^{2(x-1)}>0$ $⇒2^{2x}\left[1+\frac{1}{16}-\frac{1}{4}\right]>52⇒2^{2x}\left[\frac{16+1-4}{16}\right]>52$ so $2^{2x}\left[\frac{13}{16}\right]>52$ $2^{2x}>64$ so $x ≥3 ⇒x∈[3, \infty)$ Hence (3) is the correct answer. |