If $x=\frac{(7+5 \sqrt{2})^{\frac{1}{3}}}{2+2 \sqrt{2}}$, then x belongs to:
Answer & explanation
Correct answer: option 2
Cube both the sides, we get
$x^3=\frac{7+5 \sqrt{2}}{8(7+5 \sqrt{2})}=\frac{1}{8}$
$\Rightarrow x=\frac{1}{2}$. Cleary $x$ lies between 0 and 1 .
Hence (2) is the correct answer.