MCQ
If $x = \sqrt {7 + 4\sqrt 3 } ,$ then $x + \frac{1}{x} = $
- ✓$4$
- B$6$
- C$3$
- D$2$
$\therefore$ $\frac{1}{x} = \frac{1}{{\sqrt {7 + 4\sqrt 3 } }} = \frac{{\sqrt {7 - 4\sqrt 3 } }}{{\sqrt {7 + 4\sqrt 3 } .\sqrt {7 - 4\sqrt 3 } }} = \sqrt {7 - 4\sqrt 3 } $
$\therefore$ $x + \frac{1}{x} = \sqrt {7 + 4\sqrt 3 } + \sqrt {7 - 4\sqrt 3 } $
$ = (\sqrt 3 + 2) + (2 - \sqrt 3 ) = 4$.
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Then, which of the following statements are true?
$I.$ $g$ has exactly two distinct real roots.
$II$. $g$ can have more than two distinct real roots.
$III$. There exists a real number $\alpha$ such that $g(x) \geq \alpha$ for all real $x$