7．A student was attempting to prove that \(x = \frac { 1 } { 2 }\) is the only real root of $$x ^ { 3 } + \frac { 3 } { 4 } x - \frac { 1 } { 2 } = 0$$ The attempted solution was as follows． $$\begin{array} { r l r } & x ^ { 3 } + \frac { 3 } { 4 } x & = \frac { 1 } { 2 }
\therefore & x \left( x ^ { 2 } + \frac { 3 } { 4 } \right) & = \frac { 1 } { 2 }
\therefore & x & = \frac { 1 } { 2 }
\text { or } & x ^ { 2 } + \frac { 3 } { 4 } & = \frac { 1 } { 2 }
\text { i.e. } & x ^ { 2 } & = - \frac { 1 } { 4 } \quad \text { no solution }
\therefore & \text { only real root is } x & = \frac { 1 } { 2 } \end{array}$$ （a）Explain clearly the error in the above attempt．
（b）Give a correct proof that \(x = \frac { 1 } { 2 }\) is the only real root of \(x ^ { 3 } + \frac { 3 } { 4 } x - \frac { 1 } { 2 } = 0\) ． The equation $$x ^ { 3 } + \beta x - \alpha = 0$$ where \(\alpha , \beta\) are real，\(\alpha \neq 0\) ，has a real root at \(x = \alpha\) ．
（c）Find and simplify an expression for \(\beta\) in terms of \(\alpha\) and prove that \(\alpha\) is the only real root provided \(| \alpha | < 2\) ．
（6）
An examiner chooses a positive number \(\alpha\) so that \(\alpha\) is the only real root of equation（I） but the incorrect method used by the student produces 3 distinct real＂roots＂．
（d）Find the range of possible values for \(\alpha\) ． Marks for style，clarity and presentation： 7