Quadratic EquationHard

Question

The possible set of real values of ' $a$ ' for which, for all $x$ not exceeding unity in absolute value, the inequality $\frac{ax - a(1 - a)}{a^{2} - ax - 1} > 0$ is valid

Options

A.$(1,\infty)$
B.$\left( - \infty,\frac{- 1 + \sqrt{5}}{2} \right)$
C.$\left( - \infty,\frac{- 1 - \sqrt{5}}{2} \right)$
D.$(2,\infty)$

Solution

$(x - (1 - a))\left( x - \frac{a^{2} - 1}{a} \right) < 0\ \forall x \in \lbrack - 1,1\rbrack$

Let $f(x) = (x - (1 - a))\left( x - \frac{a^{2} - 1}{a} \right)$

$$\begin{matrix} f( - 1) < 0 & \Rightarrow \ \frac{(a - 2)\left( a^{2} + a - 1 \right) > 0}{a} \\ & \Rightarrow \ a \in \left( - \infty,\frac{- 1 - \sqrt{5}}{2} \right) \cup \left( 0,\frac{- 1 + \sqrt{5}}{2} \right) \cup (2,\infty) \\ & \text{~and~}f(1) < 0\ \Rightarrow \ a\left( \frac{a^{2} - a - 1}{a} \right) > 0\frac{+ \frac{-}{\frac{1 - \sqrt{5}}{2}} +}{\frac{1 + \sqrt{5}}{2}} \\ \Rightarrow & a \in \left( - \infty,\frac{1 - \sqrt{5}}{2} \right) \cup \left( \frac{1 + \sqrt{5}}{2},\infty \right) \\ \text{~Hence,~} & a \in \left( - \infty,\frac{- 1 - \sqrt{5}}{2} \right) \cup (2,\infty). \end{matrix}$$

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