Quadratic EquationHard
Question
The set of all real values of ' $a$ ' for which both the roots of the equation $x^{2} - 1 = 0$ lie between the roots of the equation $x^{2} + \left( 3a - a^{2} \right)x - 3a^{3} = 0$ is equal to
Options
A.$( - \infty, - 1)$
B.$\left( - 1, - \frac{1}{3} \right) \cup (1,\infty)$
C.$(1,\infty)$
D.$\left( - \frac{1}{3},1 \right)$
Solution
$f( - 1) < 0 \Rightarrow a \in \left( \frac{1}{3},\infty \right)$
$$f(1) < 0 \Rightarrow \ a \in \left( - 1, - \frac{1}{3} \right) \cup (1,\infty)\ - 1\ 1/y = f(x) = x^{2} + \left( 3a - a^{2} \right)x - 3a^{3}$$
Hence, $a \in (1,\infty)$
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