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Q.

If the two roots of equation $(a - 1) (x^{4} + x^{2} + 1) + (a + 1) {(x^{2} + x + 1)}^{2} = 0$ are real and distinct then the set of all values of ‘a’ is

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a

$(- \frac{1}{2}, 0)$

b

$(- \infty, - 2)$

c

$(2, \infty)$

d

$(- \frac{1}{2}, 0) \cup (0, \frac{1}{2})$

answer is D.

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Detailed Solution

$(x^{2} + x + 1)$
$((a - 1) (x^{2} - x + 1) + (a + 1) (x^{2} + x + 1)) = 0 \Rightarrow 2 a x^{2} + 2 x + 2 a = 0 \Rightarrow a x^{2} + x + a = 0$
has distinct roots $\Rightarrow 1 - 4 a^{2} > 0$
$\Rightarrow a \in (- \frac{1}{2}, \frac{1}{2})$ and $a \neq 0$

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