The values of $a$ which makes the expression $x^{2} - a x + 1 - 2 a^{2}$ always positive for real values of $x$ are

- \frac{2}{3} < a < \frac{2}{3}

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- \frac{2}{3} \leq a \leq \frac{2}{3}

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- \frac{2}{3} \leq a \leq 1

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0 < a < \frac{2}{3}

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Solution

The correct option is A $- \frac{2}{3} < a < \frac{2}{3}$
Given, $y = x^{2} - a x + 1 - 2 a^{2} > 0$
So, $D < 0$
$a^{2} - 4 + 8 a^{2} < 0$
$\Rightarrow 9 a^{2} - 4 < 0$
$\Rightarrow (3 a + 2) (3 a - 2) < 0$
$- \frac{2}{3} < a < \frac{2}{3}$

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x^{2} - a x + 1 - 2 a^{2}

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- 3 < \frac{x^{2} + a x - 2}{x^{2} + x + 1} < 2 \forall x \in R

a_{0}, a_{1}, a_{2}, a_{3}

a_{0} + \frac{a_{1}}{2} + \frac{a_{2}}{3} + \frac{a_{3}}{4} = 0

a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3} = 0

y = \sqrt{(\frac{(x + 1) (x - 3)}{x - 2})}

{\frac{2}{x^{2} - x + 1} - \frac{1}{x + 1} - \frac{2 x + 1}{x^{1} + 1}}^{1 / 2}

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