If $x^{2} + 2 a x + 10 - 3 a > 0$ for all x ∈ R, then

-5 < a < 2

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a < -5

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a > 5

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2 < a < 5

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Solution

The correct option is A
-5 < a < 2

According to given condition,

$4 a^{2} - 4 (10 - 3 a) < 0$ ⇒ $a^{2} + 3 a - 10 < 0$

⇒ $(a + 5) (a - 2) < 0 \Rightarrow - 5 < a < 2.$

Suggest Corrections

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