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

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4-2.Quadratic Equations and Inequations

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

A

$ - 5 < a < 2$

B

$a < - 5$

C

$a > 5$

D

$2 < a < 5$

(IIT-2004)

Solution

(a) According to given condition,

$4{a^2} – 4(10 – 3a) < 0$

==>${a^2} + 3a – 10 < 0$

==>$(a + 5)\,(a – 2) < 0$

==> $ – 5 < a < 2$.

Standard 11

Mathematics

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