- Home
- Standard 11
- Mathematics
4-2.Quadratic Equations and Inequations
medium
જો દરેક $x \in R$ માટે,${x^2} + 2ax + 10 - 3a > 0$ તો .
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
