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4-2.Quadratic Equations and Inequations
hard
$ \alpha $ એ $x$ ની ન્યૂનતમ પૃણાંક કિમત છે કે જેથી $\frac{{x - 5}}{{{x^2} + 5x - 14}} > 0$ થાય તો .....
A
${\alpha ^2} + 3\alpha - 4 = 0$
B
${\alpha ^2} - 5\alpha + 4 = 0$
C
${\alpha ^2} - 7\alpha + 6 = 0$
D
${\alpha ^2} + 5\alpha - 6 = 0$
(JEE MAIN-2013)
Solution
$\frac{x-5}{x^{2}+5 x-14}>0$
$\Rightarrow {x^2} + 5x – 14 < x – 5$
$\Rightarrow x^{2}+4 x-9<0$
$\Rightarrow \alpha=-5,-4,-3,-2,-1,0,1$
$\alpha=-5$ does not satisfy any of the options
$\alpha=-4$ satisfy the option $(a) \alpha^{2}+3 \alpha-4=0$
Standard 11
Mathematics
