A) \[\frac{69}{70}\]
B) \[\frac{1}{70}\]
C) \[1\]
D) \[0\]
Correct Answer: A
Solution :
(a): \[{{x}^{2}}+2x-4>0\] root \[\alpha ,\beta =\frac{-2\pm \sqrt{4+16}}{2}=\frac{-2\pm 2\sqrt{5}}{2}=-1\pm \sqrt{5}\] or, \[\left\{ x-\left( -1+\sqrt{5} \right) \right\}\left\{ x-\left( 1-\sqrt{5} \right) \right\}>0\] \[\therefore \] \[x\in \left( -\infty ,-1-\sqrt{5} \right)\cup \left( -1+\sqrt{5},\infty \right)\] \[\therefore \] Within natural numbers, only 1 is not the solution \[\therefore \] \[P=\frac{69}{70}\]
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