A) \[\frac{4}{5}\]
B) \[\frac{2}{3}\]
C) \[\frac{3}{5}\]
D) none of these
Correct Answer: B
Solution :
Key Idea: If roots of the quadratic equation are real, then discriminant is always be greater than equal to zero. Given equation is \[{{x}^{2}}+px+\frac{p}{4}+\frac{1}{2}=0\] Since roots are real, therefore discriminant \[\ge 0\] \[\Rightarrow \] \[{{p}^{2}}-4\left( \frac{p}{4}+\frac{1}{2} \right)\ge 0\] \[\Rightarrow \] \[{{p}^{2}}-p-2\ge 0\] \[\Rightarrow \] \[(p-2)(p+1)\ge 0\] \[\Rightarrow \] \[p\ge 2\]or \[p\le -1\] Since, it is given \[0\le p\le 5,\]so we neglect\[p\le -1.\] The possible values of p are 2, 3, 4, 5 \[\therefore \] Required probability \[=\frac{4}{6}=\frac{2}{3}\]
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