JEE Mains · Maths · STD 12 - 13. probability
The probability of selecting integers \(a \in[-5,30]\) such that \(x^{2}+2(a+4) x-5 a+64>0\), for all \(x \in R\), is:
- A \(\frac{1}{4}\)
- B \(\frac{7}{36}\)
- C \(\frac{2}{9}\)
- D \(\frac{1}{6}\)
Answer & Solution
Correct Answer
(C) \(\frac{2}{9}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{D}<0\) \(\Rightarrow 4(a+4)^{2}-4(-5 a+64)<0\) \(\Rightarrow a^{2}+16+8 a+5 a-64<0\) \(\Rightarrow a^{2}+13 a-48<0\) \(\Rightarrow(a+16)(a-3)<0\) \(\Rightarrow a \in(-16,3)\) \(\therefore\) Possible a : \(\{-5,-4, \ldots, 2\}\) \(\therefore\) Required probability…
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