Question:
The probability of selecting integers \(a\in[-5,30]\) such that \(x^2+2(a+4)x-5a+64\gt0\), for all \(x\in\mathbb{R}\), is :
The probability of selecting integers \(a\in[-5,30]\) such that \(x^2+2(a+4)x-5a+64\gt0\), for all \(x\in\mathbb{R}\), is :
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A quadratic with positive leading coefficient is positive for all real \(x\) iff its discriminant is negative.
Updated On: Nov 4, 2026
- \(\frac{7}{36}\)
- \(\frac{2}{9}\)
- \(\frac{1}{6}\)
- \(\frac{1}{4}\)
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The Correct Option is B
Solution and Explanation
Need \(D\lt0\): \(4(a+4)^2-4(64-5a)\lt0\Rightarrow a^2+8a+16-64+5a\lt0\Rightarrow a^2+13a-48\lt0\).
\((a+16)(a-3)\lt0\Rightarrow-16\lt a\lt3\).
Integers in \([-5,30]\): 36 in total. Favourable: \(a=-5,-4,\dots,2\), which is 8 integers.
Probability \(=\frac{8}{36}=\frac29\).
Answer: (2) \(\frac{2}{9}\)
\((a+16)(a-3)\lt0\Rightarrow-16\lt a\lt3\).
Integers in \([-5,30]\): 36 in total. Favourable: \(a=-5,-4,\dots,2\), which is 8 integers.
Probability \(=\frac{8}{36}=\frac29\).
Answer: (2) \(\frac{2}{9}\)
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