The probability of selecting integers a∈[−5, 30] such that x2+2(a+4)x−5a+64>0, for all x∈R, is

f(x)=x2+2(a+4)x−5a+64f(x)>0,When D 0 4(a+4)2−4(64−5a)<0⇒ a2+16+8a−64+5a<0⇒ a2+13a−48<0⇒ (a+16)(a−3)<0a∈(−16,3)⇒a∈[−5,2] (as per question)According to the question, a∈[−5,30]n(S)=30+5+1=36n(E)=2+5+1=8∴ Probability =n(E)n(S)=836=29

The probability of selecting integers a∈[−5, 30] such that x2+2(a+4)x−5a+64>0, for all x∈R, is

f(x)=x2+2(a+4)x−5a+64f(x)>0,When D 0 4(a+4)2−4(64−5a)<0⇒ a2+16+8a−64+5a<0⇒ a2+13a−48<0⇒ (a+16)(a−3)<0a∈(−16,3)⇒a∈[−5,2] (as per question)According to the question, a∈[−5,30]n(S)=30+5+1=36n(E)=2+5+1=8∴ Probability =n(E)n(S)=836=29

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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

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$\frac{1}{6}$

$\frac{7}{36}$

$\frac{2}{9}$

$\frac{1}{4}$

answer is B.

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Detailed Solution

$\begin{array}{l} f (x) = x^{2} + 2 (a + 4) x - 5 a + 64 \\ f (x) > 0, \end{array}$

When D < 0 as coefficient of x² > 0

$\begin{array}{l} 4 (a + 4)^{2} - 4 (64 - 5 a) < 0 \\ \Rightarrow a^{2} + 16 + 8 a - 64 + 5 a < 0 \\ \Rightarrow a^{2} + 13 a - 48 < 0 \\ \Rightarrow (a + 16) (a - 3) < 0 \end{array}$