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

Question

The probability of selecting integers a ∈[-5,30] such that x2+2(a+4) x-5 a+64 > 0, for all x ∈ R, is: (A) (7/36) (B) (2/9) (C) (1/6) (D) (1/4)

Tardigrade Admin · Accepted Answer

D < 0 ⇒ 4(a+4)2-4(-5 a+64)<0 ⇒ a2+16+8 a+5 a-64<0 ⇒ a2+13 a-48<0 ⇒(a+16)(a-3)<0 ⇒ a ∈(-16,3) ∴ Possible a: -5,-4, ldots ldots, 3 ∴ Required probability =(8/36) =(2/9)

Q. 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:

$\frac{7}{36}$

$\frac{2}{9}$

$\frac{1}{6}$

$\frac{1}{4}$

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

367​

92​

61​

41​

D<0 ⇒4(a+4)2−4(−5a+64)<0 ⇒a2+16+8a+5a−64<0 ⇒a2+13a−48<0 ⇒(a+16)(a−3)<0 ⇒a∈(−16,3) ∴ Possible a : {−5,−4,……,3} ∴ Required probability =368​ =92​

Q. 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:

$\frac{7}{36}$

$\frac{2}{9}$

$\frac{1}{6}$

$\frac{1}{4}$