Suppose Y is distributed uniformly in the open interval (1, 6). The probability that the polynomial 3x2 + 6xY + 3Y + 6 has only real roots is (rounded off to 1 decimal place) _________.
Note: This was Numerical Type question.
(A) 0.80
(B) 0.17
(C) 0.20
(D) 1
Answer: (A)
Explanation: For a quadratic polynomial ax2 + bx + c = 0. There are three condition:
b2 - 4ac > 0 {real and distinct root, i.e., two real roots} b2 - 4ac = 0 {real and equal roots, i.e., only one real root} b2 - 4acPolynomial 3x2 + 6xY + 3Y + 6 has only real roots,
⇒ b2 – 4ax ≥ 0 ⇒ (6Y)2 – 4(3) (3Y+ 6) ≥ 0 ⇒ Y2 – Y + 2 ≥ 0 Y ∈ (–∞, – 1] ∩ [2, ∞) ⇒ Y ∈ [2, 6)Since y is uniformly distributed in (1, 6).
Probability distributed function,
f(Y) = (1/5), 1 6Hence,
So, answer is 0.8.
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