Question 3362 – Logical-Reasoning
December 10, 2023Question 7844 – Engineering-Mathematics
December 10, 2023Question 7813 – Engineering-Mathematics
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) _____.
Correct Answer: D
Question 7 Explanation:
Given polynomial equation is
3x2 + 6xY + 3Y + 6
= 3x2 + (6Y)x + (3Y + 6)
which is in the form: ax2 + bx + c
For real roots: b2 – 4ac ≥ 0
⇒ (6Y)2 – 4(3)(3Y + 6) ≥ 0
⇒ 36Y2 – 36Y – 72 ≥ 0
⇒ Y2 – Y – 2 ≥ 0
⇒ (Y+1)(Y-2) ≥ 0
Y = -1 (or) 2
The given interval is (1,6).
So, we need to consider the range (2,6).
The probability = (1/(6-1)) * (6-2) = 1/5 * 4 = 0.8
3x2 + 6xY + 3Y + 6
= 3x2 + (6Y)x + (3Y + 6)
which is in the form: ax2 + bx + c
For real roots: b2 – 4ac ≥ 0
⇒ (6Y)2 – 4(3)(3Y + 6) ≥ 0
⇒ 36Y2 – 36Y – 72 ≥ 0
⇒ Y2 – Y – 2 ≥ 0
⇒ (Y+1)(Y-2) ≥ 0
Y = -1 (or) 2
The given interval is (1,6).
So, we need to consider the range (2,6).
The probability = (1/(6-1)) * (6-2) = 1/5 * 4 = 0.8
0.3
0.9
0.1
0.8
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