Let M be the maximum value of the product of two positive integers when their sum is 66. Let the sample space S = { x ∈ ℤ:x(66 - x) ≥ 5/9M } and the event A = { x ∈ S:x is a multiple of 3}. Then P(A) is equal to :

Solution

Maths sir · 5 steps

Find M first, then the inequality becomes a quadratic that traps x between 11 and 55.

1. Find M. For a fixed sum, the product is largest when the two numbers are equal. So M = 33 × 33 = 1089, and 5/9M = 5 × 121 = 605.
2. Solve the inequality. x(66 - x) ≥ 605 rearranges to x² - 66x + 605 ≤ 0. The roots are (66 ± √(4356 - 2420))/2 = (66 ± 44)/2, that is 11 and 55.

   (x - 11)(x - 55) ≤ 0 ⇒ 11 ≤ x ≤ 55

3. Count S. Both ends are included because the inequality allows equality (11 × 55 = 605). So S has 55 - 11 + 1 = 45 integers.
4. Count A. The multiples of 3 from 11 to 55 run 12, 15, …, 54. That is (54 - 12)/3 + 1 = 15 numbers.
5. So. P(A) = 15/45 = 1/3, which is option (a).

The idea to keep. Fixed sum means the product peaks when the parts are equal. To count the integers from a to b, use b - a + 1. Check whether the endpoints satisfy the inequality before you count.

Answer from the NTA final answer key. Solution by PrepWiser.