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(B) The product of two integers $x$ and $66-x$ is $f(x) = x(66-x)$.This is a downward parabola with maximum at $x = 33$.Thus,$M = 33 	imes 33 = 1089$

Vedclass · Accepted Answer

$\frac{1}{3}$

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(B) The product of two integers $x$ and $66-x$ is $f(x) = x(66-x)$.This is a downward parabola with maximum at $x = 33$.Thus,$M = 33 	imes 33 = 1089$.We require $x(66-x) \geq \frac{5}{9} 	imes 1089 = 5 	imes 121 = 605$.$66x - x^2 \geq 605 \implies x^2 - 66x + 605 \leq 0$.Solving $x^2 - 66x + 605 = 0$ using the quadratic formula: $x = \frac{66 \pm \sqrt{4356 - 2420}}{2} = \frac{66 \pm \sqrt{1936}}{2} = \frac{66 \pm 44}{2}$.So,$x_1 = 11$ and $x_2 = 55$.The set $S = \{11, 12, \ldots, 55\}$,so the number of elements $n(S) = 55 - 11 + 1 = 45$.The event $A$ consists of multiples of $3$ in $S$: $A = \{12, 15, 18, \ldots, 54\}$.This is an arithmetic progression with $a = 12$,$l = 54$,and $d = 3$.$54 = 12 + (n-1)3 \implies 42 = (n-1)3 \implies n-1 = 14 \implies n = 15$.Thus,$n(A) = 15$.$P(A) = \frac{n(A)}{n(S)} = \frac{15}{45} = \frac{1}{3}$.

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

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