Verified Solution • 2025–26 Session moderate

QJEEM23Ma9509: JEE Main 2023 Mathematics - Class 12 JEEM23Ma9509 Solution: Let M be the maximum value of the product of two positive…

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)≥59M} and the event A={x∈S:xisamultipleof3}. Then P(A) is equal to :

Multiple Choice Options

  • A 13
  • B 15
  • C 722
  • D 1544

Step-by-Step Verified Solution

x+y=66x+y=66

x+y2≥xy⇒33≥xy⇒xy≤1089∴M=1089S:x(66−x)≥59⋅108966x−x2≥605⇒x2−66x+605≤0

⇒(x−55)(x−11)≤0;11≤x≤55\Rightarrow(x-55)(x-11) \leq 0 ; 11 \leq x \leq 55

Therefore S={11,12,13…55}S=\{11,12,13 \ldots 55\}

⇒n(S)=45\Rightarrow n(S)=45

Elements of SS which are multiple of 3 are

12+(n−1)3=54⇒3(n−1)=42⇒n=15n(A)=15

⇒P(A)=1545=13\Rightarrow P(A)=\frac{15}{45}=\frac{1}{3}

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