JEE Mathematics 2023 Question with Solution

Given: The sum of two positive integers is $66$, and $M$ is the maximum possible product. The sample space is $S = \{x \in \mathbb{Z} : (66-x)x \geq \frac{5}{9}M\}$. Find: $P(A)$ where $A$ is the set of multiples of $3$ in $S$.

For a fixed sum, the product is maximum when the two integers are equal.

$$M = 33 \times 33 = 1089$$

Now use the given condition:

$$x(66-x) \geq \frac{5}{9} \times 1089 = 605$$

Rearranging,

$$x^2 - 66x + 605 \leq 0$$

Solve the corresponding quadratic equation:

$$x^2 - 66x + 605 = 0$$ $$x = \frac{66 \pm \sqrt{66^2 - 4 \cdot 605}}{2}$$ $$x = \frac{66 \pm \sqrt{4356 - 2420}}{2} = \frac{66 \pm \sqrt{1936}}{2}$$ $$x = \frac{66 \pm 44}{2}$$

Hence,

$$x = 11 \quad \text{or} \quad x = 55$$

Therefore,

$$11 \leq x \leq 55$$

So the total number of integers in $S$ is

$$55 - 11 + 1 = 45$$

The multiples of $3$ in this interval are $12, 15, 18, \ldots, 54$. This arithmetic progression has number of terms

$$n = \frac{54 - 12}{3} + 1 = 15$$

Therefore,

$$P(A) = \frac{15}{45} = \frac{1}{3}$$

So, the correct option is B.

Given: $x(66-x) \geq \frac{5}{9}M$ with maximum product at sum $66$. Find: $P(A)$.

Since the sum is fixed at $66$, the maximum product occurs at the midpoint:

$$M = 33^2$$

The inequality becomes

$$x(66-x) \geq \frac{5}{9} \cdot 33^2$$

From the solution working, this gives the integer interval

$$11 \leq x \leq 55$$

So there are

$$45$$

integers in the sample space.

Now count multiples of $3$ from $12$ to $54$:

$$12, 15, 18, \ldots, 54$$

These are exactly

$$15$$

numbers.

Hence,

$$P(A) = \frac{15}{45} = \frac{1}{3}$$

Therefore, the correct option is B.