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Question 7: If x and y are positive real numbers satisfying x + y = 102, then the minimum possible value of \( 2601\left(1 + \frac{1}{x}\right)\left(1 + \frac{1}{y}\right) \) is

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Video Explanation

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Explanatory Answer

To find the minimum value, keep x and y equal.
So, x = y = 51

\( 2601\left(1 + \frac{1}{51}\right)\left(1 + \frac{1}{51}\right) \)
= \( 2601 \times \frac{52}{51} \times \frac{52}{51} \)
= \( 52 \times 52 \)
= 2704

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Hence, the answer is 2704.