Discussion

Explanation:

The sum of two numbers, when their product is known, is minimum when they are equal.

The product of two numbers, when their sum is known, is maximum when they are equal.

${\left(x+\frac{1}{x}\right)}^2+{\left(y+\frac{1}{y}\right)}^2$

= (x² + y²) + $\frac{x^2+y^2}{x^2{y}^2}+4$

This expression is minimum when (x² + y²) is minimum and x²y² is maximum.

We have x + y = 1

Squaring,

x² + 2xy + y² = 1

∴ x² + y² = 1 – 2xy

x² + y² is minimum when xy is maximum.

xy is maximum when x = y

∴ For minimum value, both x and y have to be equal.

∴ x = y = $\frac{1}{2}$

$\because {\left(x+\frac{1}{x}\right)}^2+{\left(y+\frac{1}{y}\right)}^2={\left(2+\frac{1}{2}\right)}^2+{\left(2+\frac{1}{2}\right)}^2$

= (2.5)² + (2.5)²

= 12.5

Hence, option (c).

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