Two different positive numbers a and b each differ from their reciproc

It must be true that one number is more than its reciprocal, and the other is less than its reciprocal. We can let a be the one that is more than its reciprocal and hence b is the one that is less than its reciprocal. So we have:

a - 1/a = 1 and 1/b - b = 1

Subtracting these two equations, we have:

a - 1/a - 1/b + b = 0

a + b = 1/a + 1/b

a + b = (b + a)/(ab)

ab(a + b) = a + b

Since both a and b are positive, a + b ≠ 0. So we can divide both sides of the equation by a + b and obtain:

ab = 1

Now, let’s multiply the original first equation by a. We have:

a^2 - 1 = a

Similarly, let’s multiply the original second equation by b. We have:

1 - b^2 = b

Adding the two new equations, we have:

a^2 - b^2 = a + b

(a + b)(a - b) = a + b

Again, dividing both sides of the equation by a + b, we have:

a - b = 1

Squaring the above equation, we have:

a^2 - 2ab + b^2 = 1

Since ab = 1, we have:

a^2 - 2(1) + b^2 = 1

a^2 + b^2 = 3

Now add 2ab = 2 to the above equation, we have:

a^2 + 2ab + b^2 = 5

(a + b)^2 = 5

Taking the square root of both sides, we have:

a + b = √5

Answer: C