The sum of two non – zero numbers is 4. The minimum value of the sum of their reciprocals is

1) 3 / 4

2) 6 / 5

3) 1

4) None of these

Solution: (3) 1

x + y = 4

y = 4 – x

1 / x + 1 / y = (x + y) / xy

f (x) = 4 / xy = (4 / x) (4 – x)

f (x) = 4 / (4x – x2) and f’ (x) = (-4 / (4x – x2)2) (4 – 2x)

f’ (x) = 0, 4 – 2x = 0.

x = 2 and y = 2

Minimum of (1 / x + 1 / y) = 1 / 2 + 1 / 2 = 1

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