Question

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

• A. $\frac{3}{4}$
• B. $\frac{6}{5}$
• C. 1
• D. ) None of these

Answer 1

Answer: (C)

1

Answer 2

Let $x + y = 4$ or $y = 4 - x$

$\frac{1}{x} + \frac{1}{y} = \frac{x + y}{xy}$ or $f(x) = \frac{4}{xy} = \frac{4}{x(4 - x)}$

$f(x) = \frac{4}{4x - x^{2}}$, $f^{'}(x) = \frac{- 4}{(4x - x^{2})^{2}}.(4 - 2x)$

Put $f^{'}(x) = 0$ ⇒ $4 - 2x = 0$ ⇒ $x = 2$ and $y = 2$

$\therefore$ min. $\left( \frac{1}{x} + \frac{1}{y} \right) = \frac{1}{2} + \frac{1}{2} = 1$.