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

The correct option is C

$1$

Explanation of the correct option:

Step 1. Find the function

Let two non-zero numbers are $x$ and $y$.

Given : The sum of two non–zero numbers is $4$.

$x+y=4$

$\Rightarrow$ $y=4-x$

Thus sum of the reciprocals is,

$$\begin{gathered} \frac{1}{x}+\frac{1}{y} \\ \Rightarrow \frac{x+y}{xy} \\ \Rightarrow \frac{4}{x(4-x)} \end{gathered}$$

Let $f\left(x\right)=\frac{4}{4x-x^2}$

Step 2. Find the minimum value

$f\left(x\right)=\frac{4}{4x-x^2}$

For extremum $f\text{'}\left(x\right)=0$

$$\begin{gathered} \Rightarrow f\text{'}\left(x\right)=\frac{-4(4-2x)}{{(4x-x^2)}^{2^{}}} \\ \Rightarrow \frac{-4(4-2x)}{{(4x-x^2)}^2}=0 \\ \Rightarrow 4-2x=0 \\ \Rightarrow x=2 \end{gathered}$$

$\therefore y=2$

Therefore, minimum of $\left(\frac{1}{x}+\frac{1}{y}\right)$$=\frac{1}{2}+\frac{1}{2}$

minimum of $\left(\frac{1}{x}+\frac{1}{y}\right)$$=1$

Hence, option (C) is the correct option.