GATE CSE 2024 | Set 2 | GA Question: 4

Question

For positive non-zero real variables $x$ and $y$, if
\[
\ln \left(\frac{x+y}{2}\right)=\frac{1}{2}[\ln (x)+\ln (y)]
\]
then, the value of $\frac{x}{y}+\frac{y}{x}$ is

• A. $1$
• B. $1 / 2$
• C. $2$
• D. $4$

Answer 1

$\log \frac{(x+y)}{2}=1/2(\log x+\log y)$
$\implies \log \frac{(x+y)}{2}= \frac{1}{2}(\log xy)$
$\implies \log \frac{(x+y)}{2}= \log \sqrt {xy}$
$\implies \frac{(x+y)}{2}= \sqrt {xy}$
Squaring both side we get;
$\implies x^2+y^2+2xy=4xy$
$\implies x^2+y^2=4xy-2xy$
$\implies x^2+y^2=2xy$
$\implies \frac{x^2+y^2}{xy}=2 $
$\implies \frac{x^2}{xy}+\frac{y^2}{xy}=2$
$\implies \frac{x}{y}+\frac{y}{x}=2$

So $2$ is the correct answer.

Answer 2

Quick method:

Note that $x = 1, y=1$ satisfies the given equation. So, $x/y + y/x = 1+1 = 2.$

Answer 3

C. 2 

Answer 4

Given:

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

Step 1: Rewrite the equation.

$$\ln \left(\frac{x+y}{2}\right) = \frac{1}{2}\ln(xy)$$

Step 2: Use the property of logarithms.

$$\ln\left(\frac{x+y}{2}\right) = \ln\left(\sqrt{xy}\right)$$

Step 3: Drop the logarithms.

$$\frac{x+y}{2} = \sqrt{xy}$$

Step 4: Square both sides to eliminate the square root.

$$(x+y)^2 = 4xy$$

$$x^2 + 2xy + y^2 = 4xy$$

$$x^2 + y^2 = 2xy$$

Step 5: Divide both sides by $xy$.

$$\frac{x^2}{xy} + \frac{y^2}{xy} = 2$$

$$\frac{x}{y} + \frac{y}{x} = 2$$

So, the solution to the given equation is $\frac{x}{y} + \frac{y}{x} = 2$. Thus, the correct answer is $2$.

Alternate approach:

Given:

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

We observe that when $x = 1$ and $y = 1$, the equation becomes:

$$\ln \left(\frac{1+1}{2}\right) = \frac{1}{2}(\ln(1) + \ln(1))$$

$$\ln(1) = \frac{1}{2}(0 + 0)$$

$$0 = 0$$

So, $x = 1$ and $y = 1$ satisfies the given equation. Substituting these values into $\frac{x}{y} + \frac{y}{x}$, we get:

$$\frac{1}{1} + \frac{1}{1} = 1 + 1 = 2$$

Therefore, according to this alternate method, the answer is $2$.