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

Given: $$\ln\left(\frac{x+y}{2}\right)=\frac{1}{2}(\ln x+\ln y)$$ Rewrite using log properties: $$\ln\left(\frac{x+y}{2}\right)=\frac{1}{2}\ln(xy)=\ln(\sqrt{xy})$$ Drop logarithms: $$\frac{x+y}{2}=\sqrt{xy}$$ Square both sides: $$(x+y)^2=4xy \;\Rightarrow\; x^2+2xy+y^2=4xy \;\Rightarrow\; x^2+y^2=2xy$$ Divide both sides by $xy$: $$\frac{x^2}{xy}+\frac{y^2}{xy}=2 \;\Rightarrow\; \frac{x}{y}+\frac{y}{x}=2$$ Hence, the solution is $\frac{x}{y}+\frac{y}{x}=2$. Therefore, the correct answer is 2.

Alternate approach:
Given $$\ln\left(\frac{x+y}{2}\right)=\frac{1}{2}(\ln x+\ln y)$$ Take $x=1,\;y=1$: $$\ln\left(\frac{1+1}{2}\right)=\frac{1}{2}(\ln1+\ln1)\;\Rightarrow\;\ln1=0\;\Rightarrow\;0=0$$ Thus, the equation is satisfied. Then $$\frac{x}{y}+\frac{y}{x}=\frac{1}{1}+\frac{1}{1}=2$$ So, by this method also, the answer is 2.

Answer 3

Quick method:

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

Answer 4

C. 2 

Answer 5

Ans 2

 

The question itself says arithematic mean and Geometric mean are same for x and y. then x = y. so x/y=y/x=1. so sum=2

Answer 6

2