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Official PYQ · 2024

GATE CS 2024 Set 2 Q4 — General Aptitude

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For positive non-zero real variables xx and yy, if
ln⁡(x+y2)=12[ln⁡(x)+ln⁡(y)]\ln \left(\frac{x+y}{2}\right)=\frac{1}{2}[\ln (x)+\ln (y)]
then, the value of xy+yx\frac{x}{y}+\frac{y}{x} is

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Answer: C

Using the properties of logarithms, we simplify the given equation:

ln⁡(x+y2)=12[ln⁡(x)+ln⁡(y)]=ln⁡(xy)\ln \left(\frac{x+y}{2}\right) = \frac{1}{2} [\ln(x) + \ln(y)] = \ln(\sqrt{xy})

Taking the exponential on both sides:

x+y2=xy  ⟹  x+y=2xy\frac{x+y}{2} = \sqrt{xy} \implies x + y = 2\sqrt{xy}

Squaring both sides:

(x+y)2=4xy  ⟹  x2−2xy+y2=0  ⟹  (x−y)2=0(x+y)^2 = 4xy \implies x^2 - 2xy + y^2 = 0 \implies (x-y)^2 = 0

This gives x=yx = y. Substituting this into the required expression:

xy+yx=1+1=2\frac{x}{y} + \frac{y}{x} = 1 + 1 = 2

Hence, the correct option is C.

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