GATE CS 2024 Set 2 — Question 4

MCQ+1 / -0.33EasyRatios, Percentages, Powers, Exponents & LogarithmsQuantitative AptitudeGeneral Aptitude

General Aptitude → Quantitative Aptitude → Ratios, Percentages, Powers, Exponents & Logarithms

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Question

For positive non-zero real variables xx and yy, ifln(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
A.
1
B.
1/2
C.
2
D.
4

Correct answer

(C) 2

Solution

Given equation:ln(x+y2)=12[ln(x)+ln(y)]\ln \left(\frac{x+y}{2}\right) = \frac{1}{2}[\ln (x) + \ln (y)]Using logarithmic properties ln(a)+ln(b)=ln(ab)\ln(a) + \ln(b) = \ln(ab) and kln(a)=ln(ak)k \ln(a) = \ln(a^k):ln(x+y2)=12ln(xy)=ln((xy)1/2)=ln(xy)\ln \left(\frac{x+y}{2}\right) = \frac{1}{2} \ln (xy) = \ln ((xy)^{1/2}) = \ln (\sqrt{xy})Taking antilog on both sides:x+y2=xy\frac{x+y}{2} = \sqrt{xy}x+y=2xyx+y = 2\sqrt{xy}Squaring both sides:(x+y)2=(2xy)2(x+y)^2 = (2\sqrt{xy})^2x2+y2+2xy=4xyx^2 + y^2 + 2xy = 4xyx2+y22xy=0x^2 + y^2 - 2xy = 0(xy)2=0(x-y)^2 = 0x=yx = yWe need to find the value of xy+yx\frac{x}{y} + \frac{y}{x}. Since x=yx=y:xx+xx=1+1=2\frac{x}{x} + \frac{x}{x} = 1 + 1 = 2

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