If x-\frac{1}{x}=y, y-\frac{1}{y}=z, z-\frac{1}{z}=x, there which of the following equations are true? A. \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0 B. \frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=8 C. \frac{1}{x y}+\frac{1}{y z}+\frac{1}{z x}=-3 Choose the correct answer from the options given below:

Correct Answer is : A and C only

CMAT Exam 2023 Shift 2 Solved Paper

Question : 18 of 100

Marks: +1, -0

x-\;{1}/{x}=y, y-\;{1}/{y}=z, z-\;{1}/{z}=x

, there which of the following equations are true?
A.

\;{1}/{x}+\;{1}/{y}+\;{1}/{z}=0

\;{1}/{x^2}+\;{1}/{y^2}+\;{1}/{z^2}=8

\;{1}/{x y}+\;{1}/{y z}+\;{1}/{z x}=-3

Choose the correct answer from the options given below:

Solution:

It is given,

x-\;{1}/{x}=y, y-\;{1}/{y}=z

and

z-\;{1}/{z}=x

Adding all the given equations, we get

\;x+y+z-\;{1}/{x}-\;{1}/{y}-\;{1}/{z}=y+z+x

\;\;{1}/{x}+\;{1}/{y}+\;{1}/{z}=0

Therefore, equation

A

is correct.

x-\;{1}/{x}=y, y-\;{1}/{y}=z \;\text " and "\; z-\;{1}/{z}=x

Squaring on both the sides and adding three resultant equations, we get

\;x^2+\;{1}/{x^2}-2+y^2+\;{1}/{y^2}-2+z^2+\;{1}/{z^2}-2=y^2+z^2+x^2

\;\;{1}/{x^2}+\;{1}/{y^2}+\;{1}/{z^2}=6

Therefore, equation B is incorrect.
The answer is option B

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