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UPSC CDS II 2024 Math Paper

Question : 78 of 100

Marks: +1, -0

(x+\frac{1}{yz})-(y+\frac{1}{zx})=

(y+\frac{1}{zx})-(z+\frac{1}{xy})

and x + z ≠ 2y, then what is xyz equal to?

Solution:

Concept:

Simplify the given equation by grouping like terms and factoring to isolate the condition

xyz = -1

Explanation:

Start with the equation:

(x+\frac{1}{yz})-(y+\frac{1}{zx}) = (y+\frac{1}{zx})-(z+\frac{1}{xy})

Rewrite each fraction with a common denominator

xyz

\frac{1}{yz} = \frac{x}{xyz}

\frac{1}{zx} = \frac{y}{xyz}

\frac{1}{xy} = \frac{z}{xyz}

Substitute and simplify both sides:

Left:

(x-y) + \left(\frac{x}{xyz} - \frac{y}{xyz}\right) = (x-y) + \frac{x-y}{xyz}

Right:

(y-z) + \left(\frac{y}{xyz} - \frac{z}{xyz}\right) = (y-z) + \frac{y-z}{xyz}

Set them equal:

(x-y)\left(1 + \frac{1}{xyz}\right) = (y-z)\left(1 + \frac{1}{xyz}\right)

Bring all terms to one side:

(x-y - (y-z))\left(1 + \frac{1}{xyz}\right) = 0

(x - 2y + z)\left(1 + \frac{1}{xyz}\right) = 0

Given that

x + z \ne 2y

, we have

x - 2y + z \ne 0

. Therefore:

1 + \frac{1}{xyz} = 0 \quad \Rightarrow \quad \frac{1}{xyz} = -1 \quad \Rightarrow \quad xyz = -1

Answer:

xyz = -1

(Option B)

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