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UPSC CDS I 2019 Math Paper

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Question : 47 of 100

Marks: +1, -0

If

a^{x} = b^{y} = c^{z}

b^{2} = ac

\frac{1}{x}+\frac{1}{z}

equal to?

$\frac{1}{y}$
$-\frac{1}{y}$
$\frac{2}{y}$
$-\frac{2}{y}$

Solution:

Concept:

Use the common value

k

such that

a^{x}=b^{y}=c^{z}=k

, then express each base in terms of

k

and substitute into

b^{2}=ac

.

Explanation:

Let

a^{x}=b^{y}=c^{z}=k

.

Then

a=k^{\frac{1}{x}}

,

b=k^{\frac{1}{y}}

,

c=k^{\frac{1}{z}}

.

Given

b^{2}=ac

.

Substitute:

\left(k^{\frac{1}{y}}\right)^{2}=k^{\frac{1}{x}} \cdot k^{\frac{1}{z}}

.

Simplify:

k^{\frac{2}{y}}=k^{\frac{1}{x}+\frac{1}{z}}

.

Since bases are equal, equate exponents:

\frac{2}{y}=\frac{1}{x}+\frac{1}{z}

.

Thus

\frac{1}{x}+\frac{1}{z}=\frac{2}{y}

.

Answer:

\frac{2}{y}

, which corresponds to option C.

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