Test Index

UPSC CDS 2 2023 Math Paper

Question : 16 of 100

Marks: +1, -0

x(a-b+\frac{ab}{a-b})=y(a+b-\frac{ab}{a+b})

and x + y = 2a3, then what is x-y equal to ?

Solution:

Concept:

Use algebraic identities to simplify the given ratios and solve for

x

and

y

Explanation:

We are given:

x\left(a-b+\frac{ab}{a-b}\right) = y\left(a+b-\frac{ab}{a+b}\right)

and

x + y = 2a^{3}

First, simplify each bracket:

a-b+\frac{ab}{a-b} = \frac{(a-b)^{2}+ab}{a-b} = \frac{a^{2}+b^{2}-ab}{a-b}

Similarly,

a+b-\frac{ab}{a+b} = \frac{(a+b)^{2}-ab}{a+b} = \frac{a^{2}+b^{2}+ab}{a+b}

Using identities:

a^{3}+b^{3} = (a+b)(a^{2}+b^{2}-ab)

and

a^{3}-b^{3} = (a-b)(a^{2}+b^{2}+ab)

Thus,

\frac{a^{2}+b^{2}-ab}{a-b} = \frac{a^{3}+b^{3}}{(a+b)(a-b)}

and

\frac{a^{2}+b^{2}+ab}{a+b} = \frac{a^{3}-b^{3}}{(a+b)(a-b)}

So the given equation becomes:

x \cdot \frac{a^{3}+b^{3}}{(a+b)(a-b)} = y \cdot \frac{a^{3}-b^{3}}{(a+b)(a-b)}

Cancel the common denominator:

x(a^{3}+b^{3}) = y(a^{3}-b^{3})

Hence,

\frac{x}{y} = \frac{a^{3}-b^{3}}{a^{3}+b^{3}}

Let

x = k(a^{3}-b^{3})

and

y = k(a^{3}+b^{3})

for some constant

k

Using

x+y = 2a^{3}

k(a^{3}-b^{3} + a^{3}+b^{3}) = k(2a^{3}) = 2a^{3}

→

k=1

Therefore,

x = a^{3}-b^{3}

and

y = a^{3}+b^{3}

Thus,

x-y = (a^{3}-b^{3}) - (a^{3}+b^{3}) = -2b^{3}

Answer:

-2b^{3}

(Option A)

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