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UPSC CDS 1 2022 Math Paper

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Question : 26 of 100

Marks: +1, -0

If

a + b = 2

\frac{1}{a} + \frac{1}{b} = 2

then what is the value of

a^{3} + b^{3}

?

2
4
6
8

Solution:

Concept:

Given

a+b=2

and

\frac{1}{a}+\frac{1}{b}=2

, use the identity

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

and find

ab

and

a^{2}+b^{2}

from the given equations.

Explanation:

From

\frac{1}{a}+\frac{1}{b}=2

, rewrite as

\frac{a+b}{ab}=2

.

Thus

a+b=2ab

.

Substitute

a+b=2

:

2=2ab

, so

ab=1

.

Square

a+b=2

:

(a+b)^{2}=4 \Rightarrow a^{2}+b^{2}+2ab=4

.

Plug

ab=1

:

a^{2}+b^{2}+2=4 \Rightarrow a^{2}+b^{2}=2

.

Now apply the identity:

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

.

Substitute

a+b=2

,

a^{2}+b^{2}=2

,

ab=1

:

a^{3}+b^{3}=2 \times (2-1)=2 \times 1=2

.

Therefore the required value is

2

.

Answer:

2

(Option A)

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