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UPSC NDA I 2026 Mathematics Paper

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Question Numbers: 46-50

For the next five (05) items that follow:

Let

u

be a positive integer and

f

be a real number lying between 0 and 1 .

Further,

(\sqrt{2}+1)^{10}=\mathrm{u}+\mathrm{f}

and

(\sqrt{2}-1)^{10}=v

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Question : 47 of 120

Marks: +1, -0

What is the multiplicative inverse of

(\sqrt{2}+1)^{20}

?

v
$\mathrm{v}^{2}-1$
$\mathrm{v}^{2}$
$\mathrm{v}^{2}+1$

Solution:

Concept:

Use the identity

(\sqrt{2}+1)(\sqrt{2}-1)=1

.

Explanation:

Given

(\sqrt{2}-1)^{10}=v

.

Since

(\sqrt{2}+1)(\sqrt{2}-1)=1

,

(\sqrt{2}+1)^{10}(\sqrt{2}-1)^{10}=1

So

(\sqrt{2}+1)^{10}=\frac{1}{v}

.

Therefore, the multiplicative inverse of

(\sqrt{2}+1)^{20}

is:

(\sqrt{2}+1)^{-20}=(\sqrt{2}-1)^{20}=v^2

Answer:

v^{2}

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