Passage: Let u be a positive integer and f be a real number lying between 0 and 1 . Further, ( 2 +1)^ 10 =u+f and ( 2 -1)^ 10 =v . Question: What is the multiplicative inverse of ( 2 +1)^ 20 ?

Correct answer: C. v^2 The multiplicative inverse of ( 2 +1)^ 20 is 1 ( 2 +1)^ 20 . Rationalizing the base gives 1 2 +1 = 2 -1 ( 2 +1)( 2 -1) = 2 -1 . Thus, 1 ( 2 +1)^ 20 = ( 2 -1)^ 20 . Given that v = ( 2 -1)^ 10 . Squaring both sides yields v^2 = ( 2 -1)^ 20 . Therefore, the multiplicative inverse of ( 2 +1)^ 20 is v^2 . Answer: v^2

NDA 2026

Exam: NDA
Year: 2026
Subject: Mathematics
Chapter: Binomial Theorem

NDA 2026 · Mathematics · Binomial Theorem

Passage: Let u be a positive integer and f be a real number lying...

Passage: Let

u

be a positive integer and

f

be a real number lying between

0 and 1

. Further,

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

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

.<br><br><strong>Question:</strong> What is the multiplicative inverse of \left(\sqrt{2}+1\right)^{20}

Options

A $v$
B $v^2-1$
C $v^2$
D $v^2+1$

Correct answer

C. $v^2$

Step-by-step solution

The multiplicative inverse of

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

\dfrac{1}{\left(\sqrt{2}+1\right)^{20}}

.

Rationalizing the base gives

\dfrac{1}{\sqrt{2}+1} = \dfrac{\sqrt{2}-1}{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)} = \sqrt{2}-1

.

Thus,

\dfrac{1}{\left(\sqrt{2}+1\right)^{20}} = \left(\sqrt{2}-1\right)^{20}

.

Given that

v = \left(\sqrt{2}-1\right)^{10}.<br><br>Squaring both sides yields

v^2 = \left(\sqrt{2}-1\right)^{20}.

Therefore,

the multiplicative inverse of \left(\sqrt{2}+1\right)^{20}

v^2

.

Answer:

v^2

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