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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 uu be a positive integer and ff be a real number lying between 0and10 and 1. Further, \left(\sqrt{2}+1\right)^{10}=u+fand(2−1)10=vand \left(\sqrt{2}-1\right)^{10}=v.

Question: Consider the following statements :
I. (u+v+f)(u+v+f) is an integer.
II. (f+v)(f+v) is an integer.
Which of the statements given above is/are correct ?

Options

  1. AI only
  2. BII only
  3. CBoth I and II
  4. DNeither I nor II

Correct answer

C. Both I and II

Step-by-step solution

Given (2+1)10=u+f(\sqrt{2}+1)^{10} = u+f and (2−1)10=v(\sqrt{2}-1)^{10} = v.

Adding both equations, we get:

u+f+v=(2+1)10+(2−1)10u+f+v = (\sqrt{2}+1)^{10} + (\sqrt{2}-1)^{10}

Using the binomial expansion (x+y)n+(x−y)n=2(nC0xn+nC2xn−2y2+… )(x+y)^n + (x-y)^n = 2(^{n}C_{0}x^n + ^{n}C_{2}x^{n-2}y^2 + \dots), we have:

u+f+v=2(10C0(2)10+10C2(2)8+⋯+10C10)u+f+v = 2(^{10}C_{0}(\sqrt{2})^{10} + ^{10}C_{2}(\sqrt{2})^8 + \dots + ^{10}C_{10})

Since all powers of 2areeven,theright−handsideisaneveninteger.Letthisintegerbe\sqrt{2} are even, the right-hand side is an even integer. Let this integer beI.<br><br>Therefore,.<br><br>Therefore,u+f+v = I,whichmeans, which means(u+v+f)isaninteger.Thus,StatementIiscorrect.<br><br>Sinceis an integer. Thus, Statement I is correct.<br><br>Sinceuisgivenasapositiveintegerandis given as a positive integer andu+f+v = I,itfollowsthat, it follows thatf+v = I - u.<br><br>Thedifferenceoftwointegersisaninteger,so.<br><br>The difference of two integers is an integer, so(f+v)isaninteger.Thus,StatementIIisalsocorrect.<br><br>Answer:BothIandIIis an integer. Thus, Statement II is also correct.<br><br>Answer: Both I and II
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