NDA 2026 Mathematics PYQ — Let be a positive integer and be a real number lying between and … | Mathem Solvex | Mathem Solvex
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NDA 2026 — Mathematics PYQ
NDA | Mathematics | 2026
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
What is the value of (v+f)?
मान लीजिए u एक धनात्मक पूर्णांक है तथा 0 और 1 के बीच में f एक वास्तविक संख्या है। इसके अतिरिक्त, (2+1)10=u+f और (2−1)10=v है।
(v+f)का मान क्या है ?
Choose the correct answer:
A.
2
B.
1
(Correct Answer)
C.
0.5
D.
0.25
Correct Answer:
1
Explanation
Let X=(2+1)10=u+f.
Let v=(2−1)10.
Expanding these using the binomial theorem, we observe that for any even power n, the sum of the expansions of (2+1)n and (2−1)n results in an even integer because the terms containing odd powers of 2 cancel out.
Since all terms in the summation are integers, (X+v) is an even integer. Let this integer be I.
Substituting X=u+f:
(u+f)+v=I
f+v=I−u
Since u is an integer and I is an integer, (f+v) must be an integer. Given that 0 < f < 1 and 0 < v < 1 (as (2−1)≈0.414, which when raised to power 10 is a small fraction), their sum must satisfy:
0 < f + v < 2
The only integer satisfying this condition is 1. Therefore:
f+v=1
Conclusion: The value of (v+f) is 1.
Correct Option: (b) 1
Explanation
Let X=(2+1)10=u+f.
Let v=(2−1)10.
Expanding these using the binomial theorem, we observe that for any even power n, the sum of the expansions of (2+1)n and (2−1)n results in an even integer because the terms containing odd powers of 2 cancel out.
Since all terms in the summation are integers, (X+v) is an even integer. Let this integer be I.
Substituting X=u+f:
(u+f)+v=I
f+v=I−u
Since u is an integer and I is an integer, (f+v) must be an integer. Given that 0 < f < 1 and 0 < v < 1 (as (2−1)≈0.414, which when raised to power 10 is a small fraction), their sum must satisfy:
0 < f + v < 2
The only integer satisfying this condition is 1. Therefore:
