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.
Consider the following statements:
I. (u+v+f) is an integer.
II. (f+v) is an integer.
Which of the statements given above is/are correct?
मान लीजिए u एक धनात्मक पूर्णांक है तथा 0 और 1 के बीच मेंf एक वास्तविक संख्या है। इसके अतिरिक्त, (2+1)10=u+fऔर(2−1)10=vहै।
निम्नलिखित कथनों पर विचार कीजिए:
I. (u+v+f)एक पूर्णांक है
II. (f+v)एक पूर्णांक है
उपर्युक्त कथनों में से कौन-सा/कौन-से सही है/हैं ?
Choose the correct answer:
A.
I only
B.
II only
C.
Both I and II
(Correct Answer)
D.
Neither I nor II
Correct Answer:
Both I and II
Explanation
Let X=(2+1)10=u+f.
Let v=(2−1)10.
Since 2≈1.414, then 0 < \sqrt{2} - 1 < 0.5. Raising a number between 0 and 1 to the power of 10 results in a very small positive number, so 0 < v < 1.
Now, consider the expansion using the Binomial Theorem:
(2+1)10=k=0∑10(k10)(2)10−k(1)k
(2−1)10=k=0∑10(k10)(2)10−k(−1)k
When we add these two expressions:
X+v=(2+1)10+(2−1)10
The terms with odd powers of 2 cancel out, leaving:
Since all powers of 2 here are even, they result in integers. Thus, (X+v) is an integer.
Since X=u+f:
(u+f)+v=Integer
f+v=Integer−u
Since u is an integer, (f+v) must also be an integer. Because 0 < f < 1 and 0 < v < 1, the sum 0 < f + v < 2. The only integer value in this range is 1. Therefore, f+v=1.
Evaluating Statement I:(u+v+f)=u+(v+f)=u+1. Since u is an integer, (u+1) is an integer. Statement I is correct.
Evaluating Statement II:(f+v)=1, which is an integer. Statement II is correct.
Conclusion: Both statements I and II are correct.
Correct Option: (c) Both I and II
Explanation
Let X=(2+1)10=u+f.
Let v=(2−1)10.
Since 2≈1.414, then 0 < \sqrt{2} - 1 < 0.5. Raising a number between 0 and 1 to the power of 10 results in a very small positive number, so 0 < v < 1.
Now, consider the expansion using the Binomial Theorem:
(2+1)10=k=0∑10(k10)(2)10−k(1)k
(2−1)10=k=0∑10(k10)(2)10−k(−1)k
When we add these two expressions:
X+v=(2+1)10+(2−1)10
The terms with odd powers of 2 cancel out, leaving:
Since all powers of 2 here are even, they result in integers. Thus, (X+v) is an integer.
Since X=u+f:
(u+f)+v=Integer
f+v=Integer−u
Since u is an integer, (f+v) must also be an integer. Because 0 < f < 1 and 0 < v < 1, the sum 0 < f + v < 2. The only integer value in this range is 1. Therefore, f+v=1.
Evaluating Statement I:(u+v+f)=u+(v+f)=u+1. Since u is an integer, (u+1) is an integer. Statement I is correct.
Evaluating Statement II:(f+v)=1, which is an integer. Statement II is correct.
