Q200 / 270 · NDA 2026 (Phase 1) · Mathematics
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 $\left(\sqrt{2}-1\right)^{10}=v$.
Question: What is the value of $uv$ ?
Question: What is the value of $uv$ ?
- A$1$
- B$2$
- Correct answerC$0<uv<1$
- D$1<uv<2$
Solution
Given $\left(\sqrt{2}+1\right)^{10} = u+f$ and $\left(\sqrt{2}-1\right)^{10} = v$.
Multiplying these two equations, we get:
$(u+f)v = \left(\sqrt{2}+1\right)^{10}\left(\sqrt{2}-1\right)^{10}$
$(u+f)v = \left((\sqrt{2})^2 - 1^2\right)^{10} = 1^{10} = 1$
$uv + fv = 1 \Rightarrow uv = 1 - fv$
Since $f$ is the fractional part, $0 < f < 1$.
Also, $v = \left(\sqrt{2}-1\right)^{10}$, and since $0 < \sqrt{2}-1 < 1$, we have $0 < v < 1$.
Because both $f$ and $v$ lie strictly between $0$ and $1$, their product $fv$ also lies strictly between $0$ and $1$, i.e., $0 < fv < 1$.
Therefore, $uv = 1 - fv$ must satisfy:
$0 < uv < 1$
Answer: $0<uv<1$
Multiplying these two equations, we get:
$(u+f)v = \left(\sqrt{2}+1\right)^{10}\left(\sqrt{2}-1\right)^{10}$
$(u+f)v = \left((\sqrt{2})^2 - 1^2\right)^{10} = 1^{10} = 1$
$uv + fv = 1 \Rightarrow uv = 1 - fv$
Since $f$ is the fractional part, $0 < f < 1$.
Also, $v = \left(\sqrt{2}-1\right)^{10}$, and since $0 < \sqrt{2}-1 < 1$, we have $0 < v < 1$.
Because both $f$ and $v$ lie strictly between $0$ and $1$, their product $fv$ also lies strictly between $0$ and $1$, i.e., $0 < fv < 1$.
Therefore, $uv = 1 - fv$ must satisfy:
$0 < uv < 1$
Answer: $0<uv<1$
Hence, the correct answer is option: 'C'.