Q197 / 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 multiplicative inverse of $\left(\sqrt{2}+1\right)^{20}$ ?
Question: What is the multiplicative inverse of $\left(\sqrt{2}+1\right)^{20}$ ?
- A$v$
- B$v^2-1$
- Correct answerC$v^2$
- D$v^2+1$
Solution
The multiplicative inverse of $\left(\sqrt{2}+1\right)^{20}$ is $\dfrac{1}{\left(\sqrt{2}+1\right)^{20}}$.
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 $v = \left(\sqrt{2}-1\right)^{10}$.
Squaring both sides yields $v^2 = \left(\sqrt{2}-1\right)^{20}$.
Therefore, the multiplicative inverse of $\left(\sqrt{2}+1\right)^{20}$ is $v^2$.
Answer: $v^2$
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 $v = \left(\sqrt{2}-1\right)^{10}$.
Squaring both sides yields $v^2 = \left(\sqrt{2}-1\right)^{20}$.
Therefore, the multiplicative inverse of $\left(\sqrt{2}+1\right)^{20}$ is $v^2$.
Answer: $v^2$
Hence, the correct answer is option: 'C'.