Q46 / 120 · NDA 2024 (Phase 2) · Mathematics
Paragraph:
Let $(8+3 \sqrt{7})^{20}=\mathrm{U}+\mathrm{V}$ and $(8-3 \sqrt{7})^{20}=\mathrm{W}$, where U is an integer and $0 < \mathrm{V} < 1$.
Question:
What is the value of $(\mathrm{U}+\mathrm{V}) \mathrm{W}$ ?
Question: What is the value of $(\mathrm{U}+\mathrm{V}) \mathrm{W}$ ?
- A$\frac{1}{2}$
- Correct answerB$1$
- C$\frac{3}{2}$
- D$2$
Solution
We have, $(8+3 \sqrt{7})^{20}=\mathrm{U}+\mathrm{V}$....(i)
and $(8-3 \sqrt{7})^{20}=\mathrm{W}$....(ii)
Here, $\mathrm{o} < \mathrm{W} < 1$
Now, adding equation (i) \& (ii), we have,
$\begin{aligned}
& \mathrm{U}+\mathrm{V}+\mathrm{W}=(8+3 \sqrt{7})^{20}+(8-3 \sqrt{7})^{20} \\
& =2\left[{ }^{20} \mathrm{C}_0 \cdot 8^{20}+{ }^{20} \mathrm{C}_2 8^{18} \cdot(3 \sqrt{7})^2+\ldots .+(3 \sqrt{7})^{20}\right]
\end{aligned}$
R.H.S. is an even number.
Also, $\mathrm{o} < \mathrm{V} < 1, \mathrm{o} < \mathrm{W} < 1$ and U is an integer
$\Rightarrow \mathrm{V}+\mathrm{W}$ is an integer $\Rightarrow \mathrm{V}+\mathrm{W}=1$
$\begin{aligned} & (U+V) W=(8+3 \sqrt{7})^{20}(8-3 \sqrt{7})^{20} \\ = & {[(8+3 \sqrt{7})(8-3 \sqrt{7})]^{20} } \\ = & (64-63)^{20}=1^{20}=1\end{aligned}$
Hence, the correct answer is option: 'B'.