MCQ
If $\sqrt{2}=1.4142$, then $\sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}}$ is equal to
- ✓2.4142
- B5.8282
- C0.4142
- D0.1718
✓
Answer
Correct option: A.
2.4142
(a)
We find that $\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{\sqrt{2}+1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{(\sqrt{2}+1)^2}{2-1}=(\sqrt{2}+1)^2$
$\therefore \quad \sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}}=\sqrt{2}+1=1.4142+1=2.4142$
We find that $\frac{\sqrt{2}+1}{\sqrt{2}-1}=\frac{\sqrt{2}+1}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1}=\frac{(\sqrt{2}+1)^2}{2-1}=(\sqrt{2}+1)^2$
$\therefore \quad \sqrt{\frac{\sqrt{2}+1}{\sqrt{2}-1}}=\sqrt{2}+1=1.4142+1=2.4142$
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