ABC26GN1962 · Surds and Indices
Subject: General Aptitude · Chapter: Surds and Indices · Exam: 2009 · Marks: · Difficulty:
If $\mathrm{N}=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}-\sqrt{3-2 \sqrt{2}}$, then the value of N is
(a)$2 \sqrt{2}-1$
(b)3
(c)1
(d)2
Answer
Explanation
Let $X=\frac{\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2}}{\sqrt{\sqrt{5}+1}}$. Then $X^{2}=\frac{(\sqrt{\sqrt{5}+2}+\sqrt{\sqrt{5}-2})^{2}}{(\sqrt{\sqrt{5}+1})^{2}}$ $$\begin{aligned} & =\frac{(\sqrt{5}+2)+(\sqrt{5}-2)+2 \sqrt{(\sqrt{5}+2)(\sqrt{5}-2)}}{(\sqrt{5}+1)} \\ & =\frac{2 \sqrt{5}+2 \sqrt{(\sqrt{5})^{2}-(2)^{2}}}{\sqrt{5}+1}=\frac{2 \sqrt{5}+2}{\sqrt{5}+1} \\ & =\frac{2(\sqrt{5}+1)}{(\sqrt{5}+1)}=2 \end{aligned} \begin{aligned} \Rightarrow & X & =\sqrt{2} . \\ \therefore & N & =\sqrt{2}-\sqrt{3-2 \sqrt{2}}=\sqrt{2}-\sqrt{(\sqrt{2})^{2}+1^{2}-2 \times \sqrt{2} \times 1} \\ & & =\sqrt{2}-\sqrt{(\sqrt{2}-1)^{2}}=\sqrt{2}-(\sqrt{2}-1)=1 . \end{aligned}$$
Explanation as extracted from the printed page; notation may be imperfect.
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