Question

Let $f:\left[1,\infty \right)\to \left[2,\infty \right)$ be defined by $f\left(x\right)=x+\frac{1}{x}$ . If $f^{-1}\left(\sqrt{5}\right)=n_1$ , then $2\left(\frac{n_1}{2}-\sin {18}^0\right)=$

• A. 1
• B. 2
• C. 3
• D. 4

Answer 1

Answer: (A)

1

$Given:f:[1,\infty )\to [2,\infty )f\left(x\right)=x+\frac{1}{2}{f}^{-1}\left(\sqrt{5}\right)=n12(\frac{n1}{2}-sin{18}^{\circ })=?\because f^{-1}\left(\sqrt{5}\right)=n1f\left(n1\right)=\sqrt{5}\Rightarrow n1+\frac{1}{n1}=\sqrt{5}\Rightarrow {n1}^2-\sqrt{5}n1+1=0n1=\frac{\sqrt{5}\pm \sqrt{5-4}}{2}=\frac{\sqrt{5}\pm 1}{2}=\frac{\sqrt{5}+1}{2}\therefore 2(\frac{n1}{2}-sin{18}^{\circ })=2(\frac{\sqrt{5}+1}{4}-\frac{\sqrt{5}-1}{4})=\frac{\sqrt{5}}{2}+\frac{1}{2}-\frac{\sqrt{5}}{2}+\frac{1}{2}=\frac{1}{2}+\frac{1}{2}=1$