A) \[\frac{3}{4}\]
B) \[\frac{4}{3}\]
C) \[\frac{3}{5}\]
D) \[\frac{5}{3}\]
Correct Answer: B
Solution :
\[a=\frac{\sqrt{5}+1}{\sqrt{5}-1}=\frac{{{(\sqrt{5}-1)}^{2}}}{4}=\frac{6+2\sqrt{5}}{4}=\frac{3+\sqrt{5}}{2}\] \[b=\frac{\sqrt{5}-1}{\sqrt{5}+1}=\frac{{{(\sqrt{5}-1)}^{2}}}{4}=\frac{6-2\sqrt{5}}{4}=\frac{3-\sqrt{5}}{2}\] \[{{a}^{2}}={{\left( 3+\frac{\sqrt{5}}{2} \right)}^{2}}=\frac{14+6\sqrt{5}}{4}\] \[{{b}^{2}}={{\left( \frac{3-\sqrt{5}}{2} \right)}^{2}}=\frac{14-6\sqrt{5}}{4}\] \[ab=\left( \frac{\sqrt{5}+1}{\sqrt{5}-1} \right)\,\left( \frac{\sqrt{5}-1}{\sqrt{5}+1} \right)=\frac{4}{4}=1\] \[\therefore \] \[\frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}-ab+{{b}^{2}}}=\frac{32}{24}=\frac{4}{3}\]
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