| If \[a=\frac{\sqrt{3}+1}{\sqrt{3}-1}\]and \[b=\frac{\sqrt{3}-1}{\sqrt{3}+1},\]the value of \[\left( \frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}-ab+{{b}^{2}}} \right)\]is |
A) \[\frac{15}{13}\]
B) \[\frac{16}{13}\]
C) \[\frac{11}{13}\]
D) \[\frac{12}{13}\]
Correct Answer: A
Solution :
| \[a=\frac{(\sqrt{3}+1)}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}=\frac{{{(\sqrt{3}+1)}^{2}}}{{{(\sqrt{3})}^{2}}-{{(1)}^{2}}}\] |
| [by rationalisation] |
| \[=\frac{3+1+2\sqrt{3}}{3-1}=\frac{4+2\sqrt{3}}{2}=2+\sqrt{3}\] |
| \[{{a}^{2}}={{(2+\sqrt{3})}^{2}}=4+3+4\sqrt{3}=(7+4\sqrt{3})\] |
| \[{{b}^{2}}={{(2-\sqrt{3})}^{2}}=4+3-4\sqrt{3}=(7-4\sqrt{3})\] |
| \[ab=\frac{(\sqrt{3}+1)}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}-1)}{(\sqrt{3}+1)}=1\] |
| Now, \[\frac{{{a}^{2}}+ab+{{b}^{2}}}{{{a}^{2}}-ab+{{b}^{2}}}=\frac{{{a}^{2}}+{{b}^{2}}+ab}{{{a}^{2}}+{{b}^{2}}-ab}\] |
| \[=\frac{(7+4\sqrt{3})+(7-4\sqrt{3})+1}{(7+4\sqrt{3})+(7-4\sqrt{3})-1}=\frac{15}{13}\] |
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