A) \[2\]
B) \[\frac{1}{2}\]
C) \[1\]
D) \[\frac{1}{3}\]
Correct Answer: C
Solution :
| [c] \[SA=\sqrt{{{({{a}^{2}}-1)}^{2}}+{{(2a)}^{2}}}\] |
| \[=\sqrt{{{a}^{4}}+1-2{{a}^{2}}+4{{a}^{2}}}=\sqrt{{{({{a}^{2}}+1)}^{2}}}={{a}^{2}}+1\] |
| \[SB=\sqrt{{{\left( \frac{1}{{{a}^{2}}}-1 \right)}^{2}}+{{\left( \frac{-2}{a}-0 \right)}^{2}}}\] |
| \[=\sqrt{\frac{1}{{{a}^{4}}}+1-\frac{2}{{{a}^{2}}}+\frac{4}{{{a}^{2}}}}=\sqrt{{{\left( \frac{1}{{{a}^{2}}}+1 \right)}^{2}}}\] |
| \[=\frac{1}{{{a}^{2}}}+1=\frac{1+{{a}^{2}}}{{{a}^{2}}}\] |
| \[\therefore \,\,\,\,\,\frac{1}{SA}+\frac{1}{SB}=\frac{1}{{{a}^{2}}+1}+\frac{1}{\frac{1+{{a}^{2}}}{{{a}^{2}}}}=1\] |
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