A) 5
B) 4
C) 3
D) 2
Correct Answer: D
Solution :
[d] ?a? is a natural number. \[\therefore \] \[{{a}^{2}}+\frac{1}{{{a}^{2}}}={{a}^{2}}+\frac{1}{{{a}^{2}}}-2+2\] \[=\,\,{{a}^{2}}+\frac{1}{{{a}^{2}}}-2.a.\frac{1}{a}+2\] \[{{a}^{2}}+\frac{1}{{{a}^{2}}}={{\left( a-\frac{1}{a} \right)}^{2}}+2\] Now, \[{{\left( a-\frac{1}{a} \right)}^{2}}\] is always greater than or equal to zero. \[\therefore \] \[{{a}^{2}}+\frac{1}{{{a}^{2}}}\ge 2\]You need to login to perform this action.
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