A) 0
B) \[\pm \]1
C) \[\pm \]b
D) \[\pm \]\[\frac{1}{b}\]
Correct Answer: A
Solution :
[b] Given expression is \[{{b}^{2a+2}}+{{b}^{2a}}={{b}^{3a+1}}+{{b}^{a+1}}\] \[\Rightarrow \,{{b}^{2a+2}}-{{b}^{a+1}}={{b}^{3a+1}}-{{b}^{2}}a\] \[\Rightarrow \,{{b}^{a+1}}({{b}^{a+1}}-1)={{b}^{2a}}({{b}^{a+1}}-1)\] \[\Rightarrow \,({{b}^{a+1}}-1)({{b}^{a+1}}-{{b}^{2a}})=0\] If \[{{b}^{a+1}}-{{b}^{2a}}=0\Rightarrow a+1=2a\,\,or\,\,a=1\] Thus \[a=\underline{+}\,1.\] So the answer is [b].
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