A) 0
B) 1
C) 2
D) \[~-1\]
Correct Answer: B
Solution :
\[\frac{1}{1+{{4}^{a-b}}}+\frac{1}{1+{{2}^{2b-2a}}}\] \[=\,\,\,\,\frac{1}{1+{{4}^{(a-b)}}}+\frac{1}{1+{{4}^{(b-a)}}}\] \[=\,\,\,\frac{1}{1+\frac{{{4}^{a}}}{{{4}^{b}}}}+\frac{1}{1+\frac{{{4}^{b}}}{{{4}^{a}}}}\] \[=\,\,\,\frac{{{4}^{b}}}{{{4}^{a}}+{{4}^{b}}}+\frac{{{4}^{a}}}{{{4}^{a}}+{{4}^{b}}}=\underline{\mathbf{1}}\]You need to login to perform this action.
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