A) 1
B) 48
C) \[-48\]
D) \[-1\]
Correct Answer: B
Solution :
Since, \[f(x)=\frac{{{4}^{x}}}{{{4}^{x}}+2}\] \[\therefore \] \[f(1-x)=\frac{{{4}^{1-x}}}{{{4}^{1-x}}+2}\] \[=\frac{4}{4+{{2.4}^{x}}}=\frac{2}{2+{{4}^{x}}}\] \[\Rightarrow \] \[f(x)+f(1-x)=1\] On putting\[x=\frac{1}{97},\frac{2}{97},.....,\frac{48}{97},\]we get \[f\left( \frac{1}{97} \right)+f\left( \frac{2}{97} \right)+....+f\left( \frac{96}{97} \right)=48\]You need to login to perform this action.
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