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Yes - put the control bit in the state $H\lvert b\rangle$, with $H$ the Hadamard transform. Then, the controlled-$U_f$ gate will transform $(H\lvert b\rangle)\otimes\lvert i\rangle$ into $(H\lvert b\oplus f(i)\rangle)\otimes\lvert i\rangle$, as can be easily seen by checking the cases $b=0$ and $b=1$ separately. Thus, the desired action is obtained by ...

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