= Solution
When $a=0$, the first two <qubits> after the <Boolean quantum oracle> are $(-1)^d Z^b|+\rangle\otimes Z^c|+\rangle$. The <Hadamard gate> obeys $HZ^t|+\rangle=|t\rangle$ for $t\in\{0,1\}$. Applying the <Walsh-Hadamard transform> therefore gives
$$
(H\otimes H\otimes I)|\psi_{0bcd}\rangle=(-1)^d|bc\rangle\otimes|-\rangle.
$$
Consequently the <quantum measurement in the computational basis> has the deterministic outcomes
$$
\begin{array}{c|cccc}
\text{oracle}&f_{0001}&f_{0010}&f_{0100}&f_{0110}\\\hline
\text{outcome}&00&01&10&11
\end{array}
$$
These four distinct answers identify the four promised <Boolean quantum oracles> with \b[one query and certainty]. The constant term does not affect the <quantum measurement>, because a <global phase> does not change its <probabilities>.
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