For the given logic circuit, which of the following inputs combination will make both LED\(-1\) and LED\(-2\) to glow?

Let the output of the OR gate be \(Y_1\).

\(Y_1 = A + B\)

For LED-1 to glow, \(Y_1\) must be \(1\), which means \(A + B = 1\).

Let the output of the first AND gate be \(Y_2\).

\(Y_2 = Y_1 \cdot C = (A + B) \cdot C\)

Let the output of the second AND gate be \(Y_3\).

\(Y_3 = A \cdot Y_2 = A \cdot (A + B) \cdot C\)

For LED-2 to glow, \(Y_3\) must be \(1\).

Simplifying the Boolean expression for \(Y_3\):

\(Y_3 = (A \cdot A + A \cdot B) \cdot C\)

\(Y_3 = (A + AB) \cdot C\)

\(Y_3 = A(1 + B) \cdot C\)

\(Y_3 = A \cdot C\)

For \(Y_3 = 1\), we require \(A = 1\) and \(C = 1\).

If \(A = 1\), then \(Y_1 = 1 + B = 1\) for any value of \(B\), satisfying the condition for LED-1 to glow.

From the given options, the combination \(A = 1, B = 0, C = 1\) satisfies these conditions.

Answer: \(A = 1, B = 0, C = 1\)