JEE Main 2026 — Semiconductors Question with Solution

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: $A=1,B=0,C=1$