Which of the following input binary conditions will produce an ambiguous state in the flip flop?

Question

For ambiguous state, both outputs x and y must be zero(for active low) to be considered invalid i.e. of both PRESET and CLR are logic "0". Now, analysing the circuit, $$ x=a \oplus D_0 $$ and $$ y= D_2 \oplus MSB $$ that means either

$$a=D_0=0 \\ or \\ a=D_0=1 $$ and $$ D_2=MSB=0 \\ or \\ D_2=MSB=1 $$

if $$a=0, b=0, and \ c=1$$ then LSB = 0 and MSB = 1 implies $$D_1 = 1$$ and $$D_0,D_2,D_3=0$$ but since $$MSB = 1$$ output of XOR gate will give Y=1.

Why is the correct answer a=0,b=0,c=1 ?

Answer 1

Answer D) is correct: a=0, b=0, c=1.

You correctly determined that LSB = 0 and MSB = 1, but this means that D2=1. You incorrectly concluded that D1=1.

The input of the decoder is decimal 2 (binary 10), which means the D2 output is set.

Answer 2

Procedure is correct, $$ LSB = 0 $$ and $$ MSB = 1$$ hence, $$ D_2 = 1 $$ and $$ X=a \oplus D_0=0$$ $$ Y=D_2 \oplus MSB =0$$

I just misread MSB and LSB. Thanks anyways!