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Consider the circuit shown below. The output of a $2:1$ MUX is given by the function $(ac' + bc)$.

Which of the following is true?

1. $f=X_1'+X_2$
2. $f=X_1'X_2+X_1X_2'$
3. $f=X_1X_2+X_1'X_2'$
4. $f=X_1+X_2'$

g = (a and x1′) or (b and x1)
g = (1 and x1’) or (0 and x1)
g = x1’

f = ac’ + bc
f = (a and x2′) or (b and x2)
f = (g and x2′) or (x1 and x2)
f = x1’x2’ + x1x2

Ref: Geeksforgeeks.org
the diagram doesn't match with GO 2020 soft copy
this question diagram is still wrong in Go pdf 2020.please correct in the next issue

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$g = X_1'$
So, $f = ac' + bc$

$= X_1'X_2' + X_1X_2$

So, (C).
by

What is the significance of ac'+bc here?

We don't need it to get the answer. So why it is mentioned in the question?
here , f = ac'+bc where for first mux c = x1 and for second mux c = x2.

Now substitute for g = ac' + bc => a x1' + b x , now for f , f= ac'+bc => g x2' + x1 x2 => x1' x2' + x1 x2.
Why anyone is not noticing that a should be equal to 1 and b should be equal to 0 in the question
This is so weird. The solution is different from what I'm getting. I think the diagram is wrong.
Image was wrong. Fixed now.

Out of the two inputs how come we assume a is I0(Input 0) and b is I1 (Input 1). Unless we assume this is there anyway to get into the answer?

The image is wrong in GO PDF.
image is still wrong in Hard copy of GO PYQ, (page no – 471), Digital logic part.
Given output of the mux is $a{c}' + bc$

For the first mux, $g$ is the output and $a=0,b=1,c = x_{1}$

Hence,

$g=1.{x_{1}}'+0.x_{1} = {x_{1}}'$  $........(1)$

For the second mux,

$a=g={x_{1}}' [from (1)],b=x_{1},c=x_{2}$

$f={x_{1}}'.{x_{2}}'+x_{1}.x_{2}$

Hence, option (c).

$\underline{\textbf{Answer:}\Rightarrow}$

$\underline{\textbf{Explanation:}\Rightarrow}$

A multiplexer has $\mathbf 3$ things:

1. Selection Line.
2. Input Line.
3. Output Lines.
4. For $\mathbf{2:1}$ Multiplexer, there are $\mathbf 2$ inputs and $\mathbf 1$ output and $\mathbf 1$ selection line.

So, when Selection Line, $\mathbf {X_1}$ is set $\mathbf 0$, then $\mathbf {a}$ will be selected and when $\mathbf {X_1 = 1}$, then $\mathbf b$ will be selected.

$\therefore$ Equation will be: $\mathbf{g = 1.X_1' + 0.X_1 = X_1'}$

Now, this $\mathbf{g = X_1}$ will be given as input to second multiplexer.

Following the same procedure for second multiplexer and taking one input as $\mathbf{X_1'}$, we will get:

$\mathrm{f = X_2'X_1' + X_2X_1}$

$\therefore \mathbf C$ is the correct answer.

by

edited by
A silly doubt: How do we consider multiplying the select line inputs like, why X1’ *1 and not X1’*0?

As input is given as 0 in a and 1 in b. Corresponding to that the select line should be taken, right? Please help me with this doubt.

Just I need to know that why didn’t you multiplied X1’ with 0. As input is zero so select line has to be complemented, please correct me if I’m wrong.

same doubt!!

$f=ac'+bc$

Now, solve without considering $X1,X2,1,0$ and try to get function $f.$

Output of $mux_{1}:$ $g=c'a+cb$ and

Output of $mux_{2}:$ $c'g+cb\Rightarrow c'(c'a+cb)+cb=c'a+cb=$ $ac'+bc$

With this we are clear that with $X1'$ we have to consider $a's$ input and with $X1$ we have to consider $b's$ input.

So,

Output of $mux_{1}:$ $g=X1'.1+X1.0=X1'$

Output of $mux_{2}:$ $X2'g+X2.X1=X2'(X1')+X2.X1=$ $X1X2+X1'X2'$

### 1 comment

which among a and b is msb/lsb? how to know when I0/I1 is not given...

g = (X1' * 1) + (X1 * 0) =  X1'

a = g from 2nd MUX

f = (ac' + bc) = (X2' * a)+ ( X2 * b)

X2' = c' , b= X1 (given)

so, f = (X1' * X2')+(X1*X2)