For option $A$
$$f(x_1,x_2,…,x_n)=x_1’f(x_1,x_2,…,x_n)+x_1f(x_1,x_2,…,x_n) =(x_1’+x_1)f(x_1,x_2,…,x_n)=1.f(x_1,x_2,…,x_n)=f(x_1,x_2,…,x_n) $$
So option $A$ is correct.
Similarly for option $B$
$$f(x_1,x_2,…,x_n)=x_2’f(x_1,x_2,…,x_n)+x_2f(x_1,x_2,…,x_n) =(x_2’+x_2)f(x_1,x_2,…,x_n)=1.f(x_1,x_2,…,x_n)=f(x_1,x_2,…,x_n) $$
So option $B$ is correct.
For option $C$,
Now the given function, we can have two possibilities: either $x_n$ is true or $x_n$ is false.
(1) when $x_n$ is false, the function becomes $f(x_1,x_2,…,0)$ and this situation as a whole is represented conditionally as : $x_n’f(x_1,x_2,…,0)$
(2) when $x_n$ is true, the function becomes $f(x_1,x_2,…,1)$ and this situation as a whole is represented conditionally as : $x_nf(x_1,x_2,…,0)$
Since our function is a logical “or” of the above two situations, so we have:
$$f(x_1,x_2,…,x_n)=x_n’f(x_1,x_2,…,0)+x_nf(x_1,x_2,…,1)$$
So option $C$ is true.
From this we can say that option $D$ is the wrong option. But why is it so?
This so because if you think logically the function is given as:
$$f(x_1,x_2,…,x_n)=f(0,x_2,…,x_n)+f(1,x_2,…,x_n)$$
Whatever function we consider, the function given on the RHS is independent of $x_1$ which is a problem.