# Aptitude | GATE IT 2006 | Question 7

Given a boolean function f (x1, x2, …, xn), which of the following equations is NOT true

(A) f (x1, x2, …, xn) = x1’f(x1, x2, …, xn) + x1f(x1, x2, …, xn)
(B) f (x1, x2, …, xn) = x2f(x1, x2, …, xn) + x2’f(x1, x2, …,xn)
(C) f (x1, x2, …, xn) = xn’f(x1, x2, …, 0) + xnf(x1, x2, …,1)
(D) f (x1, x2, …, xn) = f(0, x2, …, xn) + f(1, x2, .., xn)

Explanation: Option A: f (x1, x2, …, xn) = x1’f(x1, x2, …, xn) + x1f(x1, x2, …, xn)
Case 1: taking x1=0
RHS = 1.f(x1, x2, …, xn) + 0.f(x1, x2, …, xn)
RHS =f(x1, x2, …, xn).

Case 2: taking x1=1
RHS = 0.f(x1, x2, …, xn) + 1.f(x1, x2, …, xn)
RHS =f(x1, x2, …, xn).
In both cases RHS=LHS, so, (A) is true

Option B: f (x1, x2, …, xn) = x2f(x1, x2, …, xn) + x2’f(x1, x2, …, xn)
Case 1: taking x2=0
RHS= 0.f(x1, x2, …, xn) + 1.f(x1, x2…,xn)
RHS =f(x1, x2, …, xn).

Case 2: taking x2=1
RHS = 1.f(x1, x2, …, xn) + 0.f(x1, x2, …, xn)
RHS =f(x1, x2, …, xn).

In both cases RHS=LHS, so, (B) is true.

Option C: f (x1, x2, …, xn) = xn’f(x1, x2, …, 0) + xnf(x1, x2, …,1)
Case 1: taking xn=0
RHS= 1.f(x1, x2, …, 0) + 0.f(x1, x2, …, 1)
RHS =f(x1, x2, …, 0)

Case 2: taking xn=1
RHS = 0.f(x1, x2, …, 0) + 1.f(x1, x2, …, 1)
RHS =f(x1, x2, …, 1)In both cases RHS=LHS, so, (C) is true.

Option D: f (x1, x2, …, xn) = f(0, x2, …, xn) + f(1, x2, .., xn)
Here, no way to equate LHS and RHS so ‘NOT true’. NO term depends on value of ‘x1’.

This solution is contributed by Sandeep pandey.

Quiz of this Question

My Personal Notes arrow_drop_up
Article Tags :

1

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.