# Aptitude | GATE IT 2006 | Question 7

• Difficulty Level : Medium
• Last Updated : 28 Jun, 2021

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.

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