Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

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)


Answer: (D)

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).

Attention reader! Don’t stop learning now.  Practice GATE exam well before the actual exam with the subject-wise and overall quizzes available in GATE Test Series Course.

Learn all GATE CS concepts with Free Live Classes on our youtube channel.

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
Recommended Articles
Page :

Start Your Coding Journey Now!