PREVIOUS YEAR SOLVED PAPERS - GATE 2017 Shift 1
- Option : B
- Explanation : ρ(A) = n|A| = 0 ⇒ one eigen value must be ‘0’ ∈ [-5, 5]
∴ (I) is true
but eigen values of A are 0, −5,5
∴ The eigen value with the largest magnitude is not greater than 5
For and Let 
⇒ eigen values = 5,5
∴ One eigen value must be in [−5,5] and largest eigen value magnitude is not greater than 5
∴ (II) is false
- Option : B
- Explanation : The Context free grammar given over alphabets ∑{a, b, c} with S and T as non terminals are: G1: S → aSb|T,T → cT|∈ G2: S → bSa|T,T → cT|∈ Lets L(G1) is the language for grammar G1 and L(G2) is the language for grammar G2 L(G1) = {ancmbn|n,m ≥ 0} L(G2) = {bncman|n,m ≥ 0} L(G1) ∩ L(G1) = {cman|m ≥ 0}; which is infinite and regular
- Option : A
- Explanation : ‘X’ is Gaussian random variable
⇒ X ∼ N(0, σ2) for -∞ < x < ∞
Given
Since median is positional average
Therefore, median of Y is '0'.
- Option : B
- Explanation : From K-map simplification we get the min-term as CA' . So We can simplify it for NOR gate expression i.e. C' NOR A = (C'+A)' = CA' Now complemented inputs are also given to us so for 2 input NOR gate we need only 1 NOR gate. 1 is correct answer
