PREVIOUS YEAR SOLVED PAPERS MCQ - GATE 2017 Shift 1
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
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 falseThe 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
‘X’ is Gaussian random variable
⇒ X ∼ N(0, σ2) for -∞ < x < ∞
Given
Since median is positional average
Therefore, median of Y is '0'.
Since median is positional average
Therefore, median of Y is '0'.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




