# GATE Solved Paper 2017-19 - GATE 2017 Shift 1

>>>>>>>>GATE 2017 Shift 1

• A

abab  • B

aaab  • C

abbaa  • D

babba  • Option : D
• Explanation :
1) S → SS → aSbS → abS → abaSb → abab
2) S → aSb → aSaSb → aaaSb → aaab
3) S → SS → aSbS → abS → abaSb → abab → abbSaSa → abbaa
4) can't not drive

• A

is 0  • B

is -1  • C

is 1  • D

does not exist  • Option : C
• Explanation : • A

(I) and (II) only  • B

(I) and (III) only  • C

(II) and (III) only  • D

(I), (II) and (III)  • Option : D
• Explanation : Given, DepId can be permitted to be NULL
I. {t | ∃ u ∈ EMP (t[EMPName] = u[EmpName] ∧ ∀ v ∈ DEPT (t[DeptId] ≠ DeptId]))} : Gives empnames who donot belong to any department
II. {t | ∃ u ∈ EMP (t[EMPName] = u[EmpName] ∧ ∃ v ∈ DEPT (t[DeptId] ≠ DeptId]))} : Gives empnames who donot belong to some department
III. {t | ∃ u ∈ EMP (t[EMPName] = u[EmpName] ∧ ∃ v ∈ DEPT (t[DeptId] = DeptId]))}: Gives empnames who donot belong to same department
All of these queries are giving some results which are finite and thus all are safe expressions.
Therefore, option D is correct.

• A

S1 is true, S2 is true  • B

S1 is true, S2 is false  • C

S1 is false, S2 is true  • D

S1 is false, S2 is false  • Option : B
• Explanation :
Statement 1 is “TRUE”. Because there can be a case when page selected to be replaced is by FIFO policy.
Statement 2 is “FALSE”. Because LRU page replacement algorithm does not suffers from Belady’s Anomaly. Only FIFO page replacement algorithm suffers from Belady’s Anomaly.

• A

12  • B

23  • C

31  • D

44  • Option : B
• Explanation : Count in the function total is static.  i Count Table i 5 0 2 4 2 3(2+1) 3 3 5(3+2) 2 5 6(5+1) 1 6 7(6+1) = 23
Related Quiz.
GATE 2017 Shift 1