PREVIOUS YEAR SOLVED PAPERS - GATE 2017 Shift 1
- 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
- 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.
- 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.
- 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
