Q16,p3,j15. which of the following is no the basic primitive of the graphics kernel system.

Options.
A) polyline.       B)polydraw.              c) Fillarea              D)Polymarker.

Answer B.

Explanation.
Basic primitive of GKS are Polyline,polymarker, fillarea,text.

Q15,p3,j15. The process of dividing an analog signal into a string of discrete outputs,each of constant amplitude is called.

Options

A)     Strobing               B)Amplification                 C)Conditioning                  D)Quantization.

Answer D.

Q13,p3,j15. Give the number of principal vanishing point along with their direction for the standard perspective transformation.

A)     Only one in direction K.

B)      Two in direction I,J.

C)      Three in direction I,J,k.

D)     Only two in direction j,K

Answer A.

Explanation.

As things get further away, from us, they seems smaller and closer together. When they get far enough away, distances become ever tinier and so form a  single point.

Q12,p3,j15. In the indexed scheme of blocks to a file, the maximum possible size of the file depends on.

Options

A)     The number of blocks used for index and size of index.

B)      Size of blocks and size of address.

C)      Size of index.

D)     Size of blocks.

Answer B.

Q11,p3,j15. The relation schema R1 and R2 form a lossless join decomposition of R if and only if.

1. R1 intersection R2 -->>(R1-R2)
2. R1->R2.
3. R1 intersection  R2 ->>(R2-R1).
4.R2->R1 intersection R2.

codes
A) 1 and 2 happens.   B)1 and 4 happens.    C) 1 and 3 happens     D) 2 and 3 happens.

Answer C.

Explanation.
for lossless decomposition .Atleast one of the following dependencies should be in F+.

R1 intersction R2->R1.                     Eq. 1.
or
R1 intersection R2->R2.                    Eq2.

when a decompostion satisfies  eq 1. given above then that will solve the  R1 intersection R2 -->>(R1-R2).
when a  decomposition satisfies eq2 given above then that will solve the  R1 intersection R2 -->>(R2-R1).

Q10,P3,J15. The Relation

vendor order(v_no,v_ord_no,v_name,qty_sup,unit_price) is in 2nf because:
options
A) Non key attribute v_name  is dependent on v_no ,which is part of composite key.
B)Non key attribute v_name is dependent on qty_sup.
C)key attribute qty_sup is dependent on primary key unit_price.
D)Key attribute v_ord_no is dependent on primary key unit_price.
Answer A.

Q9,p3,J15.select the False statement from the following statements about Normal forms.

A) Loss less  preserving decomposition in to 3nf is always possible.
B)Loss less preserving decomposition in to BCNF is always possible.
C)Any relation with two attributes is in BCNF.
D)BCNF is stronger than 3NF.
Answer B.
explanation.
Table 1(A,B,C) with FD  AB->C and C->B  cannot be preserved  when decomposition is done.