2014年8月22日星期五

C2090-546 A2040-911 A2090-731 dernières questions d'examen certification IBM et réponses publiés

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Code d'Examen: C2090-546
Nom d'Examen: IBM (DB2 9.7 Database Administrator for Linux UNIX or Windows Upgrade)
Questions et réponses: 78 Q&As

Code d'Examen: A2040-911
Nom d'Examen: IBM (Assessment: IBM WebSphere Portal 8.0 Solution Development)
Questions et réponses: 119 Q&As

Code d'Examen: A2090-731
Nom d'Examen: IBM (DB2 9 DBA for Linux,UNIX and Windows)
Questions et réponses: 138 Q&As

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NO.1 Index INDEX1 has been created as follows: CREATE INDEX index1 ON table_x (cola) Which of the
following actions can be done through an ALTER INDEX statement?
A. Add a column to the index key.
B. Change the PCTFREE specification.
C. Make the COMPRESS attribute YES or NO.
D. Make this index the clustering index.
Answer: C

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NO.2 Which of the following SQL statements will return monitoring information of tables in the DB2USER
schema that begin with the letter 'A'?
A. SELECT * FROM mon_get_table('DB2USER','A%',-2)
B. SELECT * FROM TABLE(mon_get_table('DB2USER','A%',-2))
C. SELECT * FROM mon_get_table('DB2USER',",-2) WHERE TABNAME LIKE 'A%'
D. SELECT * FROM TABLE(mon_get_table('DB2USER','',-2)) WHERE TABNAME LIKE 'A%'
Answer: D

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NO.3 Which of following is true when decomposing multiple XML documents?
A. It is possible to decompose multiple XML documents stored in a binary column or in an XML column.
B. It is not possible to decompose multiple XML documents stored in a binary column or in an XML
column.
C. It is possible to decompose multiple XML documents stored in a binary column but not in an XML
column.
D. It is possible to decompose multiple XML documents stored in a XML column but not in an binary
column.
Answer: A

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NO.4 Click on the Exhibit button
Given the following DDL statements:
If COL2 contains XML documents similar to the one shown in the scenario, what is the end result of the
CREATE INDEX statement?
A. An error will be returned because it is not possible to create an index for an XML column like COL2.
B. TBSP11 and TBSP12 will each contain one local index for XML_INDEX since the local indexes will
physically reside in the same table space as the related data.
C. An error will be returned because it is not possible to include more than one column as part of the
CREATE INDEX statement when one of the columns is of type XML.
D. A non-partitioned index will be created in table space TBSP21 since the INDEX IN clause of the table
definition is ignored and XML_INDEX will automatically be created within the first index partition listed in
the create table statement.
Answer: C

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NO.5 If table TAB_A has just been created as follows:
Which command will create a compression dictionary for table TAB_A?
A. DELETE FROM tab_a
B. REORG TABLE tab_a
C. UPDATE tab_a SET c1 = c1*1.1
D. LOAD FROM data.del OF DEL INSERT INTO tab_a
Answer: D

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NO.6 Which is true regarding collecting distribution statistics for XML data?
A. XML distribution statistics are collected for indexes over XML data of type VARCHAR.
B. XML distribution statistics are collected for indexes over XML data of type VARCHAR HASHED.
C. XML distribution statistics are collected when collecting index statistics during index creation.
D. XML distribution statistics are collected for partitioned indexes over XML data defined on a data
partitioned table.
Answer: A

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NO.7 When storing XML data in a DB2 database, which statement is valid.?
A. A table with an XML column can be defined in a non-Unicode database.
B. A table with an XML column can only be defined in a Unicode database.
C. A table with an XML column can be defined in a non-Unicode database but the table containing the
XML column will be stored in Unicode.
D. A table with an XML column can be defined in a non-Unicode database but the database must be
converted to Unicode before the XML columns can be used.
Answer: A

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NO.8 If table TAB_A is created as follows:
Assuming the cardinality of the columns is the same, which statement will create an index that will benefit
the most from compression?
A. CREATE INDEX taba_idx ON tab_a (c2)
B. CREATE INDEX taba_idx ON tab_a (c4)
C. ALTER TABLE tab_a ADD PRIMARY KEY (c1)
D. CREATE UNIQUE INDEX taba_idx ON tab_a (c2)
Answer: A

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