True False[1 Marks ]

Question 11 Mark

Let sets R and T be defined as
R = {x $\in$ Z | x is divisible by 2}
T = {x $\in$ Z | x is divisible by 6}. Then $\text{T} \subset \text{R}.$

Answer

True'.
Solution:
We can written the given sets is Roster from
R = {....., -8, -6, -4, -2, 0, 2, 4, 6, 8, .....}
And T = {....., -18, -12, -6, 0, 6, 12, 18, .....}
Since every element of T is present in R. so, $\text{T}\subset\text{R}.$
Hence, the statement is 'True'.

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Question 21 Mark

If A is any set, then $\text{A} \subset \text{A}.$

Answer

True'.
Solution:
Since every set is a subset of it self. So, it is 'True'.

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Question 31 Mark

Given A = {0, 1, 2}, $\text{B} = {\text{x} \in \text{R} | 0 \leq \text{x} \leq 2}.$ Then A = B

Answer

False'.
Solution:
Here A = {0, 1, 2}, B is a set having all real numbers from 0 to 2. So $\text{A}\neq\text{B}.$
Hence, the given statement is 'False'.

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Question 41 Mark

$\text{Q} \cup \text{Z} = \text{Q},$ where Q is the set of rational numbers and Z is the set of integers.

Answer

True'.
Solution:
True, since every integer is a rational number.
$\therefore \text{Z}\subset\text{Q},$ so $\text{Q}\cup\text{Z}=\text{Q}.$

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Question 51 Mark

Given that M = {1, 2, 3, 4, 5, 6, 7, 8, 9} and if B = {1, 2, 3, 4, 5, 6, 7, 8, 9}, then $\text{B} \not\subset \text{M}.$

Answer

False'.
Solution:
M = {1, 2, 3, 4, 5, 6, 7, 8, 9}
And B = {1, 2, 3, 4, 5, 6, 7, 8, 9}
Since M and B ahs same elements
$\therefore \text{M}=\text{B},$ so $\text{B}\not\subset \text{M}$ is 'False'.

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Question 61 Mark

The sets {1, 2, 3, 4} and {3, 4, 5, 6} are equal.

Answer

False'.
Solution:
False, since the two sets do not contain the same elements.

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MATHS True False[1 Marks ]

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