True False[1 Marks ]

Question 11 Mark

State True or False for the statements.
Every relation which is symmetric and transitive is also reflexive.

Answer

False.Solution:
Let R be a relation defined by,
R = {(1, 2), (2, 1), (1, 1), (2, 2)} on the set A = {1, 2, 3}
It is clear that $(3,3)\notin\text{R}.$ So, it is not reflexive.

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Question 21 Mark

State True or False for the statements.
An integer m is said to be related to another integer n if m is a integral multiple of n. This relation in Z is reflexive, symmetric and transitive.

Answer

False.Solution:
We know that if m is a integral multiple of n, then n can’t be the integral multiple of m.
Hence, the given relation is reflexive and transitive but not symmetric.

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Question 31 Mark

State True or False for the statements.
Every function is invertible.

Answer

False.Solution:
Only bijective functions are invertible.

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Question 41 Mark

State True or False for the statements.
The composition of functions is associative.

Answer

True.Solution:
Let us suppose f(x) = x, g(x) = x + 1 and h(x) = 2x - 1
Now, fo{goh(x)} = f[g{h(x)}]
= f{g(2x - 1)} $[\because$ h(x) = 2x - 1$]$
= f(2x - 1 + 1)
f(2x) = 2x
Also, (fog)oh(x) = (fog){h(x)}
= (fog)(2x - 1)
= f{g(2x - 1)}
= f(2x - 1 + 1)
= f(2x) = 2x

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Question 51 Mark

State True or False for the statements.
The relation R on the set A = {1, 2, 3} defined as R = {(1, 1), (1, 2), (2, 1), (3, 3)} is reflexive, symmetric and transitive.

Answer

False.
Solution:
Given that, R = {(1, 1), (1, 2), (2, 1), (3, 3)}
$(2,2)\notin\text{R}$
So, R is not reflexive.

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Question 61 Mark

State True or False for the statements.
Let A = {0, 1} and N be the set of natural numbers. Then the mapping f : N → A defined by $\text{f}(2\text{n}-1)=0,\ \text{f}(2\text{n})=1,\ \forall\ \text{n}\in\text{N},$ is onto.

Answer

True.
Solution:
Given, A = {0, 1}
$\text{f}(2\text{n}-1)=0,\ \text{f}(2\text{n})=1,\ \forall\ \text{n}\in\text{N},$
So, the mapping f : N → A is onto.

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Question 71 Mark

State True or False for the statements.
Let f : R → R be the function defined by $\text{f}(\text{x})=\sin(3\text{x}+2)\ \forall\ \text{x}\in\text{R}.$ Then f is invertible.

Answer

False.
Solution:
The first condition for any function to be invertible is that it should be one-one function.
The given function i.e., $\text{f}(\text{x})=\sin(3\text{x}+2)\ \forall\ \text{x}\in\text{R}$ is not one-one function for all $\text{x}\in\text{R}$
Hence, f is not invertible.

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Question 81 Mark

State True or False for the statements. The composition of functions is commutative.

Answer

Let $f(x) = x^2$
and $g(x) = x + 1$
$fog(x) = f(g(x))$
$= f(x + 1)$
$= (x + 1)^2$
$= x^2 + 2x + 1$
$gof(x) $
$= g(f(x))$
$= g(x^2)$
$ = x^2 + 1$
Thus, $\text{fog}(\text{x})\neq\text{gof}(\text{x})$

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Question 91 Mark

State True or False for the statements.
A binary operation on a set has always the identity element.

Answer

False.
Solution:
‘+’ is a binary operation on the set N but it has no identity element.

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Question 101 Mark

State True or False for the statements.
Let R = {(3, 1), (1, 3), (3, 3)} be a relation defined on the set A = {1, 2, 3}. Then R is symmetric, transitive but not reflexive.

Answer

False.
Solution:
We are given the relation R = {(3, 1), (1, 3), (3, 3)} which is defined on the set A = {1, 2, 3}
Since, $(1,1), (2,2)\notin\text{R}$
Hence, R is not reflexive.
Since, $(3,1), (1,3)\in\text{R}$
Hence, R is symmetric.
Since, $(3,1)\in\text{R}, (1,3)\in\text{R}$ but $(1,1)\notin\text{R}$
Hence, R is not transitive.

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MATHS True False[1 Marks ]

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