1 Marks Question

Question 11 Mark

$f(x)=2 x, f: N \rightarrow N$, show that $f(x)$ is not onto.

Answer

$f$ is not onto. For $1 \in N$ (co-domain) there is no existence of any $x$ in domain N became $f(x)=2 x$.

Question 21 Mark

By $f(x)=2 x,$ defined a function $f : A \rightarrow B$ is one $-$ one and onto both. If $A =\{1,2,3,4\}$ then find set $B$.

Answer

$ f(x)=2 x$
$f(1)=2 \times 1=2, $
$f(2)=2 \times 2=4,$
$f(3)=2 \times 3=6,$
$f(4)=2 \times 4=8,$
$\therefore B=\{2,4,6,8\}$

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Question 31 Mark

If $A=\{1,2\}$ and $B=\{3,4\}$, Find the number of relations in A and B ?

Answer

$n(A)=2, n(B)=2$
$n(A \times B )=2 \times 2=4$
then number of relations from A to $B =$ total number of subsets of $A \times B$$
=2^4=16
$

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Question 41 Mark

Is $f: Z \rightarrow Z, f(x)=x^2$ is one-one function?

Answer

No, $f: Z \rightarrow Z, f(x)=x^2$ is not one-one function because $3,-3 \in Z$ but $3 \neq-3$ and $f(-3)=f(3)=$ 9 or -3 and 3 has same image 9 .

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Question 51 Mark

In set $A=\{0,1,2,3,4,5\} R$ is equivalence relation where $R =\{(a, b):(a-b)$ is divisible by 2$\}$. Write equivalence class [0].

Answer

Equivalence class [0] is set of those elements of A which is related to zero.
i.e.$
[0]=\{(a, 0) \in R: a \in A\}
$
Now, $(a, 0) \in R$
$\Rightarrow a-0$ is divisible by 2 and $a \in A$
$\Rightarrow$$
a=0,2,4
$
thus$
[0]=\{0,2,4\}
$

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Question 61 Mark

If $f: Z \rightarrow Z, f(x)=x+5$ onto function?

Answer

Yes, here $f$ is onto function, because $f( Z )=f( Z )$ or range of $f=$ co-domain.

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Question 71 Mark

Is $f: R \rightarrow R , f(x)=\sin x$ is many-one function?

Answer

$f: R \rightarrow R , f(x)=\sin x$ is many-one function because $\sin x$ is recurring function or for more than one angles $\sin x$ has same value.

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Question 81 Mark

Is there any relation which is symmetric and also antisymmetric ?

Answer

Yes, is set of natural number ' $=$ ' relation is symmetric as well as antisymmetric.

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Question 91 Mark

In set of integers I , if a relation R is defined as $x R y$ $\Leftrightarrow x>y$ then is it a transitive relation? If yes then why ?

Answer

R is transitive relation because $x, y, z \in I$ and $x>y$ and $y>z$ then $x>y>z \Rightarrow x>z$ or $(x, y) \in R ,(y$, $z) \in R \Rightarrow(x, z) \in R \forall x, y, z \in I$.

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Question 101 Mark

A relation R is defined as $x R y \Rightarrow x$ is parallel to $y$ is set A of straight lines lie in a plane. What relations in R ?

Answer

R will reflexive because every line is parallel to itself. R will symmetric because $x$ is parallel to $y$ then $y$ is also parallel to $x$.
R will be transitive because $x\|y, y\| z \Leftrightarrow x \| z$

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Question 111 Mark

If $A=\{1,2,3\}$ then find the number of all possible non-empty relations defined in A ?

Answer

$A =\{1,2,3\}$
No. of element in $A =3$
then number of elements in $A \times A =3 \times 3=3^2=9$ hence number of non-empty relations defined in $A =$ $2^9-1=512-1=511$.

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Question 121 Mark

Is 'congruent' relation reflexive is any set of triangles ?

Answer

Yes, because every triangle is congruent to itself.

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Question 131 Mark

Is $x R y, y R z \Rightarrow x R z$ true is transitive relation?

Answer

Yes.

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Question 141 Mark

Is relation $x R y \Leftrightarrow x \perp y$ is set of straight lines defined as symmetric ? If yes, then why? Write reason.

Answer

Yes, because $x \perp y$ then $y \perp x$ will always.
So, $x R y \Rightarrow y R x$, this is symmetric.

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Question 151 Mark

C and R represent set of complex numbers and real numbers. Show that $f: C \rightarrow R , f(z)=|z| \forall z \in C$ is neither one-one nor onto.

Answer

Suppose $z_1=1-i$ and $z_2=1+i$ and $z_1 \neq z_2$
but$
\left|z_1\right|=\left|z_2\right|=\sqrt{1^2+(1)^2}=\sqrt{2}
$
$
\text { i.e. } \quad f\left(z_1\right)=f\left(z_2\right)
$
hence two different elements of C have same image. So, $f$ is not one-one. $f$ in not also onto because negative numbers in R have not pre-image in C . or range of $f R ^{+} \cup\{0\} \neq R$ (co-domain).

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Question 161 Mark

A relation $R$ is defined as $A R B \Leftrightarrow A$ is subset of $B$ is sets of set S . Is this relation R will anti symmetric?

Answer

R will antisymmetric because any two sets $A$ and $B$ such as
$A \subseteq B$ and $B \subseteq A \Rightarrow A = B$

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