(Each question 4 marks)

Question 14 Marks

Let f = {(1, 1), (2, 3), (0, –1), (–1, –3)} be a function from Z to Z defined by f(x) = ax + b, for some integers a, b. Determine a, b.

Answer

f = {(1, 1), (2, 3), (0, –1), (-1, -3)}
f(x) = ax + b
(1, 1) ∈ f
⇒ f(1) = 1
⇒ a × 1 + b = 1
⇒ a + b = 1
(0, -1) ∈ f
⇒ f(0) = -1
⇒ a × 0 + b = -1
⇒ b = -1
On substituting b =-1 in a + b = 1, we obtain a + (-1) =1 ⇒ a = 1 + 1 = 2.
Thus, the respective values of a and b are 2 and -1.

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Question 24 Marks

The relation f is defined by $\text{f(x)}=\begin{cases}\text{x}^2,0\leq\text{x}\leq3\\3\text{x},3\leq\text{x}\leq10\end{cases}$ The relation g is defined by $\text{g(x)}=\begin{cases}\text{x}^2,0\leq\text{x}\leq2\\3\text{x},2\leq\text{x}\leq10\end{cases}$ Show that f is a function and g is not a function.

Answer

It is observed that for $0\leq\text{x}<3,\text{f(x)}=\text{x}^2$ $3<\text{x}\leq10,\text{f(x)}=3\text{x}$ Also, at $\text{x}=3,\text{f(x)}=3^2=9$ or $\text{f(x)}=3\times3=9$ i.e., at $\text{x}=3,\text{f(x)}=9$
Therefore, for $0\leq\text{x}\leq10,$ the images of f(x) are unique. Thus, the given relation is a function.
The relation g is defined as, $\text{g(x)}=\begin{cases}\text{x}^2,0\leq\text{x}\leq2\\3\text{x},2\leq\text{x}\leq10\end{cases}$ It can be observed that for $x = 2, g(x) = 2^2= 4$ and $g(x) = 3 × 2 = 6$
Hence, element 2 of the domain of the relation g corresponds to two different images i.e., 4 and 6. Hence, this relation is not a function.

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Question 34 Marks

Let A = {9, 10, 11, 12, 13} and let f : A → N be defined by f(n) = the highest prime factor of n. Find the range of f.

Answer

Here A = {9, 10, 11, 12, 13} For n = 9, f(9) = 3 [$\because$ 9 = 3 × 3 and 3 is highest prime factor of 9] For n = 10, f(10) = 5 [$\because$ 10 = 2 × 5] For n = 11, f(11) = 11 [$\because$ 11 = 1 × 11] For n = 12, f(12) = 3 [$\because$ 12 = 3 × 3 × 2] For n = 13, f(13) = 13 [$\because$ 13 = 1 × 13] $\therefore$ Range of f ={5, 11, 3, 13} = {3, 5, 11, 13}

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Question 44 Marks

Let $A = {1, 2}$ and $B = {3, 4}$. Write $A × B$. How many subsets will $A × B$ have? List them.

Answer

$A=\{1,2\} \text { and } B=\{3,4\}$
$\therefore A \times B=\{(1,3),(1,4),(2,3),(2,4)\}$
$=n(A \times B)=4$
We know that if $C$ is a set with $n(C)=m$, then $n[P(C)]=2^m$.
Therefore, the set $\mathrm{A} \times \mathrm{B}$ has $2^4=16$ subsets. These are
$\phi,\{(1,3)\},\{(1,4)\},\{(2,3)\},\{(2,4)\},\{(1,3),(1,4)\},\{(1,3),(2,3)\},\{(1,3),(2,4)\},\{(1,4),(2,3)\},\{(1,4),(2,4)\},\{(2,3),(2,4)\},\{(1,3),(1,4),(2,3)\},\{(1,3),(1,4),(2,4)\},\{(1,3),(2,3),(2,4)\},\{(1,4),(2,3),(2,4)\},\{(1,3),(1,4),(2,3),(2,4)\}$

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Question 54 Marks

Let A ={1, 2, 3, 4}, B = {1, 5, 9, 11, 15, 16} and f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)} Are the following true?

f is a relation from A to B.
f is a function from A to B.

Answer

A = {1, 2, 3, 4} and B = {1, 5, 9, 11, 15, 16} $\therefore$ A × B = {(1, 1), (1, 5), (1, 9), (1, 11), (1, 15), (1, 16), (2, 1), (2, 5), (2, 9), (2, 11), (2, 15), (2, 16), (3, 1), (3, 5), (3, 9), (3, 11), (3, 15), (3, 16), (4, 1), (4, 5), (4, 9), (4, 11), (4, 15), (4, 16)} It is given that f = {(1, 5), (2, 9), (3, 1), (4, 5), (2, 11)}

A relation from a non-empty set A to a non-empty set B is a subset of the Cartesian product A × B.

It is observed that f is a subset of A × B.
Thus, f is a relation from A to B.

Since the same first element i.e., 2 corresponds to two different images i.e., 9 and 11, relation f is not a function.

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Question 64 Marks

Define a relation R on the set N of natural numbers by R = {(x, y) : y = x + 5, x is a natural number less than 4; x, y ∈ N}. Depict this relationship using roster form. Write down the domain and the range.

Answer

R = {(x, y): y = x + 5, x is a natural number less than 4, x, y ∈ N}
The natural numbers less than 4 are 1, 2, and 3.
$\therefore$ R = {(1, 6), (2, 7), (3, 8)}
The domain of R is the set of all first elements of the ordered pairs in the relation.
$\therefore$ Domain of R = {1, 2, 3}
The range of R is the set of all second elements of the ordered pairs in the relation.
$\therefore$ Range of R = {6, 7, 8}

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Question 74 Marks

Let A = {1, 2, 3, ... ,14}. Define a relation R from A to A by R = {(x, y) : 3x – y = 0, where x, y ∈ A}. Write down its domain, codomain and range.

Answer

The relation R from A to A is given as R = {(x, y): 3x – y = 0, where x, y ∈ A} i.e., R = {(x, y): 3x = y, where x, y ∈ A} $\therefore$ R = {(1, 3), (2, 6), (3, 9), (4, 12)} The domain of R is the set of all first elements of the ordered pairs in the relation. $\therefore$ Domain of R = {1, 2, 3, 4} The whole set A is the codomainof the relation R. $\therefore$ Codomain of R = A = {1, 2, 3, …, 14} The range of R is the set of all second elements of the ordered pairs in the relation. $\therefore$ Range of R = {3, 6, 9, 12}

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