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Maths Case study (4 Marks)

RELATIONS AND FUNCTIONS · Uttarakhand Board (UBSE) STD 12 Science (Uttarakhand - English Medium). 2 questions.

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Case study (4 Marks)

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Question 14 Marks
A relation R on a set A is said to be an equivalence relation on A iff it is:
  1. Reflexive i.e., $(\text{a, a})\in\ \text{R} \ \forall \ \text{a}\in\text{A}.$
  2. Symmetric i.e., $(\text{a, b})\in\ \text{R} \Rightarrow \text{(b, a) } \in\text{R}\ \forall \ \text{a, b}\in\text{A}.$
  3. Transitive i.e., $(\text{a, b})\in\ \text{R} \ \text{and}\ \text{(b, c) } \in\text{R}\Rightarrow\text{(a, c)}\in\text{R}\ \forall \ \text{a, b, c}\in\text{A}.$
Based on the above information, answer the following questions.
  1. If the relation R = {(1, 1), (1, 2), (1, 3), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)} defined on the set A = {1, 2, 3}, then R is:
  1. Reflexive
  2. Symmetric
  3. Transitive
  4. Equivalence
  1. If the relation R = {(1, 2), (2, 1), (1, 3), (3, 1)} defined on the set A = {1, 2, 3}, then R is:
  1. Reflexive
  2. Symmetric
  3. Transitive
  4. Equivalence
  1. If the relation R on the set N of all natural numbers defined as R = {(x, y): y = x + 5 and x < 4}, then R is:
  1. Reflexive
  2. Symmetric
  3. Transitive
  4. Equivalence
  1. If the relation R on the set A = {1, 2, 3, ........., 13, 14} defined as R = {(x, y): 3x - y = O}, then R is:
  1. Reflexive
  2. Symmetric
  3. Transitive
  4. Equivalence
  1. If the relation R on the set A = {I, 2, 3} defined as R = {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}, then R is:
  1. Reflexive only
  2. Symmetric only
  3. Transitive only
  4. Equivalence
Answer
  1. (a) Reflexive
Solution:

Clearly, (1, 1), (2, 2), (3, 3), $\in$ R. So, R is reflexive on A.

Since, (1, 2) $\in$ R but (2, 1) $\notin$ R. So, R is not symmetric on A.

Since, (2, 3), $\in$ R and (3, 1) $\in$ R but (2, 1) $\notin$ R. So, R is not transitive on A.
  1. (b) Symmetric
Solution:

Since, (1, 1), (2, 2) and (3, 3) are not in R. So, R is not reflexive on A.

Now, (1, 2) $\in$ R ⇒ (2, 1) $\in$ R and (1, 3) $\in$ R ⇒ (3, 1) $\in$ R. So, R is symmetric,

Clearly, (1, 2) $\in$ R and (2, 1) $\in$ R but (1, 1) $\notin$ R. So, R is not transitive on A.
  1. (c) Transitive
Solution:

We have, R = {(x, y): y = x + 5 and x < 4}, where x, y $\in$ N.

$\therefore$ R = {(1, 6), (2, 7), (3, 8)}

Clearly, (1, 1), (2, 2) etc. are not in R. So, R is not reflexive.

Since, (1, 6) $\in$ R but (6, 1) $\notin$ R. So, R is not symmetric.

Since, (1, 6) $\in$ R and there is no order pair in R which has 6 as the first element.

Same is the case for (2, 7) and (3, 8). So, R is transitive.
  1. (d) Equivalence
Solution:

We have, R = {(x, y): 3x - y = 0}, where x, y $\in$ A = {1,2, ......, 14}.

$\therefore$ R = {(1, 3), (2, 6), (3, 9), (4, 12)}

Clearly, (1, 1) $\notin$ R. So, R is not reflexive on A.

Since, (1, 3) $\in$ R but (3, 1) $\notin$ R. So, R is not symmetric on A.

Since, (1, 3) $\in$ Rand (3, 9) $\in$ R but (1, 9) $\notin$ R. So, R is not transitive on A.
  1. (d) Equivalence
Solution:

Clearly, (1, 1), (2, 2), (3, 3) $\in$ R. So, R is reflexive on A.

We find that the ordered pairs obtained by interchanging the components of ordered pairs in R are also in R. So, R is symmetric on A. For 1, 2, 3 $\in$ A such that (1, 2) and (2, 3) are in R implies that (1, 3) is also, in R. So, R is transitive on A. Thus, R is an equivalence relation.
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Question 24 Marks
Consider the mapping $f: A \rightarrow B$ is defined by $f(x) = x - $1 such that $f$ is a bijection.
Based on the above information, answer the following questions.
  1. Domain of $f$ is:
  1. $R - \{2\}$
  2. $R$
  3. $R -\{1, 2\}$
  4. $R - \{0\}$
  1. Range of $f$ is:
  1. $R$
  2. $R - \{2\}$
  3. $R -\{0\}$
  4. $R - \{1, 2\}$
  1. If $g: R - \{2\} \rightarrow R - \{1\}$ is defined by $g(x) = 2f(x) - 1,$ then $g(x)$ in terms of $x$ is:
  1. $\frac{\text{x}+2}{\text{x}}$
  2. $\frac{\text{x}+1}{\text{x}-2}$
  3. $\frac{\text{x}-2}{\text{x}}$
  4. $\frac{\text{x}}{\text{x}-2}$
  1. The function g defined above, is:
  1. One-one
  2. Many-one
  3. into
  4. None of these
  1. A function $f(x)$ is said to be one-one iff.
  1. $f(x_1) = f(x_2) \Rightarrow -x_1= x_2$ 
  2. $f(-x_1) = f(-x_2) \Rightarrow -x_1 = x_2$ 
  3. $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$ 
  4. None of these
Answer
  1. (a) $R - \{2\}$
Solution:

For $f(x)$ to be defined $x - 2; ≠ 0$ i.e., $x; ≠ 2.$

$\therefore$ Domain of $f = R - \{2\}$
  1. (b) $R - \{2\}$
Solution:

Let $y = f(x)$, then $\text{y}=\frac{\text{x}-1}{\text{x}-2}$

$\Rightarrow xy - 2y = x - 1 \Rightarrow xy - x = 2y -$
$\Rightarrow\text{x}=\frac{2\text{y}-1}{\text{y}-1}$

Since, $x \in R - \{2\},$ therefore $y ≠ 1$

Hence, range of $f = R - \{1\}$
  1. (d) $\frac{\text{x}}{\text{x}-2}$
Solution:

We have, $g(x) = 2f(x) - 1$

$=2\Big(\frac{\text{x}-1}{\text{x}-2}\Big)-1=\frac{2\text{x}-2-\text{x}+2}{\text{x}-2}=\frac{\text{x}}{\text{x}-2}$
  1. (a) One-one
Solution:

We have, $g(x)  =\frac{\text{x}}{\text{x}-2}$

Let $g(x_1) = g(x_2) \Rightarrow\frac{\text{x}_1}{\text{x}_{1}-2}=\frac{\text{x}_2}{\text{x}_{2}-2}$

$\Rightarrow x_1x_2 - 2x_1 = x_1x_2 - 2x_2 \Rightarrow 2x_1 = 2x_2 \Rightarrow x_1 = x_2$

Thus, $g(x_1) = g(x_2) \Rightarrow x_1= x_2$​​​​​​​

Hence, $g(x)$ is one-one.
  1. (c) $f(x_1) = f(x_2) \Rightarrow x_1 = x_2$​​​​​​​​​​​​​​
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