Maths Case study (4 Marks)

1. (a) R - {2}

Solution:

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

$\therefore$ Domain of f = R - {2}

2. (b) R - {2}

Solution:

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

 ⇒ xy - 2y = x - 1 ⇒ 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}

3. (d) $\frac{\text{x}}{\text{x}-2}$

Solution:

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

$=2\Big(\frac{\text{x}-1}{\text{x}-2}\Big)-1=\frac{2\text{x}-2-\text{x}+2}{\text{x}-2}=\frac{\text{x}}{\text{x}-2}$

4. (a) One-one

Solution:

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

Let g(x₁) = g(x₂) $\Rightarrow\frac{\text{x}_1}{\text{x}_{1}-2}=\frac{\text{x}_2}{\text{x}_{2}-2}$

⇒ x₁x₂ - 2x₁ = x₁x₂ - 2x₂ ⇒ 2x₁ = 2x₂ ⇒ x₁ = x₂

Thus, g(x₁) = g(x₂) ⇒ x₁ = x₂

Hence, g(x) is one-one.

5. (c) f(x₁) = f(x₂) ⇒ x₁ = x₂

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.

2.  (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.

3. (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.

4. (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.

5. (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.