A function f is defined on the set of real numbers as follows :
`f(x)={:{(1+x,1lexlt2),(2x-1,2lexlt4),(3x-5,4lexlt6):}`
(i) Find the domain of the function.
(ii) Find the range of the function.
(iii) Find f(4).
(iv) Is the function one-one ? Justify.

Answer

Let's solve the given problem step by step.

### Given Function:
The function \( f \) is defined as follows:
\[ f(x) = \begin{cases} 1 + x & \text{if } 1 \leq x < 2 \\ 2x - 1 & \text{if } 2 \leq x < 4 \\ 3x - 5 & \text{if } 4 \leq x < 6 \end{cases} \]

### (i) Find the domain of the function.

**Step 1:** Identify the intervals for \( x \) in the piecewise function.

- The first piece is defined for \( 1 \leq x < 2 \).
- The second piece is defined for \( 2 \leq x < 4 \).
- The third piece is defined for \( 4 \leq x < 6 \).

**Step 2:** Combine the intervals to find the overall domain.

The domain of \( f \) is the union of these intervals:
\[ \text{Domain} = [1, 2) \cup [2, 4) \cup [4, 6) \]

Thus, the domain of the function is:
\[ \text{Domain} = [1, 6) \]

### (ii) Find the range of the function.

**Step 1:** Calculate the outputs for the endpoints of each interval.

- For \( x = 1 \):
\[ f(1) = 1 + 1 = 2 \]
- For \( x = 2 \):
\[ f(2) = 2(2) - 1 = 3 \]
- For \( x = 4 \):
\[ f(4) = 3(4) - 5 = 7 \]
- For \( x = 6 \):
\[ f(6) = 3(6) - 5 = 13 \]

**Step 2:** Determine the range from the calculated outputs.

- The first piece \( 1 + x \) gives values from \( 2 \) to just below \( 3 \).
- The second piece \( 2x - 1 \) gives values from \( 3 \) to just below \( 7 \).
- The third piece \( 3x - 5 \) gives values from \( 7 \) to \( 13 \).

Thus, the range of the function is:
\[ \text{Range} = [2, 13) \]

### (iii) Find \( f(4) \).

**Step 1:** Identify which piece of the function to use for \( x = 4 \).

Since \( 4 \) falls in the interval \( [4, 6) \), we use the third piece:
\[ f(x) = 3x - 5 \]

**Step 2:** Calculate \( f(4) \):
\[ f(4) = 3(4) - 5 = 12 - 5 = 7 \]

Thus, \( f(4) = 7 \).

### (iv) Is the function one-one? Justify.

**Step 1:** A function is one-one (injective) if different inputs produce different outputs.

**Step 2:** Check the outputs for each piece:
- For \( 1 \leq x < 2 \): The function \( f(x) = 1 + x \) is strictly increasing.
- For \( 2 \leq x < 4 \): The function \( f(x) = 2x - 1 \) is also strictly increasing.
- For \( 4 \leq x < 6 \): The function \( f(x) = 3x - 5 \) is again strictly increasing.

**Step 3:** Since all pieces are strictly increasing and there are no overlaps in the outputs from different pieces, the function is one-one.

Thus, the function is one-one.

### Summary of Answers:
- (i) Domain: \([1, 6)\)
- (ii) Range: \([2, 13)\)
- (iii) \( f(4) = 7 \)
- (iv) The function is one-one.