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If B={x:x=(2n-1)/(n+2), n in W and n lt ...

If `B={x:x=(2n-1)/(n+2), n in W and n lt 4} ` then B can be wirtten as

A

`{6,(1)/(3),(2)/(4),(-1)/(2)}`

B

`{(-1)/(2),(1)/(3),(3)/(4),1}`

C

`{3,(-1)/(2),(3)/(4),(6)/(5)}`

D

`{(1)/(5),(1)/(3),(-1)/(2),1}`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to find the set \( B \) defined by the expression \( x = \frac{2n - 1}{n + 2} \) for whole numbers \( n \) that are less than 4. ### Step-by-step Solution: 1. **Identify the values of \( n \)**: Since \( n \) is a whole number and \( n < 4 \), the possible values for \( n \) are: \[ n = 0, 1, 2, 3 \] 2. **Calculate \( x \) for each value of \( n \)**: - For \( n = 0 \): \[ x = \frac{2(0) - 1}{0 + 2} = \frac{-1}{2} \] - For \( n = 1 \): \[ x = \frac{2(1) - 1}{1 + 2} = \frac{2 - 1}{3} = \frac{1}{3} \] - For \( n = 2 \): \[ x = \frac{2(2) - 1}{2 + 2} = \frac{4 - 1}{4} = \frac{3}{4} \] - For \( n = 3 \): \[ x = \frac{2(3) - 1}{3 + 2} = \frac{6 - 1}{5} = \frac{5}{5} = 1 \] 3. **Compile the values of \( x \)**: Now we have calculated \( x \) for all values of \( n \): - For \( n = 0 \): \( x = -\frac{1}{2} \) - For \( n = 1 \): \( x = \frac{1}{3} \) - For \( n = 2 \): \( x = \frac{3}{4} \) - For \( n = 3 \): \( x = 1 \) 4. **Write the set \( B \)**: Therefore, the set \( B \) can be written as: \[ B = \left\{ -\frac{1}{2}, \frac{1}{3}, \frac{3}{4}, 1 \right\} \] ### Final Answer: The set \( B \) can be expressed as: \[ B = \left\{ -\frac{1}{2}, \frac{1}{3}, \frac{3}{4}, 1 \right\} \]
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