Let `f(x)={{:(1+(2x)/(a)", "0lexlt1),(ax", "1lexlt2):}."If" lim_(xto1) "f(x) exists, then a is "`

Options

A. `underset(xto5^(-))f(x)=0`
B. `underset(xto5^(+))f(x)=1`
C. `underset(xto5)lim f(x)` does not exist
D. none of these

Explanation

To solve the problem, we need to find the value of \( a \) such that the limit of the function \( f(x) \) exists as \( x \) approaches 1. The function is defined as follows:

\[ f(x) = \begin{cases} 1 + \frac{2x}{a} & \text{if } 0 \leq x < 1 \\ ax & \text{if } 1 \leq x < 2 \end{cases} \]

### Step 1: Find the left-hand limit as \( x \) approaches 1

The left-hand limit is given by:

\[ \lim_{x \to 1^-} f(x) = \lim_{x \to 1^-} \left(1 + \frac{2x}{a}\right) \]

Substituting \( x = 1 \):

\[ \lim_{x \to 1^-} f(x) = 1 + \frac{2 \cdot 1}{a} = 1 + \frac{2}{a} \]

### Step 2: Find the right-hand limit as \( x \) approaches 1

The right-hand limit is given by:

\[ \lim_{x \to 1^+} f(x) = \lim_{x \to 1^+} (ax) \]

Substituting \( x = 1 \):

\[ \lim_{x \to 1^+} f(x) = a \cdot 1 = a \]

### Step 3: Set the left-hand limit equal to the right-hand limit

For the limit to exist, the left-hand limit must equal the right-hand limit:

\[ 1 + \frac{2}{a} = a \]

### Step 4: Solve for \( a \)

Rearranging the equation gives:

\[ 1 + \frac{2}{a} = a \]

Multiplying through by \( a \) (assuming \( a \neq 0 \)):

\[ a + 2 = a^2 \]

Rearranging this gives us a quadratic equation:

\[ a^2 - a - 2 = 0 \]

### Step 5: Factor the quadratic equation

We can factor the quadratic equation:

\[ (a - 2)(a + 1) = 0 \]

### Step 6: Find the roots

Setting each factor to zero gives us:

\[ a - 2 = 0 \quad \Rightarrow \quad a = 2 \]
\[ a + 1 = 0 \quad \Rightarrow \quad a = -1 \]

### Conclusion

Thus, the values of \( a \) for which the limit exists are:

\[ a = 2 \quad \text{and} \quad a = -1 \]