The tangent to the curve `y=e^(k x)` at a point (0,1) meets the x-axis at (a,0), where `a in [-2,-1]` . Then `k in ` `[-1/2,0]` (b) `[-1,-1/2]` `[0,1]` (d) `[1/2,1]`

Options

A. `[-(1)/(2),0]`
B. `[-1-(1)/(2)]`
C. `[0,1]`
D. `[(1)/(2), 1]`

Answer

To solve the problem, we need to find the values of \( k \) such that the tangent to the curve \( y = e^{kx} \) at the point \( (0, 1) \) meets the x-axis at a point \( (a, 0) \), where \( a \in [-2, -1] \).

### Step-by-step Solution:

1. **Find the derivative of the curve**:
The given curve is \( y = e^{kx} \). We differentiate this with respect to \( x \):
\[ \frac{dy}{dx} = k e^{kx} \]

2. **Evaluate the derivative at the point (0, 1)**:
At the point \( (0, 1) \), we substitute \( x = 0 \):
\[ \frac{dy}{dx} \bigg|_{x=0} = k e^{k \cdot 0} = k \cdot 1 = k \]
So, the slope of the tangent at this point is \( k \).

3. **Write the equation of the tangent line**:
The equation of the tangent line at the point \( (0, 1) \) can be expressed using the point-slope form:
\[ y - 1 = k(x - 0) \quad \Rightarrow \quad y = kx + 1 \]

4. **Find where the tangent meets the x-axis**:
The tangent line meets the x-axis when \( y = 0 \):
\[ 0 = kx + 1 \quad \Rightarrow \quad kx = -1 \quad \Rightarrow \quad x = -\frac{1}{k} \]
Thus, the point of intersection with the x-axis is \( \left(-\frac{1}{k}, 0\right) \).

5. **Set the condition for \( a \)**:
According to the problem, \( a = -\frac{1}{k} \) must lie within the interval \( [-2, -1] \):
\[ -2 \leq -\frac{1}{k} \leq -1 \]

6. **Solve the inequalities**:
- For the left inequality:
\[ -2 \leq -\frac{1}{k} \quad \Rightarrow \quad 2k \leq 1 \quad \Rightarrow \quad k \leq \frac{1}{2} \]
- For the right inequality:
\[ -\frac{1}{k} \leq -1 \quad \Rightarrow \quad \frac{1}{k} \geq 1 \quad \Rightarrow \quad k \leq 1 \quad \text{(since k must be negative)} \]

7. **Combine the results**:
From the inequalities, we have:
\[ k \in \left(-\infty, -1\right) \cup \left(-1, \frac{1}{2}\right) \]
However, since \( k \) must be negative, we only consider:
\[ k \in \left(-1, \frac{1}{2}\right) \]
This means \( k \) can take values in the range \( \left[-\frac{1}{2}, 0\right] \).

### Conclusion:
The correct option is:
**(a)** \( k \in [-\frac{1}{2}, 0] \)