f(x)=-1+kx+k neither touches nor intecepts the curve f(x)= Iog x, then minimum value of k `in`

Options

A. `((1)/(e),(1)/sqrt(e))`
B. `(e,e^(2))`
C. `((1)/sqrt(e),e)`
D. None of these

Answer

To solve the problem, we need to find the minimum value of \( k \) such that the line \( f(x) = -1 + kx + k \) neither touches nor intersects the curve \( f(x) = \log x \).

### Step 1: Set up the equations
The line can be rewritten as:
\[ f(x) = kx + (k - 1) \]
The curve is given by:
\[ g(x) = \log x \]

### Step 2: Find the condition for tangency
For the line to neither touch nor intersect the curve, the line must always be above the curve. This means that the equation:
\[ kx + (k - 1) > \log x \]
must hold for all \( x > 0 \).

### Step 3: Find the derivative of the curve
The derivative of \( g(x) = \log x \) is:
\[ g'(x) = \frac{1}{x} \]
At any point \( x_0 \), the slope of the tangent to the curve is \( \frac{1}{x_0} \).

### Step 4: Set the slope of the line
The slope of the line \( f(x) \) is \( k \). For the line to not touch the curve, we need:
\[ k < \frac{1}{x_0} \]
This implies:
\[ x_0 < \frac{1}{k} \]

### Step 5: Substitute \( x_0 \) into the inequality
We also need to ensure that:
\[ kx_0 + (k - 1) > \log x_0 \]
Substituting \( k = \frac{1}{x_0} \):
\[ \frac{1}{x_0} x_0 + \left(\frac{1}{x_0} - 1\right) > \log x_0 \]
This simplifies to:
\[ 1 + \frac{1}{x_0} - 1 > \log x_0 \]
Thus:
\[ \frac{1}{x_0} > \log x_0 \]

### Step 6: Analyze the inequality
The inequality \( \frac{1}{x_0} > \log x_0 \) holds true for \( x_0 > e \). Therefore, we can say:
\[ x_0 > e \implies k < \frac{1}{e} \]

### Step 7: Find the minimum value of \( k \)
From our analysis, we conclude that \( k \) must be greater than 1 for the line to neither touch nor intersect the curve. The minimum value of \( k \) that satisfies this condition is:
\[ k \geq 1 \]

### Step 8: Conclusion
Thus, the minimum value of \( k \) such that the line \( f(x) \) neither touches nor intersects the curve \( g(x) \) is:
\[ \text{Minimum value of } k = 1 \]