To find the point of contact `P (x_1, y_1)` of a tangent to the graph of `y = f(x)` passing through origin `O`, we equate the slope of tangent to `y = f(x)` at `P` to the slope of `OP`. Hence we solve the equation `f' (x) = f(x_1)/x_1` to get `x_1` and `y_1`.Now answer the following questions (7 -9): The equation `|lnmx|= px` where m is a positive constant has a single root for

Options

A. `0ltplt(m)/(e)`
B. `plt(e)/(m)`
C. `0ltplt(e)/(m)`
D. `pgt(m)/(e)`

Answer

To solve the given problem, we need to analyze the equation \(|\ln(mx)| = px\) and determine the condition under which it has a single root. Let's break down the solution step by step.

### Step 1: Understand the Equation
The equation we are dealing with is \(|\ln(mx)| = px\). This can be split into two cases based on the absolute value.

### Step 2: Case Analysis
1. **Case 1:** \(\ln(mx) = px\)
2. **Case 2:** \(-\ln(mx) = px\)

### Step 3: Solve Case 1
For Case 1, we have:
\[ \ln(mx) = px \]
Exponentiating both sides gives:
\[ mx = e^{px} \]
Rearranging gives:
\[ x = \frac{e^{px}}{m} \]
This equation is transcendental and generally does not have a simple analytical solution. However, we can analyze its behavior graphically.

### Step 4: Solve Case 2
For Case 2, we have:
\[ -\ln(mx) = px \]
This leads to:
\[ \ln(mx) = -px \]
Exponentiating both sides gives:
\[ mx = e^{-px} \]
Rearranging gives:
\[ x = \frac{e^{-px}}{m} \]
Similar to Case 1, this equation is also transcendental.

### Step 5: Graphical Interpretation
To find the conditions for which the original equation has a single root, we can analyze the graphs of the functions involved:
- The left-hand side, \(|\ln(mx)|\), is a curve that approaches \(-\infty\) as \(x\) approaches 0 and increases without bound as \(x\) increases.
- The right-hand side, \(px\), is a straight line that passes through the origin with slope \(p\).

### Step 6: Condition for a Single Root
For the equation to have a single root, the line \(y = px\) must be tangent to the curve of \(|\ln(mx)|\). This occurs when the derivative of \(|\ln(mx)|\) at the point of tangency equals the slope \(p\).

### Step 7: Find the Derivative
The derivative of \(|\ln(mx)|\) is:
\[ \frac{d}{dx}|\ln(mx)| = \frac{m}{mx} = \frac{1}{x} \]
Setting this equal to \(p\) gives:
\[ \frac{1}{x} = p \implies x = \frac{1}{p} \]

### Step 8: Substitute Back
Substituting \(x = \frac{1}{p}\) back into the original equation gives:
\[ |\ln(m \cdot \frac{1}{p})| = p \cdot \frac{1}{p} = 1 \]
This leads to:
\[ |\ln(\frac{m}{p})| = 1 \]
This means:
\[ \frac{m}{p} = e \quad \text{or} \quad \frac{m}{p} = \frac{1}{e} \]

### Step 9: Solve for \(p\)
From \(\frac{m}{p} = e\), we get:
\[ p = \frac{m}{e} \]
From \(\frac{m}{p} = \frac{1}{e}\), we get:
\[ p = me \]

### Conclusion
For the equation \(|\ln(mx)| = px\) to have a single root, the value of \(p\) must satisfy:
\[ p > \frac{m}{e} \]