The area bounded by the axes of reference and the normal to `y=log_(e)x` at (1,0), is

Options

A. 1 sq units
B. 2 sq units
C. `(1)/(2)` sq units
D. None of these

Answer

To find the area bounded by the axes of reference and the normal to the curve \( y = \log_e x \) at the point \( (1, 0) \), we can follow these steps:

### Step 1: Find the slope of the tangent to the curve at the point (1, 0).
The first step is to differentiate the function \( y = \log_e x \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{x} \]
At the point \( (1, 0) \), the slope of the tangent is:
\[ \frac{dy}{dx} \bigg|_{x=1} = \frac{1}{1} = 1 \]

### Step 2: Determine the slope of the normal.
The slope of the normal line is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = -1 \]

### Step 3: Write the equation of the normal line.
Using the point-slope form of the equation of a line, we can write the equation of the normal line at the point \( (1, 0) \):
\[ y - 0 = -1(x - 1) \]
Simplifying this, we get:
\[ y = -x + 1 \]
or rearranging gives:
\[ x + y = 1 \]

### Step 4: Find the points where the normal intersects the axes.
To find the x-intercept, set \( y = 0 \):
\[ x + 0 = 1 \implies x = 1 \quad \text{(Point: (1, 0))} \]
To find the y-intercept, set \( x = 0 \):
\[ 0 + y = 1 \implies y = 1 \quad \text{(Point: (0, 1))} \]

### Step 5: Sketch the area bounded by the axes and the normal.
The area we need to find is the triangle formed by the points \( (0, 0) \), \( (1, 0) \), and \( (0, 1) \).

### Step 6: Calculate the area of the triangle.
The area \( A \) of a triangle can be calculated using the formula:
\[ A = \frac{1}{2} \times \text{base} \times \text{height} \]
In this case, the base and height are both equal to 1:
\[ A = \frac{1}{2} \times 1 \times 1 = \frac{1}{2} \]

### Final Answer:
The area bounded by the axes of reference and the normal to \( y = \log_e x \) at the point \( (1, 0) \) is \( \frac{1}{2} \) square units.

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