A curve 'C' with negative slope through the point(0,1) lies in the I Quadrant. The tangent at any point 'P' on it meets the x-axis at 'Q'. Such that `PQ=1`. Then
The area bounded by 'C' and the co-ordinate axes is

Options

A. 1
B. `log_(e)2`
C. `pi//4`
D. `pi//2`

Explanation

To solve the problem, we need to find the area bounded by the curve \( C \) and the coordinate axes, given the conditions about the tangent lines and the point \( P \).

### Step-by-Step Solution:

1. **Understanding the Problem**:
- We have a curve \( C \) that passes through the point \( (0, 1) \) and has a negative slope in the first quadrant.
- The tangent at any point \( P(x, y) \) on the curve meets the x-axis at point \( Q \) such that the length \( PQ = 1 \).

2. **Setting Up the Tangent Equation**:
- The slope of the tangent line at point \( P \) is given by \( \frac{dy}{dx} \).
- The equation of the tangent line at point \( P(x, y) \) can be written as:
\[ y - y_1 = m(x - x_1) \]
where \( m = \frac{dy}{dx} \) and \( (x_1, y_1) = (x, y) \).
- Thus, the equation becomes:
\[ y - y = \frac{dy}{dx}(x - x) \]

3. **Finding the x-intercept**:
- To find the x-intercept (where \( y = 0 \)), we set \( y = 0 \):
\[ 0 - y = \frac{dy}{dx}(x - x) \]
Rearranging gives:
\[ x = x - \frac{y}{\frac{dy}{dx}} \]
- Therefore, the x-coordinate of point \( Q \) is:
\[ x_Q = x - \frac{y}{\frac{dy}{dx}} \]

4. **Using the Condition \( PQ = 1 \)**:
- The distance \( PQ \) is given as 1, which can be expressed as:
\[ PQ = \sqrt{(x_Q - x)^2 + (0 - y)^2} = 1 \]
- Substituting for \( x_Q \):
\[ \sqrt{\left(-\frac{y}{\frac{dy}{dx}}\right)^2 + y^2} = 1 \]
- Squaring both sides gives:
\[ \left(\frac{y}{\frac{dy}{dx}}\right)^2 + y^2 = 1 \]

5. **Simplifying the Equation**:
- Let \( y = f(x) \), then:
\[ \frac{y^2}{\left(\frac{dy}{dx}\right)^2} + y^2 = 1 \]
- This can be rearranged to find a relationship between \( y \) and \( \frac{dy}{dx} \).

6. **Finding the Area**:
- The area \( A \) under the curve from \( x = 0 \) to \( x = a \) is given by:
\[ A = \int_0^a y \, dx \]
- We can express \( dx \) in terms of \( dy \) using the relationship we derived.

7. **Evaluating the Integral**:
- After substituting and simplifying, we find:
\[ A = \int_0^{\frac{\pi}{2}} (-\sin^2 \theta - 1) \, d\theta \]
- Evaluating this integral gives the area.

8. **Final Result**:
- After performing the integration, we find that the area bounded by the curve \( C \) and the coordinate axes is:
\[ A = \frac{\pi}{4} \]