Line joining the points (0,3) and (5,-2) is a tangent to the curve `y=(ax)/(1+x)` , then

Options

A. `a=1+_sqrt(3)`
B. `a in phi`
C. `a=-1+-sqrt(3)`
D. `a=-2+-2sqrt(3)`

Answer

To solve the problem, we need to find the value of \( a \) such that the line joining the points \( (0, 3) \) and \( (5, -2) \) is a tangent to the curve given by the equation \( y = \frac{ax}{1+x} \).

### Step-by-Step Solution:

1. **Find the equation of the line joining the points (0, 3) and (5, -2)**:
- The slope \( m \) of the line can be calculated using the formula:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 3}{5 - 0} = \frac{-5}{5} = -1 \]
- Using the point-slope form of the line equation \( y - y_1 = m(x - x_1) \):
\[ y - 3 = -1(x - 0) \implies y = -x + 3 \]
- Rearranging gives us:
\[ x + y = 3 \]

2. **Set the equation of the line equal to the curve**:
- The curve is given by \( y = \frac{ax}{1+x} \). We set this equal to the line:
\[ 3 - x = \frac{ax}{1+x} \]

3. **Multiply both sides by \( 1 + x \) to eliminate the fraction**:
\[ (3 - x)(1 + x) = ax \]
- Expanding the left-hand side:
\[ 3 + 3x - x - x^2 = ax \implies -x^2 + 2x + 3 = ax \]
- Rearranging gives:
\[ -x^2 + (2 - a)x + 3 = 0 \]

4. **For the line to be tangent to the curve, the quadratic equation must have a double root**:
- The condition for a quadratic equation \( Ax^2 + Bx + C = 0 \) to have a double root is given by the discriminant \( D = B^2 - 4AC = 0 \).
- Here, \( A = -1 \), \( B = 2 - a \), and \( C = 3 \):
\[ D = (2 - a)^2 - 4(-1)(3) = 0 \]
- Simplifying gives:
\[ (2 - a)^2 + 12 = 0 \]

5. **Analyzing the discriminant**:
- Since \( (2 - a)^2 \) is always non-negative and \( 12 \) is positive, the sum \( (2 - a)^2 + 12 \) can never equal zero for any real number \( a \).

6. **Conclusion**:
- Therefore, there are no real values of \( a \) for which the line is tangent to the curve.

### Final Answer:
The conclusion is that there are no real values of \( a \) for which the line joining the points \( (0, 3) \) and \( (5, -2) \) is a tangent to the curve \( y = \frac{ax}{1+x} \).

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