If the line joining the points `(0,3)a n d(5,-2)` is a tangent to the curve `y=C/(x+1)` , then the value of `c` is 1 (b) `-2` (c) 4 (d) none of these

Options

A. 2
B. 3
C. 4
D. 5

Answer

To find the value of \( C \) such that the line joining the points \( (0, 3) \) and \( (5, -2) \) is a tangent to the curve \( y = \frac{C}{x + 1} \), we will follow these steps:

### Step 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:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} \]
Substituting the points \( (0, 3) \) and \( (5, -2) \):
\[ m = \frac{-2 - 3}{5 - 0} = \frac{-5}{5} = -1 \]

Using the point-slope form of the equation of a line:
\[ y - y_1 = m(x - x_1) \]
Taking \( (x_1, y_1) = (0, 3) \):
\[ y - 3 = -1(x - 0) \]
This simplifies to:
\[ y = 3 - x \]

### Step 2: Find the derivative of the curve \( y = \frac{C}{x + 1} \).

To find the slope of the tangent to the curve, we differentiate:
\[ y = \frac{C}{x + 1} \]
Using the quotient rule:
\[ \frac{dy}{dx} = \frac{0 \cdot (x + 1) - C \cdot 1}{(x + 1)^2} = \frac{-C}{(x + 1)^2} \]

### Step 3: Set the slope of the tangent equal to the slope of the line.

Since the slope of the line is \( -1 \), we set:
\[ \frac{-C}{(x + 1)^2} = -1 \]
This simplifies to:
\[ \frac{C}{(x + 1)^2} = 1 \]
Thus, we have:
\[ C = (x + 1)^2 \]

### Step 4: Find the point of tangency.

At the point of tangency, the \( y \)-coordinate of the line can be expressed as:
\[ y = 3 - x \]
We also know:
\[ y = \frac{C}{x + 1} \]
Setting these equal gives:
\[ 3 - x = \frac{C}{x + 1} \]

### Step 5: Substitute \( C \) from the previous equation.

Substituting \( C = (x + 1)^2 \) into the equation:
\[ 3 - x = \frac{(x + 1)^2}{x + 1} \]
This simplifies to:
\[ 3 - x = x + 1 \]
Rearranging gives:
\[ 3 - 1 = x + x \]
\[ 2 = 2x \implies x = 1 \]

### Step 6: Find \( C \).

Substituting \( x = 1 \) back into \( C = (x + 1)^2 \):
\[ C = (1 + 1)^2 = 2^2 = 4 \]

### Conclusion

Thus, the value of \( C \) is \( 4 \).