Find value of c such that line joining the points (0, 3) and (5, -2) becomes tangent to curve `y=c/(x+1)`

Options

A. 1
B. `-2`
C. 4
D. none of these

Answer

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

### Step 1: Find the equation of the line
The slope \( m \) of the line joining the points \( (0, 3) \) and \( (5, -2) \) can be calculated as:
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-2 - 3}{5 - 0} = \frac{-5}{5} = -1 \]
Using the point-slope form of the line equation, we can write:
\[ y - y_1 = m(x - x_1) \]
Substituting \( (x_1, y_1) = (0, 3) \) and \( m = -1 \):
\[ y - 3 = -1(x - 0) \implies y - 3 = -x \implies x + y - 3 = 0 \]

### Step 2: Differentiate the curve
The given curve is:
\[ y = \frac{c}{x + 1} \]
Differentiating this with respect to \( x \):
\[ \frac{dy}{dx} = -\frac{c}{(x + 1)^2} \]

### Step 3: Set up the tangent condition
Let \( (α, β) \) be the point of tangency. Since the line is tangent to the curve at this point, the slope of the line must equal the slope of the curve at that point:
\[ -\frac{c}{(α + 1)^2} = -1 \implies \frac{c}{(α + 1)^2} = 1 \implies c = (α + 1)^2 \]

### Step 4: Substitute \( β \) in terms of \( α \)
From the equation of the curve, we have:
\[ β = \frac{c}{α + 1} \]
Substituting \( c = (α + 1)^2 \):
\[ β = \frac{(α + 1)^2}{α + 1} = α + 1 \]

### Step 5: Use the line equation
Since the line equation is \( x + y - 3 = 0 \), we can express \( β \) in terms of \( α \):
\[ β = 3 - α \]
Now we have two expressions for \( β \):
1. \( β = α + 1 \)
2. \( β = 3 - α \)

### Step 6: Set the equations equal to each other
Setting the two expressions for \( β \) equal:
\[ α + 1 = 3 - α \]
Solving for \( α \):
\[ 2α = 2 \implies α = 1 \]

### Step 7: Find \( c \)
Now substituting \( α = 1 \) back to find \( c \):
\[ c = (α + 1)^2 = (1 + 1)^2 = 2^2 = 4 \]

### Final Answer
Thus, the value of \( c \) is:
\[ \boxed{4} \]