Question

The value of \( c \) such that the straight line joining the points \[ (0,3) \quad \text{and} \quad (5,-2) \] is tangent to the curve \[ y = \frac{c}{x+1} \] is:

• A. \( 3 \)
• B. \( 4 \)
• C. \( 5 \)
• D. \( 2 \)

Hint:

For tangency conditions, equate the derivative of the function to the slope of the given line.

Answer 1

The Correct Option is B

Step 1: Finding the equation of the line The equation of the line through \( (0,3) \) and \( (5,-2) \): \[ y - 3 = \frac{-2 - 3}{5 - 0} (x - 0). \] Simplifying, \[ y = -x + 3. \] Step 2: Condition for tangency The given curve is \( y = \frac{c}{x+1} \). For tangency, we equate slopes: \[ \frac{d}{dx} \left( \frac{c}{x+1} \right) = -1. \] Solving for \( c \), we get: \[ c = 4. \]

Answer 2

Given points:
\[ (0, 3) \quad \text{and} \quad (5, -2) \] The line joining these points is tangent to the curve:
\[ y = \frac{c}{x+1} \] Find the value of \( c \).

Step 1: Find the equation of the line joining the points.
Slope \( m \):
\[ m = \frac{-2 - 3}{5 - 0} = \frac{-5}{5} = -1 \] Using point-slope form with point \((0,3)\):
\[ y - 3 = -1(x - 0) \implies y = -x + 3 \]

Step 2: The line is tangent to the curve, so the curve and line intersect at exactly one point.
Set:
\[ \frac{c}{x + 1} = -x + 3 \] Multiply both sides by \( x + 1 \):
\[ c = (-x + 3)(x + 1) = -x(x+1) + 3(x+1) = -x^2 - x + 3x + 3 = -x^2 + 2x + 3 \]

Step 3: Rearrange as:
\[ -x^2 + 2x + 3 - c = 0 \] or \[ -x^2 + 2x + (3 - c) = 0 \] Multiply both sides by \(-1\) for convenience:
\[ x^2 - 2x - (3 - c) = 0 \] \[ x^2 - 2x - 3 + c = 0 \]

Step 4: For tangency, the quadratic has exactly one root, so discriminant \( D = 0 \):
\[ D = (-2)^2 - 4 \times 1 \times (-3 + c) = 4 + 4(3 - c) = 4 + 12 - 4c = 16 - 4c \] Set discriminant to zero:
\[ 16 - 4c = 0 \implies 4c = 16 \implies c = 4 \]

Therefore, the value of \( c \) is:
\[ \boxed{4} \]