Question

The line joining the points $ (0, 3) $ and $ (5, -2) $ is a tangent to the curve: $$ y = \frac{C}{x + 1} $$ Then find $ C $.

• A. 1
• B. -2
• C. 4
• D. 5

Hint:

Set the derivative equal to the line slope to find point of tangency. Then equate the curve and line at that point.

Answer 1

The Correct Option is C

Step 1: Find slope of the line joining the points: \[ m = \frac{-2 - 3}{5 - 0} = \frac{-5}{5} = -1 \] The curve is: \[ y = \frac{C}{x + 1} \Rightarrow \text{Differentiate: } \frac{dy}{dx} = -\frac{C}{(x + 1)^2} \] Let the point of tangency be \( (x_0, y_0) \) on the curve. We are told the line with slope -1 is tangent to the curve, so: \[ -\frac{C}{(x_0 + 1)^2} = -1 \Rightarrow C = (x_0 + 1)^2 \quad \text{(1)} \] Also, point lies on curve: \[ y_0 = \frac{C}{x_0 + 1} \quad \text{(2)} \] Now since line is tangent to curve and also passes through both (0, 3) and (5, -2), every point on the curve and tangent must satisfy: \[ y = -x + 3 \quad \text{(line equation from two points)} \] So set: \[ \frac{C}{x + 1} = -x + 3 \Rightarrow C = (x + 1)(-x + 3) = -x^2 + 2x + 3 \] Max value of C occurs at vertex \( x = -\frac{b}{2a} = -\frac{2}{-2} = 1 \) At \( x = 1 \): \[ C = -(1)^2 + 2(1) + 3 = -1 + 2 + 3 = \boxed{4} \]