Question

Determine the value of $c$ such that the line joining $(0,3),(5,-2)$ is a tangent to the curve $y=\frac{c}{x+1}$.

• A. $4$
• B. $3$
• C. $23$
• D. $1$

Answer 1

The correct option is A

$4$

Determine the value of $c$:

$$\begin{gathered} y=\frac{c}{x+1} \\ \Rightarrow \frac{dy}{dx}=\frac{-c}{{(x+1)}^2} \end{gathered}$$

Calculate the slope using given points $(0,3),(5,-2)$:

$$\begin{gathered} m=\frac{(y_2-y_1)}{(x_2-x_1)} \\ =\frac{(-2-3)}{5} \\ =\frac{-5}{5} \\ =-1 \end{gathered}$$

Equation of the line joining $(0,3)and(5,-2)$ is $y-y_1=m(x-x_1)$

$$\begin{gathered} y-3=-1(x-0) \\ \Rightarrow y=-x+3.................\left(i\right) \\ -1=-\frac{c}{{(x+1)}^2}(Slopeoflienandcurvearesameatpointofcontact) \\ \Rightarrow c={(x+1)}^2 \end{gathered}$$

Now $y=\frac{c}{x+1}$.

$$\begin{gathered} \Rightarrow y=\frac{{(x+1)}^2}{x+1} \\ \Rightarrow y=(x+1) \end{gathered}$$

$$\begin{gathered} -x+3=x+1(useequation....(i\left)\right) \\ \Rightarrow 2=2x \\ \Rightarrow x=1 \end{gathered}$$

So,

$$\begin{gathered} c={(1+1)}^2 \\ =2^2 \\ =4 \end{gathered}$$

Hence, the value of $c$ is $4$ that is “Option A” .