The value of c such that the straight line jo | vedclass

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(B) The slope of the line passing through $(0,3)$ and $(5,-2)$ is $m = \frac{-2-3}{5-0} = \frac{-5}{5} = -1$. The equation of the line is $y - 3 = -1(

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$4$

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(B) The slope of the line passing through $(0,3)$ and $(5,-2)$ is $m = \frac{-2-3}{5-0} = \frac{-5}{5} = -1$. The equation of the line is $y - 3 = -1(x - 0)$,which simplifies to $x + y - 3 = 0$ or $y = -x + 3$. For the line to be tangent to the curve $y = \frac{c}{x+1}$,the derivative of the curve must equal the slope of the line: $\frac{dy}{dx} = \frac{-c}{(x+1)^2} = -1$. This implies $c = (x+1)^2$. Substituting $y = \frac{c}{x+1}$ into the line equation: $\frac{c}{x+1} = -x + 3$. Since $c = (x+1)^2$,we have $\frac{(x+1)^2}{x+1} = -x + 3$,which simplifies to $x + 1 = -x + 3$. Solving for $x$: $2x = 2$,so $x = 1$. Substituting $x = 1$ into $c = (x+1)^2$,we get $c = (1+1)^2 = 4$.

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

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