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

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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(

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(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

Similar Questions

What is the angle between the curves $y = x^2$ and $x = y^2$ at the point $(1, 1)$?

$A(-2,9)$ and $B(1,6)$ are two points on the curve $y=x^2+5$. The coordinates of the point $C$ on the curve such that the tangent drawn at $A$ is parallel to the chord $BC$,are

If the equation of a tangent drawn to the curve $y=\cos (x+y), -1 \leq x \leq 1+\pi$ is $x+2y=k$,then $k=$

What is the angle of intersection of the curves $y = 4 - x^2$ and $y = x^2$?

What is the equation of the tangent to the curve $y = 2 \sin x + \sin 2x$ at the point $x = \pi / 3$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module