If the line joining the points ( 0 , 3 ) and ( 5 , − 2 ) is a tangent to the curve y = x + 1 c , then value of c is

Eqn. of the line joining given points is ( y + 2 ) = 5 − 0 − 2 − 3 ( x − 5 ) ⇒ y + x = 3 Solving with y = x + 1 c : ( 3 − x ) ( x + 1 ) = c ⇒ x 2 − 2 x + ( c − 3 ) = 0 ; tangency ⇒ 4 − 4 ( c − 3 ) = 0 ⇒ c = 4

Start Free Mock LoginAlready registered? Login here

Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeTangents & NormalsMedium2 minQB

MathematicsMediumsingle choice

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

Options:

Answer:

Solution:

Eqn. of the line joining given points is $(y + 2) = \frac{- 2 - 3}{5 - 0} (x - 5) \Rightarrow y + x = 3$ Solving with $y = \frac{c}{x + 1} : (3 - x) (x + 1) = c \Rightarrow x^{2} - 2 x + (c - 3) = 0$ ; tangency $\Rightarrow 4 - 4 (c - 3) = 0 \Rightarrow c = 4$

Stream:JEESubject:MathematicsTopic:Application of DerivativeSubtopic:Tangents & Normals

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...