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Mathematics - Application of Derivative Question with Solution | TestHub

MathematicsApplication of DerivativeTangents & NormalsMedium2 minQB

MathematicsMediumsingle choice

Passage / Comprehension

A curve passing through the origin is such that the slope of the tangent at any point is the reciprocal of the sum of the co-ordinates of the point of tangency.

The slope of the tangent at ordinate $ln 3 is$

Options:

Answer:

Solution:

$\frac{d y}{d x} = \frac{1}{x + y}$ , so $\frac{d x}{d y} - x = y$ . Solving with $x (0) = 0 : x = e^{y} - y - 1$ (equivalently $y = ln (x + y + 1)$ ). $\frac{d y}{d x} = \frac{1}{e ^{y} - 1};$ at $y = ln 3$ this is $\frac{1}{3 - 1} = \frac{1}{2}$ .

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

⏱ 2mℹ️ Source: QB

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