The equation of the tangent to the curve y = eˣ at the point where it crosses the y-axis is: A. x − y + 1 = 0 B. x + y = 1 C. x − y = 1 D. x + y + 1 = 0 Answer: A) x − y + 1 = 0 Explanation: Crosses y-axis at x = 0, y = 1. dy/dx = eˣ, at x = 0 slope = 1. Tangent: y − 1 = 1(x − 0) → y = x + 1 → x − y + 1 = 0.

imo class 12 application of derivatives

The equation of the tangent to the curve y = eˣ at the point where it crosses the y-axis is:

VAVidaara Admin Asked 6d ago 0 views 0 answers

The equation of the tangent to the curve y = eˣ at the point where it crosses the y-axis is:

Answer: A) x − y + 1 = 0

Explanation: Crosses y-axis at x = 0, y = 1. dy/dx = eˣ, at x = 0 slope = 1. Tangent: y − 1 = 1(x − 0) → y = x + 1 → x − y + 1 = 0.