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Exponentials and Logarithms

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Question 6

A curve has the equation y=e2xy = e^{2x}y=e2x

a.

The equation of the tangent to the curve at the point (a,e2a)(a, e^{2a})(a,e2a) is

y=2e2ax−2ae2a+e2a y = 2e^{2a}x - 2ae^{2a} + e^{2a} y=2e2ax−2ae2a+e2a

Verify this result.

[3]
b.

Verify that this tangent passes through the origin when a=0.5a=0.5a=0.5.

[2]
c.

Hence, show that the equation e2x=mxe^{2x} = mxe2x=mx has no real solutions when 0≤m<2e0 \leq m < 2e0≤m<2e.

[3]
Markscheme

Exponentials and Logarithms Questions

  1. A Level
  2. /Maths
  3. /Exponentials and Logarithms

277 exam-style questions on Edexcel A Level Maths Exponentials and Logarithms, covering 14.1 Exponential Functions, 14.2 y = e^x, 14.3 Exponential Modelling, 14.4 Logarithms, 14.5 Laws of Logarithms, 14.6 Solving Equations using Logarithms, 14.7 Working with Natural Logarithms, and 14.8 Logarithms and Non-Linear Data. Each one has a worked solution and a mark scheme showing where the marks go.

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