A-Level Edexcel Maths Pure Solving Equations: A curve C has equation y = e^{2

A curve C has equation y = e^{2x} an x, x eq (2n + 1) rac{ ext{π}}{2} - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 6

Question 3

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A curve C has equation y = e^{2x} an x, x eq (2n + 1) rac{ ext{π}}{2}. (a) Show that the turning points on C occur where $ \tan x = -1 $. (b) Find an equ... show full transcript

Worked Solution & Example Answer:A curve C has equation y = e^{2x} an x, x eq (2n + 1) rac{ ext{π}}{2} - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 6

Step 1

(a) Show that the turning points on C occur where $ \tan x = -1 $

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Answer

To find the turning points of the curve, we need to differentiate the function.

Start with the equation:

y = e^{2x} an x

Now, apply the product rule for differentiation:

\frac{dy}{dx} = e^{2x} \frac{d}{dx} (\tan x) + \tan x \frac{d}{dx} (e^{2x})

We know:

\frac{d}{dx} (\tan x) = \sec^2 x

and

\frac{d}{dx} (e^{2x}) = 2e^{2x}

Substituting these into our derivative gives:

\frac{dy}{dx} = e^{2x} \sec^2 x + \tan x (2e^{2x})

Set the derivative equal to zero to find turning points:

\frac{dy}{dx} = 0 \Rightarrow e^{2x} \sec^2 x + 2e^{2x} \tan x = 0

Simplifying:

e^{2x} (\sec^2 x + 2 \tan x) = 0

Since ( e^{2x} ) is never zero, we can set the inside to zero:

\sec^2 x + 2 \tan x = 0

Rearranging gives:

2 \tan x + 1 = 0\Rightarrow \tan x = -\frac{1}{2}

Since we are looking for where ( \tan x = -1), we can look at the specific values of x that yield this value. The turning points occur when ( \tan x = -1 ).

Step 2

(b) Find an equation of the tangent to C at the point where $ x = 0 $

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Answer

To find the equation of the tangent line at the point where ( x = 0 ), we first need to determine the point on the curve. Substitute ( x = 0 ) into the original equation:

y = e^{2(0)} \tan(0) = 1 \cdot 0 = 0

So, the point is ( (0, 0) ).

Next, we need the slope of the tangent line, which is given by ( \frac{dy}{dx} ) at ( x = 0 ). We have:

\frac{dy}{dx} = e^{2x} \sec^2 x + 2e^{2x} \tan x

Evaluate this at ( x = 0 ):

\frac{dy}{dx} \bigg|_{x=0} = e^{0} \sec^2(0) + 2e^{0} \tan(0) = 1 + 0 = 1

This tells us the slope (m) of the tangent line is 1. Therefore, using the point-slope form of the equation for the tangent line:

y - y_1 = m(x - x_1)

Substituting ( (x_1, y_1) = (0, 0) ) and slope ( m = 1 ) gives:

y - 0 = 1(x - 0) \Rightarrow y = x

Thus, the equation of the tangent to C at the point where ( x = 0 ) is:

y = x.

A curve C has equation y = e^{2x} an x, x eq (2n + 1) rac{ ext{π}}{2} - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 6

Worked Solution & Example Answer:A curve C has equation y = e^{2x} an x, x eq (2n + 1) rac{ ext{π}}{2} - Edexcel - A-Level Maths Pure - Question 3 - 2008 - Paper 6

(a) Show that the turning points on C occur where \( \tan x = -1 \)

(b) Find an equation of the tangent to C at the point where \( x = 0 \)

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