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

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A curve C has equation y = e^{2x} an x, x ≠ (2n + 1) rac{ au}{2}. (a) Show that the turning points on C occur where tan x = -1. (b) Find an equation of the ta... show full transcript

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

Step 1

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 C, we first differentiate the given equation.

  1. Differentiate the function: The derivative of y with respect to x is computed using the product rule: dydx=e2xtanx+e2xsec2xdxdx\frac{dy}{dx} = e^{2x} \tan x + e^{2x} \sec^2 x \cdot \frac{dx}{dx} Therefore, we get: dydx=e2xtanx+2e2xsec2x. \frac{dy}{dx} = e^{2x} \tan x + 2e^{2x} \sec^2 x.

  2. Set the derivative to zero: To find the turning points, we set the derivative to zero: dydx=02e2xtanx+e2xsec2x=0. \frac{dy}{dx} = 0 \Rightarrow 2e^{2x} \tan x + e^{2x} \sec^2 x = 0. We can simplify this expression: 2tanx+1sec2x=0. 2 \tan x + 1 \sec^2 x = 0.

  3. Solve the equation: Rearranging gives: 2tanx+1=0tanx=1.2 \tan x + 1 = 0 \Rightarrow \tan x = -1.
    Hence, turning points occur when (\tan x = -1).

Step 2

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 to the curve C at the point where x = 0, follow these steps:

  1. Evaluate the derivative at x = 0: Calculate (\frac{dy}{dx}) when (x = 0): dydxx=0=e0tan(0)+2e0sec2(0)=0+21=2. \frac{dy}{dx} \bigg|_{x=0} = e^{0} \cdot \tan(0) + 2e^{0} \cdot \sec^2(0) = 0 + 2 \cdot 1 = 2.
    Thus, the slope of the tangent line at (x = 0) is (2).

  2. Determine the point on the curve: Calculate y when (x = 0): y=e0tan(0)=10=0.y = e^{0} \cdot \tan(0) = 1 \cdot 0 = 0.
    So, the point is (0, 0).

  3. Use the point-slope form to write the tangent equation: With point (0, 0) and slope 2, the equation is: y0=2(x0)y=2x. y - 0 = 2(x - 0) \Rightarrow y = 2x.
    Hence, the equation of the tangent line is (y = 2x).

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