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The curve C has equation $y = \frac{3 + \sin 2x}{2 + \cos 2x}$ (a) Show that \[ \frac{dy}{dx} = \frac{6 \sin 2x + 4 \cos 2x + 2}{(2 + \cos 2x)^{2}} \] (b) Find an equation of the tangent to C at the point on C where $x = \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 4
Question 8
![The-curve-C-has-equation-$y-=-\frac{3-+-\sin-2x}{2-+-\cos-2x}$--(a)-Show-that-\[-\frac{dy}{dx}-=-\frac{6-\sin-2x-+-4-\cos-2x-+-2}{(2-+-\cos-2x)^{2}}-\]--(b)-Find-an-equation-of-the-tangent-to-C-at-the-point-on-C-where-$x-=-\frac{\pi}{2}$-Edexcel-A-Level Maths Pure-Question 8-2011-Paper 4.png](https://cdn.simplestudy.io/assets/backend/uploads/question/MathsPure_2011_4_1_QP_7_2011.jpg)
The curve C has equation $y = \frac{3 + \sin 2x}{2 + \cos 2x}$ (a) Show that \[ \frac{dy}{dx} = \frac{6 \sin 2x + 4 \cos 2x + 2}{(2 + \cos 2x)^{2}} \] (b) Find an ... show full transcript
Worked Solution & Example Answer:The curve C has equation $y = \frac{3 + \sin 2x}{2 + \cos 2x}$ (a) Show that \[ \frac{dy}{dx} = \frac{6 \sin 2x + 4 \cos 2x + 2}{(2 + \cos 2x)^{2}} \] (b) Find an equation of the tangent to C at the point on C where $x = \frac{\pi}{2}$ - Edexcel - A-Level Maths Pure - Question 8 - 2011 - Paper 4
Step 1
Show that \( \frac{dy}{dx} = \frac{6 \sin 2x + 4 \cos 2x + 2}{(2 + \cos 2x)^{2}} \)
Answer
To differentiate the function, we apply the quotient rule:
Let:
- ( u = 3 + \sin 2x )
- ( v = 2 + \cos 2x )
The derivatives are:
- ( \frac{du}{dx} = 2 \cos 2x )
- ( \frac{dv}{dx} = -2 \sin 2x )
According to the quotient rule: [ \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2} ]
Now substituting in the values: [ \frac{dy}{dx} = \frac{(2 + \cos 2x)(2 \cos 2x) - (3 + \sin 2x)(-2 \sin 2x)}{(2 + \cos 2x)^{2}} ]
Simplifying the numerator: [ \frac{dy}{dx} = \frac{4 \cos 2x + 2 \cos^2 2x + 6 \sin 2x + 2 \sin^2 2x}{(2 + \cos 2x)^{2}} ]
By recognizing that ( \sin^2 2x + \cos^2 2x = 1 ), we can combine terms: [ \frac{dy}{dx} = \frac{6 \sin 2x + 4 \cos 2x + 2}{(2 + \cos 2x)^{2}} ]
Thus proved.
Step 2
Find an equation of the tangent to C at the point on C where $x = \frac{\pi}{2}$
Answer
First, we need to find the value of ( y ) at ( x = \frac{\pi}{2} ):
[ y = \frac{3 + \sin(\pi)}{2 + \cos(\pi)} = \frac{3 + 0}{2 - 1} = 3 ]
So, the point is ( \left( \frac{\pi}{2}, 3 \right) ).
Next, we find the derivative at this point: [ m = \frac{dy}{dx} \bigg|_{x=\frac{\pi}{2}} = \frac{6 \sin(\pi) + 4 \cos(\pi) + 2}{(2 + \cos(\pi))^2} = \frac{0 - 4 + 2}{1} = -2 ]
Using point-slope form, where ( m = -2 ):
[ y - 3 = -2\left( x - \frac{\pi}{2} \right) ]
This rearranges to: [ y = -2x + (2 + 3) = -2x + 5 ]
Thus the equation of the tangent is: [ y = -2x + 5 ]