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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 1 - 2011 - Paper 4
Question 1
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 equat... 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 1 - 2011 - Paper 4
Step 1
Show that \( \frac{dy}{dx} = \frac{6 \sin 2x + 4 \cos 2x + 2}{(2 + \cos 2x)^{2}} \)
Answer
To find ( \frac{dy}{dx} ), we will apply the quotient rule:
Let ( u = 3 + \sin 2x ) and ( v = 2 + \cos 2x ).
Using the quotient rule ( \frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^{2}} ), we first calculate ( \frac{du}{dx} ) and ( \frac{dv}{dx} ):
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For ( u = 3 + \sin 2x ):
- ( \frac{du}{dx} = 2 \cos 2x )
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For ( v = 2 + \cos 2x ):
- ( \frac{dv}{dx} = -2 \sin 2x )
Now substituting these into the quotient rule:
[ \frac{dy}{dx} = \frac{(2 + \cos 2x)(2 \cos 2x) - (3 + \sin 2x)(-2 \sin 2x)}{(2 + \cos 2x)^{2}} ]
Expanding the numerator:
[ = \frac{4 \cos 2x + 2 \cos^{2} 2x + 6 \sin 2x}{(2 + \cos 2x)^{2}} ]
Finally,
[ = \frac{6 \sin 2x + 4 \cos 2x + 2}{(2 + \cos 2x)^{2}} ]
This shows the required result.
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 compute the value of ( y ) when ( x = \frac{\pi}{2} ):
[ y = \frac{3 + \sin(\pi)}{2 + \cos(\pi)} = \frac{3 + 0}{2 - 1} = 3 ]
So, the point on the curve is ( \left( \frac{\pi}{2}, 3 \right) ).
Next, we calculate ( \frac{dy}{dx} ) at this point:
[ \frac{dy}{dx} = \frac{6 \sin(\pi) + 4 \cos(\pi) + 2}{(2 + \cos(\pi))^{2}} = \frac{0 - 4 + 2}{1} = -2 ]
The slope of the tangent at this point is ( m = -2 ).
Using the point-slope form of the line equation, we have:
[ y - y_1 = m(x - x_1) ] [ y - 3 = -2\left(x - \frac{\pi}{2}\right) ]
Simplifying this gives:
[ y = -2x + (\pi + 3) ]
Thus the equation of the tangent in the required form is:
[ y = -2x + (\frac{\pi}{2} + 3) ]