# Edexcel C3 past paper question

• January 2nd 2010, 06:38 AM
Turple
Edexcel C3 past paper question
I'm using past papers to help with revision but found this question really difficult. Hope someone can help :)

Equation of curve: y=(2x-1)tan2x, x is between 0 and pi/4 (including 0)

The curve has a minimum at the point P. The x-coordinate of P is k.

a) Show that k satisfies the equation

4k-sin4k-2=0

I wasn't sure where to start but tried substituting k into the equation of the graph which hasn't helped :s

Thanks
• January 2nd 2010, 07:14 AM
skeeter
Quote:

Originally Posted by Turple
I'm using past papers to help with revision but found this question really difficult. Hope someone can help :)

Equation of curve: y=(2x-1)tan2x, x is between 0 and pi/4 (including 0)

The curve has a minimum at the point P. The x-coordinate of P is k.

a) Show that k satisfies the equation

4k-sin4k-2=0

I wasn't sure where to start but tried substituting k into the equation of the graph which hasn't helped :s

Thanks

used a calculator and determined $k \approx 0.276515...$

this value of $k$ does not satisfy the equation $4k - \sin(4k) - 2 = 0$
• January 2nd 2010, 07:24 AM
earboth
Quote:

Originally Posted by Turple
I'm using past papers to help with revision but found this question really difficult. Hope someone can help :)

Equation of curve: y=(2x-1)tan2x, x is between 0 and pi/4 (including 0)

The curve has a minimum at the point P. The x-coordinate of P is k.

a) Show that k satisfies the equation

4k-sin4k-2=0

I wasn't sure where to start but tried substituting k into the equation of the graph which hasn't helped :s

Thanks

Differentiate the function using product rule:

$y'= 2 tan(2x)+ (2x-1) \cdot \dfrac2{(\cos(2x))^2}$

Set y ' = 0 and start to solve for x:

$2 tan(2x)+ (2x-1) \cdot \dfrac2{(\cos(2x))^2} = 0~\implies~2\sin(2x) \cdot \cos(2x) + 4x - 2=0$ $~\implies~2 \cdot \frac12 \cdot \sin(4x) + 4x -2=0$

which yields:

$\sin(4x)+4x-2=0$

Could it be that there is a typo in your text?
• January 3rd 2010, 07:20 AM
Turple
Quote:

Originally Posted by earboth
Could it be that there is a typo in your text?

Ah yeah sorry about that! Thanks for pointing it out though

It is actually the same but with 4k+sin4k-2
• January 3rd 2010, 08:02 AM
earboth
Quote:

Originally Posted by Turple
Ah yeah sorry about that! Thanks for pointing it out though

It is actually the same but with 4k+sin4k-2

I don't know a method to solve the equation algebraically. If you use an iterative method (for instance Newton's method) you'll get

$x = k \approx 0.276515039...$ which is the value skeeter posted yesterday.

I used as initial value

$x_0 = 0.5$