# Thread: Edexcel C3 past paper question

1. ## 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

2. 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 $\displaystyle k \approx 0.276515...$

this value of $\displaystyle k$ does not satisfy the equation $\displaystyle 4k - \sin(4k) - 2 = 0$

3. 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:

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

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

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

which yields:

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

Could it be that there is a typo in your text?

4. 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

5. 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

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

I used as initial value

$\displaystyle x_0 = 0.5$