# Edexcel C3 Specimen Paper... Help! Pls..

Hey has anyone done the paper yet? Cos i tried it and I'm stuck on the last question.. And i know there are answers but they miss out steps and i dunno what they're doing!
If someone could please explain it to me it would be very much appreciated!
violet
Hey has anyone done the paper yet? Cos i tried it and I'm stuck on the last question.. And i know there are answers but they miss out steps and i dunno what they're doing!
If someone could please explain it to me it would be very much appreciated!

What's the question?
Gaz031
What's the question?

Erm ok the first part is:

Given that y = tan x + 2 cos x, find the exact value of dy/dx at x = pi/4

.. I differentiated it to dy/dx = sec^2x - 2 sin x
but then i get confused with substituting pi/4 in place of x cos it wants the exact value

And also the answer is 2 - square root of 2 (if it helps!)
violet
Erm ok the first part is:

Given that y = tan x + 2 cos x, find the exact value of dy/dx at x = pi/4

.. I differentiated it to dy/dx = sec^2x - 2 sin x
but then i get confused with substituting pi/4 in place of x cos it wants the exact value

And also the answer is 2 - square root of 2 (if it helps!)

So you want (sec pi/4)^2 - 2sin(pi/4) = 1/(cos pi/4)^2 - 2sin(pi/4) = 2 - 2[0.5rt2] = 2-rt2.
If ever they want an exact value and you can't immediately recall it (eg value of sin pi/4) try putting it in on your calculator and squaring the number to see if you have a surd or dividing by constants such as pi.
7 (i) Given that y = tan x + 2 cos x, find the exact value of dy/dx at x = pi/4

dy/dx = sec^2x - 2sinx

At x = pi/4,
dy/dx = 1/(cos(pi/4))^2 - 2sin(pi/4)
= 1/(1/root2)^2 - 2root2/2
= 1/(1/2) - (2root2)/2
= 2 - root2

(ii) Given that x = tan 0.5y, prove that dy/dx = 2/(1+x^2)

dx/dy = 0.5sec^2 0.5y
dy/dx = 1/(dx/dy) = 1/(0.5sec^2 0.5y) = 2/sec^2 0.5y

sec^2(A) = 1 + tan^2(A)
dy/dx = 2/(1 + tan^2 0.5y) = 2/(1 + x^2)
Ok yeh i get the first bit but i dont get the root 2 (sorry to be really annoying) but i thought sin (pi/4) is 1/&#8730;2 so when you multiply it by 2 doesn't it become 2/&#8730;2 ? Or am i totally missing the point here..
violet
Ok yeh i get the first bit but i dont get the root 2 (sorry to be really annoying) but i thought sin (pi/4) is 1/&#8730;2 so when you multiply it by 2 doesn't it become 2/&#8730;2 ? Or am i totally missing the point here..

Rationalise the denominator.
E.g. 1/&#8730;2 = 1/&#8730;2 x &#8730;2/&#8730;2 <--- (effectively multiplying by 1 here) = (&#8730;2)/2
Try both 1/&#8730;2 and(&#8730;2)/2 on your calculator.
so 2sin(pi/4) = 2 x (&#8730;2)/2 = &#8730;2
violet
Ok yeh i get the first bit but i dont get the root 2 (sorry to be really annoying) but i thought sin (pi/4) is 1/&#8730;2 so when you multiply it by 2 doesn't it become 2/&#8730;2 ? Or am i totally missing the point here..

sinpi/4=1/rt2=rt2/2, so 2sinpi/4=rt2.
Ooooh yeh i see! Thanks

Ok and the other bit is part (iii) of that question. Does anyone have it or shall i type it up aswell? Cos it has loads of e^(-x) so it might look strange if i do.
For (iii) you use the product rule to differentiate, then pull out e^-x so you have dy/dx=e^-x[acos2x+bsin2x]. You then express acos2x+bsin2x in the form Rcos(2x+a) by setting acos2x+bsin2x=Rcos(2x+a) and expanding out Rcos(2x+a) then equating coefficients of cos2x and sin2x.
Gaz031
For (iii) you use the product rule to differentiate, then pull out e^-x so you have dy/dx=e^-x[acos2x+bsin2x]. You then express acos2x+bsin2x in the form Rcos(2x+a) by setting acos2x+bsin2x=Rcos(2x+a) and expanding out Rcos(2x+a) then equating coefficients of cos2x and sin2x.

Yep I've done the differentiating, I just don't get how i can express it in that form. Where does the sin 2x go?
Ooh wait i get the formula you're using but dont the sin and cos have to be squared to be put into it? Or not..?
(iii) Given that y = e^–x sin 2x, show that dy/dx can be expressed in the form Re^–x cos (2x + @). Find, to 3 significant figures, the values of R and &#945;, where 0 < &#945; < pi/2

y = e^-x sin2x
u = e^-x, v = sin2x
dy/dx = u*dv/dx + v*du/dx (product rule)
dy/dx = e^-x(2cos2x) + sin2x(-e^-x)
dy/dx = 2e^-x(cos2x) - e^-x(sin2x)
dy/dx = e^-x(2cos2x - sin2x)

= Re^-x cos(2x + @) [cos(A+B)]
= Re^-x(cos2xcos@ - sin2xsin@)
= Re^-x(cos2xcos@) - Re^-x(sin2xsin@)

Comparing coefficients gives us.
Rcos@ = 2 and Rsin@ = 1
(Rcos@)^2 + (Rsin@)^2 = 2^2 + 1^2 = 5 = R^2
so R = rt5 = 2.24

Rcos@ = 2
cos@ = 2/rt5
@ = cos^-1(2/rt5) = 0.464
y=(e^-x)sin2x
dy/dx=(e^-x)(2cos2x)+(-e^-x)sin2x
dy/dx=(e^-x)[2cos2x-sin2x]
Let Rcos(2x+a)=2cos2x-sin2x
R[cos2xcosa-sin2xsina]=2cos2x-sin2x
cos2x[Rcosa]-sin2x[Rsina]=2cos2x-sin2x.
Rcosa=2, Rsina=1.
tana=Rsina/Rcosa=1/2, hence a=arctan0.5.
R^2 = 1^2 + 2^2 = 5, hence R=rt5.
dy/dx=(rt5)(e^-x)cos(2x+arctan0.5)
Use your calculator to give R and a to the required accuracy.
Gaz031
y=(e^-x)sin2x
dy/dx=(e^-x)(2cos2x)+(-e^-x)sin2x
dy/dx=(e^-x)[2cos2x-sin2x]
Let Rcos(2x+a)=2cos2x-sin2x
R[cos2xcosa-sin2xsina]=2cos2x-sin2x
cos2x[Rcosa]-sin2x[Rsina]=2cos2x-sin2x.
Rcosa=2, Rsina=1.
tana=Rsina/Rcosa=1/2, hence a=arctan0.5.
R^2 = 1^2 + 2^2 = 5, hence R=rt5.
dy/dx=(rt5)(e^-x)cos(2x+arctan0.5)
Use your calculator to give R and a to the required accuracy.

nice.
You too, yours was more comprehensive
Thanks you guys! You're so.... clever! I sat at that for like an hour and got nowhere. But thank yooooo!
I have another question... It's a little one though!
You see for question 2(c) on that same paper.. Well what do you do after you've done f(2.09455) and f(2.09465) ..(the upper and lower limits) .. Do you just say that's its correct because there is a change of sign between the 2 answers? Or do you have to do another little calculation?
violet
I have another question... It's a little one though!
You see for question 2(c) on that same paper.. Well what do you do after you've done f(2.09455) and f(2.09465) ..(the upper and lower limits) .. Do you just say that's its correct because there is a change of sign between the 2 answers? Or do you have to do another little calculation?

You might also have to say f is continuous in the interval so the root is in the interval which is 2.0946 (correct to 4dp)
Gaz031
You might also have to say f is continuous in the interval so the root is in the interval which is 2.0946 (correct to 4dp)

Yup, ok, thanks dude!