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P3 Questions

f(x) = (e^2x).sin2x , 0 < x < 180

a)Find the values of x for which f(x) = 0 in terms of pie

b) use calculus to find the coords of the turning points of the graph y = f(x)

2)

f(x) = p - 2x/x+q x not equal to -q

where p and q are constant. The curve, C, with equation y = f(x) has an asymptote with equation x = 2 and passes through the point with coords (3,2)

a) write down value of q (I got -2)

b) show that p = 8 (Did this)

c) Write down an equation for the second asymptote to C (cant do this)

Cheers

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Reply 1

username9816

I get the feeling that I want to do this Edexcel P3 Module just to see how it is.

Reply 2

Barny

imasillynarb

f(x) = (e^2x).sin2x , 0 < x < 180

a)Find the values of x for which f(x) = 0 in terms of pie
b) use calculus to find the coords of the turning points of the graph y = f(x)

2)

f(x) = p - 2x/x+q x not equal to -q

where p and q are constant. The curve, C, with equation y = f(x) has an asymptote with equation x = 2 and passes through the point with coords (3,2)

a) write down value of q (I got -2)

b) show that p = 8 (Did this)

c) Write down an equation for the second asymptote to C (cant do this)

Cheers

1a) I get 0, pie/2
b) I get 3/8pie and 7/8pie

Reply 3

Barny

bono

I get the feeling that I want to do this Edexcel P3 Module just to see how it is.

What do you mean?

Reply 4

username9816

imasillynarb

What do you mean?

Seems like the majority of Maths Questions on here are Edexcel P3 ones! How hard can it be! :smile:

Reply 5

elpaw

f(x) = (e^2x).sin2x , 0 < x < 180

a)Find the values of x for which f(x) = 0 in terms of pie

e^2x =/= 0 for any x => sin2x = 0 => x = 0, pi/2 = steak and kidney pie :biggrin:

b) use calculus to find the coords of the turning points of the graph y = f(x)

f'(x) = 2(e^2x)sin2x + 2(e^2x)cos2x = 0 for stat point

sin2x + cos2x = 0

sin2x = -cos2x

2x = 3pi/4, 7pi/4

x = 3pi/8, 7pi/8

y = √2/2 e^(3pi/4), -√2/2 e^(7pi/4) respectively

Reply 6

elpaw

2)

f(x) = p - 2x/x+q x not equal to -q

where p and q are constant. The curve, C, with equation y = f(x) has an asymptote with equation x = 2 and passes through the point with coords (3,2)

a) write down value of q (I got -2)

b) show that p = 8 (Did this)

c) Write down an equation for the second asymptote to C (cant do this)

f(x) = 8 - 2x/(x-2) = 8 - 2/(1-2/x) = 6 as x-> +/- infinity

=> f(x) = 6 is the horizontal asymptote

Reply 7

makesomenoise

elpaw

sin2x = -cos2x

2x = 3pi/4, 7pi/4

x = 3pi/8, 7pi/8

y = √2/2 e^(3pi/4), -√2/2 e^(7pi/4) respectively

How can you just jump from the first step to the second?

Surely there's more simplification possible as well?

Reply 8

username9816

What was with the steak and kidney pie thing?

Reply 9

makesomenoise

XTinaA

How can you just jump from the first step to the second?

Surely there's more simplification possible as well?

Just to clarify my position I would've gone on to say tan2X = -1 and so on.

Reply 10

elpaw

XTinaA

How can you just jump from the first step to the second?

Surely there's more simplification possible as well?

yes there is, but i know that the only time sin/cos values are equal in magnitude are when they are √2/2 (i.e. (45 + 90n) degrees for natural n), so cos45==sin45, cos 135 == -sin135 etc. its just logic (you dont have to do everything in a strict mathematical formulation). but if you wanted to do the simplification, this is what you'd get: (eg)

sinx = cosx
=> sin²x = cos²x
=> sin²x = 1 - sin²x
=> 2sin²x = 1
=> sin²x = 1/2
=> sinx = +/- 1/√2 = +/- √2/2

then get the corresponding values of x (remembering they have to satisfy sinx = cosx, because some will satisfy sinx = -cosx because of the squaring)

Reply 11

elpaw

bono

What was with the steak and kidney pie thing?

sillynarb wanted the answer in terms of pie, so thats what i gave him.

Reply 12

makesomenoise

elpaw

I think I'll stick to turning it into tan :tongue:

Reply 13

elpaw

XTinaA

I think I'll stick to turning it into tan :tongue:

yeah, thats probably easier, but i hate tans.

Reply 14

Matin

elpaw:

c) Write down an equation for the second asymptote to C (cant do this)

f(x) = 8 - 2x/(x-2) = 8 - 2/(1-2/x) = 6 as x-> +/- infinity

=> f(x) = 6 is the horizontal asymptote

sorry mate but i think the horizontal asymptote is f(x)= -2
cause you didnt factorise the top correctly or
its really late for me!!!!

cheers

Reply 15

makesomenoise

elpaw

yeah, thats probably easier, but i hate tans.

Stay indoors then <bu-boom, crash!>

Reply 16

elpaw

are you saying the equation is (8-2x)/(x-2)? :mad:

brackets would help!

i took f(x) = 8 - [2x/(x-2)], where the asymptote is f(x)=6, but if f(x)=(8-2x)/(x-2), the asymptote would be f(x)=-2, as you said.

Reply 17

makesomenoise

elpaw

are you saying the equation is (8-2x)/(x-2)? :mad:

brackets would help!

i took f(x) = 8 - [2x/(x-2)], where the asymptote is f(x)=6, but if f(x)=(8-2x)/(x-2), the asymptote would be f(x)=-2, as you said.