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# P3 Questions watch

1. 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
2. I get the feeling that I want to do this Edexcel P3 Module just to see how it is.
3. (Original post by 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
4. (Original post by bono)
I get the feeling that I want to do this Edexcel P3 Module just to see how it is.
What do you mean?
5. (Original post by imasillynarb)
What do you mean?
Seems like the majority of Maths Questions on here are Edexcel P3 ones! How hard can it be!
6. 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

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
7. 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
8. (Original post by 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?
9. What was with the steak and kidney pie thing?
10. (Original post by 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.
11. (Original post by 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)
12. (Original post by bono)
What was with the steak and kidney pie thing?
sillynarb wanted the answer in terms of pie, so thats what i gave him.
13. (Original post by elpaw)
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)
I think I'll stick to turning it into tan
14. (Original post by XTinaA)
I think I'll stick to turning it into tan
yeah, thats probably easier, but i hate tans.
15. 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
16. (Original post by elpaw)
yeah, thats probably easier, but i hate tans.
Stay indoors then <bu-boom, crash!>
17. are you saying the equation is (8-2x)/(x-2)? 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.
18. (Original post by elpaw)
are you saying the equation is (8-2x)/(x-2)? 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.
From the looks of the question Paw's right.
19. I'm sorry, i didnt read it properly. if it is p - [2x/x-2] then f(x) = 6.

sorry again I said its too late for me!!!!
20. Sorry, my bad for not using brackets..(8-2x)/(x-2)

Why would it be -2 sorry?

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