# P5 Questions

Watch
Announcements
This discussion is closed.
#1
Can anyone help me with these practice questions?

1)

Given that y = x^(sinhx)

prove:

y.d²y/dx² - (dy/dx)² = y² / x

2)

Given that sinhx = tant 0 < tant t < (pi)/2

Prove:

i) tanhx = sint

ii) e^x = sect + tant

Cheers. I could do them yesterday and now i cant and can't find my notes. I hope i don't forget stuff in the exam.
0
15 years ago
#2
What does ^ mean? And what exam board are you doing? Because I've never seen questions lke that in my life and I'm doing OCR P5 on monday!
0
15 years ago
#3
it means to the power of, 2^2 = 4
0
#4
well i'm doing edexcel, but hyperbolics and differentiation/integrastion are both on the ocr p5 syllabus, so it should resemble your work...
0
15 years ago
#5
I've done a question like the second one. Have you tried squaring it and then using identities? Although the question i did was prove that coshx = sect and the second part is just using the relationship you proved in the first part.

As for the first question I'd suggest taking logs and then using implicit differentiation and the product rule.
0
15 years ago
#6
The first question is a mistake, I have come across it on another forum on which somebody said they had rung up Heinmann who confirmed the question was wrong. It should be y=x^x, rather than y=x^(sinhx).
0
15 years ago
#7
(Original post by mikesgt2)
The first question is a mistake, I have come across it on another forum on which somebody said they had rung up Heinmann who confirmed the question was wrong. It should be y=x^x, rather than y=x^(sinhx).
yup ... my maths teacher said tht too ... the answer at the backis wat it should b for x^x
0
15 years ago
#8
(Original post by SUKBarracuda)

2)

Given that sinhx = tant 0 < tant t < (pi)/2

Prove:

i) tanhx = sint

ii) e^x = sect + tant
i)
tanhx = sinhx/coshx = tant/coshx

and cosh^2 x - sinh^2 x = 1
so coshx = root(1 + sinh^2 x) = root(1+tan^2 t) = root(sec^2 t) = sect

so tanhx = sinhx/coshx = tant/coshx = tant/sect
rearrange it and you ll get sint

ii)from above, coshx = sect
so sect + tant = coshx + sinhx = e^x
0
X
new posts
Back
to top

view all
Latest
My Feed

### Oops, nobody has postedin the last few hours.

Why not re-start the conversation?

see more

### See more of what you like onThe Student Room

You can personalise what you see on TSR. Tell us a little about yourself to get started.

### Poll

Join the discussion

Yes (65)
52%
Yes, but I want to swap them (10)
8%
No, but I know who I want to choose (14)
11.2%
No, I still don't know who I want to choose (33)
26.4%
I have decided I don't want to go to uni anymore and will not be choosing (3)
2.4%