Qs need help. (1 Viewer)

[ToO LaZy ^*]

i need help on these questions..
These Qs are from the 2001 HSC Specimen paper

 

[CM_Tutor]

Q1, using the result from (i), integrand becomes A + B(2cos x - sin x) / (2sin x + cos x), which is easy to integrate as 2cos x - sin x is the derivative of 2sin x + cos x, so you get Ax + Bln|2sin x + cos x| + C, for some constant C

 

[CM_Tutor]

Q2 is an induction, pity the question specifies that you have to use induction, as it is much more easily done without induction, as LHS = (2n)! / n!. (That doesn't help with your problem, it's just an observation.) What is causing the problem with the induction?

 

[CrashOveride]

For the first one just divide both sides by 2sinx+cosx and then take the integral of the LHS once u know A and B

shud work out to be:

2x + 3ln(2sinx + cosx) + C

 

[CrashOveride]

Woops too slow, ok CM can take over now =p

PS. CM come to 4unit forum and lend me a hand with some SHM stuff

 

[CM_Tutor]

Q3(i) Use Pythagoras' theorem on ΔOFD
(ii) Angle DFE = Angle BOC = 0.4 radians
(iii) Find the length of DE by using the cosine rule in ΔDFE, and the find angle DOE by using the cosine rule in isosceles ΔODE.

 

[ToO LaZy ^*]

Q1. got it

Q2. my problem is step 3--> proving n=k+1

Q3.
i) ...but i don't have the values for OD..
ii) yep
iii) needs part i)..not done..

 

[wogboy]

Q2. my problem is step 3--> proving n=k+1

assuming that it's true for n=k,
(k+1)*(k+2)*(k+3)*...*2k = 2^k * (1*3*...*(2k-1))

It's true for n=k+1 if,
(k+2)*(k+3)*...*2k*(2k+1)*(2k+2) = 2^(k+1) *(1*3*...*(2k-1)*(2k+1))
i.e. (k+2)*(k+3)*...*(2k+1)*{2*(k+1)} = 2*2^k * (1*3*...*(2k-1)*(2k+1))
i.e. (k+1)*(k+2)*(k+3)*...*(2k+1) = 2^k * (1*3*...*(2k-1)*(2k+1))
i.e. (k+1)*(k+2)*(k+3)*...*2k = 2^k * (1*3*...*(2k-1))

which is true as per our assumption

Q3.
i) ...but i don't have the values for OD..

You can see that OD = 10cm, since any interval drawn from O to the surface of the hemisphere is a radius of that hemisphere. For iii), follow CM_Tutor's advice.

 

[ToO LaZy ^*]

assuming that it's true for n=k,
(k+1)*(k+2)*(k+3)*...*2k = 2^k * (1*3*...*(2k-1))

It's true for n=k+1 if,
(k+2)*(k+3)*...*2k*(2k+1)*(2k+2) = 2^(k+1) *(1*3*...*(2k-1)*(2k+1))
i.e. (k+2)*(k+3)*...*(2k+1)*{2*(k+1)} = 2*2^k * (1*3*...*(2k-1)*(2k+1))
i.e. (k+1)*(k+2)*(k+3)*...*(2k+1) = 2^k * (1*3*...*(2k-1)*(2k+1))
i.e. (k+1)*(k+2)*(k+3)*...*2k = 2^k * (1*3*...*(2k-1))

which is true as per our assumption

i don't get how you got to the bold bit...

 

[wogboy]

i don't get how you got to the bold bit...

On the left hand side of the equation, move the (k+1) factor all the way over to the left. Then cancel off the factor of 2 on both sides of the equation.

 

[Grey Council]

thats a strange way to solve an induction question. Never seen anyone do it like that before.

manipulating to get the assumption.

 

[ToO LaZy ^*]

yeah..that's what i thought..and i still don't quite get it...

 

[CM_Tutor]

The middle bit by the normal method:

B Let k be a value of n for which the result is true.
That is, (k + 1)(k + 2)...(2k - 1)(2k) = 2^k[1 * 3 * 5 * ... * (2k - 1)] _____ (**)

We must now prove the result for n = k + 1.
That is, we must prove:
(k + 2)(k + 3)...(2k + 1)(2k + 2) = 2^k+1[1 * 3 * 5 * ... * (2k + 1)]

LHS = (k + 2)(k + 3)...(2k + 1)(2k + 2)
= (k + 2)(k + 3)...(2k)(2k + 1)[2(k+1)]
= [(k + 1)(k + 2)(k + 3)...(2k)] * (2k+1) * 2
= 2^k[1 * 3 * 5 * ... * (2k -1)] * (2k + 1) * 2, using the induction hypothesis (**)
= 2^k+1[1 * 3 * 5 * ... * (2k - 1) * (2k + 1)]
= RHS

etc...

 

[Grey Council]

hokay, lets do it standard induction method:
To prove true for n = k+1
It is true if k is replaced by k+1 on both sides of the equation.
taking LHS
(k+2)*(k+3)*...*2k*(2k+1)*(2k+2) ........ (I just replaced k by k+1 wherever I saw a k)

(k+1)(k+2)*(k+3)*...*2k*(2k+1)*(2k+2)/(k+1) .... (times top and bottom by k+1)
using assumption (ie (k+1)*(k+2)*(k+3)*...*2k = 2^k * (1*3*...*(2k-1)) )
2^k * (1*3*...*(2k-1)) * (2k+1)*(2k+2) / (k+1)
now, you factorise the last bit (2k+2), taking the 2 out:
2.2^k * (1*3*...*(2k-1)) * (2k+1)*(k+1) / (k+1)
the (k+1)'s cancel out, leaving you with:
2^(k+1) * (1*3*...*(2k-1)(2k+1))
in this instance, k+1 has replaced all k, therefore true if assumption is true.

yada yada yada.

hrm, sorry, hard to explain on the net, try using pen and paper. If you still don't get it, please ask.

 

[Grey Council]

nnooooooooooooo!
all that hard work, and CM beats me to it. GUH!

 

[Estel]

Damn I posted only to see you both had.
Deleted it.

Too Lazy: there's a HSC induction question almost every year for 3U. If you go through and do them all, you'll find they're all rather straightforward.

Grey council: factorising the last term is much more efficient.

 

[Estel]

nnooooooooooooo!
all that hard work, and CM beats me to it. GUH!

I share your pain
Only I set out the whole thing...

 

[CM_Tutor]

nnooooooooooooo!
all that hard work, and CM beats me to it. GUH!

I share your pain
Only I set out the whole thing...

Sorry to both of you.
*runs away to snow*

 

[Grey Council]

hah!

And I tried to make it obvious. I mean, Wogboy's solution is brilliant, but kinda hard to see. :-\ Well, not really, that question is just wierd. I started to treat it like a normal sums question, then got bamboozled, and did it the long way....all on the computer.

so you'll forgive me for not being efficient.
lol, and I don't know what you mean. Factorise the last term?

 

[ToO LaZy ^*]

ahhh..i see...thanks everyone