Author Topic: Kilbaha 2010 exam 2 wtf!!! (Read 11614 times) Tweet Share

stonecold · « **on:** September 25, 2010, 06:34:51 pm »

This one was so bloody difficult. I couldn't do barely any of question 1, and the last part of q 4 I also had no idea!!!

The solutions are horrible as well. They don't really say anything. Lots of proves/show questions, which they just say this and this and this and therefore this shows whatever you had to show. There is no written/conceptual explanations...

Anyway could someone please explain Q1 a.iv, bi and Q4 b.

Whatlol · « **Reply #1 on:** September 25, 2010, 06:45:15 pm »

I gave up on math methods after doing this exam !

stonecold · « **Reply #2 on:** September 25, 2010, 06:54:57 pm »

Quote from: Whatlol on September 25, 2010, 06:45:15 pm

I gave up on math methods after doing this exam !

Wish I could give up on methods!

brightsky · « **Reply #3 on:** September 25, 2010, 07:15:58 pm »

b) i) $g'(x) = 12 sec^2(x) - 5 \tan(x)\sec(x) - \sin(x)$

$g'(x) = \frac{12}{\cos^2(x)} - \frac{5 \sin(x)}{\cos^2(x)} - \frac{\sin(x)\cos^2(x)}{\cos^2(x)}$

$g'(x) = \frac{12 - 5\sin(x) - \sin(x)\cos^2(x)}{\cos^2(x)}$

$g'(x) = \frac{12 - 5\sin(x) - \sin(x)(1 - \sin^2(x))}{\cos^2(x)}$

$g'(x) = \frac{12 - 5\sin(x) - \sin(x) + \sin^3(x)}{\cos^2(x)}$

$g'(x) = \frac{\sin^3(x) - 6\sin(x) + 12}{\cos^2(x)}$

We recall that from a), $f(x) = ax^3 - 6ax + 12$ . By inspection, we find that $g'(x) = \frac{f(\sin(x))}{\cos^2(x)}$ when we set $a = 1$ .

brightsky · « **Reply #4 on:** September 25, 2010, 07:21:29 pm »

The coordinates of P is $(a, f(a))$ , which in other words is $\left(a, \frac{1}{a}\right)$

$f'(a) = -\frac{1}{a^2}$

Hence the equation of the tangent, bearing in mind that it passes through point $\left(a, \frac{1}{a}\right)$ , is given by:

$y - \frac{1}{a} = -\frac{1}{a^2} (x - a)$

$y = -\frac{1}{a^2} x + \frac{2}{a}$

stonecold · « **Reply #5 on:** September 25, 2010, 07:24:01 pm »

Quote from: brightsky on September 25, 2010, 07:21:29 pm

The coordinates of P is $(a, f(a))$ , which in other words is $\left(a, \frac{1}{a}\right)$

$f'(a) = -\frac{1}{a^2}$

Hence the equation of the tangent, bearing in mind that it passes through point $\left(a, \frac{1}{a}\right)$ , is given by:

$y - \frac{1}{a} = -\frac{1}{a^2} (x - a)$

$y = -\frac{1}{a^2} x + \frac{2}{a}$

Thanks, I was talking about the very last page though. This q was rather straight forward.

Whatlol · « **Reply #6 on:** September 25, 2010, 07:24:54 pm »

The only thing i didnt understand is why does sin(x)cos^2(x) = (1-sin^2(x)).

brightsky · « **Reply #7 on:** September 25, 2010, 07:30:45 pm »

Quote from: Whatlol on September 25, 2010, 07:24:54 pm

The only thing i didnt understand is why does sin(x)cos^2(x) = (1-sin^2(x)).

Substitute $\cos^2(x) = 1 - \sin^2(x)$ .

Whatlol · « **Reply #8 on:** September 25, 2010, 07:31:40 pm »

Quote from: brightsky on September 25, 2010, 07:30:45 pm

Quote from: Whatlol on September 25, 2010, 07:24:54 pm
The only thing i didnt understand is why does sin(x)cos^2(x) = (1-sin^2(x)).

Substitute $\cos^2(x) = 1 - \sin^2(x)$ .

oh! ok thank you.

8039 · « **Reply #9 on:** September 25, 2010, 07:43:45 pm »

They should call it killyourmoralebaha

brightsky · « **Reply #10 on:** September 25, 2010, 08:03:37 pm »

Oh, sorry stonecold.

4. b) i) Draw the graph. Or since $n> 0$ , we can directly consider the graph given in a).

Bound the graph between $n$ and $n+1$ as required. Then $f(n) = \frac{1}{n}$ and $f(n+1) = \frac{1}{n+1}$ . Plot those points in. The right rectangle approximation (just the one block) $= \frac{1}{n+1} ((n+1) - 1) = \frac{1}{n+1} >$ real area. The left rectangle approximation (just the one block) $= \frac{1}{n} ((n+1)- 1) = \frac{1}{n} <$ real area.

Since the real area is $\int^{n+1}_{n} \frac{1}{x} dx$ , then that proves that $\frac{1}{n} < \int^{n+1}_{n} \frac{1}{x} dx < \frac{1}{n+1}$

4. b) ii) $\int^{n+1}_{n} \frac{1}{x} dx = \log_e|n+1| - \log_e|n| = \log_e\left|\frac{n+1}{n}\right| = \log_e\left(\frac{n+1}{n} \right) \text{(since n} > \text{0)} = \log_e\left(1 + \frac{1}{n} \right)$

We know that $\frac{1}{n} < \log_e\left(1 + \frac{1}{n}\right) < \frac{1}{n+1}$

Looking at $\frac{1}{n} < \log_e\left(1 + \frac{1}{n}\right)$ first:

$n\log_e\left(1 + \frac{1}{n}\right) > 1$ (remembering $n > 0$ )

$\log_e\left(1 + \frac{1}{n}\right)^n > 1$

$e > \left(1 + \frac{1}{n}\right)^n$

Now consider $\log_e\left(1 + \frac{1}{n}\right) < \frac{1}{n+1}$

$(n+1)\log_e\left(1 + \frac{1}{n}\right) < 1$

$\log_e\left(1 + \frac{1}{n}\right)^{n+1} < 1$

$e < \left(1 + \frac{1}{n}\right)^{n+1}$

Combining the two, we have the answer required.

EDIT: Cheers, m@tty!

Jdog · « **Reply #11 on:** September 25, 2010, 08:04:28 pm »

where did u guys get these 2010 ones,

any links ?

thanks heaps in advance,

8039 · « **Reply #12 on:** September 25, 2010, 08:35:40 pm »

Quote from: Jdog on September 25, 2010, 08:04:28 pm

where did u guys get these 2010 ones,

any links ?

thanks heaps in advance,

http://vcenotes.com/forum/index.php/topic,29541.0.html

m@tty · « **Reply #13 on:** September 25, 2010, 10:14:29 pm »

^^brightsky, you need to add a "[tex]" to your post; the LaTeX at the start of 4. b) ii) isn't rendered.

kyzoo · « **Reply #14 on:** September 25, 2010, 11:43:11 pm »

0.o...Now I'm really tempted to check out Kilbaha exams for Spesh

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Author Topic: Kilbaha 2010 exam 2 wtf!!! (Read 11614 times) Tweet Share

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