Author Topic: Interesting questions (spesh) (Read 13733 times) Tweet Share

Martoman · « **Reply #60 on:** October 08, 2010, 06:45:32 pm »

$h(x)$ is an increasing function and positive over the interval $0 \leq x \leq c$ where $f(0) = p, f(c) = 5$ and $\int_{p}^{5} h^{-1}(x)dx = c+p$ . The definite integral $\int_{0}^{c}h(x)dx = ?$

brightsky · « **Reply #61 on:** October 08, 2010, 07:44:03 pm »

Mostly diagram work but $\int^c_0 h(x) dx = 5c - \int^5_p h^{-1}(x) dx = 5c - c - p = 4c - p$ . Is this right?

Martoman · « **Reply #62 on:** October 08, 2010, 07:48:56 pm »

kamil9876 · « **Reply #63 on:** October 08, 2010, 08:12:19 pm »

Are complex numbers still part of the course?

1)How many complex numbers, $z$ , are there such that $z^{128}=1$ but $z^n$ is not 1 for $n=1,2,3...127$ ?

2)What is the product of $cis(0)*cis(2\pi/100)*cis(2*2\pi/100)...*cis(99*2\pi/100)$

3) Show that any polynomial of degree 100 with real coefficients can be written as a product of quadratics with real coefficients.

yep I am bored.

Martoman · « **Reply #64 on:** October 08, 2010, 11:49:34 pm »

2)

$cis(0+ \frac{2\pi}{100}(1+2+3+4...99))$

Forms an arithmetic series $\frac{n}{2}(a+l)$ with

$a = 1, l = 99, n = 99-1 +1 = 99$

$\frac{99}{2}(100) = 4950$ :smitten:

So $cis(\frac{2\pi}{100} \times 4950) = cis(99\pi)<br /><br />$

$cis(\pi)$ = -1....nice Q and thanks for Kamil9876 for giving the q and pointing out my error in a = 1, instead of a = 2 gah.

kamil9876 · « **Reply #65 on:** October 09, 2010, 12:28:40 am »

good except for the dividing by $2\pi$ wtf? and also the arithmetic sequence starts from 1.

A sneaky way is to decompose $z^{100}-1$ into a product of linear factors:

$z^{100}-1=(z-cis0)(z-cis(2pi/100) )....(z-cis(99 * 2\pi /100))$

and now just expand and equate coefficients. You get $-1$ . (notice that if u equation the $z$ coefficient you get another interesting identity, that the sum of the roots is 0 (which may be obvious for even n, like 100, but not so for odd n))

Martoman · « **Reply #66 on:** October 09, 2010, 12:31:28 am »

oh.

lol.

Mao · « **Reply #67 on:** October 09, 2010, 02:16:57 am »

No contenders for q1? The answer is 121 right?

itolduso · « **Reply #68 on:** October 09, 2010, 07:06:01 pm »

itolduso · « **Reply #69 on:** October 09, 2010, 07:12:34 pm »

3) Expand conjugate linear factors to get quadratics with real coefficients

ghadz7 · « **Reply #70 on:** October 09, 2010, 07:53:07 pm »

What is the hardest spesh exam you've done? In general which company makes the hardest exams? I personally found the itute's exam 2 easier than TSSM.

kamil9876 · « **Reply #71 on:** October 09, 2010, 09:35:06 pm »

correct on both itolduso. (could expand more on 3 though it is the right idea). For q1, which of the complex numbers have the property? can you generalise for naturals other than 128?

brightsky · « **Reply #72 on:** October 21, 2010, 12:38:20 am »

$\frac{d}{dx} (x \log_e(x)) = 1 + \log_e(x)$ So $\int \log_e(x) dx = x\log_e(x) - x + C$

For $\int \frac{xe^x}{\sqrt{1 + e^x}} dx$

Sub $u = \sqrt{1 + e^x}$ , $du = \frac{1}{2e^x\sqrt{e^x+1}} dx \implies du = \frac{1}{2(u^2 - 1)u} dx$

$\because x = \log_e(u^2 - 1)$

$\therefore 2 \int \log_e(u^2 - 1) du = 2 \left(\int \log_e(u+1) du + \int \log_e(u - 1) du\right)$

By the result found earlier,

$= 2 \left((u+1) \log_e(u+1) - u + (u - 1) \log_e(u - 1) - u\right) + C$

Substitute in and you're done.

Martoman · « **Reply #73 on:** October 21, 2010, 01:22:50 am »

*excellent* with a capital e, that one was quite difficult.

Though I haven't dedicated much thought to your problem, it seems intriguing still.

brightsky · « **Reply #74 on:** October 30, 2010, 09:49:02 pm »

As requested (solution to the definite integration question):

$\int^{1}_{0} \frac{\arctan(x)}{x+1} dx$

Integrate by parts, letting $f(x) = \arctan(x)$ and $g'(x) = \frac{1}{x+1}$

Integrand becomes: $= \frac{\pi}{4} \log_e(2) - \int^{1}_{0} \frac{\log_e(x+1)}{x^2 + 1} dx$

Let $x = \tan(u)$ , $dx = \sec^2(u) du$ , $du = \frac{dx}{x^2 + 1}$

So it becomes: $\frac{\pi}{4} \log_e(2) - \int^{\frac{\pi}{4}}_{0} \log_e(\tan(u) + 1) du$

This is where the trick comes in. For definite integrals $\int^{a}_{0} f(x) dx = \int^{a}_{0} f(a - x) dx$ .

Let $I = \int^{\frac{\pi}{4}}_{0} \log_e(\tan(u) + 1) du = \int^{\frac{\pi}{4}}_{0} \log_e\left(\tan\left(\frac{\pi}{4} - u\right) + 1\right) du$

By trig laws:

$I = \int^{\frac{\pi}{4}}_{0} \log_e\left(\frac{\tan(\frac{\pi}{4}) - \tan(u)}{1 + \tan(\frac{\pi}{4})\tan(u)} + 1\right) du<br />$

$= \int^{\frac{\pi}{4}}_{0} \log_e\left(\frac{1 - \tan(u)}{1 + \tan(u)} + 1\right) du$

$= \int^{\frac{\pi}{4}}_{0} \log_e\left(\frac{2}{\tan(u) + 1}\right) du$

$= \int^{\frac{\pi}{4}}_{0} (\log_e(2) - \log_e(tan(u) + 1)) du$

So $I = \int^{\frac{\pi}{4}}_{0} \log_e(2) du - I$

$2I = \int^{\frac{\pi}{4}}_{0} \log_e(2) du$

$2I = \frac{\pi}{4} \log_e(2)$

$I = \frac{\pi}{8} \log_e(2)$

So the whole original integral becomes:

$\frac{\pi}{4} \log_e(2) - \frac{\pi}{8} \log_e(2)$

$= \frac{\pi}{8} \log_e(2)$

Please tell me if there are any mistakes.

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