Author Topic: Recreational Problems (SM level) (Read 76998 times) Tweet Share

asa.hoshi · « **Reply #30 on:** October 22, 2007, 07:03:33 am »

8. 2 (nearest whole number)

Ahmad · « **Reply #31 on:** October 22, 2007, 07:04:11 am »

Right =)

Can you post a brief note of how you did each?

Edit: 8 is not right

asa.hoshi · « **Reply #32 on:** October 22, 2007, 07:10:57 am »

10. sin2x = 2sinxcosx
int(2sinx.cosx.sinx)dx
let u= sinx
du/dx = cosx
u-sub integration basically.
= 4/3

asa.hoshi · « **Reply #33 on:** October 22, 2007, 07:14:42 am »

11. tan^2(x) = sec^2(x) -1
int(sec^2(x) -1)dx
= -pi/2 + 2

Ahmad · « **Reply #34 on:** October 22, 2007, 07:17:03 am »

Correct.

asa.hoshi · « **Reply #35 on:** October 22, 2007, 07:26:20 am »

P(1) = 0 in x^4 + x^3 - 4x^2 + x + 1
hence (x-1) is a factor
long division, = x^3+2x^2 - 2x+1
P(1) = 0 also. in there hence (x-1)
so... (x-1)(x^2-x -1) - factorizing by recognition
so (x-1)^2(x^2-x-1)=0
x= 1, (x^2-x-1)==> quadratic formulae
hence:x=1, x= [-3+squareroot(5)]/2, x= [-3-squareroot(5)]/2

asa.hoshi · « **Reply #36 on:** October 22, 2007, 07:27:51 am »

i hate typing on here... i can't type properly :x

8. i guess. i assumed if "1" is not correct than it might be "2" LOL

Ahmad · « **Reply #37 on:** October 22, 2007, 07:31:54 am »

That's completely logical, if only 1 were not the correct answer :lol:

Defiler · « **Reply #38 on:** October 30, 2007, 01:10:48 pm »

12. Let f(x) = sin^4(x) + cos^4(x). Find the 100th derivative of f. [hide=hint](sin^2(x) + cos^2(x))^2 = 1 [/hide]

f(x) = sin^4(x) + cos^4(x)

Therefore f(x) = 1 (sin^2(theta) + cos^2(theta) = 1)

Hence f(x) = 1
f'(x) = 0
f''(x) = 0
f'''(x) = 0

See a pattern ? o_O

Hence the 100th derivate is the same as all those before it, and all equal to 0.

Ahmad · « **Reply #39 on:** October 30, 2007, 01:12:35 pm »

$\sin^4x + \cos^4x \neq 1$

Reconsider your argument :wink:

Defiler · « **Reply #40 on:** October 30, 2007, 01:13:19 pm »

oh touche.

I can simplify it down to (3+cos(4x))/4, which I assume you're probably not looking for, as, when you derive it 100 times, you get

f(100)(x) = 401734511064747568885490523085290630550748445698208825344cos(4x)

=)

Collin Li · « **Reply #41 on:** December 14, 2007, 08:27:11 pm »

Collin Li · « **Reply #42 on:** December 14, 2007, 11:41:29 pm »

I have figured out how to do questions 5 and 6 with a little bit of help. I will not post these solutions, since I didn't figure them out myself.

$\mbox{\textbf{7.} If }f\mbox{ is continuous and }f(x) + f(-x) = x^2\mbox{, what is }\displaystyle\int^1_{-1} f(x)\,dx\mbox{?}$

$\displaystyle\int^1_{-1}[f(x) + f(-x)]\,dx = \displaystyle\int^1_{-1}x^2\,dx$

$\Rightarrow \displaystyle\int^1_{-1}f(x)\,dx + \displaystyle\int^1_{-1}f(-x)\,dx = \frac{2}{3}$

$\mbox{Letting }u = -x\mbox{ yields:}$

$\Rightarrow \displaystyle\int^1_{-1}f(x)\,dx + \displaystyle\int^{-1}_1-f(u)\,du = \frac{2}{3}$

$\Rightarrow \displaystyle\int^1_{-1}f(x)\,dx + \displaystyle\int^1_{-1}f(u)\,du = \frac{2}{3}$

$\Rightarrow 2\displaystyle\int^1_{-1}f(x)\,dx = \frac{2}{3}$

$\Rightarrow \displaystyle\int^1_{-1}f(x)\,dx = \frac{1}{3}$

$\mbox{\textbf{8.} Let }f(x) = \frac{3x^4+16}{2x^4+11}\mbox{. Find the whole number nearest to }f(2^{32})$

$f'(x) = \frac{12x^3(2x^4+11) - 8x^3(3x^4 + 16)}{(2x^4+11)^2} = \frac{4x^3}{(2x^4+11)^2} > 0\mbox{ for }x > 0$

$\Rightarrow f(1) < f(2^{32}) < \lim_{x\to\infty}f(x)$

$\Rightarrow 19/13 < f(2^{32}) < 3/2$

$\mbox{Therefore, }f(2^{32})\mbox{ is closest to }1$

$\mbox{\textbf{9.} Find the smallest possible constant }A\mbox{ such that }\log_ex\leq Ax^2\mbox{ for all }x > 0$

$\mbox{The smallest possible constant }A\mbox{ is when }\log_ex\mbox{ and }Ax^2\mbox{ \emph{touch} at some point for }x>0$

$\mbox{\emph{Touching} requires that their derivatives are equal upon intersection: }\frac{d}{dx}\left(Ax^2\right) = \frac{d}{dx}\left(\log_ex\right)\mbox{ when }Ax^2 = \log_ex$

$\frac{d}{dx}\left(Ax^2\right) = \frac{d}{dx}\left(\log_ex\right) \Rightarrow 2Ax = \frac{1}{x} \Rightarrow x = \frac{1}{\sqrt{2A}}\mbox{ for }x>0$

$\mbox{When }x = \frac{1}{\sqrt{2A}}\mbox{, the intersection becomes: }A\cdot\frac{1}{2A} = \log_e\frac{1}{\sqrt{2A}}$

$\Rightarrow \frac{1}{2} = -\frac{1}{2}\log_e(2A)$

$\Rightarrow A = \frac{1}{2e}$

Collin Li · « **Reply #43 on:** December 15, 2007, 01:57:44 am »

These ones are easy integrals:

$\mbox{\textbf{10.} }\displaystyle\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \sin (x)\cdot\sin (2x)\,dx =\displaystyle\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \sin (x)\cdot 2\sin (x)\cos (x)\,dx$

$= 2\displaystyle\int^{\frac{\pi}{2}}_{-\frac{\pi}{2}} \sin^2x\cdot\cos x\,dx$

$\mbox{Letting }u = \sin x\mbox{ yields:}$

$= 2\displaystyle\int^1_{-1} u^2\,du = \frac{2}{3}\cdot(1 + 1) = \frac{4}{3}$

$\mbox{\textbf{11.} }\displaystyle\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}} \frac{\sin^2x}{\cos^2x}\,dx = \displaystyle\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}} \tan^2x\,dx = \displaystyle\int^{\frac{\pi}{4}}_{-\frac{\pi}{4}} (\sec^2x - 1)\,dx = \tan\left(\frac{\pi}{4}\right) + \tan\left(\frac{\pi}{4}\right) - \left(\frac{\pi}{4} + \frac{\pi}{4}\right) = 2 - \frac{\pi}{2}$

$\mbox{\textbf{12.} Let }f(x) = \sin^4x + \cos^4x\mbox{. Find the 100th derivative of }f\mbox{. Hint: }(\sin^2x + \cos^2x)^2 = 1$

$f(x) = \sin^4x + 2\sin^2x\cos^2x + \cos^4x - 2(\sin x\cos x)^2 = (\sin^2x + \cos^2x)^2 - 2\cdot(\frac{1}{2}\sin 2x)^2 = 1 - \frac{1}{2}\sin^2 2x$

$\mbox{Using }\cos^2x - \sin^2x = \cos 2x \Rightarrow \cos 2x = 1 - 2\sin^2x \Rightarrow \sin^2x = \frac{1}{2}(1-\cos 2x)$

$\Rightarrow f(x) = 1 - \frac{1}{4}(1-\cos 4x) = \frac{3}{4} + \frac{1}{4}\cos 4x$

$\Rightarrow f'(x) = -\sin 4x$
$\Rightarrow f''(x) = -4\cos 4x$
$\Rightarrow f^{(3)}(x) = 4^2\sin 4x$
$\Rightarrow f^{(4)}(x) = 4^3\cos 4x$

$\mbox{Guess: }f^{(2n)}(x) = (-1)^n\cdot4^{2n-1}\cos 4x\mbox{ for }n \geq 1$

$\mbox{The basis }(n=1)\mbox{: }f^{(2)}(x) = (-1)^1\cdot 4^{2-1}\cos 4x = -4\cos 4x\mbox{ as required.}$

$\mbox{The inductive step, show that: }\frac{d^2}{dx^2}\left(f^{(2n)}(x)\right) = f^{2(n+1)}(x)$

$\frac{d^2}{dx^2}\left(f^{(2n)}(x)\right) = \frac{d^2}{dx^2}\left((-1)^n\cdot4^{2n-1}\cos 4x\right) = \frac{d}{dx}\left((-1)^{n+1}\cdot4^{2n}\sin 4x\right) = (-1)^{n+1}\cdot4^{2n+1}\cos 4x$

$= (-1)^{n+1}\cdot4^{2(n+1) - 1}\cos 4x = f^{2(n+1)}(x)\mbox{ as required.}$

$\Rightarrow f^{(100)}(x) = (-1)^{50}\cdot4^{100-1}\cos 4x = 4^{99}\cos 4x$

All problems solved!

Mao · « **Reply #44 on:** December 15, 2007, 09:58:02 am »

*commits suicide*

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