Author Topic: FOUR FANTASTI-GOOD QUESTIONS! (Read 1396 times) Tweet Share

the.watchman · « **on:** February 24, 2010, 03:16:45 pm »

Here goes, #1000!

Here are some quiz questions (non-calc) for all MM-takers, PM me the answers within the next two days and I should reply to give you a score out of ten!

1) Find $\frac{d}{dx}(log_e (\tan x))$ , hence find an anti-derivative of $\frac{\sin^2 x + 1}{\sin x \cos x}$ (3 marks)

2) Find the general solution of $4\cos^2 (3x) \tan^2 (3x) = 3$ (2 marks)

3) Prove that $f(x) = e^x - log_e x$ is increasing for $x>1$ (3 marks)

4) If $f \Big(\frac{x}{2}+1 \Big) = 3x^4 + 29x^3 + 95x^2 + 137x + 71$ , find f(x) (2 marks)

Good luck!

TrueTears · « **Reply #1 on:** February 24, 2010, 03:19:22 pm »

3. f'(x) = e^x - \frac{1}{x}

e^x > \frac{1}{x} \ \forall \ x > 1

e^x - \frac{1}{x} > 0 \ \forall \ x>1

Quote from: the.watchman on February 24, 2010, 03:16:45 pm

4) If $f \Big(\frac{x}{2}+1 \Big) = 3x^4 + 29x^3 + 95x^2 + 137x + 71$ (2 marks)

q doesnt make sense

EDIT: white for no spoiler

the.watchman · « **Reply #2 on:** February 24, 2010, 03:20:49 pm »

Noooo...don't post the answers, it's a quiz

Just PM me the answers

the.watchman · « **Reply #3 on:** February 24, 2010, 07:19:30 pm »

Quote from: TrueTears on February 24, 2010, 03:19:22 pm

EDIT: white for no spoiler

Lulz, i guess that'll do

TrueTears · « **Reply #4 on:** February 24, 2010, 07:20:53 pm »

GerrySly · « **Reply #5 on:** February 24, 2010, 07:24:22 pm »

Quote from: TrueTears on February 24, 2010, 07:20:53 pm

ok.

You've lost ideas for posts now TT? Starting to burn out lol

TrueTears · « **Reply #6 on:** February 24, 2010, 07:46:33 pm »

no just trying different styles

http://vcenotes.com/forum/index.php/topic,9668.msg159631.html#msg159631

the.watchman · « **Reply #7 on:** February 24, 2010, 10:24:47 pm »

Quote from: GerrySly on February 24, 2010, 07:24:22 pm

Quote from: TrueTears on February 24, 2010, 07:20:53 pm
ok.
You've lost ideas for posts now TT? Starting to burn out lol

LOL, that's funny

crappy · « **Reply #8 on:** February 25, 2010, 05:07:49 pm »

IT HASN'T EVEN BEEN 2 DAYS YET.

YOU LIED

the.watchman · « **Reply #9 on:** February 25, 2010, 05:09:23 pm »

Quote from: crappy on February 25, 2010, 05:07:49 pm

IT HASN'T EVEN BEEN 2 DAYS YET.

YOU LIED

LOL I did say within the next two days

I'll white it out then until tomorrow

EDIT: Gah! I'll post it again tmoz, sorry for the confusion

TrueTears · « **Reply #10 on:** February 25, 2010, 05:09:30 pm »

Quote from: crappy on February 25, 2010, 05:07:49 pm

IT HASN'T EVEN BEEN 2 DAYS YET.

YOU LIED

gg

the.watchman · « **Reply #11 on:** February 28, 2010, 09:27:56 am »

Here are my solutions, did you like the questions?:

1)

$\frac{d}{dx} (log_e (\tan x)) = [\tan x]' \times \frac{1}{\tan x}$

$\frac{d}{dx} (log_e (\tan x)) = \frac{1}{\cos^2 x} \cdot \frac{\cos x}{\sin x}$

$\frac{d}{dx} (log_e (\tan x)) = \frac{1}{\sin x \cdot \cos x}$

So, $\int \frac{1}{\sin x \cdot \cos x}\, dx = log_e (\tan x) + C$ , where C is a constant

Because, $\frac{\sin^2 x + 1}{\sin x \cdot \cos x} = \frac{\sin^2 x}{\sin x \cdot \cos x} + \frac{1}{\sin x \cdot \cos x}$

$\frac{\sin^2 x + 1}{\sin x \cdot \cos x} = \tan x + \frac{1}{\sin x \cdot \cos x}$

Therefore, $\int \frac{\sin^2 x + 1}{\sin x \cdot \cos x}\, dx = \int \tan x\, dx + \int \frac{1}{\sin x \cdot \cos x}\,dx$

$\int \frac{\sin^2 x + 1}{\sin x \cdot \cos x}\, dx = -log_e (\cos x) + log_e (\tan x) + C$

$\int \frac{\sin^2 x + 1}{\sin x \cdot \cos x}\, dx = log_e (\frac{1}{\cos x}) + log_e (\tan x) + C$

$\int \frac{\sin^2 x + 1}{\sin x \cdot \cos x}\, dx = log_e (\frac{\sin x}{\cos^2 x}) + C$

So an anti-derivative of $\frac{\sin^2 x + 1}{\sin x \cdot \cos x}$ is $log_e (\frac{\sin x}{\cos^2 x})$

2)

$4\cos^2 (3x) \cdot \tan^2 (3x) = 3$

$4\cos^2 (3x) \cdot \frac{\sin^2 (3x)}{\cos^2 (3x)} = 3$

$4\sin^2 (3x) = 3$

$\sin^2 (3x) = \frac{3}{4}$

$\sin (3x) = \pm \frac{\sqrt{3}}{2}$

$3x = n\pi + \frac{\pi}{3}, n\pi - \frac{\pi}{3}$

$3x = \frac{3n\pi + \pi}{3}, \frac{3n\pi - \pi}{3}$

$x = \frac{\pi(3n + 1}{9}, \frac{\pi(3n - 1}{9}$

3)

Because $f'(x) = e^x - \frac{1}{x}$

And $e^x > \frac{1}{x}$ for $x>1$ (at least)

So $f'(x) = e^x - \frac{1}{x}>0$ , $f(x)$ is increasing for $x>1$

4)

Let $f(x) = ax^4 + bx^3 + cx^2 + dx + e$

Then $f \Big(\frac{x}{2}+1 \Big) = a(\frac{x}{2}+1)^4 + b(\frac{x}{2}+1)^3 + c(\frac{x}{2}+1)^2 + d(\frac{x}{2}+1) + e$

$f \Big(\frac{x}{2}+1 \Big) = \frac{a}{16}x^4 + \frac{4a+b}{8}x^3 + \frac{6a+3b+c}{4}x^2 + \frac{4a+3b+2c+d}{2}x + a + b + c + d + e$

Because $f \Big(\frac{x}{2}+1 \Big) = 3x^4 + 29x^3 + 95x^2 + 137x + 71$

Equating co-efficients:

$\frac{a}{16} = 3$ , $a = 48$

$\frac{4a+b}{8} = 29$ , $4(48) + b = 232$ , $b = 40$

$\frac{6a+3b+c}{4} = 95$ , $6(48) + 3(40) + c = 380$ , $c = -28$

$\frac{4a+3b+2c+d}{2} = 137$ , $4(48) + 3(40) + 2(-28) + d = 274$ , $d = 18$

$a+b+c+d+e = 71$ , $48+40-28+18+e = 71$ , $e = -7$

SO $f(x) = 48x^4 + 40x^3 - 28x^2 + 18x - 7$

Mao · « **Reply #12 on:** February 28, 2010, 10:56:33 am »

A much shorter way for #4:

Let $g(x) = f\left(\frac{x}{2} + 1\right)$ , then $f(x) = g\left(2x -2\right)$ [by applying the inverse transformations]

Also, note for #1, be very careful about the domain you work with. $\log(\tan(x))$ does not have the same domain as $\log\left(\frac{\sin(x)}{\cos^2(x)}\right)$

the.watchman · « **Reply #13 on:** February 28, 2010, 10:57:47 am »

Quote from: Mao on February 28, 2010, 10:56:33 am

Your way of doing (4) is very long and indirect.

Let $g(x) = f\left(\frac{x}{2} + 1\right)$ , then $f(x) = g\left(2x -2\right)$ [by applying the inverse transformations]

LOL ok sorry

Thanks for the tip!

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Author Topic: FOUR FANTASTI-GOOD QUESTIONS! (Read 1396 times) Tweet Share

the.watchman

FOUR FANTASTI-GOOD QUESTIONS!

TrueTears

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the.watchman

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the.watchman

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TrueTears

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GerrySly

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TrueTears

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the.watchman

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crappy

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the.watchman

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TrueTears

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the.watchman

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