4u Mathematics Marathon V 1.0 (1 Viewer)

[KeypadSDM]

Without the hint I suppose the only real method you could use is something like mine. Trying to get rid of the e^-x using manipulative algebra. Or noting the symmetry of the limits of integration, and using the fact that sin²[x] is even. All of those worked together quite nicely to make the question simpler.

 

[icycloud]

OK guys, since Keypad doesn't like posting questions, here is the next one!

Next Question:

Prove that 3^n > n^3 for all positive integers n > 3 using induction.

 

[KeypadSDM]

Next Question:

Prove that 3^n > n^3 for all positive integers n > 3 using induction.

This one's an easy one:

Spoiler

And we've gotta use the principle of mathematical induction. It's a shame, because we can do it fairly simply by using the property that the logarithm of x increases slower than any power of x:

Note, for n = 4 we have:

4 * Log[3] > 3 * Log[4], and as the logarithm increases slower than any power of x, we know that for all n >= 4 [integral or otherwise]:

n * Log[3] > 3 * Log[n] [for n > 4][strictly, as we started @ n = 4]
3ⁿ > n³ [Putting both sides to the power of e]

But alas, I need to do it the hard way:

Spoiler

RTP: 3ⁿ > n³ for all integers n > 3

For n = 4 we have:
3⁴ = 81 > 64 = 4³

Now assume the statement is true for n = k (>= 4)

3^k > k³

we now need to show that the statement is true for n = k + 1.

3^k+1 = 3*3^k > 3k³

Now we need to show that 3k³ > (k + 1)³
I.e. we need to show that 2k³ > 3k² + 3k + 1

Now, since k >= 4

k³ >= 4k² > 3k² > 12k = 3k + 9k > 3k + 36 > 3k + 1

Thus we have:
k³ > 3k² and:
k³ > 3k + 1

Adding these inequalities yields:
2k³ > 3k² + 3k + 1

I.e.
3k³ > (k + 1)³ (For all integral k >= 4)

So we have:
3^k+1 = 3*3^k > 3k³ > (k + 1)³
3^k+1 > (k + 1)³

Insert essay here.

Not such a bad question, now here's an integration question. Pretty simple, just don't look too hard for the solution. [And don't even think about doing a substitution]

Integral((cos[x] - 1)e^{sin[x] + x} + cos[x]e^sin[x])/(1 + e^x)²dx

Hint:

Spoiler

Quotient Rule

 

[icycloud]

Not such a bad question, now here's an integration question. Pretty simple, just don't look too hard for the solution. [And don't even think about doing a substitution]

Integral((cos[x] - 1)esin[x] + x + cos[x]esin[x])/(1 + ex)2dx

Spoiler

What's wrong with the substitution u = e^sin(x) / (1+e^x)?? (Yes, ok I'll admit I read your hint... couldn't resist hehe)

 

[SeDaTeD]

(e^sinx)/(1+e^x) + c

 

[KeypadSDM]

(e^sinx)/(1+e^x) + c

Come on guys, I'm trying here.

Spoiler

At least do the rearrangement before writing down "The Integrator's" answer.

 

[icycloud]

Come on guys, I'm trying here.

Spoiler

At least do the rearrangement before writing down "The Integrator's" answer.

Here it is:

Spoiler

Choose u = e^sin(x) and v = 1 + e^x
u' = cos(x) e^sin(x)
v' = e^x

Now, I = ∫{(cos(x)-1)e^sin(x)e^x + cos(x)e^sin(x)} / (1+e^x)^2 dx
= ∫{cos(x)e^sin(x)e^x - e^sin(x)e(x) + cos(x)e^sin(x)} / v^2 dx
= ∫{u'(v-1) - uv' + u'} / v^2 dx
= ∫{vu' - uv'} / v^2 dx
= u/v + C
= e^sin(x) / (1+e^x) + C
#

Is that good enough?

 

[KeypadSDM]

Fine, this is more of what I was after.

Spoiler

Using the hint, note that the bottom is v² so pick appropriate v. Then using some manipulative algebra take out some occurunces of v and v' in the top line and note the convenience of e^sin[x]

You couldn't use your "convenient" method without knowing the integral before doing it.

 

[KeypadSDM]

Ok, here's the next question. I'm making up for lost time:

∫(2cos[x] + sin[2x])/(cosec[x] + sin[x])dx

Hint 1:

Spoiler

u = sin[x]

Hint 2:

Spoiler

multiply top and bottom by sin[x] or u

 

[icycloud]

Did not use the hints. Here's my solution:

Spoiler

I = ∫(2cos[x] + sin[2x])/(cosec[x] + sin[x]) dx
= ∫(2cos[x] + 2sin[x]cos[x]) / (1/sin[x] + sin[x]) dx
= ∫2cos[x] (1+sin[x]) / (1/sin[x] + sin[x]) dx

Let u = sin[x], du = cos[x] dx

I = ∫2(1+u) / (1/u + u) du
= ∫2u(1+u) / (u² + 1) du
= ∫2u / (1+u²) du + 2∫u² / (1+u²) du
= ln(u²+1) + 2∫du - 2∫du/(1+u²)
= ln(u²+1) + 2u - 2arctan(u) + C
= ln(sin²[x] + 1) + 2sin[x] - 2arctan(sin[x]) + C
#

Next question:
Express { 1 + i * Tan[(4k + 1)/4m pi] } ^m in a+ib form, where m and k are integers.

Edit: Sorry guys, the first "i" was meant to be a "1" in the above question. I've edited it to reflect this.

 

[KeypadSDM]

Spoiler

Set @ = (4k + 1)pi/(4m)

(1 + iTan[@])^m
=sec^m[@](cis[@])^m
=sec^m[@](cis[m@])
=sec^m[@](cis[kpi + pi/4])

I'm thinking about how to simplify, give me a moment.

=sec^m[@]cos[kpi + pi/4](1 + itan[kpi + pi/4])
=sec^m[@]cos[kpi + pi/4](1 + i)

Uhh, maybe some more. Maybe.

=sec^m[@](-1)^k(1 + i)/Sqrt[2]

 

[SeDaTeD]

Wha?
I didn't use the integrator. Just don't like typing up full solutions up. But I did use the (1+e^x)^2 as v^2 and split up and regroup the top bit.

 

[Templar]

Why use the integrator when you've got Mathematica.

 

[SeDaTeD]

Of course Templar. Slight problem. I don't have Mathematica.

 

[KeypadSDM]

And everyone's got "the integrator"

 

[pLuvia]

Next Question
Find the critical points of these graphs:

1) y = x + |x|
2) y = x² - |x|
3) y = |x| + |x-2|
4) y = |x| - |x-2|

 

[Mountain.Dew]

Next Question
Find the critical points of these graphs:

1) y = x + |x|
2) y = x² - |x|
3) y = |x| + |x-2|
4) y = |x| - |x-2|

when u mean 'critical points', do u mean x-intercepts + y-intercepts?

 

[pLuvia]

Where the point on the graph is undefined. I.e the "sharp" points

 

[Porcia]

Next Question
Find the critical points of these graphs:

1) y = x + |x|
2) y = x² - |x|
3) y = |x| + |x-2|
4) y = |x| - |x-2|

<spoiler>where x=0 for all </spoiler>

 

[pLuvia]

I'll post up the answers, forget about the last 2

Spoiler

1)
y=x+|x|
y' = 1+1=2 (x>0)
or
y' = 1-1=0 (x<0)
As x-->0⁺ y'-->2
As x-->0^- y' -->0

(0,0) is a critical point

2)
y=x²-|x|
y'=2x-1 (x>0)
or
y'=2x+1 (x<0)
As x-->0⁺ y'-->2x-1
As x-->0^- y'-->2x+1

(0,0) is a critical point