4u Mathematics Marathon V 1.0 (1 Viewer)

[Yip]

Q. Obtain the value of the constant a so that the sum of the local maximum and mimumum values of the function y=e^[2x]+ae^x+2x is -11
[I think this would be relevant to the harder 3u topic]

 

[Stan..]

a = sqrt(-4/11)
Character limit is annoying

 

[Yip]

wrong stan ~.~
hint 1: apply sum and product rules

 

[Stan..]

wrong stan ~.~
hint 1: apply sum and product rules and discriminants

I had a feeling that it was wrong, I placed it back in and it didn't come out as planned.

 

[Yip]

working would be nice too

 

[Riviet]

Q. Obtain the value of the constant a so that the sum of the local maximum and mimumum values of the function y=e^[2x]+ae^x+2x is -11

I got a=22 by differentiating and using sum of roots of the resulting quadratic. Is product of roots and discriminant necessary?

 

[Yip]

actually, the discriminant is unnecessary ~.~
but u do need the product of the roots
btw riviet, ur ans is wrong...~.~
ur error is that ur thinking that the sum of the local maximum and minimum values is referring to the x-coordinates of these local extrema, local maxumum values and minimum values are understood to be the y-values of the local extreme values
ie the quadratic y=x^2+5 has a local minimum value of 5 at 0

 

[Yip]

Its been like 5 days with no one posting any attempts, so i better post the solution ~.~

Spoiler

f(x)=e^2x+ae^x+2x
f'(x)=2e^2x+ae^x+2
Since extreme values occur when f'(x)=0,
2e^2x+ae^x+2=0
Let e^x=t
2t^2+at+2=0
Now, for f(x) to have both a relative maximum and minimum value, there must be 2 solutions to f'(x) that are positive[since t=e^x>0 for all real x]
Hence,
From Discriminant>0[since there are 2 solutions], a^2-16>0
Let the roots of f'(x) be e^x1 and e^x2, where x1 and x2 are the x-coordinates of the local extrema

e^x1+e^x2=-a/2>0
[e^x1][e^x2]=e^[x1+x2]=1
x1+x2=0

Since -a/2>0 and a^2-16>0,
a<-4

Therefore,
f(x1)+f(x2)=(e^2x1+ae^x1+2x1)+(e^2x2+ae^x2+2x2)
=(e^x1+e^x2)^2-2[e^x1][e^x2]+a(e^x1+ex2)+2(x1+x2)
=[-a/2]^2-2(1)+a(-a/2)+2(0)
=-0.25a^2-2=-11
a^2=36
a=6,-6
However, from a<-4,
a=6 is extraneous, and thus
a=-6

 

[sikeveo]

what is the square root of 6i?

 

[webby234]

(a+bi)^2 = 6i
a^2 - b^2 = 0
+ or - a = + or - b

2ab = 6
ab = 3

a = +/- sqrt 3
b = +/- sqrt 3

so sqrt of 6i = +/- (sqrt 3 + sqrt3i)

 

[pLuvia]

Spoiler

z² be the sqrt of 6i (letting z = x+iy)
.: x² - y² = 0
2xy = 6
y = 3/x

x² - 9/x² = 0
x⁴ - 9 = 0
x = +sqrt 3
y = +sqrt 3

Squareroots are +(sqrt 3+sqrt 3)

 

[pLuvia]

Next question: Prove that (cos @ + isin @)ⁿ = cos n@ + isin n@

 

[klaw]

Next question: Prove that (cos @ + isin @)ⁿ = cos n@ + isin n@

Spoiler

When n=1,
LHS=cos @ + isin@=RHS
.: true for n=1

Assume true for n=k
i.e. (cos @ + isin @)^k = cos k@ + isin k@

When n=k+1,
LHS=(cos @ + isin @)^(k+1)
=(cos @ + isin @)(cos k@ + isin k@)
=[cosk@cos@-sink@sin@]+i(cosk@sin@+sink@cos@)
=cos(n+a)@+isin(n+1)@=RHS

(Insert essay here)

Given that z=cis@, show that z^n+z^-n=2cosn@. Hence, solve the equation 4z^4-4z^3+6z^2-4z+4=0

 

[pLuvia]

zⁿ+z^-n
= cos n@ + isin n@ + cos (-n@) + isin (-n@)
= 2cos n@

As req'd

4z⁴-4z³+6z²-4z+4 = 0
Divide by z²
= 4z²-4z+6-4z^-1+4z^-2 = 0
= 4(z² + z^-2) - 4(z + z^-1) +6 = 0
= 8cos 2@ - 8cos @ + 6 = 0
= 8(2cos²@ - 1) - 8cos@ + 6 = 0
= 16cos²@ - 8cos@ - 2 = 0
= 2[8cos²@ - 4cos@ - 1] = 0
Solve for cos@

@ = 46*55* and 100*32*

 

[sikeveo]

what is the square root of -6i? (my imagination is severely limited)

 

[Riviet]

what is the square root of -6i? (my imagination is severely limited)

Lol! Extremely limited XD

Spoiler

(a+bi)²=-6i

a²-b²+2abi=-6i

a²-b²=0 *

ab=-3 **

b=-3/a

sub. into *

=> a²-9/a²=0

a⁴-9=0
(a²+3)(a+sqrt3)(a-sqrt3)=0
a=+sqrt3
.: b=+3/sqrt3

.: sqrt(-6i)=+(sqrt3+3i/sqrt3)

Next question:

Find ∫6cos³xsinx dx

 

[KeypadSDM]

Find ∫6cos³xsinx dx

This one's not that hard if you see it straight away. Sometimes you can be REALLY thrown off track if you go about it the wrong way:

Spoiler

∫6.cos³[x].sin[x].dx

=-∫6.cos³[x].d/dx(cos[x]).dx

=-6.cos⁴[x]/4 + K

=- (3/2)cos⁴[x] + K

 

[Riviet]

Hmm... that is alot faster than how i had to do it. Are you sure that's right keypad?

 

[icycloud]

Hmm... that is alot faster than how i had to do it. Are you sure that's right keypad?

Mmm, that's the way I would've done it as well. Keypad's method of "direct integration" is equivalent to using the substitution u = cos[x].

I'm guessing you would've used the expansion of cos³[x] or something similar? You would get the same answer either way, differing by only a constant.

 

[Riviet]

Oh god i did it a heaps long way lol. It involved 2cos@.

Oh well, i thought it was hard because no one responded soon after.