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Any Math Geniuses Here?
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| creon444 |
Hello.
We are currently on a brief survey of parametric equations in my math class. And these two equations were used as Bonus problems on my last take-home test, but no one in the class could figure out how to do any
of the two. So my question is, how the hell can you possibly solve them? I do realize they are of much greater difficulty and such, but I'm still curious as to what methods you would need to use when trying to solve these problems. The problems are:
1. Transform the equation (8*a^2+1)*(sinx)^3-(4*a^2+1)*sinx+2*a*(cosx)^3=0 to its explicit parametric form. That is, express x in terms of a, where a is a parameter.
2. Find all values of parameter a for which the equation (a-x^2-cos(11*Pi*x/4))*sqrt(8-a*x)=0 has an odd number of solutions on the interval -2<=x<=3.
Any explanations/hints/general thoughts would be of great help! Thank you! |
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| Floorfiller |
| i'll think about it later...its too early to do math... |
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| SuperFarStucker |
| Man, I'm getting a headache looking at that. |
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| mentalbarter |
| its ing maths damnit |
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| raek |
| the answer is.......chair |
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| Floorfiller |
| quote: | Originally posted by mentalbarter
its ing maths damnit |
not in america :rolleyes: |
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| Floorfiller |
| quote: | Originally posted by creon444
Hello.
We are currently on a brief survey of parametric equations in my math class. And these two equations were used as Bonus problems on my last take-home test, but no one in the class could figure out how to do any
of the two. So my question is, how the hell can you possibly solve them? I do realize they are of much greater difficulty and such, but I'm still curious as to what methods you would need to use when trying to solve these problems. The problems are:
1. Transform the equation (8*a^2+1)*(sinx)^3-(4*a^2+1)*sinx+2*a*(cosx)^3=0 to its explicit parametric form. That is, express x in terms of a, where a is a parameter.
2. Find all values of parameter a for which the equation (a-x^2-cos(11*Pi*x/4))*sqrt(8-a*x)=0 has an odd number of solutions on the interval -2<=x<=3.
Any explanations/hints/general thoughts would be of great help! Thank you! |
can you be a little more specific in your typing of the equation...i'm having trouble distinquishing whether its
a^2 + 1
or
a^(2+1)
and
sinx + 2
or
sin(x+2)
etc... |
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| jon |
| ^^ aye, if you could "draw" the equations in paint then link to the image i may be able to help more |
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| Noisician |
| quote: | Originally posted by creon444
2. Find all values of parameter a for which the equation (a-x^2-cos(11*Pi*x/4))*sqrt(8-a*x)=0 has an odd number of solutions on the interval -2<=x<=3.
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well, here's one way to solve it:
first, by plugging in 0 in place of x, we can find all values of a that give the initial equation the solution x=0:
if x=0, we get (a-1)√8=0 ⇔ a=1
thus, x=0 will be (one of) the initial equation's solution(s) if, and only if, a=1. we can also determine whether there are an odd number of solutions on the given interval by plugging in 1 instead of a. we'll get the equation:
(1-x²-cos(11πx/4)√(8-x)=0
since √(8-x) is strictly positive on the interval x∈[-2,3], on this interval the above equation can be rewritten as
x²+cos(11πx/4)=1
since f(x)=x²+cos(11πx/4) is an even function, the latter equation will have an odd number of solutions on the interval x∈[-2,2] (because even if x1 is its solution, then (-x1) will be one just as well. besides, x=0 is also its solution, which was already determined earlier. so the total number will always be ODD). in addition, that equation obviously has no solutions on the interval x∈[2,3], since on this specific interval
x²+cos(11πx/4)≥4-1=3>1
thus, if a=1, the initial equation has an odd number of solutions on the interval x∈[-2,3]. that means a=1 is one of the answers to this problem.
next, let's see what happens when a equals 0. if a=0, the initial equation can be rewritten as
x²+cos(11πx/4)=0
as has been noted earlier, the function on the left side is even, but then it follows that, on the interval x∈[-2,2], the above equation will have either an EVEN number of solutions, since x=0 is no longer among the solutions (remember, in this case, a=0, not 1!), or none at all. in addition, the above equation has no *additional* solutions on the interval x∈[2,3] because
x²+cos(11πx/4)≥4-1=3>0
thus, if a=0, the initial equation has an even number of solutions on the given interval. so a=0 should be excluded from the final answer.
next, for each and every value of parameter a such that a≠0 and a≠1, the initial equation has the solution x=8/a. in order for this solution to be within the given interval (otherwise it would be invalid), the double inequality
-2≤8/a≤3
must be true for all such values of x. when u solve it, u'll get two possible intervals for parameter a: a≤-4 and a≥8/3.
thus, all we need to do is study 3 intervals involving a. after having done that, the initial problem can safely be considered done. the 3 intervals are:
i. a≤-4
ii. -4< a<8/3, a≠0, a≠1
iii. a≥8/3
i. if a≤-4, the initial equation has the solution x=8/a on the given interval. we now need to determine if it has any more of them. it will have additional solutions if, and only if, the equation
x²+cos(11πx/4)=a
has solutions when a≤-4. but since x²+cos(11πx/4)≥cos(11πx/4)≥-1>-4, it appears that the above equation is unsolvable on the interval a≤-4. thus, when a≤-4, the initial equation has only one solution, x=8/a. that means any a≤-4 is among the right answers to the initial problem.
ii. if -4 < a < 8/3, a ≠ 0, a ≠1, then √(8-ax)≠0 on the interval x∈[-2,3] (see above). that means on this particular interval for a, the initial equation is equivalent to the equation
x²+cos(11πx/4)=a
if x∈[-2,2], it will have either an even number of solutions or none at all (see way above). if x∈(2,3], it will have no solutions whatsoever because, on this interval,
x²+cos(11πx/4)≥x²-1≥4-1=3>8/3
thus, when -4< a<8/3, a≠0, a≠1, the initial equation either has no solutions at all or has an EVEN number of solutions. that means, all values within the above interval should be excluded from the final answer.
iii. if a≥8/3, the initial equation will have AT LEAST one solution (x=8/a) on the interval x∈[-2,3], which was determined earlier. but, unlike with case i, on this interval for a, all eligible values must satisfy not one but two conditions:
1). since we already know that, on this interval, the initial equation has x=8/a as its solution (see above), to make the total number of solutions ODD, the equation
x²+cos(11πx/4)=a
must have an EVEN number of solutions on this interval or not have any at all. however, all such solutions must also meet a second condition:
2). for all such values of a, (8-ax) must be non-negative.
so in this particular case, the initial problem can be restated as:
on the interval a∈[8/3,∞), find all values of a, such that the equation
x²+cos(11πx/4)=a
has an EVEN number of solutions on the interval x∈[-2,8/a).
iii-a. let's see what happens if a=4 (i picked 4 because it makes for a whole number in the above interval). in this case, the above equation can be rewritten as
x²+cos(11πx/4)=4, and the interval itself is x∈[-2,8/4=2)
on the interval x∈(-2,2), that equation does, in fact, have an even number of solutions. however, x=-2 is also its solution (u can check it by plugging it in in the above equation). that means, when a=4, the above equation actually has an odd number of solutions, making the grand total even. thus, a=4 must be excluded from the final answer.
iii-b. if a>4, then 8/a<2. and since the equation "x²+cos(11πx/4)=a" has an even number of solutions on x∈[-2,2], the valid answers to the initial problem can only be the values of parameter a (a>4) such that the latter equation, on the interval x∈[8/a,2], has either an even number of solutions or no solutions at all. in other words, if we can prove that it is unsolvable on x∈[8/a,2], we will prove that all a>4 belong among the right answers to the initial problem. now, since x=2 is not a solution in this case (a≠4) and since a-x²-cos(11πx/4)≥8/x-x²-cos(11πx/4), all we need to do is prove that "8/x-x²-cos(11πx/4)" is positive on the interval x∈(0,2).
1). if 18/11 < x < 2, then 9π/2 < 11πx/4 < 11π/2 and therefore cos(11πx/4)<0; meaning that on this particular interval
8/x-x²-cos(11πx/4)>8/x-x²>0 (since x<2)
2). if 0 < x ≤ 18/11, then 8/x-x²≥11·8/18-(18/11)²>44/9-3>1 (since f(x)= 8/x-x² is monotonically decreasing when x>0). hence,
8/x-x²-cos(11πx/4)>1-cos(11πx/4)≥0
iii-c. if 8/3 ≤ a < 4, then 8/a > 2. this case is almost identical to iii-b. we need to find such values of a, for which the equation "x²+cos(11πx/4)=a" will have either an even number of solutions or none on the interval x∈(2,8/a). and just like with iii-b, all we need to do is prove that it is unsolvable on x∈(2,8/a) when 8/3 ≤ a < 4. only in this case (because, on this interval, a<8/x and, therefore, a-x²-cos(11πx/4)<8/x-x²-cos(11πx/4)) we only need to prove that "8/x-x²-cos(11πx/4)" is negative on x∈(2,3).
1) if 2 < x < 26/11, then 11π/2 < 11πx/4 < 13π/2 and thus cos(11πx/4) > 0; meaning that on this particular interval
8/x-x²-cos(11πx/4)<8/x-x²<0 (since x>2)
2) if 26/11 ≤ x < 3, then 8/x-x²≤11·8/26-(26/11)²<44/13-5<-1 (since f(x)= 8/x-x² is monotonically decreasing when x>0). hence,
8/x-x²-cos(11πx/4)<-1-cos(11πx/4)≤0
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so if i haven't made any mistakes, the final answer should be: a∈(-∞,-4]∪{1}∪[8/3,4)∪(4,∞) |
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| LoCa |
...
you are God.:wtf: |
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