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August 18th, 2013, 01:03 PM   #1
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Two Questions That Have Me Stumped

Hey everyone,

Could really use help on these two problems so I can finish studying for my math placement test tomorrow.

1) I need to find the set of real solutions if (3/3-x)=(1/3)-(1/x).
Answer Choices are: A. empty set, B. {3}, C. {-3}, D. {3, -3}, E. {3i, -3i}

2) For which of the following functions is f(a+b)=f(a)+f(b) for all values of a and b?
Answer Choices are: A. f(x)=3x-1, B. f(x)=x+3, C. f(x)=2x, D. f(x)=x^2, E. x^2+1


Just completely blanked on how to solve these two. Thanks a lot in advance!!
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August 18th, 2013, 02:11 PM   #2
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Re: Two Questions That Have Me Stumped

1)

Solve 3/(3 - x) = 1/3 - 1/x for x:

Multiply through by 3x(3 - x):

9x = x(3 - x) - 3(3 - x)

Expand:

9x = 3x - x - 9 + 3x

Rearrange and simplify:

x + 3x + 9 = 0.

3 - 4(1)(9) = 9 - 36 < 0 hence this quadratic has no real roots, so answer choice A is correct.

2)
For which of the following functions is f(a + b) = f(a) + f(b)?
A: f(x) = 3x - 1. f(a + b) = 3(a + b) - 1 = 3a + 3b - 1. f(a) + f(b) = 3a - 1 + 3b - 1 = 3a + 3b - 2.
B: f(x) = x + 3. f(a + b) = a + b + 3. f(a) + f(b) = a + 3 + b + 3 = a + b + 6.
C: f(x) = 2x. f(a + b) = 2(a + b) = 2a + 2b. f(a) + f(b) = 2a + 2b.
D: f(x) = x. f(a + b) = (a + b) = a + 2ab + b. f(a) + f(b) = a + b.
E: f(x) = x + 1. f(a + b) = (a + b) + 1 = a + 2ab + b + 1. f(a) + f(b) = a + b + 2.

Only choice C has f(a + b) = f(a) + f(b).
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August 20th, 2013, 07:22 AM   #3
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Re: Two Questions That Have Me Stumped

Quote:
Originally Posted by Tyrean
Hey everyone,

Could really use help on these two problems so I can finish studying for my math placement test tomorrow.

1) I need to find the set of real solutions if (3/3-x)=(1/3)-(1/x).
I assume that on the left you mean 3/(3- x), not (3/3)- x= 1- x.

Quote:
Answer Choices are: A. empty set, B. {3}, C. {-3}, D. {3, -3}, E. {3i, -3i}
Since you are given choices, one way to answer this, and perhaps the easiest, is to try each possible answer.
If x= 3 the left side has denominator is 0 which is impossible.
If x= -3, the left side is 3/(3+ 3)= 3/6= 1/2 while the left is (1/3)- (-1/3)= 2/3, not equal to 1/2.
Since neither 3 nor -3 is a solution, {3, -3} is not a solution set.
If x= 3i, 3/(3- 3i)= 1/(1- i). "Rationalize the denominator" by multiplying both numerator and denominator by 1+ i: (1+ i)/2. On the right, (1/3)- (1/3i)= (1/3)(1- 1/i)= (1/3)(1+ i). 1/3 is not equal to 1/2 so this is not a solution set.
The only option left is "A. empty set".

Quote:
2) For which of the following functions is f(a+b)=f(a)+f(b) for all values of a and b?
Answer Choices are: A. f(x)=3x-1, B. f(x)=x+3, C. f(x)=2x, D. f(x)=x^2, E. x^2+1
Again: if f(x)= 3x- 1 then f(a+ b)= 3(a+ b)- 1= 3a+ 3b- 1 while f(a)+ f(b)= (3a+ 1)+ (3b+ 1)= 3a+ 3b+ 2. Those are not the same.
If f(x)= x+ 3, then f(a+ b)= a+ b+ 3 while f(a)+ f(b)= (a+ 3)+ (b+ 3)= a+ b+ 6. Those are not the same.
If f(x)= 2x, then f(a+ b)= 2(a+ b)= 2a+ 2b while f(a)+ f(b)= (2a)+ (2b). Yes, those are the same!
If f(x)= x^2, then f(a+ b)= (a+ b)^2= a^2+ 2ab+ b^2 while f(a)+ f(b)= a^2+ b^2. Those are not the same.
If f(x)= x^2+ 1, then f(a+ b)= (a+ b)^2+ 1= a^2+ 2ab+ b^2+ 1 while f(a)+ f(b)= a^2+ 1+ b^2+ 1= a^2+ b^2+ 2. Those are not the same.

Quote:
Just completely blanked on how to solve these two. Thanks a lot in advance!!
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August 20th, 2013, 06:18 PM   #4
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1. If you do it correctly you should arrive with but by doing the process of elimination i.e. it can't be any other correct answer but A. empty set.
2. Seriously, come on . . . just plug in x = a + b, a or b (e.g. A. 3(a + b) - 1 = 3a - 1 + 3b - 1 but this isn't correct, so it's not A.)!
Sometimes I'm just too lazy . . . with the benefit of hindsight, C. f(x) = 2x is the correct answer (because B. and E. are similar situations to A. but again D. would include 2ab which is additionally incorrect.
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August 20th, 2013, 06:25 PM   #5
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Re: Two Questions That Have Me Stumped

Question 1 asks for real solutions, so why would choice E even be considered?
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