Help in these Problems

Problems:

  1. If x + y + xy = 1, where x and y are non-zero real numbers, what is xy + 1/xy - y/x - x/y? (The answer is 4 but needing simple algebraic manipulation.)

  2. The quartic polynomial P(x) satisfies P(1) = 0 and attains its maximum value of 3 at both x = 2 and x = 3. Compute P(5). More appreciated if the solution does not require calculus at least.

  3. Let S(X) be the sum of elements of a nonempty finite set X, where X is a set of numbers. Calculate the sum of all numbers S(X) where X ranges over all nonempty subsets of the set {1, 2, 3, ..., 16}. Please show quick method.

-From PMO

Note by John Ashley Capellan
5 years, 11 months ago

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2 votes

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  1. Put y=1x1+xy = \tfrac{1-x}{1+x}. Then xy+1xy=x(1x)1+x+1+xx(1x)  =  x2(1x)2+(1+x)2x(1x2)xy+yx=x(1+x)1x+1xx(1+x)  =  x2(1+x)2+(1x)2x(1x2) \begin{array}{rcl} xy + \frac{1}{xy} & = & \frac{x(1-x)}{1+x} + \frac{1+x}{x(1-x)} \; = \; \frac{x^2(1-x)^2 + (1+x)^2}{x(1-x^2)} \\ \frac{x}{y} + \frac{y}{x} & = & \frac{x(1+x)}{1-x} + \frac{1-x}{x(1+x)} \; = \; \frac{x^2(1+x)^2 + (1-x)^2}{x(1-x^2)} \end{array} and so xy+1xyxyyx=(1+x)2(1x2)(1x)2(1x2)x(1x2)=(1+x)2(1x)2x  =  4 \begin{array}{rcl} xy + \frac{1}{xy} - \frac{x}{y} - \frac{y}{x} & = & \frac{(1+x)^2(1-x^2) - (1-x)^2(1-x^2)}{x(1-x^2)} \\ & = & \frac{(1+x)^2 - (1-x)^2}{x} \; = \; 4 \end{array}

  2. The polynomial is P(x)=334(x2)2(x3)2P(x) = 3 - \tfrac34(x-2)^2(x-3)^2, and so P(5)=24P(5) = -24.

  3. For simplicity, let S()=0S(\varnothing) = 0, so that XX can range over all subsets of {1,2,,n}\{1,2,\ldots,n\}. Each number 1jn1 \le j \le n occurs in precisely half of the 2n2^n subsets of {1,2,,n}\{1,2,\ldots,n\}, and so contributes a total of j×2n1j \times 2^{n-1} to the total sum Stotal  =  X{1,2,,n}S(X) S_\mathsf{total} \; = \; \sum_{X \subseteq\{1,2,\ldots,n\}} S(X) Thus Stot  =  2n1j=1nj  =  n(n+1)2n2 S_\mathsf{tot} \; = \; 2^{n-1}\sum_{j=1}^n j \; = \; n(n+1)2^{n-2}

Mark Hennings - 5 years, 11 months ago

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Hello.. I just noticed that For x = 1, the value is not 0... contradicting the given...

John Ashley Capellan - 5 years, 11 months ago

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So, OK, if x=1x=1 then y=0y=0. This just means that for xx and yy to be nonzero and satisfy the identity x+y+xy=1x+y+xy=1, they also have to be not equal to 11. That just means that there is an additional "hidden" restriction on the possible values of xx and yy, but nothing more. If you think about it, xx and yy cannot be equal to 1-1, either.

Mark Hennings - 5 years, 11 months ago

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@Mark Hennings Oh.. sorry.. I was talking about the second problem, not the first... Sorry for not mentioning the number...

John Ashley Capellan - 5 years, 11 months ago

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@John Ashley Capellan Good point. I have corrected the solution, which needed 34\tfrac34 instead of 14\tfrac14.

Mark Hennings - 5 years, 11 months ago

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