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# 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
3 years, 2 months ago

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

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

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

· 3 years, 2 months ago

Hello.. I just noticed that For x = 1, the value is not 0... contradicting the given... · 3 years, 2 months ago

So, OK, if $$x=1$$ then $$y=0$$. This just means that for $$x$$ and $$y$$ to be nonzero and satisfy the identity $$x+y+xy=1$$, they also have to be not equal to $$1$$. That just means that there is an additional "hidden" restriction on the possible values of $$x$$ and $$y$$, but nothing more. If you think about it, $$x$$ and $$y$$ cannot be equal to $$-1$$, either. · 3 years, 2 months ago

Oh.. sorry.. I was talking about the second problem, not the first... Sorry for not mentioning the number... · 3 years, 2 months ago

Good point. I have corrected the solution, which needed $$\tfrac34$$ instead of $$\tfrac14$$. · 3 years, 2 months ago