Help in these Problems

Problems:

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.)
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.
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

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

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 $\begin{array}{rcl} xy + \frac{1}{xy} - \frac{x}{y} - \frac{y}{x} & = & \frac{(1+x)^2(1-x^2) - (1-x)^2(1-x^2)}{x(1-x^2)} \\ & = & \frac{(1+x)^2 - (1-x)^2}{x} \; = \; 4 \end{array}$
The polynomial is $P(x) = 3 - \tfrac34(x-2)^2(x-3)^2$ , and so $P(5) = -24$ .
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 $S_\mathsf{tot} \; = \; 2^{n-1}\sum_{j=1}^n j \; = \; n(n+1)2^{n-2}$

Mark Hennings - 7 years, 7 months ago

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

John Ashley Capellan - 7 years, 7 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.

Mark Hennings - 7 years, 7 months ago

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

John Ashley Capellan - 7 years, 7 months ago

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

Mark Hennings - 7 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$