Waste less time on Facebook — follow Brilliant.
×

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, 10 months ago

No vote yet
2 votes

Comments

Sort by:

Top Newest

  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 \[ S_\mathsf{tot} \; = \; 2^{n-1}\sum_{j=1}^n j \; = \; n(n+1)2^{n-2} \]

Mark Hennings · 3 years, 10 months ago

Log in to reply

@Mark Hennings Hello.. I just noticed that For x = 1, the value is not 0... contradicting the given... John Ashley Capellan · 3 years, 10 months ago

Log in to reply

@John Ashley Capellan 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 · 3 years, 10 months ago

Log in to reply

@Mark Hennings Oh.. sorry.. I was talking about the second problem, not the first... Sorry for not mentioning the number... John Ashley Capellan · 3 years, 10 months ago

Log in to reply

@John Ashley Capellan Good point. I have corrected the solution, which needed \(\tfrac34\) instead of \(\tfrac14\). Mark Hennings · 3 years, 10 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...