Analysis for Question 1:

Two different positive real numbers $x$ and $y$ each differ from their reciprocals by $1$. Evaluate $x+y$.

This is a really good algebra challenge, as the given info provide us the following:

$1=\left|x-\dfrac{1}{x}\right|$ and $1=\left|y-\dfrac{1}{y}\right|$

Each of which could then be used to produce two quadratic equations with their roots be $x$ and $y$:

$1=(x-\dfrac{1}{x})\implies\,x^2-x-1=0$ or

$1=-(x-\dfrac{1}{x})=1\implies\,x^2+x-1=0$

Since $x^2-(\text{sum of roots})x+(\text{product of roots})=0$ and we're told $x$ and $y$ are two positive real numbers, so their sum must be another positive number and hence, $x+y=1$ and the answer is $1$.

Analysis for Question 2:

The product of three consecutive positive integers is 8 times their sum. Find the sum of their squares.

You can solve this question pretty quickly or slowly, and it all boils down to what expressions you would use to represents the three consecutive positive integers.

If you let them as $x-1,\,x,\,x+1$, then you are bound for the fast train to reach for the destination since

$(x-1)(x)(x+1)=8(x-\cancel{1}+x+x+\cancel{1})$

$(x-1)\cancel{(x)}(x+1)=8(3\cancel{(x)})$

$x^2-1=24$

$x^2=\pm 5$

$x=5$ since $x>0$ and hence $(x-1)^2+x^2+(x+1)^2=4^2+5^2+6^2=77$

But if you chose to work with $x,\,x+1,\,x+2$, you will end up with a more complicated situation. Let's see:

$(x)(x+1)(x+2)=8(x+x+1+x+2)$

$(x)(x+1)(x+2)=8(3x+3)$

$(x)\cancel{(x+1)}(x+2)=8(3)\cancel{(x+1)}$

$(x)(x+2)=24$

$x^2+2x-24=0$

$x=4$ or $x=-6$ and so $x=4>0$ and $(x)^2+(x+1)^2+(x+2)^2=4^2+5^2+6^2=77$ Yes, this choice for the three consecutive terms works as well, but it is just that you have to undergo one more step to factor to get the real answer.

Analysis for Question 3:

Question 3: Positive integers $a, b$ and $c$ are chosen such that $c>b>a$ and the system of equations $y=|x-a|+|x-b|+|x-c|$ and $2x+y=1999$ has exactly one solution. Find the least possible value of c.

The function for $y=|x-a|+|x-b|+|x-c|$ can be broken down into different functions in separate intervals:

$x\ge c$ gives

$\begin{align}y&=x-a+x-b+x-c\\&=3x-(a+b+c)\implies \text{gradient}=3\end{align}$

$b\le x < c$ gives

$\begin{align}y&=x-a+x-b-(x-c)\\&=x-(a+b-c)\implies \text{gradient}=1\end{align}$

$a\le x < b$ gives

$\begin{align}y&=x-a-(x-b)-(x-c)\\&=-x-(a-b-c)\implies \text{gradient}=-1\end{align}$

$x< a$ gives

$\begin{align}y&=-(x-a)-(x-b)-(x-c)\\&=-3x+(a+b+c)\implies \text{gradient}=-3\end{align}$

The end points at $x=a,\,b,\,c$ are $(a,\,-2a+b+c)$, $(b,\,-a+c)$ and $(c,\,2c-a-b)$.

To make sure the system of equations $y=|x-a|+|x-b|+|x-c|$ and $2x+y=1999$ has exactly one solution, we must examine the gradient of $2x+y=1999$ (which is equivalent to $y=-2x+1999$ and $\therefore m_{y=-2x+1999}=-2$) and compare it with the gradients of the system of equation $y=|x-a|+|x-b|+|x-c|$

The diagram above tells us $2x+y=1999$ must intercept the system at $(a,\,-2a+b+c)$ and hence:

$2a+-2a+b+c=1999$

$b+c=1999=999+1000$ $(c>b>a)$ (given)

We can conclude that the least value of $c$ is $1000$.

## No comments:

## Post a Comment