Limit of two sequence? And Geometry Test?

Calculus Level 3

Let ${ x }_{ 1 }$ and ${ y }_{ 1 }$ be real numbers and ${ x }_{ 1 }>{ y }_{ 1 }$ . For any positive integer $n$ , define $\begin{cases} { x }_{ n+1 }=\frac { 5 }{ 6 } { x }_{ n }+\frac { 1 }{ 6 } { y }_{ n } \\ { y }_{ n+1 }=\frac { 2 }{ 9 } { x }_{ n }+\frac { 7 }{ 9 } { y }_{ n } \end{cases}$

Given that $\displaystyle \lim _{ n\rightarrow \infty }{ { x }_{ n } }$ can be written as $\dfrac { a{ x }_{ 1 }+b{ y }_{ 1 } }{ c }$ , where $a$ , $b$ , and $c$ are positive integers. Furthermore, suppose ${ x }_{ 1 }=12$ , ${ y }_{ 1 }=5$ , and define $\displaystyle F=\lim _{ n\rightarrow \infty }{ { x }_{ n } }$ .

It is known that $\begin{cases} C:{ x }^{ 2 }+{ y }^{ 2 }-6x-4y+F=0 \\ L:2x+y-k=0 \end{cases}$ have two distinct intersection point on the Cartesian plane if and only if $p<k<q$ .

Define $D=abc-pq$ . Find the value of $D$ .

Hint: $4{ x }_{ n }+3{ y }_{ n }$ is a constant for any positive integer $n$ .

The answer is 40.

1 solution

Stephen Brown
Dec 18, 2017

Using the hint provided, we can easily see that $\lim_{n\rightarrow\infty}4x_n+3y_n=4x_1+3y_1$ . We also know that the sequence $(x_n,y_n)$ converges to a fixed point $(x,y)$ , which we can calculate as

$x=\frac{5}{6}x+\frac{1}{6}y, y=\frac{2}{9}x+\frac{7}{9}y\ \Rightarrow x=y$

Combining these pieces of information, we get that

$\lim_{x\rightarrow\infty}x_n=\frac{4x_1+3y_1}{7}$

So $a=4,b=3,c=7$ and $F=9$

Now consider the intersection of $C$ and $L$ :

$y=k-2x \Rightarrow x^2+(k-2x)^2-6x-4(k-2x)+9=0 \Rightarrow 5x^2+(2-4k)x+(k^2-4k+9)=0$ $x=\frac{2k-1\pm\sqrt{-k^2+16k-44}}{5}$

For this to have two distinct solutions, we must have $-k^2+16k-44>0 \Rightarrow (k-r_1)(k-r_2)<0 \Rightarrow r_1<k<r_2$ where $r_1<r_2$ are the roots of $k^2-16k+44$ , but this means $p=r_1$ and $q=r_2$ , and we know that $pq=r_1r_2=44$ . Thus we have

$D=abc-pq=(4)(3)(7)-44=\boxed{40}$

Limit of two sequence? And Geometry Test?

The answer is 40.

1 solution

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