Question (x_1,x_2),(x_2,x_3),(x_3,x_4)...(x_2014,x_2015) are all points on the graph y =1/(x-1) then:

• In every choice x_1,x_2,x_3...,x_2015 are all distinct.
• There are infinitely many choices in which all x_i are equal.

All x_i are real numbers.

Solution

The question has a lot of information which may be difficult to handle directly so first let us understand the question.

It says, if we take a point (x_1,x_2) on the function y=f(x)=1/(x-1). Then we can find the next point in the series by (x_2,f(x_2)) and x_3=f(x_2). Then this process continues.

Now let us look us at the options. First option says that all x_1,x_2,x_3...,x_2015 are distinct.

Now if !y=x=1/(x-1), then we get x= 1/2+-sqrt(5)/2

So if x_1=1/2+-sqrt(5)/2 then x_2=1/2+-sqrt(5)/2 and so on and so forth.

So obviously option is not correct.

Now coming to option B:

If all x_i are equal then as we looked above there are only 2 possible values of x_i so we will get only 2 series, hence even option B is incorrect.

Hope this helps.