Series of points

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.

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