Question

If a line parallel but nont identical with x-axis cuts the graph of the curve `y=(x-1)/((x-2)(x-3))` at `x=a,x=b` then (a,b)=?

Answer 1

See explanation.

Explanation:

y-intercept `= - 1/ 6`.

Asymptotes: `larr y = 0 rarr, uarr x = 2 and uarr x = 3`. .

graph{(y (x-2)(x-3)-(x-1))((x+1)^2+(y+1/6)^2-0.03)(x^2+(y+1/6)^2-0.03)=0}.

The line `y = c ne 0`, parallel to x-axis, meets the curve at

just one point `in Q_3`, if `c = - 1/6`,

at `( - 1, - 1/6 )`. See graph.

Above , `y = c > - 1/6` meets the curve again, at

`x = a, b = 1/2( 2.5 +1/c( 1+-sqrt( 1 + 6c + c^2))`.

Answer 2

`color(red)("In this question, we are asked to find the value of "`

`color(blue)((a-1)(b-1))`

It has been informed to me.

Let a line parallel to `x`-axis be `y=c` which cuts the given curve at `x=aandx=b`

Hence we have

`c=(a-1)/((a-2)(a-3))=(b-1)/((b-2)(b-3))`

`=>(a-1)/((a-2)(a-3))=(b-1)/((b-2)(b-3))`

`=> a^2b-5ab+6b-a^2+5a-6=ab^2-5ab+6a-b^2+5b-6`

`=>ab(a-b)- (a-b)-(a^2-b^2)=0`

`=>ab-1-a-b=0` [Since `a!=b`]

`=>ab-a-b+1-2=0`

`=>a(b-1)-1(b-1)=2`

`=>(a-1)(b-1)=2`