If a line parallel but nont identical with x-axis cuts the graph of the curve y=(x-1)/((x-2)(x-3))y=x1(x2)(x3) at x=a,x=bx=a,x=b then (a,b)=?

2 Answers
Jul 29, 2018

See explanation.

Explanation:

y-intercept = - 1/ 6=16.

Asymptotes: larr y = 0 rarr, uarr x = 2 and uarr x = 3y=0,x=2andx=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 0y=c0, parallel to x-axis, meets the curve at

just one point in Q_3Q3, if c = - 1/6c=16,

at ( - 1, - 1/6 )(1,16). See graph.

Above , y = c > - 1/6y=c>16 meets the curve again, at

x = a, b = 1/2( 2.5 +1/c( 1+-sqrt( 1 + 6c + c^2))x=a,b=12(2.5+1c(1±1+6c+c2).

Jul 31, 2018

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