Wavy curve method

Dear neeraj

f(x)=(x-2)²(1-x)(x-3)³(x-4)²/(x+1) ≤ 0

 (x-2)² (x-3)²(x-4)²   is always positive

at x=2,3,4   (x-2)² (x-3)²(x-4)² is  zero 

so x=2,3,4 is a solution  

now for  (1-x)(x-3)/(x+1) ≤ 0

  or        (x-1)(x-3)/(x+1) ≥ 0

so silution is [-1,1] and [3,∞) 

so final solution   [-1,1] and [3,∞)  and 2

Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you
the answer and detailed  solution very  quickly.

 We are all IITians and here to help you in your IIT JEE preparation.

All the best.
 
Regards,
Askiitians Experts
Badiuddin

(-1,1] U [3,∞) U {2}      

sir,i am unable to understand the solution for the question.please explain clearly

-(x-2)² (x-1)(x-3)³ (x-4)² /(x+1)≤0

(x-2)² (x-1)(x-3)³ (x-4)² /(x+1)≥0

equate every linear factor by 0,

x=1,2,3,4,-1

by ploting the values nof x on no line we get,

x Ε (-∞,-1]υ[1,2]υ[3,4]υ[4,∞)

........

HERE, ON WHAT BASIS IS 2 PLACED IN ANSWER.

AND HOW CLOSE AND OPEN BRACKETS ARE ADDED.

PLEASE HELP ME. FAST.

f(x)=(x-2)²(1-x)(x-3)³(x-4)²/(x+1) ≤ 0

 (x-2)² (x-3)²(x-4)²   is always positive

at x=2,3,4   (x-2)² (x-3)²(x-4)² is  zero 

so x=2,3,4 is a solution   

now for  (1-x)(x-3)/(x+1) ≤ 0

  or         (x-1)(x-3)/(x+1) ≥ 0

ans:   (-1,1] U [3,∞) U {2}

HERE, ON WHAT BASIS IS 2 PLACED IN ANSWER.

AND HOW CLOSE AND OPEN BRACKETS ARE ADDED.

PLEASE HELP ME. FAST.

explain to me: why does the wavy curve method hold?  what is the actual idea behind it?