 # Please explain in brief the wavy curve method & also solve this ques..f(x)=(x-2)²(1-x)(x-3)³(x-4)²/(x+1) ≤ 0 find the value of X. Badiuddin askIITians.ismu Expert
148 Points
13 years ago

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

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10 years ago

(-1,1] U [3,∞) U {2} 10 years ago

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

10 years ago

-(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,∞)

........

9 years ago

HERE, ON WHAT BASIS IS 2 PLACED IN ANSWER.

AND HOW CLOSE AND OPEN BRACKETS ARE ADDED.

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.