# 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.

148 Points
14 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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ROHIT RANJAN
10 Points
12 years ago

(-1,1] U [3,∞) U {2}

shakti dash
18 Points
11 years ago

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

aabha tiwari
18 Points
11 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,∞)

........

shreyash kawalkar
18 Points
11 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.

sayak chakrabarty
18 Points
11 years ago

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

anukul
8 Points
10 years ago
I can answer your question very well, if you want me to; now. Anukul Sangwan Class 10 FIITJEE South Delhi
krishna chaitanya
17 Points
10 years ago
xE(-8,-1)U[1.2]U[3,8]