```                   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.
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4 years ago

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```                    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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4 years ago
```                    (-1,1] U [3,∞) U {2}
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one year ago
```                    sir,i am unable to understand the solution for the question.please explain clearly
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one year 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,∞)
........
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one year ago
```                    explain to me: why does the wavy curve method hold?  what is the actual idea behind it?
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10 months 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.
```
one year ago
```                    I can answer your question very well, if you want me to; now.Anukul SangwanClass 10FIITJEE South Delhi
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3 months ago