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```				   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.
```

6 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
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 ExpertsBadiuddin
```
6 years ago
```										(-1,1] U [3,∞) U {2}
```
4 years ago
```										sir,i am unable to understand the solution for the question.please explain clearly
```
4 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,∞)
........
```
4 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.
```
3 years ago
```										explain to me: why does the wavy curve method hold?  what is the actual idea behind it?
```
3 years ago
```										I can answer your question very well, if you want me to; now.Anukul SangwanClass 10FIITJEE South Delhi
```
2 years ago
```										xE(-8,-1)U[1.2]U[3,8]
```
2 years ago

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