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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.`
7 years ago

147 Points
```										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
```
7 years ago
ROHIT RANJAN
10 Points
```										(-1,1] U [3,∞) U {2}
```
5 years ago
shakti dash
18 Points
```										sir,i am unable to understand the solution for the question.please explain clearly
```
5 years ago
aabha tiwari
18 Points
```										-(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,∞)
........
```
5 years ago
shreyash kawalkar
18 Points
```										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.
```
4 years ago
sayak chakrabarty
18 Points
```										explain to me: why does the wavy curve method hold?  what is the actual idea behind it?
```
4 years ago
anukul
8 Points
```										I can answer your question very well, if you want me to; now.Anukul SangwanClass 10FIITJEE South Delhi
```
3 years ago
krishna chaitanya
17 Points
```										xE(-8,-1)U[1.2]U[3,8]
```
3 years ago
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