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neeraj pawar Grade: 12
        

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

Answers : (8)

Badiuddin askIITians.ismu Expert
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




 




 


 


 


 


7357-2077_7662_untitled.JPG


so silution is [-1,1] and [3,∞) 


 


so final solution   [-1,1] and [3,∞)  and 2


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7 years ago
ROHIT RANJAN
10 Points
										

(-1,1] U [3,∞) U {2}      Smile

4 years ago
shakti dash
18 Points
										

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

4 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,∞)


........

4 years ago
shreyash kawalkar
18 Points
										

HERE, ON WHAT BASIS IS 2 PLACED IN ANSWER.


AND HOW CLOSE AND OPEN BRACKETS ARE ADDED.


PLEASE HELP ME. FAST.


 


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


 7357-2077_7662_untitled.JPG


ans:   (-1,1] U [3,∞) U {2}


HERE, ON WHAT BASIS IS 2 PLACED IN ANSWER.


AND HOW CLOSE AND OPEN BRACKETS ARE ADDED.


PLEASE HELP ME. FAST.

3 years ago
sayak chakrabarty
18 Points
										

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

3 years ago
anukul
8 Points
										I can answer your question very well, if you want me to; now.

Anukul Sangwan
Class 10
FIITJEE South Delhi
3 years ago
krishna chaitanya
12 Points
										xE(-8,-1)U[1.2]U[3,8]
										
2 years ago
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