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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.

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.

Grade:12

8 Answers

Badiuddin askIITians.ismu Expert
148 Points
11 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


 


 

 

 

 

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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ROHIT RANJAN
10 Points
9 years ago

(-1,1] U [3,∞) U {2}      Smile

shakti dash
18 Points
9 years ago

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

aabha tiwari
18 Points
9 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
8 years ago

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.

sayak chakrabarty
18 Points
8 years ago

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

anukul
8 Points
8 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
7 years ago
xE(-8,-1)U[1.2]U[3,8]

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