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.

4 years ago

Share

Answers : (8)

                                        

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


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 Experts
Badiuddin

4 years ago
                                        

(-1,1] U [3,∞) U {2}      Smile

2 years ago
                                        

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

2 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,∞)


........

2 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.

one year ago
                                        

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

one year ago
                                        I can answer your question very well, if you want me to; now.

Anukul Sangwan
Class 10
FIITJEE South Delhi
one year ago
                                        xE(-8,-1)U[1.2]U[3,8]
                                        
9 months ago

Post Your Answer

More Questions On Differential Calculus

Ask Experts

Have any Question? Ask Experts
Post Question
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!!
Click Here for details
Check monotonocity of x^(1/3) at x=0
 
 
Hii Student Just check the question and send in again. Monotonocity can be defined in a interval and at a single point it’s meaningless . So check the question again before posting. We...
  img
Sourabh Singh 4 months ago
Q. THE RANGE OF THE FUNCTION sin(inverse)( x(square) + 2x)
 
 
Ans: Hello Student, Please find answer to your question below First we need to check the domain of the function ….......(1) So the domain of f(x) is (1). Since it covers all the...
  img
Jitender Singh 5 months ago
Find doamin of f(x) g(x)=|sinx|+sinx h(x)=sinx+cosx , x is in b/w 0 to pi f(x)=(logh(x) g(x))1/2 log is in square root answer is pi/6,pi/2 pls give two solution graph and analytical. thanks
 
 
Ans: 1st Part: 2nd Part: 1st Condition: …...........(1) 2nd Condition: …...............(2) Final Solution: Intersection of (1) & (2) Thanks & Regards Jitender Singh IIT...
  img
Jitender Singh 7 months ago
x=(1/cos1°) + (1/cos2°cos3°) +(1/cos3°cos4°) +....+(1/cos44°cos45°) , then x(sin1) is ?
 
 
Hello student, first term of your question should be 1/cos1cos2 in place of 1/cos1, check the solution of your question given below: x=(1/cos1°cos2) +(1/cos2°cos3°)...
  img
Sunil Raikwar 5 months ago
cos^2a-6sinacosa+3sin^2a+2
 
 
Ans: For maxima: when Maximum value: Thanks & Regards Jitender Singh IIT Delhi askIITians Faculty
  img
Jitender Singh 7 months ago
View all Questions »