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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.
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
(-1,1] U [3,∞) U {2}
sir,i am unable to understand the solution for the question.please explain clearly
-(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,∞)
........
HERE, ON WHAT BASIS IS 2 PLACED IN ANSWER.
AND HOW CLOSE AND OPEN BRACKETS ARE ADDED.
PLEASE HELP ME. FAST.
ans: (-1,1] U [3,∞) U {2}
explain to me: why does the wavy curve method hold? what is the actual idea behind it?
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