 # 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 positiveat 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.PLEASE HELP ME. FAST.

9 years ago

we have.
(x-1)(x-3)/(x+1) ≥ 0
=>(x-1)(x-3)(x+1)
≥ 0

=>plotting x=1,-1,3 and seeing the value of the above function by putiing values...

e.g.=>x=0 satisfies it so the interval between -1 and 1 is +
Hence the given domain of x=(-1,1] U [3,infinnte)
2 is the ans as already obtained at the first part.
to inlude it also in the the ans. a U {2} is added.
Regards.

9 years ago

2 is placed because at two f(x) assumes 0 na
and in ques it is  given f(x) <= 0
dont forget that equal to sign

Now close brackets should be place when equal to sign is present
but take care not to place closed brackets there where function becomes undefined. ..

9 years ago

since 2 is the solution it can satisfy (1-x)(x-3) and it hence it  becomes positive so 2 is added in the answer with open bracket