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ย
f(x)=(x-2)ยฒ(1-x)(x-3)ยณ(x-4)ยฒ/(x+1) โค 0
ย (x-2)2ย (x-3)2(x-4)2ย ย ย is always positive
at x=2,3,4ย ย ย (x-2)2ย (x-3)2(x-4)2ย isย zeroย
so x=2,3,4 is a solution ย ย
now forย ย (1-x)(x-3)/(x+1) โค 0
ย orย ย ย ย ย ย ย ย ย (x-1)(x-3)/(x+1) โฅ 0
ans: ย (-1,1] U [3,โ) U {2}
HERE, ON WHAT BASIS IS 2 PLACED IN ANSWER.
AND HOW CLOSE AND OPEN BRACKETS ARE ADDED.
PLEASE HELP ME. FAST.
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.
2 is placed because at two f(x) assumes 0 naand in ques it is given f(x) <= 0 dont forget that equal to signNow close brackets should be place when equal to sign is presentbut take care not to place closed brackets there where function becomes undefined. ..
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
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