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f(x)= mod(x-1) + mod(x-2) + mod(x-3) FIND RANGE OF THIS?
here , x-1≥0
or,x≤1.
similarly, x≤2 andx≤3 (because this is a modulus function.)
by taking common through sign - scheme it's range is(-∞,3].
if u like my answer plz approve it by clicking on yes!!!
Hi Srichandra,
This is a question on range of a modulus function, where you need to break it into intervals and solve it.
We can clearly break this into interval where there is a sign change in a particular modulus,
ie less than 1, between 1 and 2, between 2 and 3, and greater than 3.
Let x be in [3,infinity)
Which will give f(x) = y = (x-1) + (x-2) + (x-3) = 3x - 6, so range is [3,infinity) {when x = 3, y = 3 and will take all values greater than 3 when x is greater than 3}
Next x be in [2, 3)
Which will give y = (x-1) + (x-2) - (x-3) = x, so range is [2, 3). { same as x }
Next x be in [1,2)
Which will give y = (x-1) - (x-2) - (x-3) = 4-x, so range is (2,3]
Next x is less than 1, ie x be in (-infinity,1)
which will give y = -(x-1) -(x-2) -(x-3) = 6-3x, so range in thsi interval is (3,infinity).
Hence the range of this function is the union of the ranges in all the respective intervals, which would be the interval [2,infinity) which is the range of this function.
Note that the range has to be posititve in any case, since it is sum of modulus functions, each of which is individually positive.
You can also have questions on this like what would be the minimum value of this function, which in this case would be 2, when x = 2.
Hope that helps.
All the best,
Regards,
Ashwin (IIT Madras).
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