Q. 143.9( 17 Votes )

Find the do

Answer :


We know the square root of a real number is never negative.


Clearly, f(x) takes real values only when x – 2 and 3 – x are both positive or negative.


(a) Both x – 2 and 3 – x are positive


x – 2 ≥ 0

x ≥ 2


3 – x ≥ 0

x ≤ 3


Hence, x ≥ 2 and x ≤ 3


x [2, 3]


(b) Both x – 2 and 3 – x are negative


x – 2 ≤ 0 x ≤ 2


3 – x ≤ 0 x ≥ 3


Hence, x ≤ 2 and x ≥ 3


However, the intersection of these sets in null set. Thus, this case is not possible.


In addition, f(x) is also undefined when 3 – x = 0 because the denominator will be zero and the result will be indeterminate.


3 – x = 0

x = 3


Hence, x [2, 3] – {3}


x [2, 3)


Thus, domain of f = [2, 3)

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