If HCF(a, b) = 1,then HCF(a - b, a + b) is eq | vedclass

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(C) Let $d = HCF(a - b, a + b)$.Then $d$ must divide their sum and their difference.$d$ divides $(a + b) + (a - b) = 2a$.$d$ divides $(a + b) - (a - b

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$1$ or $2$

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(C) Let $d = HCF(a - b, a + b)$.Then $d$ must divide their sum and their difference.$d$ divides $(a + b) + (a - b) = 2a$.$d$ divides $(a + b) - (a - b) = 2b$.Since $d$ divides $2a$ and $2b$,$d$ must divide $HCF(2a, 2b) = 2 	imes HCF(a, b)$.Given $HCF(a, b) = 1$,so $d$ must divide $2 	imes 1 = 2$.Therefore,$d$ can be either $1$ or $2$.For example,if $a = 3, b = 2$,$HCF(3, 2) = 1$,then $HCF(3-2, 3+2) = HCF(1, 5) = 1$.If $a = 3, b = 1$,$HCF(3, 1) = 1$,then $HCF(3-1, 3+1) = HCF(2, 4) = 2$.Thus,the result is $1$ or $2$.

If $HCF(a, b) = 1$,then $HCF(a - b, a + b)$ is equal to:

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