Suppose m = HCF of x and y and n = LCM of x and y. Consider the following :<\/p>\n
Which of the following is\/are correct ?<\/p>\n
[amp_mcq option1=”1 only” option2=”2 only” option3=”Both 1 and 2″ option4=”Neither 1 nor 2″ correct=”option1″]<\/p>\n
Consider statement 1: $m^2 n \\leq x^2 y$.
\nWe can substitute $y = mn\/x$ from the fundamental relationship into the inequality:
\n$m^2 n \\leq x^2 (mn\/x)$
\n$m^2 n \\leq xmn$
\nSince x, m, and n are positive for positive integers x, y (unless x or y is 0 or 1, but the context implies standard HCF\/LCM), we can divide both sides by mn:
\n$m \\leq x$.
\nThe HCF (m) of x and y is always a divisor of x. Therefore, for positive integers, $m \\leq x$ is always true.
\nThus, statement 1 is correct.<\/p>\n
Consider statement 2: $mn^2 \\leq x^2 y$.
\nSubstitute $y = mn\/x$ into the inequality:
\n$mn^2 \\leq x^2 (mn\/x)$
\n$mn^2 \\leq xmn$
\nSince x, m, and n are positive, we can divide both sides by mn:
\n$n \\leq x$.
\nThe LCM (n) of x and y is a multiple of x. For positive integers, this means $n \\geq x$. The inequality $n \\leq x$ is only true in specific cases (e.g., when y divides x, in which case $n=x$) but is generally false. For example, if x=2 and y=3, LCM(2,3)=6, and $6 \\not\\leq 2$.
\nThus, statement 2 is incorrect in general.<\/p>\n