Suppose m = HCF of x and y and n = LCM of x and y. Consider the following :
- 1. m²n ≤ x²y
- 2. mn² ≤ x²y
Which of the following is/are correct ?
We know the fundamental relationship between HCF (m) and LCM (n) of two numbers x and y:
$x \times y = m \times n$.
Consider statement 1: $m^2 n \leq x^2 y$.
We can substitute $y = mn/x$ from the fundamental relationship into the inequality:
$m^2 n \leq x^2 (mn/x)$
$m^2 n \leq xmn$
Since 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:
$m \leq x$.
The HCF (m) of x and y is always a divisor of x. Therefore, for positive integers, $m \leq x$ is always true.
Thus, statement 1 is correct.
Consider statement 2: $mn^2 \leq x^2 y$.
Substitute $y = mn/x$ into the inequality:
$mn^2 \leq x^2 (mn/x)$
$mn^2 \leq xmn$
Since x, m, and n are positive, we can divide both sides by mn:
$n \leq x$.
The 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 $6 \not\leq 2$.
Thus, statement 2 is incorrect in general.
Since only statement 1 is correct, the answer is A.
– HCF of x and y is a divisor of x (so $m \leq x$).
– LCM of x and y is a multiple of x (so $n \geq x$).