Bunuel wrote:
Official Solution:
If \(y(u - c) = 0\) and \(j(u - k) = 0\), which of the following must be true, given \(c < k\)?
A. \(yj \lt 0\)
B. \(yj \gt 0\)
C. \(yj = 0\)
D. \(j = 0\)
E. \(y = 0\)
From \(y(u - c) = 0\), it follows that \(u=c\) or \(y=0\) (or both).
From \(j(u - k) = 0\), it follows that \(u=k\) or \(j=0\) (or both).
Note that \(u=c\) and \(u=k\) cannot simultaneously be correct because if they were, it would imply that \(u=c=k\), but we have been given that \(c < k\). Thus, only one can be correct, which further implies that either \(y=0\) or \(j=0\), making \(yj = 0\) always true.
Answer: C
I find it tough to understand that why option d. y = 0 and option e. j = 0 are not always true. Is nt it true that from the 1st equation y must be true and from the second equation j must be true? If not, can someone please explain and elucidate the cases, when y = 0 and j = 0 are not must be true possibles?
BunuelD, \(j = 0\), is not necessarily true because we can have a case when \(y=0\) and \(u=k\) (observer, that both \(y(u - c) = 0\) and \(j(u - k) = 0\) hold true for these values). In this case j can be any number, not necessarily 0.
Similarly, E, \(y = 0\), is not necessarily true because we can have a case when \(j=0\) and\(u=c\) (observer, that both \(y(u - c) = 0\) and \(j(u - k) = 0\) hold true for these values). In this case j can be any number, not necessarily 0.