Question

If $u = \frac{x+y}{x-y}$ and $v = \frac{xy}{(x-y)^2}$ are functionally dependent, then the relation between $u$ & $v$ is

• A. \( u = 1+4v \)
• B. \( u^2 = 1+4v \)
• C. \( v^2 = 1+4u \)
• D. \( v = 1+4u \)

Hint:

When dealing with functional dependence, especially with rational functions involving $(x+y)$, $(x-y)$, and $xy$, consider algebraic identities involving these terms. The identity $(x+y)^2 = (x-y)^2 + 4xy$ is particularly useful here. Also, squaring one of the given expressions ($u$ in this case) often helps to introduce common denominators or terms that can be directly related to the other expression ($v$).

Answer 1

The Correct Option is B

Two functions $u(x,y)$ and $v(x,y)$ are functionally dependent if there exists a non-trivial relation $F(u,v)=0$. One way to find this relation is to manipulate the given expressions to eliminate $x$ and $y$. Given: $u = \frac{x+y}{x-y} \quad \ldots (1)$ $v = \frac{xy}{(x-y)^2} \quad \ldots (2)$
Step 1: Square the expression for $u$. Squaring equation (1), we get: $$ u^2 = \left(\frac{x+y}{x-y}\right)^2 = \frac{(x+y)^2}{(x-y)^2} $$
Step 2: Expand the numerator. Recall the identity $(a+b)^2 = (a-b)^2 + 4ab$. Applying this to the numerator $(x+y)^2$: $$ (x+y)^2 = (x-y)^2 + 4xy $$
Step 3: Substitute the expanded numerator into the expression for $u^2$. $$ u^2 = \frac{(x-y)^2 + 4xy}{(x-y)^2} $$
Step 4: Separate the terms in the numerator. $$ u^2 = \frac{(x-y)^2}{(x-y)^2} + \frac{4xy}{(x-y)^2} $$ $$ u^2 = 1 + \frac{4xy}{(x-y)^2} $$
Step 5: Substitute the expression for $v$ into the equation for $u^2$. From equation (2), we know that $v = \frac{xy}{(x-y)^2}$. Substitute $v$ into the equation for $u^2$: $$ u^2 = 1 + 4 \left( \frac{xy}{(x-y)^2} \right) $$ $$ u^2 = 1 + 4v $$ Thus, the relation between $u$ and $v$ is $u^2 = 1+4v$.