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Let $f(x, y) = x^{n}y^{m} = P$. If $x$ is doubled and $y$ is halved, the new value of $f$ is

1. $2^{n-m}P$
2. $2^{m-n}P$
3. $2(n - m)P$
4. $2(m - n)P$
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$f(x,y) = x^{n}y^{m}=P$

Now, $x$ is doubled so we substitute $2x$ for $x$ AND $y$ is halved so we substitute $y/2$ for $y$

$f(x,y) = (2x)^{n}(\frac{y}{2})^{m}$

We get,

$f(x,y) = (2)^{n} (x)^{n}(y)^{m}(2)^{-m}$

$f(x,y) = 2^{n-m} x^{n}y^{m} = 2^{n-m} P$

Answer A) $2^{n-m}P$
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option A is correct
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