Question

Which one of the following is homogeneous function?
This question has multiple correct options
A $f\left( x,y \right)=\dfrac{x-y}{{{x}^{2}}+{{y}^{2}}}$
B $f\left( x,y \right)={{x}^{\dfrac{1}{3}}}.{{y}^{-\dfrac{2}{3}}}{{\tan }^{-1}}\dfrac{x}{y}$
C $f\left( x,y \right)=x\left( \ln \sqrt{{{x}^{2}}+{{y}^{2}}}-\ln y \right)+y{{e}^{\dfrac{x}{y}}}$
D $f\left( x,y \right)=x\left[ \ln \dfrac{2{{x}^{2}}+{{y}^{2}}}{x}-\ln \left( x+y \right) \right]+{{y}^{2}}\tan \dfrac{x+2y}{3x-y}$

Answer 1

Hint: We solve this question by using the condition of homogeneity to find out the homogeneous functions from the above given options. A homogeneous function is a function that has the scaling property which means that if each of its terms are multiplied by a factor, then the entire function is multiplied by a power of this factor. The condition used for checking homogeneous functions is given by $f\left( hx,hy \right)={{h}^{m}}f\left( x,y \right),$ where $m\in R.$

Complete step by step answer:
In order to solve this question, let us use the condition of homogeneity to check if the given functions are homogeneous. The condition used is represented as $f\left( hx,hy \right)={{h}^{m}}f\left( x,y \right),$ where $m\in R.$ Here, it means that if every individual variable term of the function is scaled up or multiplied by a factor h, then the entire function can be represented as a multiple of this factor h or a power of it ${{h}^{m}}$ such that m is a real number.
Let us solve the first part,
$\Rightarrow f\left( x,y \right)=\dfrac{x-y}{{{x}^{2}}+{{y}^{2}}}$
Let us replace x by hx and y by hy,
$\Rightarrow f\left( hx,hy \right)=\dfrac{hx-hy}{{{\left( hx \right)}^{2}}+{{\left( hy \right)}^{2}}}$
Now we take the h common out from the numerator and square the individual terms in the denominator,
$\Rightarrow f\left( hx,hy \right)=\dfrac{h\left( x-y \right)}{{{h}^{2}}{{x}^{2}}+{{h}^{2}}{{y}^{2}}}$
Taking the ${{h}^{2}}$ common out from the denominator,
$\Rightarrow f\left( hx,hy \right)=\dfrac{h\left( x-y \right)}{{{h}^{2}}\left( {{x}^{2}}+{{y}^{2}} \right)}$
Substituting for the original function and the h terms cancel giving us a ${{h}^{-1}}$ in the numerator,
$\Rightarrow f\left( hx,hy \right)={{h}^{-1}}f\left( x,y \right)$
Hence, option A is a homogeneous function.
Let us solve the second part now,
$\Rightarrow f\left( x,y \right)={{x}^{\dfrac{1}{3}}}.{{y}^{-\dfrac{2}{3}}}{{\tan }^{-1}}\dfrac{x}{y}$
Let us replace x by hx and y by hy,
$\Rightarrow f\left( hx,hy \right)={{\left( hx \right)}^{\dfrac{1}{3}}}.{{\left( hy \right)}^{-\dfrac{2}{3}}}{{\tan }^{-1}}\left( \dfrac{hx}{hy} \right)$
Now we take the power of the individual terms and then cancelling the h from the numerator and denominator term of the tan function,
$\Rightarrow f\left( hx,hy \right)={{h}^{\dfrac{1}{3}}}{{x}^{\dfrac{1}{3}}}.{{h}^{-\dfrac{2}{3}}}{{y}^{-\dfrac{2}{3}}}{{\tan }^{-1}}\left( \dfrac{x}{y} \right)$
Taking the h terms common out,
$\Rightarrow f\left( hx,hy \right)={{h}^{\dfrac{1}{3}-\dfrac{2}{3}}}{{x}^{\dfrac{1}{3}}}.{{y}^{-\dfrac{2}{3}}}{{\tan }^{-1}}\left( \dfrac{x}{y} \right)$
Simplifying the power for h,
$\Rightarrow f\left( hx,hy \right)={{h}^{-\dfrac{1}{3}}}{{x}^{\dfrac{1}{3}}}.{{y}^{-\dfrac{2}{3}}}{{\tan }^{-1}}\left( \dfrac{x}{y} \right)$
Substituting for the original function,
$\Rightarrow f\left( hx,hy \right)={{h}^{-\dfrac{1}{3}}}f\left( x,y \right)$
Hence, option B is a homogeneous function too.
Let us solve the third part now,
$\Rightarrow f\left( x,y \right)=x\left( \ln \sqrt{{{x}^{2}}+{{y}^{2}}}-\ln y \right)+y{{e}^{\dfrac{x}{y}}}$
Let us replace x by hx and y by hy,
$\Rightarrow f\left( hx,hy \right)=hx\left( \ln \sqrt{{{\left( hx \right)}^{2}}+{{\left( hy \right)}^{2}}}-\ln hy \right)+hy{{e}^{\dfrac{hx}{hy}}}$
Now we take the ${{h}^{2}}$ term common from inside the root and it comes out of the root as h. Then, we cancel the h in the numerator and denominator in the power of the exponential term.
$\Rightarrow f\left( hx,hy \right)=hx\left( \ln h\sqrt{{{x}^{2}}+{{y}^{2}}}-\ln hy \right)+hy{{e}^{\dfrac{x}{y}}}$
We know that $\ln a-\ln b=\ln \dfrac{a}{b}.$ Using this in the above equation,
$\Rightarrow f\left( hx,hy \right)=hx\left( \ln \dfrac{h\sqrt{{{x}^{2}}+{{y}^{2}}}}{hy} \right)+hy{{e}^{\dfrac{x}{y}}}$
Cancelling the h terms from the numerator and denominator of the ln term,
$\Rightarrow f\left( hx,hy \right)=hx\left( \ln \dfrac{\sqrt{{{x}^{2}}+{{y}^{2}}}}{y} \right)+hy{{e}^{\dfrac{x}{y}}}$
Taking h common out from the first and second terms,
$\Rightarrow f\left( hx,hy \right)=h\left( x\left( \ln \dfrac{\sqrt{{{x}^{2}}+{{y}^{2}}}}{y} \right)+y{{e}^{\dfrac{x}{y}}} \right)$
Using $\ln a-\ln b=\ln \dfrac{a}{b},$ and splitting the ln terms,
$\Rightarrow f\left( hx,hy \right)=h\left( x\left( \ln \sqrt{{{x}^{2}}+{{y}^{2}}}-\ln y \right)+y{{e}^{\dfrac{x}{y}}} \right)$
Substituting the original function,
$\Rightarrow f\left( hx,hy \right)=hf\left( x,y \right)$
Hence, option C is a homogeneous function too.
Next, we consider option D.
$\Rightarrow f\left( x,y \right)=x\left[ \ln \dfrac{2{{x}^{2}}+{{y}^{2}}}{x}-\ln \left( x+y \right) \right]+{{y}^{2}}\tan \dfrac{x+2y}{3x-y}$
Let us replace x by hx and y by hy,
$\Rightarrow f\left( hx,hy \right)=hx\left[ \ln \dfrac{2{{\left( hx \right)}^{2}}+{{\left( hy \right)}^{2}}}{hx}-\ln \left( hx+hy \right) \right]+{{\left( hy \right)}^{2}}\tan \dfrac{hx+2hy}{3hx-hy}$
Taking the ${{h}^{2}}$ term common out from the numerator of the first ln function and cancelling with the one h in the denominator,
$\Rightarrow f\left( hx,hy \right)=hx\left[ \ln \dfrac{h\left( 2{{x}^{2}}+{{y}^{2}} \right)}{x}-\ln \left( hx+hy \right) \right]+{{\left( hy \right)}^{2}}\tan \dfrac{hx+2hy}{3hx-hy}$
Taking the h terms common out from the numerator and denominator of the tan function, and using the formula $\ln a-\ln b=\ln \dfrac{a}{b}$ for the two ln functions,
$\Rightarrow f\left( hx,hy \right)=hx\left[ \ln \dfrac{h\left( 2{{x}^{2}}+{{y}^{2}} \right)}{xh\left( x+y \right)} \right]+{{\left( hy \right)}^{2}}\tan \dfrac{h\left( x+2y \right)}{h\left( 3x-y \right)}$
Cancelling the h terms from the numerator and denominator of the tan function and similarly cancelling the h terms from the numerator and denominator of the ln function,
$\Rightarrow f\left( hx,hy \right)=hx\left[ \ln \dfrac{\left( 2{{x}^{2}}+{{y}^{2}} \right)}{x\left( x+y \right)} \right]+{{\left( hy \right)}^{2}}\tan \dfrac{\left( x+2y \right)}{\left( 3x-y \right)}$
Splitting the ln terms using $\ln a-\ln b=\ln \dfrac{a}{b},$ and squaring the h in the second term,
$\Rightarrow f\left( hx,hy \right)=hx\left[ \ln \dfrac{\left( 2{{x}^{2}}+{{y}^{2}} \right)}{x}-\ln \left( x+y \right) \right]+{{h}^{2}}{{y}^{2}}\tan \dfrac{\left( x+2y \right)}{\left( 3x-y \right)}$
Taking the h term common out from both the terms,
$\Rightarrow f\left( hx,hy \right)=h\left( x\left[ \ln \dfrac{\left( 2{{x}^{2}}+{{y}^{2}} \right)}{x}-\ln \left( x+y \right) \right]+h{{y}^{2}}\tan \dfrac{\left( x+2y \right)}{\left( 3x-y \right)} \right)$
We cannot replace the original function here since there is an additional h term in the second term. Therefore, option D is not a homogeneous function.

Hence, options A, B and C are homogeneous functions.

Note: We need to know the concept of homogeneity to solve such questions. Care must be taken while cancelling the terms of h and we need to use the formula $\ln a-\ln b=\ln \dfrac{a}{b},$ for further simplification of the terms. We need to note that it will not be a homogeneous function if we are not able to replace the original function back in the scaled equation.