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In a homogeneous expression, all terms will have
(a) different degree
(b) degree>2
(c) degree<2
(d) same degree

Answer Verified Verified
Hint: To solve the above question, we have to know the concept of homogeneous expression and non homogeneous expression. An expression is called homogeneous if all the terms have the same degree and non homogeneous is just opposite is just opposite of homogeneous expression that is all the terms will have different degrees.

Complete step-by-step answer:
If we go through hint part we are getting a little bit idea about of homogeneous expression and non homogeneous expression.As we can see that all terms of homogeneous expression have same degree.So, now just taking an example of homogeneous expression and that will clear our concept fully-
If we start with :$f\left( x,y \right)=x+4y$
Now we multiply each of variable by m: $f(mx,my)=mx+4my$
Now we rearrange it out by factoring out m: $f(mx,my)=m(x+4y)$
And $x+4y$ is $f(x,y)$ : $f(mx,my)=mf(x,y)$
Which we are wanted ,with $n=1$ : ${{m}^{1}}f(x,y)$
Now it is a homogeneous expression where you can see that all terms have the same degree.
Now we have to know the concept of non homogeneous expression.Now we are taking an example of non homogeneous expression that will polish our concept.
Example: ${{x}^{3}}+{{z}^{2}}$
Now we start with :$f(x,y)={{x}^{3}}+{{z}^{2}}$
Now we multiply each variable by $p$ : $f(px,py)={{\left( px \right)}^{3}}+{{\left( pz \right)}^{2}}$
That is $f(px,py)={{p}^{3}}{{x}^{3}}+{{p}^{2}}{{z}^{2}}$
Now if we factoring out expression will be: $f(px,py)={{p}^{2}}(p{{x}^{3}}+{{z}^{2}})$
But in this case we can see $p{{x}^{3}}+{{z}^{2}}$ is not $f(x,y)$ !.

So, the correct answer is “Option (D)”.

Note: Here student must take care of the concept of homogeneous expression and non homogeneous expression. Sometimes students make mistakes to understand what is a homogeneous and Non- homogeneous expression. So, for that students have to always keep in mind that all the terms of homogeneous expression have the same degree.