
Which of the following is a second degree homogeneous expression in x and y:
A. $a{{x}^{2}}+bx+c$
B. $a{{x}^{2}}+bx+cy$
C. $a{{x}^{2}}+bx+c{{y}^{2}}$
D. $a{{x}^{2}}+bxy+c{{y}^{2}}$
Answer
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Hint: Remember that a homogenous expression can be mathematically defined as: $f(kx,ky)={{k}^{n}}f(x,y)$ . So, replace x by kx and y by ky in the original expression, then try to express it according to the mathematical definition of homogeneous equation.
Complete step by step answer:
Before moving to the options, let us talk about what a homogenous expression is.
A second degree homogeneous expression is an expression for which if (x,y) is a solution, then (kx,ky) will also be one of the solutions provided k is a scalar.
A homogenous expression can be mathematically defined as:
$f(kx,ky)={{k}^{n}}f(x,y)$
Where n is the degree of homogenous expression.
Starting with option (a);
$f(x,y)=a{{x}^{2}}+bx+c$
Let x=kx and y=ky.
$f(kx,ky)=a{{k}^{2}}{{x}^{2}}+bkx+c$
$\therefore f(kx,ky){{k}^{2}}f(x,y)$
So, option (a) is not a homogenous expression.
Similarly, solving for option (b). We get;
$f(x,y)=a{{x}^{2}}+bx+cy$
Let x=kx and y=ky.
$f(kx,ky)=a{{k}^{2}}{{x}^{2}}+bkx+cky$
$\therefore f(kx,ky){{k}^{2}}f(x,y)$
So, option (b) is not a second degree homogeneous expression.
Solving for option (c). We get;
$f(x,y)=a{{x}^{2}}+bx+c{{y}^{2}}$
Let x=kx and y=ky.
$f(kx,ky)=a{{k}^{2}}{{x}^{2}}+bkx+c{{k}^{2}}y$
$\Rightarrow f(kx,ky)=k\left( ak{{x}^{2}}+bx+cky \right)$
$\therefore f(kx,ky){{k}^{2}}f(x,y)$
So, option (c) is not a second degree homogeneous expression.
Finally, solving option (d). We get;
$f(x,y)=a{{x}^{2}}+bxy+c{{y}^{2}}$
Let x=kx and y=ky.
$f(kx,ky)=a{{k}^{2}}{{x}^{2}}+b{{k}^{2}}xy+c{{k}^{2}}{{y}^{2}}$
$\Rightarrow f(kx,ky)={{k}^{2}}\left( a{{x}^{2}}+bxy+c{{y}^{2}} \right)$
$\therefore f(kx,ky)={{k}^{2}}f(x,y)$
So, option (d) is a second degree homogeneous expression.
Hence, option (d) is the correct answer.
Note: Be careful with the degree of the homogenous expression and also don’t forget to replace all the arbitrary variables in the expression for checking the homogeneity of the expression. An alternative to checking the homogeneity can be adding the powers of arbitrary variables of each term separately. If all are equal to the degree of homogeneity, then the expression is homogenous. Else it is non-homogenous. However, if terms involving trigonometric functions are present, it might be difficult to do so.
Complete step by step answer:
Before moving to the options, let us talk about what a homogenous expression is.
A second degree homogeneous expression is an expression for which if (x,y) is a solution, then (kx,ky) will also be one of the solutions provided k is a scalar.
A homogenous expression can be mathematically defined as:
$f(kx,ky)={{k}^{n}}f(x,y)$
Where n is the degree of homogenous expression.
Starting with option (a);
$f(x,y)=a{{x}^{2}}+bx+c$
Let x=kx and y=ky.
$f(kx,ky)=a{{k}^{2}}{{x}^{2}}+bkx+c$
$\therefore f(kx,ky){{k}^{2}}f(x,y)$
So, option (a) is not a homogenous expression.
Similarly, solving for option (b). We get;
$f(x,y)=a{{x}^{2}}+bx+cy$
Let x=kx and y=ky.
$f(kx,ky)=a{{k}^{2}}{{x}^{2}}+bkx+cky$
$\therefore f(kx,ky){{k}^{2}}f(x,y)$
So, option (b) is not a second degree homogeneous expression.
Solving for option (c). We get;
$f(x,y)=a{{x}^{2}}+bx+c{{y}^{2}}$
Let x=kx and y=ky.
$f(kx,ky)=a{{k}^{2}}{{x}^{2}}+bkx+c{{k}^{2}}y$
$\Rightarrow f(kx,ky)=k\left( ak{{x}^{2}}+bx+cky \right)$
$\therefore f(kx,ky){{k}^{2}}f(x,y)$
So, option (c) is not a second degree homogeneous expression.
Finally, solving option (d). We get;
$f(x,y)=a{{x}^{2}}+bxy+c{{y}^{2}}$
Let x=kx and y=ky.
$f(kx,ky)=a{{k}^{2}}{{x}^{2}}+b{{k}^{2}}xy+c{{k}^{2}}{{y}^{2}}$
$\Rightarrow f(kx,ky)={{k}^{2}}\left( a{{x}^{2}}+bxy+c{{y}^{2}} \right)$
$\therefore f(kx,ky)={{k}^{2}}f(x,y)$
So, option (d) is a second degree homogeneous expression.
Hence, option (d) is the correct answer.
Note: Be careful with the degree of the homogenous expression and also don’t forget to replace all the arbitrary variables in the expression for checking the homogeneity of the expression. An alternative to checking the homogeneity can be adding the powers of arbitrary variables of each term separately. If all are equal to the degree of homogeneity, then the expression is homogenous. Else it is non-homogenous. However, if terms involving trigonometric functions are present, it might be difficult to do so.
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