Question

Arrange the following in the decreasing order of their values:

$A$. If $f(x,y)=x^3+y^3-2x^2{y}^2$ then $\left(f_{xx}\right){|}_{(1,1)}$

$B$. If $f(x,y,z)=(x^2+y^2+z^2{)}^{-1/2}$ then $\sum \frac{{\partial }^{2}f}{\partial x^2}$

$C$. If $\mu =\frac{x^{1/4}+y^{1/4}}{x^{1/6}+y^{1/6}}$ then $\frac{1}{\mu }[x\frac{\partial \mu }{\partial x}+y\frac{\partial \mu }{\partial y}]$

• A. $A,C,B$
• B. $B,C,A$
• C. $A,B,C$
• D. $C,A,B$

Answer 1

The correct option is A $A,C,B$

$A:$ $f(x,y)=x^3+y^3-2x^2{y}^2{f}_x=3x^2-4x{y}^2{f}_{xx}=6x-4y^2{f}_{xx}{|}_{1,1}=6\left(1\right)-4\left(1\right)=2$

$B:$ $f(x,y,z)=\frac{1}{{(x^2+y^2+z^2)}^{1/2}}\therefore \frac{\partial f}{\partial x}=\frac{-2x}{(x^2+y^2+z^2)}\frac{{\partial }^{2}f}{{\partial x}^2}=\frac{(x^2+y^2+z^2)(-2)-(-2x)\left(2x\right)}{{(x^2+y^2+z^2)}^2}=\frac{2(x^2-y^2-z^2)}{{(x^2+y^2+z^2)}^2}Similarly,\frac{{\partial }^{2}f}{{\partial y}^2}=\frac{2(-x^2+y^2-z^2)}{{(x^2+y^2+z^2)}^2}\frac{{\partial }^{2}f}{{\partial z}^2}=\frac{2(-x^2-y^2+z^2)}{{(x^2+y^2+z^2)}^2}\therefore \sum \frac{{\partial }^{2}f}{{\partial x}^2}=\frac{{\partial }^{2}f}{{\partial x}^2}+\frac{{\partial }^{2}f}{{\partial y}^2}+\frac{{\partial }^{2}f}{{\partial z}^2}=-2$

$C:$ $\mu =\frac{x^{1/4}+y^{1/4}}{x^{1/6}+y^{1/6}}$

Let $x=p^{12}$ and $y=q^{12}$

$\Rightarrow x\frac{d\mu }{dx}=\frac{d\mu }{dp}\cdot \frac{dp}{dx}\times p^{12}=\frac{d\mu }{dp}\cdot \frac{p^{12}}{12p^{11}}=\frac{p}{12}\frac{d\mu }{dp}Similarly,y\frac{d\mu }{dy}=\frac{q}{12}\frac{d\mu }{dq}\therefore \mu =\frac{p^3+q^3}{p^2+q^2}\Rightarrow \frac{d\mu }{dp}=\frac{p^5+3p^3{q}^2-2p^2{q}^3}{{(p^2+q^2)}^2}\frac{d\mu }{dq}=\frac{q^5+3q^3{p}^2-2q^2{p}^3}{{(p^2+q^2)}^2}(p\frac{d\mu }{dp}+q\frac{d\mu }{dq})\times \frac{1}{\mu }=\frac{p^5+q^5+q^3{p}^2+q^2{p}^3}{(p^2+q^2)(p^3+q^3)}=1\therefore A=2,B=-2,C=1$

Hence, correct answer is $A>C>B$.