Question

Let $f:R^{+}\to R^{+}$ be a function satisfying the relation \(f(x.f(y))=f(xy)+x~for~all~x,y \epsilon R^{+}\). Then $\underset{x\to 0}{lim}\left(\frac{\left(f\right(x){)}^{\frac{1}{3}}-1}{\left(f\right(x){)}^{\frac{1}{2}}-1}\right)$=

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Given relation is $f(x.f(y\left)\right)=f\left(xy\right)+x$ (1.56)
Interchanging x and y in Eq. (1.56), we have
$f(y.f(x\left)\right)=f\left(yx\right)+y$ (1.57)
Again replacing $x$ with $f\left(x\right)$ in Eq. (1.56) we get
$f\left(f\right(x).f(y\left)\right)=f(y.f(x\left)\right)+f\left(x\right)$ (1.58)
Therefore, Eqs. (1.56) - (1.58) imply
$f\left(f\right(x).f(y\left)\right)=f\left(xy\right)+y+f\left(x\right)$ (1.59)
Again interchanging $x$ and $y$ in Eq. (1.59), we have
$f\left(f\right(y).f(x\left)\right)=f\left(yx\right)+x+f\left(y\right)$ (1.60)
Equations (1.59) and (1.60) imply
$f\left(xy\right)+y+f\left(x\right)=f\left(yx\right)+x+f\left(y\right)$ (1.61)
Suppose $f\left(x\right)-x=f\left(y\right)-y=\lambda$
Substituting $f\left(x\right)=\lambda +x$ in Eq. (1.56), we have
$x.f\left(y\right)+\lambda =(xy+\lambda )+x$
$\Rightarrow x.f\left(y\right)=xy+x$
Therefore $x(y+\lambda )=xy+x[\because f(y)=\lambda +y]$
$\Rightarrow \lambda =x$
$\Rightarrow \lambda =1(\because x>0)$
So $f\left(x\right)=x+\lambda =x+1$
Hence $\underset{x\to 0}{lim}\frac{\left(f\right(x){)}^{\frac{1}{3}}-1}{\left(f\right(x){)}^{\frac{1}{2}}-1}=\underset{x\to 0}{lim}\frac{(1+x{)}^{1/3}-1}{(1+x{)}^{1/2}-1}=\underset{x\to 0}{lim}\left(\frac{(1+x{)}^{1/3}-1}{1+x-1}\right).\left(\frac{1+x-1}{(1+x{)}^{1/2}-1}\right)=\frac{1/3}{1/2}=\frac{2}{3}$