Question

Let $f\left(x\right)$ be a real valued function not identically zero, such that

$f(x+y^n)=f\left(x\right)+{\left(f\left(y\right)\right)}^n\forall x,y\in R$

where $n\in N(n\ne 1)$ and $f^{\prime }\left(0\right)\ge 0$. We may get an explicit form of the function $f\left(x\right)$.

The value of $f′(0)$f′(0) is :

• A. $1$
• B. $n$
• C. $n+1$
• D. $2$

Answer 1

The correct option is A $1$

Putting $x=y=0$ in the given functional equation, we have
$f\left(0\right)=f(0+0)=f\left(0\right)+{\left(f\left(0\right)\right)}^n\Rightarrow f\left(0\right)=0$
$f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}=\underset{h\to 0}{lim}\frac{f\left(h\right)}{h}=l$(say)
Again $f^{\prime }\left(0\right)=\underset{h\to 0}{lim}\frac{f\left(h\right)-f\left(0\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(0\right)+{\left(f\left(h^{1/n}\right)\right)}^n-f\left(0\right)}{{\left(h^{1/n}\right)}^n}$
$=\underset{h\to 0}{lim}{\left(\frac{f\left(h^{1/n}\right)}{h^{1/n}}\right)}^n=l^n$
So $l^n=l\Rightarrow l=0,1,-1$
But $l=-1$ is ruled out as $f^{\prime }\left(0\right)\ge 0$
If $l=0$, then $f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$=\underset{h\to 0}{lim}\frac{f\left(x\right)+{\left(f\left(h^{1/n}\right)\right)}^n-f\left(x\right)}{h}=l^n=0$
$\Rightarrow$ $f^{\prime }\left(x\right)=0$ i.e., $f$ is a constant funclion
but $f\left(0\right)=0$ so $f$ is identically zero which is not the case.
Hence t=1
$\Rightarrow$ $f\left(x\right)=x+c$ but $f\left(0\right)=0$
so $f\left(x\right)=x$. Therefor, $f^{\prime }\left(0\right)=1$.