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 $f\left(x\right)$.

${\int }_0^{1}f\left(x\right)dx$ is equal to

• A. $\frac{1}{2n}$
• B. $2n$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

The correct option is C $\frac{1}{2}$

$f^{\prime }\left(x\right)=\frac{f(x+h)-f\left(x\right)}{h}$

Take $h=y^n$

$\Rightarrow f^{\prime }\left(x\right)=\frac{f(x+y^n)-f\left(x\right)}{y^n}\Rightarrow f^{\prime }\left(x\right)=\frac{\left\{f\right(y){\}}^n}{y^n}$

Take $y=x\Rightarrow f^{\prime }\left(x\right)=\frac{\left\{f\right(x){\}}^n}{x^n}={\left(\frac{f\left(x\right)}{x}\right)}^n$

$\Rightarrow \int \left\{f\right(y){\}}^{-n}{f}^{\prime }\left(x\right)dx=\int x^{-n}dx$

Solving it, we get $f\left(x\right)=x$

$\Rightarrow {\int }_0^{1}f\left(x\right)dx={\int }_0^{1}xdx={\left|\frac{x^2}{2}\right|}_0^1=\frac{1}{2}$