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 $f\left(5\right)$ is :

• A. $6$
• B. $3$
• C. $5n$
• D. $5$

Answer 1

The correct option is D $5$

Take $x=y=0$

$\Rightarrow f(0+0)=f\left(0\right)+\left\{f\right(0){\}}^n\Rightarrow f(0)=f(0)+\{f\left(0\right){\}}^n\Rightarrow \left\{f\right(0){\}}^n=0\Rightarrow f(0)=0$

Take $x=0,y=1$

$\Rightarrow f(0+{\left(1\right)}^n)=f\left(0\right)+{\left(f\right(1\left)\right)}^n\Rightarrow f\left(1\right)=f\left(0\right)+{\left\{f\right(1\left)\right\}}^n\Rightarrow f\left(1\right)={\left\{f\right(1\left)\right\}}^n\Rightarrow f\left(1\right)\{1-{\left\{f\right(1\left)\right\}}^{n-1}\}=0\Rightarrow f\left(1\right)=1$

Take $x=y=1$

$\Rightarrow f(1+1)=f\left(1\right)+{\left\{f\right(1\left)\right\}}^n\Rightarrow f\left(2\right)=1+1^n=2\Rightarrow f\left(2\right)=2x=2,y=1\Rightarrow f(2+1^n)=f\left(2\right)+{\left\{f\right(1\left)\right\}}^n\Rightarrow f\left(3\right)=f\left(2\right)+1^n=2+1=3\Rightarrow f\left(3\right)=3x=3,y=1\Rightarrow f(3+1^n)=f\left(3\right)+{\left\{f\right(1\left)\right\}}^n\Rightarrow f\left(4\right)=3+1^n=4\Rightarrow f\left(4\right)=4x=4,y=1\Rightarrow f(4+1^n)=f\left(4\right)+{\left\{f\right(1\left)\right\}}^n=4+1^n\Rightarrow f\left(5\right)=5$

Hence, correct answer is $5$