Question

Let f(x) be a real-valued differentiable function not identically zero such that $f(x+y^{2n+1})=f\left(x\right)+{\left\{f\right(y\left)\right\}}^{2n+1},nϵN$ and x, y are any real numbers and $f^{\prime }\left(0\right)\ge 0$. Find the value of f(5)

• A. $0$
• B. $1$
• C. $2$
• D. $5$

Answer 1

The correct option is D $5$

Here, $f(x+y^{2n+1})=f\left(x\right)+{\left\{f\right(y\left)\right\}}^{2n+1}$ .....(i)

Putting $x=0,y=0$, we get

$f\left(0\right)=f\left(0\right)+{\left\{f\right(0\left)\right\}}^{2n+1}\Rightarrow f\left(0\right)=0$

$f^{\prime }\left(0\right)\ge 0$ (given)

$\Rightarrow \underset{h\to 0}{lim}\frac{f\left(x\right)-f\left(0\right)}{x-0}\ge 0\Rightarrow \underset{h\to 0}{lim}\frac{f\left(x\right)}{x}\ge 0$

Now, if $x>0\Rightarrow f\left(x\right)\ge 0$ .....(ii)

Putting $x=0,y=1$ in Eq. (i),
$f\left(1\right)=f\left(0\right)+{\left\{f\right(1\left)\right\}}^{2n+1}$ or $f\left(1\right)[1-{\left\{f\right(1\left)\right\}}^{2n}]=0$

$\therefore f\left(1\right)=0$ or 1 [using Eq. (ii)]

Putting $y=1$ in Eq. (i), for all real x,
$f(x+1)=f\left(x\right)+{\left\{f\right(1\left)\right\}}^{2n+1}$ .....(iii)

Now, two cases arise either $f\left(1\right)=0$ or 1

Case I If $f\left(1\right)=0$
$\Rightarrow f(x+1)=f\left(x\right)$ [using Eq. (iii)]

$\Rightarrow f\left(1\right)=f\left(2\right)=f\left(3\right)=....=0$
$\therefore f\left(x\right)$ is identically zero. (which is not possible)

Case II If $f\left(1\right)=1$
$\Rightarrow f(x+1)=f\left(x\right)+1$
$\therefore f\left(2\right)=f1)+1=1+1=2$
$f\left(3\right)=f\left(2\right)+1=2+1=3$
$f\left(4\right)=f\left(3\right)+1=3+1=4$
$f\left(5\right)=f\left(4\right)+1=4+1=5$

Proceeding in same way, we get
$f\left(x\right)=x$ and $f^{\prime }\left(x\right)=1\Rightarrow f^{\prime }\left(10\right)=1$
Hence, $f\left(5\right)=5$ and $f^{\prime }\left(10\right)=1$