Mathematics - Continuity - Differentiability Question with Solution | TestHub

Given $f\left(x+y^{2n+1}\right)=f(x)+\{f(y){\}}^{2n+1}$ - (1)

Diff w.r.t. ' x '

$$\begin{array}{rlr}& f^{\prime }\left(x+y^{2n+1}\right)\left(1+(2n+1)\cdot y^{2n}\cdot \frac{dy}{dx}\right)& =f^{\prime }(x)+(2n+1)\cdot \{f(y){\}}^{2n}\cdot \frac{dy}{dx}\end{array}$$

$\because \mathrm{x}$ and y are independent real numbers

$$\begin{array}{rlrl}& \frac{dy}{dx}=0& \therefore f^{\prime }\left(x+y^{2n+1}\right)=f^{\prime }(x)& \Rightarrow f(x)=x^{1/2n+1}\text{ and }x=0\end{array}$$

$$\begin{array}{rlr}& f^{\prime }(x)=f^{\prime }(0)wheny=x^{1/2n+1}andx=0& \\ & integrating& \\ & f(x)=f^{\prime }(0)x+cputtingx=0,y=1& \\ & \therefore f(1)=0;c=0inthegivenequation& \\ & \because f(x)=f^{\prime }(0)x& \\ & \therefore f(x)=f^{\prime }(0)x& \\ & f(1)=f^{\prime }(0)\cdot 1\Rightarrow f^{\prime }(0)=1& \\ & \therefore f(x)=x& \end{array}$$

$Hencef(5)=5$