Mathematics - Functions Question with Solution | TestHub

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

Differentiate with respect to 'x':

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

Since x and y are independent real numbers,

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

$$\begin{array}{rlrl}& (1)f^{\prime }(x)=f^{\prime }(0)wheny=x^{1/(2n+1)}andx=0& integrating& f(x)=f^{\prime }(0)x+c\\ & puttingx=0,y=1& \therefore f(1)=0;c=0& \\ & inthegivenequation& \because f(x)=f^{\prime }(0)x& \\ & \therefore f(x)=f^{\prime }(0)x& f(1)=f^{\prime }(0)\cdot 1\Rightarrow f^{\prime }(0)=1& \\ & \therefore f(x)=xHencef(5)=5& & \end{array}$$