Let f(x) be a real valued function not identically zero in Z such that \[f\left( {x + {y^{2n + 1}}} \right) = f\left( x \right) + {\left\{ {f\left( y \right)} \right\}^{2n + 1}}\], $ n \in N,xy \in R $ . If $ f'(0) \geqslant 0 $ then f’(6) is equal to
A. 0
B. 1
C. – 1
D. None of these

Hint: Here, find some pattern of function by putting different values of x. and then apply mean value theorem for derivatives. Find the values of function at points one below 6 and one after 6 so that we can apply the mean value theorem. Then find the value derivative function at 6.

Complete step-by-step answer:
\[f\left( {x + {y^{2n + 1}}} \right) = f\left( x \right) + {\left\{ {f\left( y \right)} \right\}^{2n + 1}}\]
Putting x = 0 and y = 0
\[f\left( {0 + {0^{2n + 1}}} \right) = f\left( 0 \right) + {\left\{ {f\left( 0 \right)} \right\}^{2n + 1}}\]
\[ \Rightarrow f(0) = f\left( 0 \right) + {\left\{ {f\left( 0 \right)} \right\}^{2n + 1}}\]
\[ \Rightarrow {\left\{ {f\left( 0 \right)} \right\}^{2n + 1}} = 0 \Rightarrow f(0) = 0\]
Putting x = 1 and y = 0
\[f\left( {0 + {1^{2n + 1}}} \right) = f\left( 0 \right) + {\left\{ {f\left( 1 \right)} \right\}^{2n + 1}}\]
\[ \Rightarrow f(1) = f\left( 0 \right) + {\left\{ {f\left( 1 \right)} \right\}^{2n + 1}}\]
\[ \Rightarrow f(1)\left[ {{{\{ f(1)\} }^{2n}} - 1} \right] = 0 \Rightarrow {\{ f(1)\} ^{2n}} - 1 \Rightarrow {\{ f(1)\} ^{2n}} = 1 \Rightarrow f(1) = \pm 1\]
Now put x = 1 and y = 1
\[ \Rightarrow f(2) = f\left( 1 \right) + {\left\{ {f\left( 2 \right)} \right\}^{2n + 1}}\]
So, F(2) = 0 or 2
Similarly, we can find different values
Putting x = 2 and y = 1
F(3) = 1 or 3
F(5) = 3 or 5
F(7) = 5 or 7
Now Let's move on to the derivative function
Partially differentiate the required equation first with respect to y as a variable
Now changing the value of y > 0 for all real values and keeping x = 0(constant)
This means that the derivative of the function is constant for all values of y
So, from Mean value theorem we have \[F\left( c \right){\text{ }}\prime = \dfrac{{\left( {F\left( b \right) - F\left( a \right)} \right)}}{{\left( {b - a} \right)}}\] where c lies somewhere between b and a.
\[F\left( 6 \right){\text{ }}\prime = \dfrac{{\left( {F\left( 7 \right) - F\left( 5 \right)} \right)}}{{\left( {7 - 5} \right)}} = \dfrac{{7 - 5}}{2} = \dfrac{2}{2} = 1\]
Hence, f’(6) =1
So, the correct answer is “Option B”.

Note: In these types of questions, the derivative function is not directly given and the function given is slightly complex. So by putting random values of x and y try to find values of function with some pattern so that we can recognize the value of function at a particular point. We cannot find the derivative of function by directly differentiating with respect to x or y as both are variables.