Question

If $f:R\to R$ be a differentiable function, such that $f(x+2y)=f\left(x\right)+f\left(2y\right)+4xy$ for all $x,y\in R$ then

• A. $f^{\prime }\left(1\right)=f^{\prime }\left(0\right)+1$
• B. $f^{\prime }\left(1\right)=f^{\prime }\left(0\right)-1$
• C. $f^{\prime }\left(0\right)=f^{\prime }\left(1\right)+2$
• D. $f^{\prime }\left(1\right)=f^{\prime }\left(0\right)+2$

Answer 1

The correct option is D $f^{\prime }\left(1\right)=f^{\prime }\left(0\right)+2$

$f\left(x+2y\right)=f\left(x\right)+f\left(2y\right)+4xy$

$f^1\left(x+2y\right)=f\left(x\right)+0+4y$

$x=0,y=\frac{1}{2}$

$f^1\left(1\right)=f^1\left(0\right)+2$