A) \[f'(1)=\,f'(0)+1\]
B) \[f'(1)=\,f'(0)-1\]
C) \[f'(0)=\,f'(1)+2\]
D) \[f'(0)=\,f'(1)-2\]
Correct Answer: D
Solution :
Let \[x=y=0\Rightarrow f(0)=0\] |
Now we have\[\begin{align} & \frac{f\left( x+2y \right)-f(x)}{2y}=2x+\frac{f\left( 2y \right)}{2y} \\ & =2x+\frac{f(2y)-f(0)}{2y} \\ \end{align}\] |
Taking limit on both sides as \[~2y\to 0\]we get |
\[f'(x)=2x+f'(0)\,as\,f(0)=0\] |
\[\Rightarrow f'(0)=f'(1)-2\] |
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