A) \[2f'(c)=g'(c)\]
B) \[2f'(c)=3g'(c)\]
C) \[f'(c)=g'(c)\]
D) \[f'(c)=2g'(c)\]
Correct Answer: D
Solution :
Let \[h(f)=f(x)-2g(x)\] as \[h(0)=h(1)=2\] Hence, using Rolle?s theorem \[h'(c)=0\] \[\Rightarrow \] \[f\,'(c)=2g'(c)\]You need to login to perform this action.
You will be redirected in
3 sec