A) f'(x) =0 for all x
B) f'(x)+4g'(x)=0 for at least one x
C) f'(x)=2g'(x) for at most done x
D) none of these
Correct Answer: B
Solution :
[b] We know that there exists at least one x in (0, 1) for which \[\frac{f(1)-f(0)}{g(1)-g(0)}=\frac{f'(x)}{g'(x)}\] Or \[\frac{2-10}{4-2}=\frac{f'(x)}{g'(x)}\]or \[f'(x)+4g'(x)=0\] For at least one x in (0, 1).You need to login to perform this action.
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