If f and g are differentiable functions in [0,1] satisfying f(0)=2=g(1),g(0)=0 and f(1)=6, then for some c ϵ ]0,1[
A
f′(c)=g′(c)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
f′(c)=2g′(c)
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
C
2f′(c)=g′(c)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
2f′(c)=3g′(c)
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution
The correct option is Bf′(c)=2g′(c) Since, f and g both are continuous functions on [0,1] and differentiable on (0,1) then ∃cϵ(0,1) such that f′(c)=f(1)−f(0)1=6−21=4
and g′(c)=g(1)−g(0)1=2−01=2
Thus, we get f′(c)=2g′(c)