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Q.

If x and g(x) are differentiable functions in [0,1] such that 0=2, g())=0, ⨍(1)=6, g(1)=2, then there exists c, 0<c<1 such that ⨍'(c)=
 

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a

g'(c) 
 

b

2g'(c) 
 

c

–g'(c) 
 

d

3g'(c) 
 

answer is C.

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Detailed Solution

g1(c)f1(c)=g(1)-g(0)f(1)-f(0)            =2-06-2 =24=12 2g1(c)=f1(c)

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