Question

# Let $$f(x)$$ and $$g(x)$$ be differentiable for $$0\leq x \leq1$$, such that $$f(0)=0$$, $$g(0)=0$$, $$f(1)=6$$. Let there exists a real number $$c$$ in $$(0,1)$$ such that $$f'(c)=2g'(c)$$. Then the value of $$g(1)$$ must be

A
1
B
3
C
2
D
1

Solution

## The correct option is C $$3$$From lagrange's theorem$$\dfrac{f(b)-f(a)}{b-a}=2\times\dfrac{g(b)-g(a)}{b-a}$$$$\dfrac{f(1)-f(0)}{1-0}=2\times \dfrac{g(1)-g(0)}{1-0}$$$$\dfrac{(6-0)}{1-0}=2\times \dfrac{(g(1)-0)}{1-0}$$$$6=2g(1)$$$$g(1)=3$$Mathematics

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