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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
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B
3
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C
2
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D
1
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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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