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Let  $$f\left ( x \right )$$  and  $$g\left ( x \right )$$  be differentiable for   $$0\leq x\leq 1$$,  such that  $$f\left ( 0 \right )= 2,  g\left ( 0 \right )= 0,  f\left ( 1 \right )= 6$$.  Let these exist a real number c in  $$\left [ 0, 1 \right ]$$  such that  $$f'\left ( c \right )= 2g'\left ( c \right )$$  then  $$g\left ( 1 \right )= $$


A
1
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B
2
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C
2
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D
1
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Solution

The correct option is C $$2$$
Given that $$f(0)=2,f(1)=6,g(0)=0$$

Given $$f(x),f^{'}(x)$$ are differentiable on $$(0,1)$$.

Therefore, Lagrange's mean value theorem can be applied.

Lagrange's mean value theorem states that if $$f(x)$$ be continuous on $$[a,b]$$ and differentiable on $$(a,b)$$ then there exists some $$c$$ between $$a$$ and $$b$$ such that $$f^{'}(c)=\dfrac{f(b)-f(a)}{b-a}$$

Given that $$f^{'}(c)=2g^{'}(c)$$

$$\implies \dfrac{f(1)-f(0)}{1-0}=2(\dfrac{g(1)-g(0)}{1-0})$$

$$\implies 6-2=2(g(1)-0)$$

$$\implies 2g(1)=4$$

$$\implies g(1)=2$$

Mathematics

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