Application of Derivatives 3 Question 3
3. If $f(x)$ and $g(x)$ are differentiable functions for $0 \leq x \leq 1$, such that
$$ \begin{gathered} f(0)=2, g(0)=0 \\ f(1)=6, g(1)=2 \end{gathered} $$
Then, show that there exists $c$ satisfying $0<c<1$ and $f^{\prime}(c)=2 g^{\prime}(c)$.
(1982, 2M)
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Solution:
- Since, $f(x)$ and $g(x)$ are differentiable functions for $0 \leq x \leq 1$.
$$ \therefore \quad f^{\prime}(c)=\frac{f(1)-f(0)}{1-0} $$
Using Lagrange’s Mean Value theorem,
$$ \begin{aligned} & \frac{6-2}{1-0}=4 \\ & \text { and } \quad g^{\prime}(c)=\frac{g(1)-g(0)}{1-0} \\ & =\frac{2-0}{1-0}=2 \\ & \Rightarrow \quad f^{\prime}(c)=2 g^{\prime}(c) \end{aligned} $$
