SATHEE: Application of Derivatives 3 Question 3

Application of Derivatives 3 Question 3

####3. If $f (x)$ and $g (x)$ are differentiable functions for $0 \leq x \leq 1$ , such that

$\begin{matrix} f (0) = 2, g (0) = 0 \\ f (1) = 6, g (1) = 2 \end{matrix}$

Then, show that there exists $c$ satisfying $0 < c < 1$ and $f^{'} (c) = 2 g^{'} (c)$ .

(1982, 2M)

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Solution:

Since, $f (x)$ and $g (x)$ are differentiable functions for $0 \leq x \leq 1$ .

$∴ f^{'} (c) = \frac{f (1) - f (0)}{1 - 0}$

Using Lagrange’s Mean Value theorem,

$\begin{aligned} \frac{6 - 2}{1 - 0} = 4 \\ and g^{'} (c) = \frac{g (1) - g (0)}{1 - 0} \\ = \frac{2 - 0}{1 - 0} = 2 \\ \Rightarrow f^{'} (c) = 2 g^{'} (c) \end{aligned}$

Application of Derivatives 3 Question 3

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Application of Derivatives 3 Question 3

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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