First slide

Mean value theorems

Question

$Let f (x) satisfy all the conditions of Lagrange's mean value theorem in [0, 2] . If$ $f (0) = 0 and |f^{'} (x)| \leq \frac{1}{2} for all x in [0, 2], then$

Moderate

A

$f (x) < \frac{1}{2}$
B

$|f (x)| \leq 1$
C

$f (x) = 2 x$
D

$f (x) = 3 for at least one x in [0, 2]$

Solution

$\begin{array}{l} By using lagrange's mean value theorem, \frac{f (b) - f (a)}{b - a} = f^{'} (x) \\ \Rightarrow \frac{f (2) - f (0)}{2 - 0} = f^{'} (x) \\ \Rightarrow \frac{f (2) - 0}{2} = f^{'} (x) \\ i n t e g r a t i n g o n b o t h s i d e s \\ \Rightarrow f (x) = \frac{f (2)}{2} x + c (∵ f (0) = 0 \Rightarrow c = 0) \\ ∴ f (x) = \frac{f (2)}{2} x \dots \dots (1) \\ Also, |f^{'} (x)| \leq \frac{1}{2} \\ \Rightarrow |\frac{f (2)}{2}| \leq \frac{1}{2} \end{array}$
From Eq. (1),
$\begin{array}{l} | f (x) | = |\frac{f (2)}{2} x| = |\frac{f (2)}{2}| | x | \leq \frac{1}{2} | x | \\ In interval [0, 2], for maximum value of x take x = 2 \\ | f (2) | \leq \frac{1}{2} \cdot 2 \\ \Rightarrow | f (2) | \leq 1 \\ h e n c e |f (x)| \leq 1 \end{array}$

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