Question

For the function f (x) = x + \[\frac{1}{x}\] ∈ [1, 3], the value of c for the Lagrange's mean value theorem is 

 

पर्याय

• 1

• \[\sqrt{3}\]

• 2

• none of these

Answer 1

\[\sqrt{3}\] 

We have

\[f\left( x \right) = x + \frac{1}{x} = \frac{x^2 + 1}{x}\]

Clearly,  \[f\left( x \right)\]  is continuous on 

\[\left[ 1, 3 \right]\] and derivable on \[\left( 1, 3 \right)\] .

Thus, both the conditions of Lagrange's theorem are satisfied.
Consequently, there exists\[c \in \left( 1, 3 \right)\] such that

\[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1} = \frac{f\left( 3 \right) - f\left( 1 \right)}{2}\]

Now, \[f\left( x \right) = \frac{x^2 + 1}{x}\]

\[f'\left( x \right) = \frac{x^2 - 1}{x^2}\],\[f\left( 1 \right) = 2\] ,\[f\left( 3 \right) = \frac{10}{3}\]

∴  \[f'\left( x \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{2}\]

\[\Rightarrow \frac{x^2 - 1}{x^2} = \frac{4}{6}\]

\[ \Rightarrow \frac{x^2 - 1}{x^2} = \frac{2}{3}\]

\[ \Rightarrow 3 x^2 - 3 = 2 x^2 \]

\[ \Rightarrow x = \pm \sqrt{3}\]

Thus,\[c = \sqrt{3} \in \left( 1, 3 \right)\] such that \[f'\left( c \right) = \frac{f\left( 3 \right) - f\left( 1 \right)}{3 - 1}\] .