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Solution for For the Function F (X) = X + 1 X ∈ [1, 3], the Value of C for the Lagrange'S Mean Value Theorem is (A) 1 (B) √ 3 (C) 2 (D) None of These - CBSE (Science) Class 12 - Mathematics

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Question

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

(a) 1
(b) \[\sqrt{3}\]

(c) 2
(d) none of these

Solution

(b) \[\sqrt{3}\] 

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

  Is there an error in this question or solution?
Solution for question: For the Function F (X) = X + 1 X ∈ [1, 3], the Value of C for the Lagrange'S Mean Value Theorem is (A) 1 (B) √ 3 (C) 2 (D) None of These concept: Maximum and Minimum Values of a Function in a Closed Interval. For the courses CBSE (Science), CBSE (Commerce), CBSE (Arts), PUC Karnataka Science
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