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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 - Mathematics

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प्रश्न

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

 

विकल्प

  • 1

  • \[\sqrt{3}\]

  • 2

  • none of these

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उत्तर

\[\sqrt{3}\] 

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

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अध्याय 15: Mean Value Theorems - Exercise 15.4 [पृष्ठ १९]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 15 Mean Value Theorems
Exercise 15.4 | Q 3 | पृष्ठ १९

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