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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
पर्याय
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
\[\Rightarrow \frac{x^2 - 1}{x^2} = \frac{4}{6}\]
\[ \Rightarrow \frac{x^2 - 1}{x^2} = \frac{2}{3}\]
\[ \Rightarrow 3 x^2 - 3 = 2 x^2 \]
\[ \Rightarrow x = \pm \sqrt{3}\]
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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