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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 (Commerce) Class 12 - Mathematics

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### 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?

#### APPEARS IN

Solution 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.
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