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# Solution for Verify Lagrange'S Mean Value Theorem for the Following Function on the Indicated Intervals. Find a Point 'C' in the Indicated Interval as Stated by the Lagrange'S F(X) = Tan−1 X on [0, 1] ? - CBSE (Commerce) Class 12 - Mathematics

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ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### Question

Verify Lagrange's mean value theorem for the following function on the indicated intervals. find a point 'c' in the indicated interval as stated by the Lagrange's mean value theore  f(x) = tan1 x on [0, 1] ?

#### Solution

We have,

$f\left( x \right) = \tan^{- 1} x$

Clearly,

$f\left( x \right)$ is continuous on $\left[ 0, 1 \right]$ and derivable on $\left( 0, 1 \right)$
Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some $c \in \left( - 3, 4 \right)$ such that $f'\left( c \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0} = \frac{f\left( 1 \right) - f\left( 0 \right)}{1}$
Now,
$f\left( x \right) = \tan^{- 1} x$
$f'\left( x \right) = \frac{1}{1 + x^2}$ ,
$f\left( 1 \right) = \frac{\pi}{4}$ ,
$f\left( 0 \right) = 0$
∴  $f'\left( x \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}$

$\Rightarrow \frac{1}{1 + x^2} = \frac{\pi}{4} - 0$

$\Rightarrow \left( \frac{4}{\pi} - 1 \right) = x^2$

$\Rightarrow x = \pm \sqrt{\frac{4 - \pi}{\pi}}$

Thus,

$c = \sqrt{\frac{4 - \pi}{\pi}} \in \left( 0, 1 \right)$ such that

$f'\left( c \right) = \frac{f\left( 1 \right) - f\left( 0 \right)}{1 - 0}$ .
Hence, Lagrange's theorem is verified.

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#### APPEARS IN

Solution Verify Lagrange'S Mean Value Theorem for the Following Function on the Indicated Intervals. Find a Point 'C' in the Indicated Interval as Stated by the Lagrange'S F(X) = Tan−1 X on [0, 1] ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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