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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 F(X) = Sin X − Sin 2x − X on [0, π] ? - CBSE (Science) Class 12 - Mathematics

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 theorem  f(x) = sin x − sin 2x − x on [0, π] ?

#### Solution

We have ,

$f\left( x \right) = \sin x - \sin2x - x$

Since

$\sin x, \sin2x \text { & }x$ are everywhere continuous and differentiable]

Therefore,

$f\left( x \right)$ is continuous on $\left[ 0, \pi \right]$ and differentiable on $\left( 0, \pi \right)$
Thus, both the conditions of lagrange's theorem are satisfied.
Consequently, there exists some
$c \in \left( 0, \pi \right)$  such that
$f'\left( c \right) = \frac{f\left( \pi \right) - f\left( 0 \right)}{\pi - 0} = \frac{f\left( \pi \right) - f\left( 0 \right)}{\pi}$
Now,
$f\left( x \right) = \sin x - \sin2x - x$
$f'\left( x \right) = \cos x - 2\cos2x - 1$,
$f\left( \pi \right) = - \pi$,
$f\left( 0 \right) = 0$
∴ $f'\left( x \right) = \frac{f\left( \pi \right) - f\left( 0 \right)}{\pi - 0}$

$\Rightarrow \cos x - 2\cos2x - 1 = - 1$

$\Rightarrow \cos x - 2\cos2x = 0$

$\Rightarrow \cos x - 4 \cos^2 x = - 2$

$\Rightarrow 4 \cos^2 x - \cos x - 2 = 0$

$\Rightarrow \cos x = \frac{1}{8}\left( 1 \pm \sqrt{33} \right)$

$\Rightarrow x = \cos^{- 1} \left[ \frac{1}{8}\left( 1 \pm \sqrt{33} \right) \right]$

Thus,

$c = \cos^{- 1} \left( \frac{1 \pm \sqrt{33}}{8} \right) \in \left( 0, \pi \right)$ such that

$f'\left( c \right) = \frac{f\left( \pi \right) - f\left( 0 \right)}{\pi - 0}$.

Hence, Lagrange's theorem is verified.

Is there an error in this question or solution?

#### 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 F(X) = Sin X − Sin 2x − X on [0, π] ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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