PUC Karnataka Science Class 12Department of Pre-University Education, Karnataka
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# Solution for Verify Rolle'S Theorem for the Following Function on the Indicated Interval F ( X ) = 6 X π − 4 Sin 2 X on [ 0 , π / 6 ] ? - PUC Karnataka Science Class 12 - Mathematics

ConceptMaximum and Minimum Values of a Function in a Closed Interval

#### Question

Verify Rolle's theorem for the following function on the indicated interval $f\left( x \right) = \frac{6x}{\pi} - 4 \sin^2 x \text { on } [0, \pi/6]$ ?

#### Solution

The given function is $f\left( x \right) = \frac{6x}{\pi} - 4 \sin^2 x$ .

Since

$\sin^2 x \text { & }x$ are everywhere continuous and differentiable, $f\left( x \right)$ is continuous on $\left[ 0, \frac{\pi}{6} \right]$ and differentiable on $\left( 0, \frac{\pi}{6} \right)$ .

Also,

$f\left( \frac{\pi}{6} \right) = f\left( 0 \right) = 0$

Thus,

$f\left( x \right)$ satisfies all the conditions of Rolle's theorem.

Now, we have to show that there exists

$c \in \left( 0, \frac{\pi}{6} \right)$ such that $f'\left( c \right) = 0$ .
We have

$f\left( x \right) = \frac{6x}{\pi} - 4 \sin^2 x$

$\Rightarrow f'\left( x \right) = \frac{6}{\pi} - 8 \sin x \cos x$

$\therefore f'\left( x \right) = 0$

$\Rightarrow \frac{6}{\pi} - 8\sin x\cos x = 0$

$\Rightarrow \sin2x = \frac{3}{2\pi}$

$\Rightarrow x = \frac{1}{2} \sin^{- 1} \left( \frac{3}{2\pi} \right)$

Thus,

$c = \frac{1}{2} \sin^{- 1} \left( \frac{3}{2\pi} \right) \in \left( 0, \frac{\pi}{6} \right)$ such that $f'\left( c \right) = 0$ .
​Hence, Rolle's theorem is verified.
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#### APPEARS IN

Solution Verify Rolle'S Theorem for the Following Function on the Indicated Interval F ( X ) = 6 X π − 4 Sin 2 X on [ 0 , π / 6 ] ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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