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# Solution for Verify Rolle'S Theorem for the Following Function on the Indicated Interval F(X) = 2 Sin X + Sin 2x on [0, π] ? - CBSE (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(x) = 2 sin x + sin 2x on [0, π] ?

#### Solution

The given function is $f\left( x \right) = 2\sin x + \sin2x$ .

Since

$\sin x \text { and }\sin2x$ are everywhere continuous and differentiable,
$f\left( x \right)$ is continuous on $\left[ 0, \pi \right]$ and differentiable on $\left( 0, \pi \right)$ .
Also,
$f\left( \pi \right) = f\left( 0 \right) = 0$
Thus,
$f\left( x \right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists $c \in \left( 0, \pi \right)$  such that $f'\left( c \right) = 0$ .
We have

$f\left( x \right) = 2\sin x + \sin2x$

$\Rightarrow f'\left( x \right) = 2\cos x + 2\cos2x$

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

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

$\Rightarrow \cos x + \cos2x = 0$

$\Rightarrow \cos x + 2 \cos^2 x - 1 = 0$

$\Rightarrow 2 \cos^2 x + \cos x - 1 = 0$

$\Rightarrow \left( \cos x + 1 \right) \left( 2\cos x - 1 \right) = 0$

$\Rightarrow \cos x = - 1, \cos x = \frac{1}{2}$

$\Rightarrow \cos x = cos\pi, \cos x = \frac{\pi}{3}$

$\Rightarrow x = \pi, \frac{\pi}{3}$

Thus,

$c = \frac{\pi}{3} \in \left( 0, \pi \right)$ such that $f'\left( c \right) = 0$ .
​Hence, Rolle's theorem is verified.
Is there an error in this question or solution?

#### APPEARS IN

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