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# Solution for Verify Rolle'S Theorem for the Following Function on the Indicated Interval F(X) = Sin4 X + Cos4 X on [ 0 , π 2 ] ? - CBSE (Commerce) 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) = sin4 x + cos4 x on $\left[ 0, \frac{\pi}{2} \right]$ ?

#### Solution

The given function is $f\left( x \right) = \sin^4 x + \cos^4 x$ .

Since

$\sin x \text { and } \cos x$ are everywhere continuous and differentiable,

$f\left( x \right) = \sin^4 x + \cos^4 x$ is continuous on $\left[ 0, \frac{\pi}{2} \right]$ and differentiable on $\left( 0, \frac{\pi}{2} \right)$ .
Also,
$f\left( \frac{\pi}{2} \right) = f\left( 0 \right) = 1$
Thus,
$f\left( x \right)$ satisfies all the conditions of Rolle's theorem.
Now, we have to show that there exists $c \in \left( 0, \frac{\pi}{2} \right)$ such that $f'\left( c \right) = 0$ .
We have

$f\left( x \right) = \sin^4 x + \cos^4 x$

$\Rightarrow f'\left( x \right) = 4 \sin^3 x\cos x - 4 \cos^3 x\sin x$

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

$\Rightarrow 4 \sin^3 x\cos x - 4 \cos^3 x\sin x = 0$

$\Rightarrow \sin^3 x\cos x - \cos^3 x\sin x = 0$

$\Rightarrow \tan^3 x - \tan x = 0$

$\Rightarrow \tan x\left( \tan^2 x - 1 \right) = 0$

$\Rightarrow \tan x = 0, \tan^2 x = 1$

$\Rightarrow \tan x = 0, \tan x = \pm 1$

$\Rightarrow x = 0, x = \frac{\pi}{4}, \frac{3\pi}{4}$

Thus

$c = \frac{\pi}{4} \in \left( 0, \frac{\pi}{2} \right)$ 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) = Sin4 X + Cos4 X on [ 0 , π 2 ] ? Concept: Maximum and Minimum Values of a Function in a Closed Interval.
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