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  • Verify the Rolle’s theorem for each of the functions f(x) = sin^4x+cos^4x

Verify the Rolle’s theorem for each of the functions f(x)=\sin ^{4} x+\cos ^{4} x \text { in }\left[0, \frac{\pi}{2}\right]

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Solution

Given: f(x)=\sin ^{4} x+\cos ^{4} x \text { in }\left[0, \frac{\pi}{2}\right]

Now, we have to show that f(x) verify the Rolle’s Theorem

First of all, Conditions of Rolle’s theorem are:

a) f(x) is continuous at (a,b)

b) f(x) is derivable at (a,b)

c) f(a) = f(b)

If all three conditions are satisfied then there exist some ‘c’ in (a,b) such that f’(c) = 0

Condition 1:

f(x)=\sin ^{4} x+\cos ^{4} x

Since, f(x) is a trigonometric function and trigonometric function is continuous everywhere

\\ \Rightarrow f(x) = sin\textsuperscript{4}x + cos\textsuperscript{4}x \ is \ continuous \ at \ x \in \left [ 0,\pi/2 \right ]

Hence, condition 1 is satisfied.

Condition 2:

f(x)=\sin ^{4} x+\cos ^{4} x

On differentiating above with respect to x, we get

f'(x) = 4 \times sin\textsuperscript{3} (x) \times cos x + 4 \times cos\textsuperscript{3} x \times (- sin x)

\because\left[\frac{\mathrm{d}}{\mathrm{dx}} \sin \mathrm{x}=\cos \mathrm{x} \& \frac{\mathrm{d}}{\mathrm{dx}} \cos \mathrm{x}=-\sin \mathrm{x}\right]

\\ \Rightarrow f'(x) = 4sin\textsuperscript{3} x cos x - 4 cos\textsuperscript{3} x sinx

\Rightarrow f'(x) = 4sin x cos x [sin\textsuperscript{2}x - cos\textsuperscript{2} x]

 \Rightarrow f'(x) = 2 sin2x [sin\textsuperscript{2}x - cos\textsuperscript{2} x] [\because 2 sin x cos x = sin 2x]

 \Rightarrow f'(x) = 2 sin 2x [-cos 2x]

[\because cos\textsuperscript{2} x - sin\textsuperscript{2} x = cos 2x]

[\because cos\textsuperscript{2} x - sin\textsuperscript{2} x = cos 2x]

\Rightarrow  f(x) is differentiable at  \left [ 0, \pi/2 \right ]

Hence, condition 2 is satisfied.

Condition 3:

\\f(x) = sin\textsuperscript{4}x + cos\textsuperscript{4}x \\\\f(0) = sin\textsuperscript{4}(0) + cos\textsuperscript{4}(0) = 1

\begin{aligned} \mathrm{f}\left(\frac{\pi}{2}\right)=& \sin ^{4}\left(\frac{\pi}{2}\right)+\cos ^{4}\left(\frac{\pi}{2}\right)=1 \\ & \therefore \mathrm{f}(0)=\mathrm{f}\left(\frac{\pi}{2}\right) \end{aligned}

Hence, condition 3 is also satisfied.

Now, let us show that c ∈ \left [ 0, \pi/2 \right ] such that f’(c) = 0
\\f(x) = sin\textsuperscript{4}x + cos\textsuperscript{4}x \\ \Rightarrow f'(x) = - 2 sin 2x cos 2x

Put x = c in above equation, we get

\Rightarrow f'(c) = - 2 \ sin 2c \ cos 2c

\becauseall the three conditions of Rolle’s theorem are satisfied

f’(c) = 0
\\ \Rightarrow - 2 sin 2c cos 2c = 0 \\ \Rightarrow sin 2c cos 2c = 0 \\ \Rightarrow sin 2c = 0 \\ \Rightarrow 2c = 0\\ \\ \Rightarrow c = 0

or Now, cos 2c = 0

\\ \Rightarrow \cos 2 c=\cos \frac{\pi}{2} \\ \Rightarrow 2 c=\frac{\pi}{2} \\ \Rightarrow c=\frac{\pi}{4} \\ \Rightarrow c=\left\{0, \frac{\pi}{4}\right\} \in\left[0, \frac{\pi}{2}\right]

Thus, Rolle’s theorem is verified.

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