# Verify the Rolle’s theorem for each of the functions Given: f(x) = sin4x + cos4x

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) = sin4x + cos4x

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

f(x) = sin4x + cos4x is continuous at Hence, condition 1 is satisfied.

Condition 2:

f(x) = sin4x + cos4x

On differentiating above with respect to x, we get

f’(x) = 4 × sin3 (x) × cos x + 4 × cos3 x × (- sin x) f’(x) = 4sin3 x cos x – 4 cos3 x sinx

f’(x) = 4sin x cos x [sin2x – cos2 x]

f’(x) = 2 sin2x [sin2x – cos2 x]

[ 2 sin x cos x = sin 2x]

f’(x) = 2 sin 2x [- cos 2x]

[ cos2 x – sin2 x = cos 2x]

f’(x) = - 2 sin 2x cos 2x

f(x) is differentiable at Hence, condition 2 is satisfied.

Condition 3:

f(x) = sin4x + cos4x

f(0) = sin4(0) + cos4(0) = 1  Hence, condition 3 is also satisfied.

Now, let us show that c ( ) such that f’(c) = 0

f(x) = sin4x + cos4x

f’(x) = - 2 sin 2x cos 2x

Put x = c in above equation, we get

f’(c) = - 2 sin 2c cos 2c

, all the three conditions of Rolle’s theorem are satisfied

f’(c) = 0

- 2 sin 2c cos 2c = 0

sin 2c cos 2c = 0

sin 2c = 0

2c = 0

c = 0

Now, cos 2c = 0    Thus, Rolle’s theorem is verified.

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