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.