Question

f(x) = `sin^4x + cos^4x` in `[0, pi/2]`

Answer 1

We have, f(x) = `sin^4x + cos^4x` in `[0, pi/2]`

We know that sin x and cos x are conditions and differentiable

∴ sin4x and cos4x and hence sin4x + cos4x is continuous and differentiable

Now f(0) = 0 + 1 = 1 and `"f"(pi/2)` = 1 + 0 = 1

⇒ f(0) = `"f"(pi/2)`

So, conditions of Rolle's theorem are satisfied.

Hence, there exists atleast one `"c" ∈ (0, pi/2)`  such that f'(c) = 0

∴ `4sin^3"c" cos "c" - 4cos^3"c" sin"c"` = 0

⇒ `4sin"c" cos"c" (sin^2"c" - cos^2"c")` = 0

⇒ `4sin"c" cos"c"(-cos 2"c")` = 0

⇒ `-2 sin 2"c" * cos 2"c"` = 0

⇒ sin 4c = 0

⇒ 4c = π

⇒ c = `pi/4 ∈ (0, pi/2)`.

Hence, Rolle's theorem has been verified.