Question

Let f(x) = sin³x + λsin²x, – π/2 < x < π/2 Find the interval in which l should lies in order that f(x) has exactly one minimum and one maximum.

Answer 1

Given f(x) = sin³x + λsin²x

⇒ f'(x) = 3sin²x cos x + λsin2x

⇒ f'(x) = sinx cos x (3sinx + 2λ)

⇒ f'(x) = 0 gives

⇒ sin x cos x (3sinx + 2λ) = 0

⇒ sin x = 0, cos x = 0, (3sinx + 2λ) = 0

But if λ = 0, then sin x = 0 has only one solution.

Thus, λ ∈ (– 3/2 , 3/2) – {0}

For these value of l, there are two distinct solutions.

Since f(x) is continuous function, these solutions give one maxima and one minima. Beacuse for a continuous function, local max and min occur alternate way