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