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Let $f(x) = \sin^3 x + \lambda\sin^2 x$ , $-\frac{\pi}{2} < x < \frac{\pi}{2}$. Find '$\lambda$' such that the function $f$ has (i) no points of extremum, (ii) exactly two points of extremum: one of local maximum and other of local minimum, (iii) only a local maximum, (iv) only a local minimum.

I tried differentiating the function but could not find any proper logic.

I tried plotting the graphs of $\sin^3 x$ and $sin^2 x$ separately and observed that when $\lambda$ is $0$ then the function will have no maxima or minima in the given conditions. But I could not find similar logic for the other cases.

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    What do you know so far? – 2017-02-23

1 Answers 1

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$$f'(x) = \sin x \cos x (3\sin x + 2\lambda)$$ Since $-\frac{\pi}{2} < x < \frac{\pi}{2}$, $\cos x \neq 0$. Hence $f'(x) = 0$ if $x = 0, \sin x = -\frac{2\lambda}{3}$. When $|\lambda| \geq \frac{3}{2}$, $f'(x) = 0$ has only one zero, $x=0$. When $|\lambda| < \frac{3}{2}$, and $\lambda \neq 0$, $f'(x)$ vanishes at one more point in the given interval.

$$f''(x) = (\cos^2 x - \sin^2 x)(3\sin x + 2\lambda) + 3\sin x \cos^2 x$$ At $x= 0$, $f''(0) = 2\lambda$. Thus $x=0$ is a maximum when $\lambda < 0$, minimum when $\lambda > 0$ and a point of inflection when $\lambda = 0$. When $\sin x = -\frac{2\lambda}{3}$, $f''(x) > 0$ if $\lambda < 0$ and negative when $\lambda > 0$. The corresponding point is a maximum or minimum in these cases.

Hence $f$ has no points of extremum when $\lambda = 0$, one minimum and one maximum when $|\lambda| < \frac{3}{2}$ and $\lambda \neq 0$.

When $|\lambda| > \frac{3}{2}$, $x=0$ is a maximum when $\lambda < 0$ and minimum when $\lambda > 0$.