Question

Let $f\left(x\right)={\sin }^{3}x+\lambda {\sin }^{2}x,-\left(\frac{\pi }{2}\right)<x<\left(\frac{\pi }{2}\right)$. Then the number of integral values of $\lambda$ in which $\lambda$ lies in order than $f\left(x\right)$ has exactly one maximum are _____________.

Answer 1

$f^{\prime }\left(x\right)=\cos x\left(3{\sin }^2\right(x)+2\lambda \sin x)=0\Rightarrow x=0,\pm \pi /2$ or $3\sin \left(x\right)+2\lambda =0$

It is easy to see that 0 is a local minimum of the function and the values $\pm \pi /2$ are not included in the domain of definition of the function.

So the function can have an extrema only at values of x where $3\sin \left(x\right)+2\lambda =0\Rightarrow \sin \left(x\right)=-\frac{2\lambda }{3}$

Now, we know that $-1\le \sin \left(x\right)\le 1\Rightarrow -1\le -\frac{2\lambda }{3}\le 1\Rightarrow -\frac{3}{2}\le \lambda \le \frac{3}{2}$.

So the possible integral values that $\lambda$ can take are $-1,0,1$.

But $\lambda =0$ gives back $x=0$ which is an inflection point when $\lambda =0$ as $f\left(x\right)={\sin }^{3}x\Rightarrow f^{\prime }\left(x\right)=3{\sin }^{2}x\cos x$.

Observe that when $\lambda =0$, $f\left(0\right)=f^{\prime }\left(0\right)=f^{\prime \prime }\left(0\right)=0$, hence $x=0$ is an inflection point in that case, not a maxima.

Thus, there are 2 integral values of $\lambda$ for which the given function has exactly one maxima in the given domain and those values are $\pm 1$