Animated Solution for Mathematics - Differentiation: Let $f(x)={\sin }^{3}x+\lambda {\sin }^{2}x,-\pi /2<x<\pi /2$. Find the intervals in which $\lambda$ should lie in order that $f(x)$ has exactly one minimum and exactly one maximum.

Imagine you are standing on the edge of a mathematical landscape, looking at the function $f(x)={\sin }^{3}x+\lambda {\sin }^{2}x$. You are tasked with a mission: find the values of $\lambda$ that force this function to have exactly one maximum and one minimum within the interval $(-\frac{\pi }{2},\frac{\pi }{2})$.

The interval $(-\frac{\pi }{2},\frac{\pi }{2})$ is special. In this region, the cosine function $\cos x$ is always strictly positive. This is a gift from the geometry of the unit circle, and it will save us from division-by-zero nightmares later.

To find the peaks and valleys of any function, we must look at its rate of change. We need the derivative $f^{\prime }(x)$. Applying the chain rule, we differentiate $f(x)={\sin }^{3}x+\lambda {\sin }^{2}x$.

The first term, ${\sin }^{3}x$, yields $3{\sin }^{2}x\cos x$. The second term, $\lambda {\sin }^{2}x$, yields $2\lambda \sin x\cos x$. Putting them together, we get:

Now, look at this expression. It is begging to be factored. Both terms share $\sin x\cos x$. Pulling that out, we arrive at the elegant form:

This is our engine. The roots of this equation are the critical points where the function might turn.

We set $f^{\prime }(x)=0$. Since $\cos x$ is never zero in our interval, we are left with two possibilities: either $\sin x=0$ or $3\sin x+2\lambda =0$.

The first case, $\sin x=0$, gives us $x=0$. This is a constant, a fixed point in our landscape regardless of what $\lambda$ does. It is always there.

But the second case, $\sin x=-\frac{2\lambda }{3}$, is where the magic happens. This is the variable root. For the function to have exactly one maximum and one minimum, we need two distinct critical points where the derivative changes sign.

Here is where many students stumble. We need the second root to be distinct from the first. If $\sin x=-\frac{2\lambda }{3}$ happens to be $0$, then $\lambda =0$.

If $\lambda =0$, our derivative becomes $f^{\prime }(x)=3{\sin }^{2}x\cos x$. At $x=0$, this derivative does not change sign; it just touches zero and continues. That is an inflection point, not an extremum. So, we must strictly forbid $\lambda =0$.

Furthermore, for the root to exist within our interval, the value of $\sin x$ must be strictly between $-1$ and $1$. Thus, we set the inequality:

Multiplying by $-3$ (and flipping the inequality signs, as we must!), we get $3>2\lambda >-3$. Dividing by $2$, we find $-\frac{3}{2}<\lambda <\frac{3}{2}$.

Combining everything, we have the range $(-\frac{3}{2},\frac{3}{2})$, but we must carve out the hole at $\lambda =0$.

Our final answer is $\lambda \in (-\frac{3}{2},0)\cup (0,\frac{3}{2})$. You have navigated the trap, respected the domain, and solved the puzzle. This is the essence of JEE Advanced mathematics: not just calculating, but understanding the soul of the function.