IIT-JEE 1985Question 1
Let $f(x)=\sin^{3}x+\lambda\sin^{2}x$, $-\dfrac\pi2<x<\dfrac\pi2$. Find the intervals in which $\lambda$ should lie in order that $f(x)$ has exactly one minimum and exactly one maximum.
Differentiation and its Application
Answer$\lambda\in\left(-\dfrac32,0\right)\cup\left(0,\dfrac32\right)$
Solution
Concept. $t=\sin x$ is increasing on $\left(-\frac\pi2,\frac\pi2\right)$ with $t\in(-1,1)$, so extrema in $x$ match those of $g(t)=t^3+\lambda t^2$. Formulas used. (1) $g'(t)=3t^2+2\lambda t=t(3t+2\lambda)$Step 1 - Critical points. By (1): $t=0$ and $t=-\frac{2\lambda}{3}$. Step 2 - Condition. One max and one min need two distinct simple roots in $(-1,1)$: $$-\frac{2\lambda}{3}\ne0,\qquad\left|\frac{2\lambda}{3}\right|<1.$$ Step 3 - Interval. $\lambda\ne0$ and $|\lambda|<\frac32$. Answer: $\lambda\in\left(-\dfrac32,0\right)\cup\left(0,\dfrac32\right)$