Question

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

• A. $(-3,0)\cup (0,3).$
• B. $(-3/2,0)\cup (0,3/2).$
• C. $(-2,0)\cup (0,2).$
• D. $(-2,0)\cup (0,3).$

Answer 1

The correct option is B $(-3/2,0)\cup (0,3/2).$

$f\left(x\right)={\sin }^{3}x+\lambda {\sin }^{2}x$
$f^{\prime }\left(x\right)=3{\sin }^{2}x\cos x+2\lambda \sin x\cos x$
For maxima or minima,
$f^{\prime }\left(x\right)=0$
$\Rightarrow 3{\sin }^{2}x\cos x+2\lambda \sin x\cos x=0$
$\therefore \sin x\cos x(3\sin x+2\lambda )=0$ ......(1)
$\therefore \sin x=0or\cos x=0or\sin x=-2\lambda /3$
$\cos x=0$ is not possible as $-\pi /2<x<\pi /2$
$\therefore \sin x=0i.ex=0$
$\sin x=-2\lambda /3$
$-1<-\frac{2\lambda }{3}<1$
$-3<-2\lambda <3$
$\Rightarrow 3>2\lambda >-3$
$-\frac{3}{2}<\lambda <\frac{3}{2}$
$\therefore \lambda ϵ(-\frac{3}{2},0)\cup (0,\frac{3}{2}).$ ($\lambda$ cannot be 0 as the solution are to be distinct. )
When $\lambda$ lies in the above intervals, there are two distinct solutions, one gives maximum and the other minimum.