Question

Let f(x) $={\sin }^{3}x+\lambda {\sin }^{2}x,-\frac{\pi }{2}<x<\frac{\pi }{2}$ Find the intervals in which $\lambda$ should lie in order that f(x) has exactly one minimum and exactly one maximum.
If the solution is $\lambda \in (-C,0)\cup (0,C)$
Find $12C$?

• A. $9$
• B. $36$
• C. $12$
• D. $18$

Answer 1

The correct option is D $18$

$f\left(x\right)={\sin }^{3}x+\lambda {\sin }^{2}x$
$f^{\prime }\left(x\right)=\sin x\cos x(3\sin x+2\lambda )$
$f"\left(x\right)=6\sin x{\cos }^{2}x-3{\sin }^{3}x+2\lambda \cos 2x$
$f^{\prime }\left(x\right)=0\Rightarrow \sin x=0$ or $\cos x=0$ or $\sin x=\frac{-2\lambda }{3}$
$\cos x\ne 0$ if $-\frac{\pi }{2}<x<\frac{\pi }{2}$
$\sin x=0\Rightarrow x=0\Rightarrow \sin x=\frac{-2\lambda }{3}$
$-1<\sin x<1\Rightarrow -1<\frac{-2\lambda }{3}<1\Rightarrow \frac{-3}{2}<\lambda <\frac{3}{2}$
$\lambda \ne 0$ otherwise there is only one critical point If
$\lambda >0$ then $f"\left(0\right)>0$ $\Rightarrow x=0$ point of minima & $f^{\prime }\left(x\right)$ changes sign from positive to negative for
$x={\sin }^{-1}\left(\frac{-2\lambda }{3}\right)$ (point of maxima)
If $\lambda <0$ then $x=0$ is a point of maxima while $x={\sin }^{-1}\left(\frac{-2\lambda }{3}\right)$ is a point of minima
Thus for $\lambda \in (-\frac{3}{2}.\frac{3}{2})-\left\{0\right\}$
function has exactly one maxima & exactly one minima