Let $f(x)=\sin ^{3} x+\lambda \sin ^{2} x,-\frac{\pi}{2}

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.

Professor PC Thomas Classes · Answer

$f(x)=\sin ^{3} x+\lambda \sin ^{2} x,-\frac{\pi}{2}0$\ or $\lambda<0$\ when $x=0, f(x)$ has a maximum if $\lambda<0$ and minimum if $\lambda>0$\ Also when $\sin x=\frac{-2 \lambda}{3}, \frac{-3}{2}<\lambda<\frac{3}{2}$\ $f^{\prime \prime}(x)=\left(1-2 \sin ^{2} x ight)(3 \sin x+2 \lambda)+3 \sin x$ $\left(1-\sin ^{2} x ight)$\ $=-\frac{8 \lambda}{9}\left(\frac{9}{4}-\lambda^{2} ight)$\ $ herefore f^{\prime \prime}(x)$ is +ve if $\frac{-3}{2}<\lambda<0$, in this case $f(x)$ is minimum and $f^{\prime \prime}(x)$ is -ve and if $0<\lambda<\frac{3}{2}$ , in this case $f(x)$ is maximum.\ Hence, $f(x)$ has exactly one minimum and exactly one maximum if\ $-\frac{3}{2}<\lambda<0$ or $0<\lambda<\frac{3}{2}$.\