Question

Let the function $f:(0,\pi )\to R$ be defined by $f\left(\theta \right)=(\sin \theta +\cos \theta {)}^2+(\sin \theta -\cos \theta {)}^4.$ Suppose the function $f$ has a local minimum at $\theta$ precisely when $\theta \in \{{\lambda }_1\pi ,\dots ,{\lambda }_r\pi \},$ where $0<{\lambda }_1<\cdots <{\lambda }_r<1.$ Then the value of ${\lambda }_1+\cdots +{\lambda }_r$ is

Answer 1

$f\left(\theta \right)=(\sin \theta +\cos \theta {)}^2+(\sin \theta -\cos \theta {)}^4$
$={\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta +\left(\right(\sin \theta -\cos \theta {)}^2{)}^2$
$=1+\sin 2\theta +({\sin }^2\theta +{\cos }^2\theta -2\sin \theta \cos \theta {)}^2$
$=1+\sin 2\theta +(1-\sin 2\theta {)}^2$
$=1+\sin 2\theta +1+{\sin }^{2}2\theta -2\sin 2\theta$
$={\sin }^{2}2\theta -\sin 2\theta +2$
$\Rightarrow f\left(\theta \right)={(\sin 2\theta -\frac{1}{2})}^2+\frac{7}{4}$

Since $\theta \in (0,\pi ),$
$\therefore 2\theta \in (0,2\pi )$
$f\left(\theta \right)$ is minimum when $\sin 2\theta =\frac{1}{2}$
$\therefore 2\theta =\frac{\pi }{6},\frac{5\pi }{6}$
$\Rightarrow \theta =\frac{\pi }{12},\frac{5\pi }{12}$
${\lambda }_1=\frac{1}{12},{\lambda }_2=\frac{5}{12}$
${\lambda }_1+{\lambda }_2=\frac{1}{12}+\frac{5}{12}=\frac{1}{2}=0.50$