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 $\theta \in \{{\lambda }_1\pi ,...,{\lambda }_r\pi \}$, where $0<{\lambda }_1<\cdots <{\lambda }_r<1$. Then the value of ${\lambda }_1+\cdots +{\lambda }_r$ is _____.

Answer 1

Step 1: Simplifying $f\left(\theta \right)$

Given,$f\left(\theta \right)={(\sin \theta +\cos \theta )}^2+{(\sin \theta –\cos \theta )}^4$

$$\begin{gathered} f\left(\theta \right)={(\sin \theta +\cos \theta )}^2+{(\sin \theta –\cos \theta )}^4 \\ =({\sin }^2\theta +{\cos }^2\theta )+2\sin \theta \cos \theta +({(\sin \theta –\cos \theta )}^2{)}^2\mathbf{}\left[\because {(a+b)}^2=a^2+b^2\right] \\ =1+\sin 2\theta +{({\sin }^2\theta +{\cos }^2\theta –2\sin \theta \cos \theta )}^2\mathbf{}\left[\because {(a-b)}^2=a^2-b^2\right] \\ =1+\sin 2\theta +{(1–\sin 2\theta )}^2 \\ =1+\sin 2\theta +1+{\sin }^{2}2\theta –2\sin 2\theta \\ ={\sin }^{2}2\theta –\sin 2\theta +2 \\ f(\theta )={\left(\sin 2\theta –\frac{1}{2}\right)}^2-\frac{1}{4}+2 \\ f(\theta )={\left(\sin 2\theta –\frac{1}{2}\right)}^2+\frac{7}{4} \end{gathered}$$

Step 2: Finding $\theta$

Because $\theta \in [0,\pi ]$

Therefore, $2\theta \in [0,2\mathrm{\pi }]$

$f\left(\theta \right)$min. when $\sin \left(2\theta \right)=\frac{1}{2}$

Therefore,$2\theta =\frac{\pi }{6},\frac{5\pi }{6}$

$\theta =\frac{\pi }{12},\frac{5\pi }{12}$

Step 3: Finding the value of ${\lambda }_1+{\lambda }_2$

$$\begin{gathered} {\lambda }_1=\frac{1}{12};{\lambda }_2=\frac{5}{12} \\ {\lambda }_1+{\lambda }_2=\frac{1+5}{12} \\ {\lambda }_1+{\lambda }_2=\frac{6}{12} \\ {\lambda }_1+{\lambda }_2=\frac{1}{2} \\ {\lambda }_1+{\lambda }_2=0.5 \end{gathered}$$

Therefore, the value of ${\lambda }_1+\cdots +{\lambda }_r$, i.e., ${\lambda }_1+{\lambda }_2$ is $0.5$.