- Home
- Standard 11
- Mathematics
Let the function $:(0, \pi) \rightarrow R$ be defined by
$f (\theta)=(\sin \theta+\cos \theta)^2+(\sin \theta-\cos \theta)^4$
Suppose the function $f$ has a local minimum at $\theta$ precisely when $\theta \in\left\{\lambda_1 \pi, \ldots, \lambda_{ T } \pi\right\}$, where $0<\lambda_1<\cdots<\lambda_r<1$. Then the value of $\lambda_1+\cdots+\lambda_r$ is. . . . .
$0.40$
$0.50$
$0.60$
$0.70$
Solution
$f(\theta) =(\sin \theta+\cos \theta)^2+(\sin \theta-\cos \theta)^4$
$f(\theta) =\sin ^2 2 \theta-\sin 2 \theta+2$
$f^{\prime}(\theta) =2(\sin 2 \theta) \cdot(2 \cos 2 \theta)-2 \cos 2 \theta$
$ =2 \cos 2 \theta(2 \sin 2 \theta-1)$
critical points
so, minimum at $\theta=\frac{\pi}{12}, \frac{5 \pi}{12}$
$\lambda_1+\lambda_2=\frac{1}{12}+\frac{5}{12}=\frac{6}{12}=\frac{1}{2}$
