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Mathematics · Application of Derivatives

JEE Advanced 2020 — Paper 2 — Question 51

Let the function f:(0,π)→R\mathrm{f}:(0, \pi) \rightarrow \mathbb{R} be defined by f(θ)=(sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)4.f(\theta)=(\sin \theta+\cos \theta)^{2}+(\sin \theta-\cos \theta)^{4} . Suppose the function f has a local minimum at θ\theta precisely when θ∈{λ1π,…,λrπ}\theta \in\left\{\lambda_{1} \pi, \ldots, \lambda_{\mathrm{r}} \pi\right\}, where 0<λ1<…<λr0<\lambda_{1}<\ldots<\lambda_{\mathrm{r}} <1<1. Then the value of λ1+…+λr\lambda_{1}+\ldots+\lambda_{\mathrm{r}} is ____\_\_\_\_

Answer: 0.5

Numerical answer — enter this value.

Step-by-step solution

f(θ)=(sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)4f(\theta)=(\sin \theta+\cos \theta)^{2}+(\sin \theta-\cos \theta)^{4}

f′(θ)=2(sin⁡θ+cos⁡θ)(cos⁡θ−sin⁡θ)+4(sin⁡θ−cos⁡θ)3(cos⁡θ+sin⁡θ)\mathrm{f}^{\prime}(\theta)=2(\sin \theta+\cos \theta)(\cos \theta-\sin \theta)+4(\sin \theta-\cos \theta)^{3}(\cos \theta+\sin \theta)

=2(sin⁡θ+cos⁡θ)(sin⁡θ−cos⁡θ)(1−2sin⁡2θ)=2(\sin \theta+\cos \theta)(\sin \theta-\cos \theta)(1-2 \sin 2 \theta)

f′(θ)=0⇒θ=π12,π4,5π12,3π4f^{\prime}(\theta)=0 \Rightarrow \theta=\frac{\pi}{12}, \frac{\pi}{4}, \frac{5 \pi}{12}, \frac{3 \pi}{4}

Pts of minima at x=π12,5π12\mathrm{x}=\frac{\pi}{12}, \frac{5 \pi}{12}

λ1+λ2=112+512=12=0.5\lambda_{1}+\lambda_{2}=\frac{1}{12}+\frac{5}{12}=\frac{1}{2}=0.5

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Exam
JEE Advanced 2020
Paper
Paper 2
Subject
Mathematics
Chapter
Application of Derivatives
Topic
Local, Global extremum

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