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JEE Advanced 2020Paper 2
Exam
JEE Advanced
Year
2020
Paper
Paper 2
Subject
Mathematics
Chapter
Application of Derivatives

JEE Advanced 2020 · Paper 2 · Mathematics · Application of Derivatives

Let the function f: (0, π) → ℝ be defined by f(θ) = (sinθ + cosθ) ^...

Let the function f:(0,π)→Rf: \left(0, \pi\right) \to \mathbb{R} be defined by f(θ)=(sinθ+cosθ)2+(sinθ−cosθ)4f\left(\theta\right) = {\left(sin\theta + cos\theta\right)}^{2} + {\left(sin\theta - cos\theta\right)}^{4}. Suppose the function ff has a local minimum at θ\theta precisely when θ∈{λ1π,…,λrπ},\theta \in \left\{{\lambda}_{1}\pi, \ldots , {\lambda}_{r}\pi\right\}, where 0<0 <λ1<⋯<λr<1.{\lambda}_{1} < ⋯ < {\lambda}_{r} < 1 . Then the value of λ1+⋯+λr{\lambda}_{1} + ⋯ + {\lambda}_{r} is_______

Correct answer

0

Step-by-step solution

f(θ)=1+sin⁡2θ+(1−sin⁡2θ)2f\left(\theta\right) = 1 + \sin 2\theta + {\left(1 - \sin 2\theta\right)}^{2}

=sin⁡22θ−sin⁡2θ+2= \sin^{2}2\theta - \sin 2\theta + 2

=(sin⁡2θ−12)2+74= {\left(\sin 2\theta - \dfrac{1}{2}\right)}^{2} + \dfrac{7}{4}

It has minima when sin⁡2θ=12=sinπ6\sin 2\theta = \dfrac{1}{2} = sin\dfrac{\pi}{6}

θ∈(0,π)\theta \in \left(0, \pi\right)

2θ=π6,5π62\theta = \dfrac{\pi}{6}, \dfrac{5\pi}{6}

θ=π12,5π12\theta = \dfrac{\pi}{12}, \dfrac{5\pi}{12}

λ1=112,λ2=512{\lambda}_{1} = \dfrac{1}{12}, {\lambda}_{2} = \dfrac{5}{12}

sum=12sum = \dfrac{1}{2}.

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