Let the function \(f:(0, \pi) \rightarrow \mathbb{R}\) 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 \(\theta \in\{\lambda_{1} \pi, \ldots, \lambda_{r} \pi\}\), where \(0<\lambda_{1}<\cdots<\lambda_{r}<1\). Then the value of \(\lambda_{1}+\cdots+\lambda_{r}\) is ____

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Application of Derivatives

Let the function

f : (0, π) \to R

be defined by

f (θ) = (sin θ + cos θ)^{2} + (sin θ - cos θ)^{4}

Suppose the function

f

has a local minimum at

θ

precisely when

θ \in {λ_{1} π, \dots, λ_{r} π}

, where

0 < λ_{1} < \dots < λ_{r} < 1

. Then the value of

λ_{1} + \dots + λ_{r}

is ____

Answer: 0.5

Simplify

f (θ)

(sin θ + cos θ)^{2} = 1 + sin 2 θ

and

(sin θ - cos θ)^{4} = (1 - sin 2 θ)^{2} = 1 - 2 sin 2 θ + sin^{2} 2 θ

. So

f (θ) = 1 + sin 2 θ + 1 - 2 sin 2 θ + sin^{2} 2 θ = sin^{2} 2 θ - sin 2 θ + 2

. Taking the derivative:

f^{'} (θ) = 2 sin 2 θ \cdot 2 cos 2 θ - 2 cos 2 θ = 2 cos 2 θ (2 sin 2 θ - 1)

. Critical points:

cos 2 θ = 0

(giving

θ = π /4, 3 π /4

) or

sin 2 θ = 1/2

(giving

2 θ = π /6, 5 π /6

, so

θ = π /12, 5 π /12

). Checking the second derivative or sign changes, the local minima occur at

θ = π /12

and

θ = 5 π /12

. Thus

λ_{1} = 1/12

and

λ_{2} = 5/12

, giving

λ_{1} + λ_{2} = 6/12 = 0.50

JEE Advanced · 2020 · 27 Sep 2020 · Paper 2 · Numerical answer