Let the function f:(0, π) → 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 ∈ {λ₁π, ..., λᵣπ}, where 0 < λ₁ < ... < λᵣ < 1. Then the value of λ₁ + ... + λᵣ is ______

Let the function f:(0, π) → 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 ∈ {λ₁π, ..., λᵣπ}, where 0 < λ₁ < ... < λᵣ < 1. Then the value of λ₁ + ... + λᵣ is ______ f(θ) = (sinθ + cosθ)² + (sinθ - cosθ)⁴ θ ∈ (0,π) Simplify f ⇒ find f' ⇒ critical points and nature (sinθ+cosθ)² = 1 + sin 2θ (sinθ-cosθ)² = 1 - sin 2θ u = sin 2θ ⇒ f = (1+u) + (1-u)² = u² - u + 2 f'(θ) = (2u-1)· 2cos 2θ = 2 (2sin 2θ - 1)cos 2θ sin 2θ = 1/2 ⇒ θ = π/12, 5π/12 cos 2θ = 0 ⇒ θ = π/4, 3π/4 f': (0,π/12) -, (π/12,π/4) +, (π/4,5π/12) - (5π/12,3π/4) +, (3π/4,π) - ⇒ local minima at θ = π/12, 5π/12 θ = λπ ⇒ λ₁ = 1/12, λ₂ = 5/12 λ₁ + λ₂ = 1/12 + 5/12 = 6/12 = 1/2

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

Answer: $7$

Application of Derivatives Let the function

f:(0, \pi) \to \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, ..., \lambda_{r}\pi\}

, where

0 < \lambda_{1} < ... < \lambda_{r} < 1

. Then the value of

\lambda_{1} + ... + \lambda_{r}

is ______

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Written solution

$f(\theta) = (\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^4$ $\theta \in (0,\pi)$ $\text{Simplify } f \Rightarrow \text{find } f' \Rightarrow \text{critical points and nature}$ $(\sin\theta+\cos\theta)^2 = 1 + \sin 2\theta$ $(\sin\theta-\cos\theta)^2 = 1 - \sin 2\theta$ $u = \sin 2\theta \Rightarrow f = (1+u) + (1-u)^2$ $= u^2 - u + 2$ $f'(\theta) = (2u-1)\cdot 2\cos 2\theta$ $= 2\,(2\sin 2\theta - 1)\cos 2\theta$ $\sin 2\theta = \frac12 \Rightarrow \theta = \frac{\pi}{12},\ \frac{5\pi}{12}$ $\cos 2\theta = 0 \Rightarrow \theta = \frac{\pi}{4},\ \frac{3\pi}{4}$ $f':\ (0,\tfrac{\pi}{12})\,-,\quad (\tfrac{\pi}{12},\tfrac{\pi}{4})\,+,\quad (\tfrac{\pi}{4},\tfrac{5\pi}{12})\,-$ $(\tfrac{5\pi}{12},\tfrac{3\pi}{4})\,+,\quad (\tfrac{3\pi}{4},\pi)\,-$ $\Rightarrow \text{local minima at } \theta = \frac{\pi}{12},\ \frac{5\pi}{12}$ $\theta = \lambda\pi \Rightarrow \lambda_1 = \frac{1}{12},\ \lambda_2 = \frac{5}{12}$ $\lambda_1 + \lambda_2 = \frac{1}{12} + \frac{5}{12} = \frac{6}{12}$ $= \frac{1}{2}$

Written solution

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