JEE Challenger
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MathMedAdv 2020 • Paper 2

Sum of Coefficients of Local Minima Points of Trigonometric Function

Consider the function f:(0,π)→Rf: (0, \pi) \rightarrow \mathbb{R} given 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. Assume the function ff attains a local minimum at θ\theta precisely when θ∈{λ1π,…,λrπ}\theta \in \{\lambda_1 \pi, \dots, \lambda_r \pi\}, where 0<λ1<⋯<λr<10 < \lambda_1 < \dots < \lambda_r < 1. Determine the value of λ1+⋯+λr\lambda_1 + \dots + \lambda_r.

Official Numerical Answer0.5

Topics & Concepts

Application of DerivativesMaxima and Minima

Step-by-Step Solution

To find the points of local minima for the function f:(0,π)→Rf: (0, \pi) \rightarrow \mathbb{R} given 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,

we first simplify the trigonometric expression. Using the identity sin⁡2θ=2sin⁡θcos⁡θ\sin 2\theta = 2\sin\theta\cos\theta, we have: (sin⁡θ+cos⁡θ)2=sin⁡2θ+cos⁡2θ+2sin⁡θcos⁡θ=1+sin⁡2θ(\sin \theta + \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta + 2\sin \theta \cos \theta = 1 + \sin 2\theta (sin⁡θ−cos⁡θ)2=sin⁡2θ+cos⁡2θ−2sin⁡θcos⁡θ=1−sin⁡2θ(\sin \theta - \cos \theta)^2 = \sin^2 \theta + \cos^2 \theta - 2\sin \theta \cos \theta = 1 - \sin 2\theta

Substituting these into f(θ)f(\theta), we get: f(θ)=(1+sin⁡2θ)+(1−sin⁡2θ)2f(\theta) = (1 + \sin 2\theta) + (1 - \sin 2\theta)^2 f(θ)=1+sin⁡2θ+1−2sin⁡2θ+sin⁡22θ=sin⁡22θ−sin⁡2θ+2f(\theta) = 1 + \sin 2\theta + 1 - 2\sin 2\theta + \sin^2 2\theta = \sin^2 2\theta - \sin 2\theta + 2

To find the critical points, we differentiate f(θ)f(\theta) with respect to θ\theta: f′(θ)=ddθ(sin⁡22θ−sin⁡2θ+2)=(2sin⁡2θ−1)⋅(2cos⁡2θ)=2(2sin⁡2θ−1)cos⁡2θf'(\theta) = \frac{d}{d\theta} \left(\sin^2 2\theta - \sin 2\theta + 2\right) = (2\sin 2\theta - 1) \cdot (2\cos 2\theta) = 2(2\sin 2\theta - 1)\cos 2\theta

For θ∈(0,π)\theta \in (0, \pi), let t=2θ∈(0,2π)t = 2\theta \in (0, 2\pi). Setting f′(θ)=0f'(\theta) = 0 yields:

  1. 2sin⁡t−1=0  ⟹  sin⁡t=12  ⟹  t=π6,5π62\sin t - 1 = 0 \implies \sin t = \frac{1}{2} \implies t = \frac{\pi}{6}, \frac{5\pi}{6}
  2. cos⁡t=0  ⟹  t=π2,3π2\cos t = 0 \implies t = \frac{\pi}{2}, \frac{3\pi}{2}

Thus, the critical points in θ∈(0,π)\theta \in (0, \pi) are: θ=π12,π4,5π12,3π4\theta = \frac{\pi}{12}, \frac{\pi}{4}, \frac{5\pi}{12}, \frac{3\pi}{4}

Now, we analyze the sign change of f′(θ)=2(2sin⁡2θ−1)cos⁡2θf'(\theta) = 2(2\sin 2\theta - 1)\cos 2\theta across these critical points:

  1. For θ∈(0,π12)\theta \in \left(0, \frac{\pi}{12}\right) (2θ∈(0,π6)2\theta \in \left(0, \frac{\pi}{6}\right)): sin⁡2θ<12  ⟹  2sin⁡2θ−1<0\sin 2\theta < \frac{1}{2} \implies 2\sin 2\theta - 1 < 0 and cos⁡2θ>0\cos 2\theta > 0, so f′(θ)<0f'(\theta) < 0.

  2. For θ∈(π12,π4)\theta \in \left(\frac{\pi}{12}, \frac{\pi}{4}\right) (2θ∈(π6,π2)2\theta \in \left(\frac{\pi}{6}, \frac{\pi}{2}\right)): sin⁡2θ>12  ⟹  2sin⁡2θ−1>0\sin 2\theta > \frac{1}{2} \implies 2\sin 2\theta - 1 > 0 and cos⁡2θ>0\cos 2\theta > 0, so f′(θ)>0f'(\theta) > 0.

    Since f′(θ)f'(\theta) changes sign from negative to positive at θ=π12\theta = \frac{\pi}{12}, the function attains a local minimum at θ=π12\theta = \frac{\pi}{12}.

  3. For θ∈(π4,5π12)\theta \in \left(\frac{\pi}{4}, \frac{5\pi}{12}\right) (2θ∈(π2,5π6)2\theta \in \left(\frac{\pi}{2}, \frac{5\pi}{6}\right)): sin⁡2θ>12  ⟹  2sin⁡2θ−1>0\sin 2\theta > \frac{1}{2} \implies 2\sin 2\theta - 1 > 0 and cos⁡2θ<0\cos 2\theta < 0, so f′(θ)<0f'(\theta) < 0.

    Thus, θ=π4\theta = \frac{\pi}{4} is a local maximum.

  4. For θ∈(5π12,3π4)\theta \in \left(\frac{5\pi}{12}, \frac{3\pi}{4}\right) (2θ∈(5π6,3π2)2\theta \in \left(\frac{5\pi}{6}, \frac{3\pi}{2}\right)): sin⁡2θ<12  ⟹  2sin⁡2θ−1<0\sin 2\theta < \frac{1}{2} \implies 2\sin 2\theta - 1 < 0 and cos⁡2θ<0\cos 2\theta < 0, so f′(θ)>0f'(\theta) > 0.

    Since f′(θ)f'(\theta) changes sign from negative to positive at θ=5π12\theta = \frac{5\pi}{12}, the function attains a local minimum at θ=5π12\theta = \frac{5\pi}{12}.

  5. For θ∈(3π4,π)\theta \in \left(\frac{3\pi}{4}, \pi\right) (2θ∈(3π2,2π)2\theta \in \left(\frac{3\pi}{2}, 2\pi\right)): sin⁡2θ<12  ⟹  2sin⁡2θ−1<0\sin 2\theta < \frac{1}{2} \implies 2\sin 2\theta - 1 < 0 and cos⁡2θ>0\cos 2\theta > 0, so f′(θ)<0f'(\theta) < 0.

    Thus, θ=3π4\theta = \frac{3\pi}{4} is a local maximum.

The points of local minima are precisely: θ∈{π12,5π12}\theta \in \left\{ \frac{\pi}{12}, \frac{5\pi}{12} \right\}

Comparing this with the given set {λ1π,…,λrπ}\{\lambda_1 \pi, \dots, \lambda_r \pi\}, we have: λ1=112,λ2=512(r=2)\lambda_1 = \frac{1}{12}, \quad \lambda_2 = \frac{5}{12} \quad (r = 2)

Therefore, the required sum is: λ1+λ2=112+512=612=12=0.5\lambda_1 + \lambda_2 = \frac{1}{12} + \frac{5}{12} = \frac{6}{12} = \frac{1}{2} = 0.5

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