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JEE Advanced 2020 · Paper 2, Mathematics question 18

Let the function f:(0,π)→Rf:(0,\pi)\to\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 ff has a local minimum at θ\theta precisely when θ∈{λ1π,…,λrπ}\theta\in\{\lambda_1\pi, \ldots, \lambda_r\pi\}, where 0<λ1<⋯<λr<10 < \lambda_1 < \cdots < \lambda_r < 1. Then the value of λ1+⋯+λr\lambda_1 + \cdots + \lambda_r is ______

Answer and worked solution

Official answer: 0.5

As a quadratic. Let s=sin⁡2θs=\sin2\theta. Then (sin⁡θ+cos⁡θ)2=1+s(\sin\theta+\cos\theta)^2=1+s and (sin⁡θ−cos⁡θ)4=(1−s)2(\sin\theta-\cos\theta)^4=(1-s)^2, so
f=s2−s+2,f=s^2-s+2,
which is smallest at s=12s=\frac12.

Local minima.

  • s=12s=\frac12 when 2θ=π62\theta=\frac\pi6 or 5π6\frac{5\pi}6, i.e. θ=π12\theta=\frac\pi{12} and 5π12\frac{5\pi}{12}. There ss passes through 12\frac12, so these are local minima.
  • At θ=π4\theta=\frac\pi4 (s=1s=1) and θ=3π4\theta=\frac{3\pi}4 (s=−1s=-1), ff has local maxima.

Sum.
λ1+λ2=112+512=12.\lambda_1+\lambda_2=\frac1{12}+\frac5{12}=\frac12.
Answer: 0.5

Topics:Maxima and minimaTrigonometric equations
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