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← JEE Advanced 2019 · Paper 1
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JEE Advanced 2019 · Paper 1, Mathematics question 2

Let

M=[sin⁡4θ−1−sin⁡2θ1+cos⁡2θcos⁡4θ]=αI+βM−1,M = \begin{bmatrix}\sin^4\theta & -1-\sin^2\theta\\ 1+\cos^2\theta & \cos^4\theta\end{bmatrix} = \alpha I + \beta M^{-1},

where α=α(θ)\alpha = \alpha(\theta) and β=β(θ)\beta = \beta(\theta) are real numbers, and II is the 2×22\times2 identity matrix. If

α∗\alpha^* is the minimum of the set {α(θ):θ∈[0,2π)}\{\alpha(\theta) : \theta\in[0,2\pi)\} and

β∗\beta^* is the minimum of the set {β(θ):θ∈[0,2π)}\{\beta(\theta) : \theta\in[0,2\pi)\},

then the value of α∗+β∗\alpha^* + \beta^* is

Choose one option
Answer and worked solution

Official answer: C

Find α and β. Multiplying M=αI+βM−1M=\alpha I+\beta M^{-1} by MM gives M2−αM−βI=0M^2-\alpha M-\beta I=0. Compare with Cayley-Hamilton, M2−(tr⁡M)M+(det⁡M)I=0M^2-(\operatorname{tr}M)M+(\det M)I=0:
α=tr⁡M,β=−det⁡M.\alpha=\operatorname{tr}M,\qquad\beta=-\det M.

In terms of t=sin⁡2θcos⁡2θ∈[0,14]t=\sin^2\theta\cos^2\theta\in[0,\frac14].
α=sin⁡4θ+cos⁡4θ=1−2t.\alpha=\sin^4\theta+\cos^4\theta=1-2t.
det⁡M=sin⁡4θcos⁡4θ+(1+sin⁡2θ)(1+cos⁡2θ)=t2+t+2.\det M=\sin^4\theta\cos^4\theta+(1+\sin^2\theta)(1+\cos^2\theta)=t^2+t+2.

Minima. Both are smallest at t=14t=\frac14:
α∗=12,β∗=−(116+14+2)=−3716.\alpha^*=\tfrac12,\qquad\beta^*=-\left(\tfrac1{16}+\tfrac14+2\right)=-\tfrac{37}{16}.
So α∗+β∗=−2916\alpha^*+\beta^*=-\frac{29}{16}.

(C)

Topics:Matrix algebraDeterminants and propertiesTrigonometric ratios and identities
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