Let $\mathrm{M}=\left[\begin{array}{cc}\sin ^{4} \theta & -1-\sin ^{2} \theta \\ 1+\cos ^{2} \theta & \cos ^{4} \theta\end{array}\right]=\alpha \mathrm{I}+\beta \mathrm{M}^{-1}$, where $\alpha=\alpha(\theta)$ and $\beta=\beta(\theta)$ are real number, and I is the $2 \times 2$ identity matrix. If $$\begin{aligned} & \alpha^{*} \text { is the minimum of the set }\{\alpha(\theta): \theta \in[0,2 \pi)\} \text { and } \\ & \beta^{*} \text { is the minimum of the set }\{\beta(\theta): \theta \in[0,2 \pi)\}, \end{aligned}$$ then the value of $\alpha^{*}+\beta^{*}$ is

Matrices — JEE Advanced · 2019 · 27 May 2019 · Paper 1 Question with Solution

Let

M = [sin^{4} θ 1 + cos^{2} θ - 1 - sin^{2} θ cos^{4} θ] = α I + β M^{- 1}

,
where

α = α (θ)

and

β = β (θ)

are real number, and I is the

2 \times 2

identity matrix. If

α^{*} is the minimum of the set {α (θ) : θ \in [0, 2 π)} and β^{*} is the minimum of the set {β (θ) : θ \in [0, 2 π)},

then the value of

α^{*} + β^{*}

Answer: (B)

Given

M = α I + β M^{- 1}

, multiply both sides by M to get

M^{2} - α M - β I = O

.

Computing

M^{2}

and comparing coefficients:

α (θ) = 1 - 2 sin^{2} θ cos^{2} θ = 1 - \frac{sin ^{2} 2 θ}{2} \geq \frac{1}{2}

For

β (θ)

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

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

Let

t = \frac{s i n ^{2} 2 θ}{4} \in [0, \frac{1}{4}]

. Then:

β (θ) = - (t^{2} + t + 2) \geq - \frac{37}{16}

Therefore

α^{*} = \frac{1}{2}

and

β^{*} = - \frac{37}{16}

, giving

α^{*} + β^{*} = \frac{1}{2} - \frac{37}{16} = - \frac{29}{16}

JEE Advanced · 2019 · 27 May 2019 · Paper 1 · Single correct option