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Mathematics · Matrices

JEE Advanced 2019 — Paper 1 — Question 29

Let M=[sin⁡4θ−1−sin⁡2θ1+cos⁡2θcos⁡4θ]=α1+βM−1M=\left[\begin{array}{cc}\sin ^{4} \theta & -1-\sin ^{2}\\ \theta 1+\cos ^{2} \theta & \cos ^{4} \theta\end{array}\right]=\alpha 1+\beta M^{-1}, where α=α(θ)\alpha=\alpha(\theta) and β=β(θ)\beta=\beta(\theta) are

real numbers, and 1 is the 2×22 \times 2 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

  1. Option A:

    −3716-\frac{37}{16}

  2. Option B:

    −3116-\frac{31}{16}

  3. Option C:

    −1716-\frac{17}{16}

  4. Option D:

    −2916-\frac{29}{16}

    Correct

Answer: D

Step-by-step solution

M=α[1001]+β[cos⁡4θ1+sin⁡2θ−1−cos⁡2θsin⁡4θ]M=\alpha\left[\begin{array}{ll}1 & 0\\ 0 & 1\end{array}\right]+\beta\left[\begin{array}{cc}\cos ^{4} \theta & 1+\sin ^{2}\\ \theta -1-\cos ^{2} \theta & \sin ^{4} \theta\end{array}\right]

on comparing we have

sin⁡4θ=α+βcos⁡4θ∣M∣−1−sin⁡2θ=α+β(1+sin⁡2θ)∣M∣α=sin⁡4θ+cos⁡4θ Now α=(sin⁡2θ+cos⁡2θ)2−2sin⁡2θcos⁡2θ=1−sin⁡22θ2⇒α∗=12\begin{aligned} & \sin ^{4} \theta=\alpha+\frac{\beta \cos ^{4} \theta}{|\mathrm{M}|} \\& -1-\sin ^{2} \theta=\alpha+\frac{\beta\left(1+\sin ^{2} \theta\right)}{|\mathrm{M}|} & \quad \alpha=\sin ^{4} \theta+\cos ^{4} \theta \\& \text { Now } \alpha=\left(\sin ^{2} \theta+\cos ^{2} \theta\right)^{2}-2 \sin ^{2} \theta \cos ^{2} \theta & =1-\frac{\sin ^{2} 2 \theta}{2} \\& \Rightarrow & \alpha^{*}=\frac{1}{2} \end{aligned}

We have, ∣M∣=sin⁡4θcos⁡4θ+(1+sin⁡2θ)(1+cos⁡2θ)|\mathrm{M}|=\sin ^{4} \theta \cos ^{4} \theta+\left(1+\sin ^{2} \theta\right)\left(1+\cos ^{2} \theta\right)

=2+sin⁡2θcos⁡2θ+sin⁡4θcos⁡4θ=(sin⁡2θcos⁡2θ+1/2)2+7/4⇒β=−∣M∣=−74−(sin⁡2θcos⁡2θ+12)2⇒β∗=−74−(14+12)2=−74−916=−3716α∗+β∗=−2916.\begin{aligned} & =2+\sin ^{2} \theta \cos ^{2} \theta+\sin ^{4} \theta \cos ^{4} \theta \\& = \left(\sin ^{2} \theta \cos ^{2} \theta+1 / 2\right)^{2}+7 / 4 \\& \Rightarrow \beta=-|M|=-\frac{7}{4}-\left(\sin ^{2} \theta \cos ^{2} \theta+\frac{1}{2}\right)^{2} \\& \Rightarrow \beta^{*}=-\frac{7}{4}-\left(\frac{1}{4}+\frac{1}{2}\right)^{2}=-\frac{7}{4}-\frac{9}{16}=-\frac{37}{16} \\& \alpha^{*}+\beta^{*}=-\frac{29}{16} . \end{aligned}

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2019
Paper
Paper 1
Subject
Mathematics
Chapter
Matrices
Topic
Inverse of a Matrix

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