MCQ
Let $M=\left[\begin{array}{cc}\sin ^4 \theta & -1-\sin ^2 \theta \\ 1+\cos ^2 \theta & \cos ^4 \theta\end{array}\right]=\alpha I +\beta M ^{-1}$, where $\alpha=\alpha(\theta)$ and $\beta=\beta(\theta)$ are real number, and $I$ is the $2 \times 2$ identity matrix. If $\alpha^*$ is the minimum of the set $\{\alpha(\theta): \theta \in[0,2 \pi)\}$ and $\beta^*$ is the minimum of the set $\{\beta(\theta): \theta \in[0,2 \pi)\}$, then the value of $\alpha^*+\beta^*$ is
  • A
     $-\frac{37}{16}$
  • $-\frac{29}{16}$
  • C
    $-\frac{31}{16}$
  • D
    $-\frac{17}{16}$

Answer

Correct option: B.
$-\frac{29}{16}$
b
Given $M =\alpha I +\beta M ^{-1}$

$\Rightarrow M ^2-\alpha M -\beta I = O$

By putting values of $M$ and $M^2$, we get

$\alpha(\theta)=1-2 \sin ^2 \theta \cos ^2 \theta=1-\frac{\sin ^2 2 \theta}{2} \geq \frac{1}{2}$

Also,$\beta(\theta)=-\left(\sin ^4 \theta \cos ^4 \theta+\left(1+\cos ^2 \theta\right)\left(1+\sin ^2 \theta\right)\right)$

$=-\left(\sin ^4 \theta \cos ^4 \theta+1+\cos ^2 \theta+\sin ^2 \theta+\sin ^2 \theta \cos ^2 \theta\right)$

$=-\left(t^2+t+2\right), t=\frac{\sin ^2 2 \theta}{4} \in\left[0, \frac{1}{4}\right]$

$\Rightarrow \quad \beta(\theta) \geq-\frac{37}{16}$

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