JEE Advanced 2019Paper 1
- Exam
- JEE Advanced
- Year
- 2019
- Paper
- Paper 1
- Subject
- Mathematics
- Chapter
- Matrices
JEE Advanced 2019 · Paper 1 · Mathematics · Matrices
Let M = [ cases ^ 4 θ & - 1 - ^ 2 θ 1 + ^ 2 θ & ^ 4 θ cases ] = + β...
Let M = \left[\begin{cases}\sin^{4} \theta & - 1 - \sin^{2} \theta \\ 1 + \cos^{2} \theta & \cos^{4} \theta\end{cases}\right] = \alphaI + \beta{M}^{- 1}, where and are real number, and is the identity matrix. If {\beta}^{*} is the minimum of the set \left\{\beta\left(\theta\right): \theta \in \left[0, 2\pi\right)\right\},{\alpha}^{*} + {\beta}^{*} is
Options
- A
- B
- C
- D
Correct answer
C.
Step-by-step solution
Method I
As given, M = \left[\begin{cases}\sin^{4}\theta & - 1 - \sin^{2}\theta \\ 1 + \cos^{2}\theta & \cos^{4}\theta\end{cases}\right] = \alphaI + \beta{M}^{- 1}
adj\left(M\right) = \left[\begin{cases}\cos^{4}\theta & 1 + \sin^{2}\theta \\ - 1 - \cos^{2}\theta & \sin^{4}\theta\end{cases}\right]
{M}^{- 1} = \dfrac{1}{| M |}\left[\begin{cases}\cos^{4}\theta & 1 + \sin^{2}\theta \\ - 1 - \cos^{2}\theta & \sin^{4}\theta\end{cases}\right]
\left[\right] = \alpha\left[\right] + \dfrac{\beta}{\left|M\right|}\left[\right]{a}_{12}- 1 - \sin^{2}\theta = 0 + \dfrac{\beta}{\left|M\right|}\left(1 + \sin^{2}\theta\right)\beta = | M |{a}_{11}\sin^{4}\theta = \alpha + \dfrac{\beta}{\left|M\right|}\left(\cos^{4}\theta\right)\sin^{4}\theta = \alpha - \cos^{4}\theta\alpha = \sin^{4}\theta + \cos^{4}\theta\alpha = \sin^{4}\theta + \cos^{4}\theta\alpha = {\left(\sin^{2}\theta + \cos^{2}\theta\right)}^{2} - 2\sin^{2}\theta\cos^{2}\theta\alpha = 1 - \dfrac{\sin^{2}2\theta}{2}0 \le \sin^{2}2\theta \le 1- \dfrac{1}{2} \le \dfrac{- \sin^{2}2\theta}{2} \le 0\dfrac{1}{2} \le 1 - \dfrac{\sin^{2}2\theta}{2} \le 1\alpha \in , {\alpha}_\min = \dfrac{1}{2} = \alpha *
For , consider
\left|M\right| = \sin^{4}\theta\cos^{4}\theta + \sin^{2}\theta\cos^{2}\theta + 2
\left|M\right|\left\{{\left(\dfrac{\sin^{2}2\theta}{4} + \dfrac{1}{2}\right)}^{2} + \dfrac{7}{4}\right\}0 \le \sin^{2}2\theta \le 1
{\beta}_\min = - \dfrac{37}{16} = \beta *
Method II
Multiply with both side
...(i)
Equation (i) represents characteristic equation of matrix
Hence, Trace
\left|M\right| = - \beta∴ \alpha = \sin^{4}\theta + \cos^{4}\theta\beta = - \left(\sin^{4}\theta\cos^{4}\theta + \sin^{2}\theta\cos^{2}\theta + 2\right)
Now proceed as Method 1
As given, M = \left[\begin{cases}\sin^{4}\theta & - 1 - \sin^{2}\theta \\ 1 + \cos^{2}\theta & \cos^{4}\theta\end{cases}\right] = \alphaI + \beta{M}^{- 1}
adj\left(M\right) = \left[\begin{cases}\cos^{4}\theta & 1 + \sin^{2}\theta \\ - 1 - \cos^{2}\theta & \sin^{4}\theta\end{cases}\right]
{M}^{- 1} = \dfrac{1}{| M |}\left[\begin{cases}\cos^{4}\theta & 1 + \sin^{2}\theta \\ - 1 - \cos^{2}\theta & \sin^{4}\theta\end{cases}\right]
\left[\right] = \alpha\left[\right] + \dfrac{\beta}{\left|M\right|}\left[\right]{a}_{12}- 1 - \sin^{2}\theta = 0 + \dfrac{\beta}{\left|M\right|}\left(1 + \sin^{2}\theta\right)\beta = | M |{a}_{11}\sin^{4}\theta = \alpha + \dfrac{\beta}{\left|M\right|}\left(\cos^{4}\theta\right)\sin^{4}\theta = \alpha - \cos^{4}\theta\alpha = \sin^{4}\theta + \cos^{4}\theta\alpha = \sin^{4}\theta + \cos^{4}\theta\alpha = {\left(\sin^{2}\theta + \cos^{2}\theta\right)}^{2} - 2\sin^{2}\theta\cos^{2}\theta\alpha = 1 - \dfrac{\sin^{2}2\theta}{2}0 \le \sin^{2}2\theta \le 1- \dfrac{1}{2} \le \dfrac{- \sin^{2}2\theta}{2} \le 0\dfrac{1}{2} \le 1 - \dfrac{\sin^{2}2\theta}{2} \le 1\alpha \in , {\alpha}_\min = \dfrac{1}{2} = \alpha *
For , consider
\left|M\right| = \sin^{4}\theta\cos^{4}\theta + \sin^{2}\theta\cos^{2}\theta + 2
\left|M\right|\left\{{\left(\dfrac{\sin^{2}2\theta}{4} + \dfrac{1}{2}\right)}^{2} + \dfrac{7}{4}\right\}0 \le \sin^{2}2\theta \le 1
{\beta}_\min = - \dfrac{37}{16} = \beta *
Method II
Multiply with both side
...(i)
Equation (i) represents characteristic equation of matrix
Hence, Trace
\left|M\right| = - \beta∴ \alpha = \sin^{4}\theta + \cos^{4}\theta\beta = - \left(\sin^{4}\theta\cos^{4}\theta + \sin^{2}\theta\cos^{2}\theta + 2\right)
Now proceed as Method 1