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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 α=α(θ)\alpha = \alpha\left(\theta\right) and β=β(θ)\beta = \beta\left(\theta\right) are real number, and II is the 2×22 \times 2 identity matrix. If α∗istheminimumofthesetα(θ):θ∈[0,2π){\alpha}^{*} is the minimum of the set { \alpha\left(\theta\right): \theta \in \left[0, 2\pi\right) }andand{\beta}^{*} is the minimum of the set \left\{\beta\left(\theta\right): \theta \in \left[0, 2\pi\right)\right\},thenthevalueofthen the value of{\alpha}^{*} + {\beta}^{*} is

Options

  1. A−3716- \dfrac{37}{16}
  2. B−3116- \dfrac{31}{16}
  3. C−2916- \dfrac{29}{16}
  4. D−1716- \dfrac{17}{16}

Correct answer

C. −2916- \dfrac{29}{16}

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[{sin⁡4θ−1−sin⁡2θ1+cos⁡2θcos⁡4θ\begin{cases}\sin^{4}\theta & - 1 - \sin^{2}\theta \\ 1 + \cos^{2}\theta & \cos^{4}\theta\end{cases}\right] = \alpha\left[{1001\begin{cases}1 & 0 \\ 0 & 1\end{cases}\right] + \dfrac{\beta}{\left|M\right|}\left[{cos⁡4θ1+sin⁡2θ−1−cos⁡4θsin⁡4θ\begin{cases}\cos^{4}\theta & 1 + \sin^{2}\theta \\ - 1 - \cos^{4}\theta & \sin^{4}\theta\end{cases}\right]<br/>Compareelement<br />Compare element{a}_{12}inL.H.S.andR.H.S.<br/>in L.H.S. and R.H.S.<br />- 1 - \sin^{2}\theta = 0 + \dfrac{\beta}{\left|M\right|}\left(1 + \sin^{2}\theta\right)<br/><br />\beta = | M |...(iii)<br/>Compareelement...(iii)<br />Compare element{a}_{11}inLHS.andR.H.S.<br/>in LHS. and R.H.S.<br />\sin^{4}\theta = \alpha + \dfrac{\beta}{\left|M\right|}\left(\cos^{4}\theta\right)<br/>Fromequation(iii)<br/><br />From equation (iii)<br />\sin^{4}\theta = \alpha - \cos^{4}\theta<br/><br />\alpha = \sin^{4}\theta + \cos^{4}\theta...(iv)<br/>Nowfromequation(iv)<br/>...(iv)<br />Now from equation (iv)<br />\alpha = \sin^{4}\theta + \cos^{4}\theta<br/><br />\alpha = {\left(\sin^{2}\theta + \cos^{2}\theta\right)}^{2} - 2\sin^{2}\theta\cos^{2}\theta<br/><br />\alpha = 1 - \dfrac{\sin^{2}2\theta}{2}<br/><br />0 \le \sin^{2}2\theta \le 1<br/><br />- \dfrac{1}{2} \le \dfrac{- \sin^{2}2\theta}{2} \le 0<br/><br />\dfrac{1}{2} \le 1 - \dfrac{\sin^{2}2\theta}{2} \le 1<br/><br />\alpha \in [12,1]\left[\dfrac{1}{2}, 1\right] , {\alpha}_\min = \dfrac{1}{2} = \alpha *
For β\beta , consider
\left|M\right| = \sin^{4}\theta\cos^{4}\theta + \sin^{2}\theta\cos^{2}\theta + 2<br/>∣M∣{(sin⁡2θcos⁡2θ+12)2+74}<br />\left|M\right|\left\{{\left(\sin^{2}\theta\cos^{2}\theta + \dfrac{1}{2}\right)}^{2} + \dfrac{7}{4}\right\}
\left|M\right|\left\{{\left(\dfrac{\sin^{2}2\theta}{4} + \dfrac{1}{2}\right)}^{2} + \dfrac{7}{4}\right\}<br/><br />0 \le \sin^{2}2\theta \le 1<br/>Hence,∣M∣∈[2,3716]<br />Hence, \left|M\right| \in \left[2, \dfrac{37}{16}\right]
β=−∣M∣\beta = - \left|M\right|
β∈[−3716,−2]\beta \in \left[- \dfrac{37}{16}, - 2\right]
{\beta}_\min = - \dfrac{37}{16} = \beta *
α∗+β∗=12−3716=−2916\alpha* + \beta* = \dfrac{1}{2} - \dfrac{37}{16} = - \dfrac{29}{16}
Method II
M=\alphaI+βM−1M = \alphaI + \beta{M}^{- 1}
Multiply with MM both side
M2=\alphaM+\betaI{M}^{2} = \alphaM + \betaI
M2−\alphaM−BI=0{M}^{2} - \alphaM - BI = 0 ...(i)
Equation (i) represents characteristic equation of matrix ‘M’‘M’
Hence, Trace (M)=α( M ) = \alpha
\left|M\right| = - \beta<br/>Sumofrootsofcharacteristicequation=Traceofmatrix<br/>Productofrootsofcharacteristicequation=determinedofmatrix<br/><br />Sum of roots of characteristic equation = Trace of matrix<br />Product of roots of characteristic equation = determined of matrix<br />∴ \alpha = \sin^{4}\theta + \cos^{4}\theta<br/><br />\beta = - \left(\sin^{4}\theta\cos^{4}\theta + \sin^{2}\theta\cos^{2}\theta + 2\right)
Now proceed as Method 1
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